Manufacturing Planning and Control for Supply Chain Management 6th Edition Jacobs Solutions Manual

Type:

Solution Manual

Manufacturing Planning and Resource: Control for Supply Chain Management Edition:

6th Edition

Author(s):

F. Robert Jacobs William Lee Berry David Clay Whybark Thomas E. Vollmann

Chapter 1 Solution Manual

CHAPTER 1 Solution Manual Discussion Questions 1. The intent of this question is to get the student to think about timing issues. The first of these issues is that the process of direction setting, planning, and execution all go on simultaneously.

The distinction is that the direction setting and future planning may be for products that have not yet even been designed, while the detailed material and capacity planning is for products that will be made in the very near future, and the execution activities are for producing products that were planned some time ago. It is very difficult for the students to get that timing sequence down. Their examples should clearly illustrate they understand that the planning taking place today is for products that will be produced in the future.

2. Taken from the student's perspective, the breakdown is: major (front end), courses and sequence (engine) and performance (back end). Once the major is planned, a schedule of courses is required. This MPS is exploded into classes by term (frozen in the short run). The shop floor system monitors and adjusts for changes like failing, dropping courses, or having courses canceled.

3. This is obviously a question designed to let the class run loose. Among the things that they may come up with are: the population explosion, advances in information technology, new production technology, the emergence of China, developments in theory, and new organizational innovations.

4. Before asking questions, lash her soundly. She should be focusing on what type of business the company is in and what processes are used. Why is she going straight to software and system choice? This chapter stresses the need to fit the system to the needs of the firm, market and supply chain partners. Has she determined the specific tasks that need to be performed to do any of this? Only after she has worked on defining the MPC needs of the firm can she select the software and system choice.

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Chapter 01A Solution Manual

CHAPTER 1A Solution Manual Discussion Questions 1. Essential attributes of ERP systems are the following: (1) multifunctional in scope, (2) integrated in that when a transaction or piece of data representing an activity of the business is entered by one of the function, data regarding the other related function is changed as well, (3) the software is modular in structure, and (4) the software much facilitate manufacturing planning and control activities including forecasting, production planning, and inventory management.

2. Processes probably do need to be reengineered accommodate a new ERP system since functions can change significantly due to the processes needed to support the new system. It is important, though, that the competitive impact of making the change be considered. Is the new system actually an improvement over the old system? This may or may not be true depending on the case at hand.

3. Real-time data allows the exact status of customer orders and the activities critical to the success of the firm to be determined at any time. It’s important to recognize, though, that reacting to short term changes can often be quite disruptive. A good plan should be robust relative to short term changes in demand and supply and this should be a real goal in making the best use of information from an ERP system 4. Comparing the offerings from SAP, Oracle and PeopleSoft is a good exercise for the students. These websites change often based on the marketing emphasis that each company is pursuing. Actually, major changes in the underlying software does not occur very frequently, possibly on every 10 years. The software is constantly being repackaged, though. These companies are also continually developed very specialized applications often directed at specific industries. These applications will be advertised on the websites as well.

5. i2 Technologies is well known for advanced forecasting, bottleneck scheduling (shop level) and advanced logistics applications (truck loading, vehicle routing).

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Chapter 01A Solution Manual

Chapter 1A Problem Summaries Problem Difficulty 1 2 3 4 5

Easy Hard Moderate Easy Easy

6 7 8 9 10 11 12 13 14 15

Moderate Moderate Moderate Moderate Moderate Moderate Moderate Easy Easy Moderate

Problem Type Cash-to-cash cycle time calculation.

Cash-to-cash cycle comparison using data from the Internet.

Cash-to-cash cycle time sensitivity analysis.

Labor requirements analysis.

Sensitivity analysis using ERP data. Problems 6-12 are extensions that explore the use of ERP type data.

Extension of #5.

Labor requirements – extension of #4 and #5.

Extension of #7.

Extension of #8.

Extension of #9.

Extension of #10.

Extension of #11.

Benchmarking problem.

Benchmarking comparison between firms.

Eli Lilly analysis of the value of reducing inventory level through the use of an ERP system.

1.

Sd = S/d ARd = AR/Sd Cd = Sd * (COS/S) Id = I/Cd APd = AP/Cd

= 8028 / (13 weeks * 7 days) = 2689 / 88.22 = 88.22 * (6580 / 8028) = (126 + 224) / 72.31 = 4326 / 72.31

= 88.22 M per day = 30.48 days = 72.31M = 4.84 days = 59.83 days

Cash to Cash Cycle Time = 30.48 + 4.84 – 59.83 = -24.51 days

2. Calculations are similar to what is given in problem 1. This is an interesting comparisons that can generate good discussion about how these companies operate differently.

3. Reduction in cost-of-goods sold. This should improve cash-to-cash cycle time. More frequent deliveries from suppliers. Improve cash-to-cash cycle time. Reductions in time customers are allowed to pay for goods. Improve cash-to-cash cycle time. Change from paying suppliers on receipt of goods to waiting 60 days to pay suppliers. This should improve cash-to-cash cycle time. Write-off of obsolete inventory. Improves cash-to-cash cycle time. Reduction in labor content in a production process. Improves cash-to-cash cycle time. Outsourcing the production of a major product. Improves cash-to-cash cycle time.

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Chapter 01A Solution Manual

4.

2300/60 = (38.33 hours) / 5 hours per student ≈ 8 part-time college students

5.

(2100 x $2.00) – (2300 x $1.25) = $1325.

(1500 x $2.00) – (2300 x $1.25) = $125.

6.

There would be no discarded sandwiches in either case, since the next day all could be sold. But it does require the sellers to use up the old inventory first. Ajax would need a linkage to each selling site to know how many less sandwiches to send each day (i.e., gross-to-net).

7.

500 x .9 x .9 = 405 roast beef plus 500 + (500 – 405) * .5 or 547.5 tuna – in round numbers 400 roast beef and 550 tuna or 950 total sandwiches.

8.

About 50 (1000 – 950) sandwiches at a margin of $0.75 ($2.00 - $1.25) = $37.50

9.

Monthly delivery savings = 3 x $20 = $60 ($720/year). Cost of capital tied up in customer inventories is about 1 to 3 weeks of stock (say 2 on average) =$200 x 15% = $30.

10.

The master production schedule for assembling sandwiches could be used to determine the forward need (forecast) of peanut butter requirements. It could also be used for payments – assuming the schedule is actually achieved. Another potential source of information for payments could be the actual build results – perhaps coming from labor reporting in the HR systems. Additionally, there would probably be information in the sales units for shipments to the various outlets that could be used. If the payment was based on actual end item sandwich sales then it would be necessary to connect to the selling unit to the detailed sales analysis. A problem here is that the supplier should also be paid for sandwiches not sold. It would also be necessary to periodically reconcile the inventories in both companies.

11.

There are half of the varieties (12) that have 5% leftover on any day so 5% (leftover) * 20 (sandwiches per day) * 12 (varieties) * 50% (half that have excess) = 6 leftover sandwiches each day (tossed out?). Additionally the 2 best sellers could have 10% (.10 x 20) more = 2 each, so we could sell 4 more sandwiches (2 sandwiches * 2 varieties).

These four could be made from the read of the 6 leftovers loaves – only tossing out 2 (6 4) loaves of bread.

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Chapter 01A Solution Manual

12.

The MRP portion of ERP (using actual sales data from the stores) could do a continual gross-to-net set of calculations as to which sandwiches were selling which could be used to determine the estimated inventory levels of all materials in the system – and where they are located. From these, very accurate inventory replenishment needs could be calculated.

13.

A difference of about 30 days (84 days - 55 days). If say a 1% saving for each 10 days then (30/10) * 1% = 3% extra return on sales.

14.

1.7 turns versus 4.7 turns would indicate that the best in class company should be able to operate with about 1/3 (36%) of the investment in assets of the average company. When one thinks about those that are below average – the comparison becomes even more pronounced: Good supply chain performance is not just something that is “nice to have.”

15.

Currently, $1,000.000 = 78 days of inventory, therefore reduction of 10 days = $128,205 reduction in inventory.

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Chapter 02 Solution Manual

CHAPTER 2 Solution Manual Discussion Questions 1. The customer order decoupling point is the point at which the customer does his or her “shopping.” If it is at finished goods, they buy finished goods at their timing and quantities.

However, from finished goods back into the company, management has control of the demands on upstream departments and materials, again, both in terms of timing and quantity.

The advantage of dependent demand is that it can be planned for as opposed to being forecast.

This reduces uncertainty and associated costs as well as enabling more economic production.

2. Your students should be able to provide several examples of each. Almost all the goods at a Wal-mart are in the make-to-stock category, for example. For assemble to order, they might think of the computer kiosks that are found in many stores, making up a plate at a cafeteria or a car purchase. The make-to-order category might include contact lenses, architect designed houses and the buildings on campus. For the advantages, they might start with the increased knowledge upon which the products are manufactured (as opposed to forecasts). Additional advantages might include reduced inventory, space, and transportation needs.

Chapter 2 Problem Summaries Problem Difficulty 1

Moderate

Problem Type Assemble to order vs make to stock comparison.

1. Northland Computer Shop a. The number of potential finished products is the product of all the components and options (assuming that all are viable, but the problem says “potential” number, so we’re okay).

The total number of finished products possible is: 7 * 6 * 5 * 3 * 4 = 2520 potential finished products b. To get the difference we need the total number of components and options which is: 7 + 6 + 5 + 3 + 4 = 25 components At $10 per forecast we can save $2,495/week (100*(2520-25)).

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Chapter 03 Solution Manual

CHAPTER 3 Solution Manual Discussion Questions 1. The response to this question really centers on the preparation of the organization and continued involvement of the people in the process of rolling up and forcing down the reconciliations. There must be an ongoing feedback mechanism to provide for exceptions if the organization is going to "own" the forecast number.

2. While it's easy for a grocery store to capture sales data, it's nearly impossible to capture demand data since customers get their own products and may change their minds, use a substitute, or simply not buy any product. In a warehouse, on the other hand, the demand is received by phone, mail, or a customer at the counter. Thus, it's easier to capture the demand data right at the source.

3. The primary reason for this argument is that it may be more important to be right "on the average" than to be consistently wrong. For example, if the forecasts are consistently high, excess capacity, extra inventories, and too much manpower may be provided to produce the product actually required. On the other hand, if there is substantial error but on the average capacity is right, buffering techniques can be used to mitigate the forecast error.

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Chapter 3 Problem Summaries Problem Difficulty

Problem Type

1

Moderate

Least squares regression analysis (intercept, slope, and standard error) and creating a forecast using the data from the analysis.

2

Moderate

Creating several forecast on the same data using weighted moving average, moving average, exponential smoothing and linear regression.

3

Easy

Calculating an exponential smoothing forecast and the MAD.

4

Hard

Forecasting using seasonal factors with linear regression to determine the trend. Have to deseasonalize the data, determine the trend line, then create a seasonal forecast.

5

Hard

Creating a seasonal forecast using regression to determine the trend line and seasonal factors.

6

Moderate

Moving average and weighted moving average forecasting problem.

7

Moderate

Exponential smoothing problem, allows comparison to #6.

8

Moderate

Exponential smoothing problem.

9

Hard

Spreadsheet comparison of moving averages and exponential smoothing models.,,

10

Hard

Test two exponential smoothing models using a data set.

11

Moderate

Evaluate the value of reducing forecast error

12

Moderate

Impact of aggregating demand.

13

Hard

Extension of problem 12

14

Hard

Roll-up and roll-down forecast exercise

15

Moderate

Individual item versus aggregate forecast.

16

Moderate

Roll-up and roll-down forecast exercise.

17

Moderate

Extension of problem #14 to consider “what-if” analysis.

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Chapter 03 Solution Manual

1. NINJA INDUSTRIES Month

x

y

xy

x2

Y

January February March April May June July August September October November December

1 2 3 4 5 6 7 8 9 10 11 12

4,200 4,300 4,000 4,400 5,000 4,700 5,300 4,900 5,400 5,700 6,300 6,000

4,200 8,600 12,000 17,600 25,000 28,200 37,100 39,200 48,600 57,000 69,300 72,000

1 4 9 16 25 36 49 64 81 100 121 144

3,959 4,151 4,344 4,536 4,728 4,921 5,113 5,305 5,497 5,690 5,882 6,074

78

60,200

418,800

650

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58,081 22,201 118,336 18,496 73,984 48,841 34,969 164,025 9,409 100 174,724 5,476 728,642

a.

Month

x

January February March April May June July August September October November December

Y 13 14 15 16 17 18 19 20 21 22 23 24

6,267 6,459 6,651 6,843 7,036 7,228 7,420 7,613 7,805 7,997 8,189 8,382

b. The standard error of the forecast

She must carry 270 x 3 = 810 extra units of for safety

2. HISTORICAL DEMAND Month

Demand

Jan Feb Mar Apr May Jun a.

12 11 15 12 16 15

Weighted moving average forecast Weights are .60, .30, .10 WMAFJuly = .60 * 15 +.3 *16 + .10 * 12 = 15

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Chapter 03 Solution Manual

b. 3-period moving average forecast MAFJuly = (15 + 16 + 12)/3 = 43/3 = 14.33 c.

Exponential smoothing forecast α = .20, ESFJune = 13 ESFJuly = .20 * 15 + (1-.20) * 13 = 13.4

d. Simple linear regression

e.

Month

x

y

xy

x2

Jan Feb Mar Apr May Jun

1 2 3 4 5 6 21

12 11 15 12 16 15 81

12 22 45 48 80 90 297

1 4 9 16 25 36 91

Regression forecast x = 7 Y7 = 10.8 + .7714 * 7 = 16.2

3. JANUARY FORECAST a. Exponential Forecast

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January February March April May June

Actual 100 94 106 80 68 94

Exponential Absolute Forecast Error error 80 20 20 84 10 10 86 20 20 90 -10 10 88 -20 20 84 10 10 30 90

b. Mean Absolute Deviation MAD = sum of absolute error/n = 90/6 = 15

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Chapter 03 Solution Manual

4. ZEUS COMPUTER CHIPS Deseasonalize the data

Year 2007

2008

2009

Quarter I II III IV I II III IV I II III IV

Actual Period Demand (x) (y) 1 2 3 4 5 6 7 8 9 10 11 12 78

4800 3500 4300 3000 3500 2700 3500 2400 3200 2100 2700 1700 37400

Average of the same quarter each year

Seasonal Factor

Deseasonalized Demand (Yd)

x2

x * Yd

1.23 0.89 1.12 0.76 1.23 0.89 1.12 0.76 1.23 0.89 1.12 0.76 12.00

3902.4 3932.6 3839.3 3947.4 2845.5 3033.7 3125.0 3157.9 2601.6 2359.6 2410.7 2236.8 37392.5

1 4 9 16 25 36 49 64 81 100 121 144 650

3902.4 7865.2 11517.9 15789.6 14227.5 18202.2 21875 25263.2 23414.4 23596 26517.7 26841.6 219013

3833.3 2766.7 3500.0 2366.7

Calculate the seasonal factors and then determine the regression trend line

Calculate the forecast for 2010

Year

Quarter

2010 I II III IV

Period

Y from Regression Line

Seasonal Factor

Forecast (Y * seasonal factor)

13 14 15 16

2023.4 1855.3 1687.2 1519.1

1.23 0.89 1.12 0.76

2488.78 1651.22 1889.66 1154.52

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5. BI-MONTHLY SALES DATA a. Plot the data

b. Fit simple linear regression model to the data Month

x

January–February March–April May–June July–August September–October November–December January–February March–April May–June July–August September–October November–December

1 2 3 4 5 6 7 8 9 10 11 12 78

3-8

y

xy

x2

109 104 150 170 120 100 115 112 159 182 126 106 1,553

109 208 450 680 600 600 805 896 1,431 1,820 1,386 1,272 10,257

1 4 9 16 25 36 49 64 81 100 121 144 650

Chapter 03 Solution Manual

c. Determine seasonal factors using the regression model

Month January–February March–April May–June July–August September–October November–December January–February March–April May–June July–August September–October November–December

y

Y

109 104 150 170 120 100 115 112 159 182 126 106

123 124 125 127 128 129 130 131 132 133 135 136

Seasonal Average Factor Seasonal (yi/Yi) Factor 0.89 0.84 1.2 1.34 0.94 0.78 0.88 0.85 1.2 1.37 0.93 0.78

0.89 0.85 1.2 1.36 0.94 0.78

d. Prepare the seasonalized forecast for next year

Month January–February March–April May–June July–August September–October November–December

x

Y 13 14 15 16 17 18

2,218 2,282 2,346 2,410 2,474 2,538

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Seasonal Factor

Seasonalized Forecast

0.89 0.85 1.2 1.36 0.94 0.78

1,974 1,940 2,815 3,278 2,326 1,980

6.

MAVERICK a.

Month 1 2 3 4 5 6 7 8 9 10 11 12 1

Demand Forecast

Error

Absolute Error

20 21 23 25 24 26 25 24 24 24

5 3 4 -3 6 -3 -5 5 -2

5 3 4 3 6 3 5 5 2

SUM MEAN

10 1.11

36 4

Demand Forecast

Error

Absolute Error

20 20 23 25 25 26 25 27 23 23

5 4 4 -3 5 -3 -5 2 -1

5 4 4 3 5 3 5 2 1

SUM MEAN

8 0.9

32 3.6

20 18 21 25 24 27 22 30 23 20 29 22

b.

Month 1 2 3 4 5 6 7 8 9 10 11 12 1

20 18 21 25 24 27 22 30 23 20 29 22

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Chapter 03 Solution Manual

c. The three month weighted moving average performed better than the three month moving average on both measures of forecast error. Sometimes there can be contradictions in these two measures , but in this case the weighted moving average is the best on both measures.

Bias

MAD

3-month moving avg.

1.11

4

Weighted 3-mo.

MA

0.9

3.6

7. MAVERICK (CONTINUED) a. .

Month 1 2 3 4 5 6 7 8 9 10 11 12 1 SUM MEAN

Demand Forecast Error Absolute Forecast Error Absolute Forecast Error Absolute 0.2 Error 0.5 Error 0.8 Error 20 18 21 25 20 5 5 20 5 5 20 5 5 24 21 3 3 22.5 1.5 1.5 24 0 0 27 21.6 5.4 5.4 23.3 3.7 3.7 24 3 3 22 22.7 -0.7 0.7 25.2 -3.2 3.2 26.4 -4.4 4.4 30 22.6 7.4 7.4 23.6 6.4 6.4 22.9 7.1 7.1 23 24.1 -1.1 1.1 26.8 -3.8 3.8 28.6 -5.6 5.6 20 23.9 -3.9 3.9 24.9 -4.9 4.9 24.1 -4.1 4.1 29 23.1 5.9 5.9 22.5 6.5 6.5 20.8 8.2 8.2 22 24.3 -2.3 2.3 25.8 -3.8 3.8 27.4 -5.4 5.4 23.8 23.9 23.1 18.7 2.08

34.7 3.86

7.4 0.82

3-11

38.8 4.31

3.8 0.42

42.8 4.76

b. The MAD values are very much the same for all three smoothing constants. The bias, however, decreases as the Alfa increases. When compared to the weighted average results from problem 6, the MAD values are all about the same. The bias for the weighted moving average is near the values for the higher smoothing constants. That gives some credibility to the argument of weighting the current information more heavily than old information. However, the lowest MAD was created with the lowest smoothing constant, even though the bias was the highest with that forecast.

8. TALBOT PUBLISHING COMPANY a.

5 6 7

Actual Sales 18,000 20,000 17,000

Forecast 20,200 20,200 20,200

Forecast Error -2,200 -200 -3,200 -5,600

Absolute Deviation 2,200 200 3,200 5,600

b.

Bias (average error) = -5600/3 MAD = 5600/3 Standard Deviation = 1867 * 1.25

= = =

-1867 Units 1867 Units 2333.75 Units

c.

5 6 7

Actual Sales 18,000 20,000 17,000

Forecast 20,200 19,760 19,808

Absolute Deviation 2,200 240 2,808 5,248

MAD = 5,248/3 = 1,749 units. There has been a slight improvement in forecast accuracy by updating the forecast each period.

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Chapter 03 Solution Manual

9.

COMPARISON a. The following spreadsheet is one way of setting up to do the problem. If you set  = .3 it provides a lower MAD (8.24 vs.8.46) for the exponential smoothing model versus the three-period moving average approach.

1 2 3 4 5 6 7 8 9 10

b.

Esponential Moving Absolute Absolute Smoothing Average Error Error Demand Forecast Forecast ESF MAF 27 26 32 41 28 35 30.8 33.7 4.2 1.3 43 32.1 34.7 10.9 8.3 47 35.4 35.3 11.6 11.7 28 38.9 41.7 10.9 13.7 32 35.6 39.3 3.6 7.3 34.5 35.7 Sum 41.2 42.3 Mean 8.24 8.46

When the starting value of the exponential smoothing model (actually the forecast for period 6) is changed to the average of periods 3, 4, and 5 the MAD for exponential smoothing drops to 7.32 ( = .3).

6 7 8 9 10

Demand 35 43 47 28 32

Esponential Moving Absolute Absolute Smoothing Average Error Error Forecast Forecast ESF MAF 33.7 33.7 1.3 1.3 34.1 34.7 8.9 8.3 36.8 35.3 10.2 11.7 39.9 41.7 11.9 13.7 36.3 39.3 4.3 7.3 35 35.7

Sum Mean

36.6 7.32

10.

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42.3 8.46

11. TWO DEMAND SETS The results indicate that data set number 2 has a much larger level of random variation than data set number 1. Furthermore, in both data sets the value of MAD resulting from the use of an  value of .5 exceeds the value of MAD resulting from the use of an  value of .1. The average error values of .06 and .09 in data set 1 indicate that the forecasts produced by both the exponential smoothing models have very little bias. However, the forecasts produced by both exponential smoothing models on data set number 2 exceed actual demand by 1 to 2 units per period. Overall, one would recommend the use of a smoothing constant value of .1 in forecasting the demand with exponential smoothing for both data sets.

Period 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Average =

Demand Set Number 1

 = 0.10

Demand 51 46 49 55 52 47 51 48 56 51 45 52 49 48 43 46 55 53 54 49

Error 1.00 -4.10 -0.69 5.38 1.84 -3.34 0.99 -2.11 6.10 0.49 -5.56 2.00 -1.20 -2.08 -6.87 -3.19 -6.87 3.52 4.17 -1.25 0.06

Forecast 50.00 50.10 49.69 49.62 50.16 50.34 50.01 50.11 49.90 50.51 50.56 50.00 50.20 50.08 49.87 49.19 48.87 49.48 49.83 50.25

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ABS(Error) 1.00 4.10 0.69 5.38 1.84 3.34 0.99 2.11 610 0.49 5.56 2.00 1.20 2.08 6.87 3.19 6.87 3.52 4.17 1.25 3.10

Chapter 03 Solution Manual

Demand Set Number 1 Period 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Average =

Period 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Demand 51 46 49 55 52 47 51 48 56 51 45 52 49 48 43 46 55 53 54 49

 = 0.50

Forecast 50.00 50.50 48.25 48.63 51.81 51.91 49.45 50.23 49.11 52.56 51.78 48.39 50.19 49.60 48.80 45.90 45.95 50.47 51.74 52.87

Error 1.00 -4.50 0.75 6.37 0.19 -4.91 1.55 -2.23 6.89 -1.56 -6.78 3.61 -1.19 -1.60 -5.80 0.10 9.05 2.53 2.26 -3.87 0.09

Demand Set Number 2

 = 0.10

Demand 77 83 90 22 10 80 16 19 27 79 73 88 15 21 85 22 88

Forecast 50.00 52.70 55.73 59.16 55.44 50.90 53.81 50.03 46.92 44.93 48.34 50.80 54.52 50.57 47.61 51.35 48.42 3-15

Error 27.00 30.30 34.27 -37.16 -45.44 29.10 -37.81 -31.03 -19.92 34.07 24.66 37.20 -39.52 -29.57 37.39 29.35 39.58

ABS(Error) 1.00 4.50 0.75 6.37 0.19 4.91 1.55 2.23 6.89 1.56 6.78 3.61 1.19 1.60 5.80 0.10 9.05 2.53 2.26 3.87 3.34

ABS(Error) 27.00 30.30 34.27 37.16 45.44 29.10 37.81 31.03 19.92 34.07 24.66 37.20 39.52 29.57 37.39 -29.35 39.58

18 19 20 Average =

Period 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Average =

75 14 16

52.38 54.64 50.57

22.62 -40.64 -34.57 -1.44

Demand Set Number 2

 = 0.50

Demand 77 83 90 22 10 80 16 19 27 79 73 88 15 21 85 22 88 75 14 16

Forecast 50.00 63.50 73.25 81.63 51.81 30.91 55.45 35.73 27.36 27.18 53.09 63.05 75.52 45.26 33.13 59.07 40.53 64.27 69.63 41.82

Error 27.00 19.50 16.75 -59.63 -41.81 49.09 -39.45 -16.73 -0.36 51.82 19.91 24.95 -60.52 -24.26 51.87 -37.07 47.47 10.73 -55.63 -25.82 -2.11

22.62 40.64 34.57 33.06

ABS(Error) 27.00 19.50 16.75 59.63 41.81 49.09 39.45 16.73 0.36 51.82 19.91 24.95 60.52 24.26 51.87 37.07 47.47 10.73 55.63 25.82 34.02

12. THANSKAVEL a. Before the initiatives the average demand per week was 200 units and the standard deviation is 22 units for Eggsbar. This means the distribution over the manufacturing lead time had a mean of 1200 (6*200), a variance of 2904 ((222)*6) and a standard deviation of 53.89 ( ). After the investments, the mean was 800 (4*200), the variance 1936 (4*484) and the standard deviation 44 ( ). b. With the company policy of holding 2.5 standard deviations of safety stock and a standard deviation of 53.89, the company was carrying 135 (53.89 * 2.5) units of safety stock. This amounts to $74.07 (10000/135) per unit. After the investments, the amount of safety stock inventory was reduced to 110 (44*2.5) units or $8148 (74.07*110). The savings, therefore, was $1852 (10000 – 8148).

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Chapter 03 Solution Manual

13. CUMBERLAND a. Yearly distribution of each product: Average = 1,200/ year

 = 12  10 2 = 34.64 b. Monthly distribution of all products together: Average = 500/ month

 = 5  10 2 = 22.36 c. Yearly distribution of all products together: Average = 6,000/year

 = 5  12  10 2 = 77.46 14. COMPARISON WITH CUMBERLAND 1Month 3 in%

Cumberl and

1Year 3 in%

Cumberl and

Family

3*(22.36)/500=13.42%

12%

3*(77.46)/6000=3.87%

8%

SKU

3*(10)/100=30.00%

30%

3*(34.64)/1200=8.66%

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15. MACRONALD’S a.

Family Burgers

Chicken Hoagies

Pizza

Product Regular Super Super-Duper Regular Cajun Italian French American Cheese Pepperoni

Forecast Forecast Of units $/unit Of sales 1200 $1.00 $1,200.00 2700 $1.50 $4,050.00 2100 $1.80 $3,780.00 1800 $2.50 $4,500.00 2700 $2.75 $7,425.00 2250 $3.50 $7,875.00 1650 $3.00 $4,950.00 1350 $3.25 $4,387.50 750 $1.75 $1,312.50 1200 $2.25 $2,700.00 $42,180.00

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Chapter 03 Solution Manual

b.

Family Burgers Chicken Hoagies Pizza

Manager’s Rolled Up Forecast Rolled Down Forecast Forecast $10,000.00 $9,030.00 $13,000.00 $15,000.00 $11,925.00 $19,500.00 $20,000.00 $17,212.50 $26,000.00 $5,000.00 $4,012.50 $6,500.00 $50,000.00 $42,180.00 $65,000.00

Family Burgers

Chicken Hoagies

Pizza

Rolled Down Forecast Product $Sales units Regular $1,727.57 Super $5,830.56 Super-Duper $5,441.86 Regular $7,358.49 Cajun $12,141.51 Italian $11,895.42 French $7,477.12 American $6,627.45 Cheese $2,126.17 Pepperoni $4,373.83 $65,000.00

3-19

1727 3887 3023 2943 4414 3398 2492 2039 1215 1944

16. KNOX PRODUCTS CORPORATION a.

Rolled-up Forecast Average Price

Business Forecast ($)

$11,000 $8,400

Rolled-up Forecast ($)

Eastern 589 Units 450 Units $7.17

Western 982 Units 750 Units $6.90

Forced Forecast

Forced Forecast Initial Forecast Price

Less Knox 196 Units 150 Units $4.50

Item Level Forced Forecast (Units) Initial Forecast Units) Disparity (Units) b. .

More Knox 393 Units 300 Units $8.50

Less Knox 393 Units 300 Units $4.50

Eastern Less Knox More Knox 196 393 150 300 46 93

3-20

More Knox 589 Units 450 Units $8.50

Western Less Knox More Knox 393 589 300 450 93 149

Chapter 03 Solution Manual

Rolled-up Forecast Average Price

Business Forecast ($)

$9,000 $8,400

Rolled-up Forecast ($)

Eastern 482 Units 450 Units $7.17

Western 803 Units 750 Units $6.90

Forced Forecast

Forced Forecast Initial Forecast Price

Less Knox 161 Units 150 Units $4.50

Item Level Final Forecast (Units)

More Knox 231 Units 300 Units $8.50

Less Knox 321 Units 300 Units $4.50

Eastern Less Knox More Knox 161 321

More Knox 482 Units 450 Units $8.50

Western Less Knox More Knox 321 482

17. GONZALES ELECTRIC COMPANY a. Family Forecast 250,000 Product Family Forecast Family Rolled-Up Family $ Forecast $ Forecast 1 50,000 46,000 2 50,000 64,000 3 75,000 65,000 4 75,000 80,000 250,000 255,000

3-21

Item Forecasts

Product A B C

Roll Up Unit Forecast 10 15 20

D E F

5 3 2

G H 1

100 180 220

J K L

2 4 3

b.

Roll Down Dollars Dollar Dollar Unit /Unit Forecast Forecast Forecast 1,000 10,000 10,870 11 1,200 18,000 19,565 16 900 18,000 19,565 22 46,000 50,000 5,000 25,000 19,531 4 7,000 21,000 16,406 2 9,000 18,000 14,063 2 64,000 50,000 250 25,000 28,846 115 100 18,000 20,769 208 100 22,000 25,385 254 65,000 75,000 10,000 20,000 18,750 2 9,000 36,000 33,750 4 8,000 24,000 22,500 3 80,000 75,000

Sales Goal 300,000

Family 1 2 3 4

Product Family Forecast Family Rolled-Up Rolled-Down $ Forecast $ Forecast $ Sales Goal 50,000 46,000 60,000 50,000 64,000 60,000 75,000 65,000 90,000 75,000 80,000 90,000 250,000 255,000 300,000 Item Forecasts

Product A B C D E F

Roll Up Unit Forecast 10 15 20 5 3 2

Dollars Dollar /Unit Forecast 1,000 10,000 1,200 18,000 900 18,000 46,000 5,000 25,000 7,000 21,000 9,000 18,000 64,000

3-22

Roll Down Dollar Unit Goal Goal 13,043 13 23,478 20 23,478 26 60,000 23,438 5 19,688 3 16,875 2 60,000

Chapter 03 Solution Manual

Product G H I

Roll Up Unit Forecast 100 180 220

J K L

2 4 3

Roll Down Dollars Dollar Dollar Unit /Unit Forecast Goal Goal 250 25,000 34,615 138 100 18,000 24,923 249 100 22,000 30,462 305 65,000 90,000 10,000 20,000 22,500 2 9,000 36,000 40,500 5 8,000 24,000 27,000 3 80,000 90,000

18. GONZALES ELECTRIC COMPANY (continued) a. a. Sales Goal 300,000 Family 3 Forecast 75,000

Family 1 2 3 4

Product A B C

Product Family Forecast Family Rolled-Up Rolled-Down $ Forecast $ Forecast $ Sales Goal 50,000 46,000 64,286 50,000 64,000 64,286 75,000 65,000 75,000 75,000 80,000 96.429 250,000 255,000 300,000

Unit Forecast 10 15 20

D E F

5 3 2

G H I

100 180 220

Item Forecasts Roll Up Roll Down Dollars Dollar Dollar /Unit Forecast Goal 1,000 10,000 13.975 1,200 18,000 25,155 900 18,000 25,155 46,000 64,286 5,000 25,000 25,112 7,000 21,000 21,094 9,000 18,000 18,080 64,000 64,286 250 25,000 28,846 100 18,000 20,769 100 22,000 25,385 65,000 75,000

3-23

Unit Goal 14 21 28 5 3 2 115 208 254

Product J K L

Unit Forecast 2 4 3

Roll Up Roll Down Dollars Dollar Dollar /Unit Forecast Goal 10,000 20,000 24,107 9,000 36,000 43,393 8,000 24,000 28,929 80,000 96,429

Unit Goal 2 5 4

b. Sales Goal 300,000 Family Forecast largest of family or sum of product forecasts

Family 1 2 3 4

Product Family Forecast Family Rolled-Up Rolled-Down $ Forecast $ Forecast $ Sales Goal 50,000 46,000 55,762 64,000 64,000 71,375 75,000 65,000 83,643 80,000 80,000 89,219 269,000 255,000 300,000 Item Forecasts

Product A B C

Roll Up Unit Forecast 10 15 20

D E F

5 3 2

G H I

100 180 220

J K L

2 4 3

Roll Down Dollars Dollar Dollar Unit /Unit Forecast Goal Goal 1,000 10,000 12,122 12 1,200 18,000 21,820 18 900 18,000 21,820 24 46,000 55,762 5,000 25,000 27,881 6 7,000 21,000 23,420 3 9,000 18,000 20,074 2 64,000 71,375 250 25,000 32,170 129 100 18,000 23,163 232 100 22,000 28,310 283 65,000 83,643 10,000 20,000 22,304 2 9,000 36,000 40,149 4 8,000 24,000 26,766 3 80,000 89,219

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Chapter 03 Solution Manual

c. Sales Goal 300,000 Family Forecast 350,000 (windfall of sales of J)

Family 1 2 3 4

Product A B C

Product Family Forecast Family Rolled-Up Rolled-Down $ Forecast $ Forecast $ Sales Goal 50,000 46,000 42,857 64,000 64,000 42,857 75,000 65,000 64,286 175,000 180,000 150,000 350,000 355,000 300,000

Unit Forecast 10 15 20

D E F

5 3 2

G H I

100 180 220

J K L

2 4 3

Item Forecasts Roll Up Roll Down Dollars Dollar Dollar Unit /Unit Forecast Goal Goal 1,000 10,000 9,317 9 1,200 18,000 16,770 14 900 18,000 16,770 19 46,000 42,857 5,000 25,000 16,741 3 7,000 21,000 14,063 2 9,000 18,000 12,054 1 64,000 42,857 250 25,000 24,725 99 100 18,000 17,802 178 100 22,000 21,758 218 65,000 64,286 10,000 120,000 100,000 10 9,000 36,000 30,000 3 8,000 24,000 20,000 3 180,000 150,000

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Chapter 04 Solution Manual

CHAPTER 4 Solution Manual Discussion Questions 1. Figure 4.1 makes this relatively clear. The consistency of the sales and operations plan assures top management communication with manufacturing. If manufacturing then complies with the operations plan, the sales and operations plan can be realized. Any changes made in the strategy of the firm must be passed on to manufacturing through the operations plan.

2. The student here should recognize that the terminology may be different at different levels in the organization. The common language at the top is likely to be students or dollars; in the computer center; computer hours; in buildings and grounds, rooms or seats; and for an individual major, probably credit hours. The student should also see that there is a transfer back and forth implied between these. A number of students implies a certain number of credit hours each term, etc.

3. Sales and operations planning is directed at the strategic linking of the firm's functions. Once the production levels to support those strategies have been defined, the financial part of the sales and operations planning process ought to result in the budgets necessary to accomplish the plans. This implies that when conditions suggest changing the sales and operations plans the budgets ought to change as well (or that flexible budgets be used).

4. The student here should think through carefully the merits of the argument that forcing a replanning on a periodic basis may bring to the attention of management alternatives that may not be included in the current plan and, therefore, not subject to exception conditions. In addition, to replan on the exception, means that there is an effective measurement system in place. On the other hand, if a periodic replanning is coming up soon, a needed replanning for taking care of an exception may be postponed because "we'll be doing it again shortly."

5. Some of the benefits of the tabular charts are the ease of arithmetic operations, the ease of incorporating accounting or management target data, explicit numerical statements of capacity limitations, and ease of computer adaptation. On the other hand, advocates of the cumulative chart point out its smoothing effects, its easy visualization, the ease of exploring feasible options and the clear representation of the interrelationship between marketing and manufacturing.

6. This is an open invitation to the informal system again. If there is not an explicit decision on the use of manufacturing capacity, manufacturing will resolve it for efficiency and marketing will desire inventory and flexibility. The resulting conflicts will be resolved informally at levels in the organization where the information to make intelligent trade-off decisions may not be available.

4-1

7. For the commodity type businesses the strategic implications are likely to be low costs, so high production efficiencies and low levels of flexibility would be incorporated in the sales and operations plan. In one sense this would mean that marketing would "lose" many of the sales and operations planning battles. Virtually the opposite would be true of the high fashion or cosmetic firm where quick reaction to changes in the marketplace are essential, implying high degrees of flexibility in manufacturing. In the production of an industrial product, combinations of customer service and costs are important and so a position someplace between the other two would be likely.

Chapter 4 - Problem Summaries Problem

Difficulty

Problem Type

1

Easy

This problem should be assigned as a starting point for the chapter.

The student can learn to analyze actual business performance against the sales and operations plan.

2

Moderate

This problem enables the student to analyze actual business performance against the plan, and to develop recommendations for changes in the sales and operations plan, using the S&OP spreadsheet framework.

3

Moderate

This problem continues on with problem 2. It is concerned with the analysis of sales forecast errors and their impact on the sales and operations plan, using the S&OP spreadsheet framework.

4

Easy

Problems 4-7 are linked together and should be assigned as a starting assignment for the chapter. The students learn about trade-offs, chase, level, overtime, hiring and laying off by doing these four problems. The first problem is to do a level plan and then create a cumulative chart.

5

Easy

This problem increases the forecast from problem 4 but still requires a level plan. It provides a look at the trade-offs between the different alternatives for satisfying demand.

6

Moderate

This continues on with problem 4 and has the student learn how to develop a level plan with no backorders.

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Chapter 04 Solution Manual

Chapter 4 - Problem Summaries Problem

Difficulty

Problem Type

7

Hard

This continues on with problem 4 and has the student do a basic chase plan. The second part has them develop a chase plan when they can not change the size of the work force and must use overtime, subcontracting, and part-time workers.

8

Easy

Has the student determine inventory and backorders of a given production plan.

9

Hard

Creating a no backorder level production plan but there are varying number of days per quarter. It is a little more difficult that the other problem but also more realistic.

10

Easy

Level production planning (with and without backorders) and cumulative charting.

11

Easy

Three period level plan with cumulative charting problem.

12

Moderate

Calculating a level production plan using cumulative charting. They need to convert the production plan into the correct number of employees to calculate labor cost.

4-3

13

Hard

This problem requires the determination of the daily production rate and the number of workers. Each quarter has a different number of working days.

14

Hard

This problem is concerned with the analysis of business performance against the sales and operations plan, and the development of changes to the sales and operations plan to improve company performance. Options concerning improvements in inventory performance, and employment levels are examined.

15

Hard

The problem concerns the analysis of business performance against the sales and operations plan, and the modification of the sales and operations plan to meet changed business conditions. Further, this problem examines the implications of pursuing different operations strategies, i.e. chase sales and level production strategies.

16

Moderate

Problem 16, 17, and 18 are linked together. This is a production planning problem to determine a chase plan and its cost.

17

Moderate

Uses the information from #16 but now the student must create a level plan.

18

Moderate

Expanded version of #16 with increased demand and no stockouts.

Student must complete both a level and a chase plan.

4-4

Chapter 04 Solution Manual

1. ELM COMPANY Completed Table

Elm Co. Sales and Operations Planning Spreadsheet Sales Forecast Actual Diff: Month Cumulative

(in Million $) (in units) (in units)

History October 0.80 800 826 26

Operations Plan

(in units) (in # employees) Number Working Days/Mo. Actual (in units) Diff: Month Cumulative

November December 0.85 0.90 850 900 851 949 1 49 27 76

800 6 23 798 -2 -2

800 8 19 802 +2 0

800 8 19 800 0 0

(in units) (in 000 $)

150 105

100 70

0 0

Actual (in units) Days of Supply

122

73 1.6

-76 -1.5

Inventory Plan

3.4

4-5

2. TRAPPER LAWN EQUIPMENT COMPANY Revised plan: Trapper Lawn Equipment Company Sales and Op's Planning Spreadsheet - Riding Mowers Product Group (Make-to-Stock)

Sales Forecast

(M$) (units) (units)

Actual Diff: Month Cumulative Operations Plan (units) (# employ) # Work Days/Mo. Actual (units) Diff: Month Cumulative Inventory Plan

History Oct 12.50 5000 4384 -616

Plan Nov 10.00 4000 3626 -374 -990

Dec 16.25 6500 6065 -435 -1425

Jan 5.00 2000

Feb 5.00 2000

Mar 7.50 3000

4000 70 19 4091 91 740 Target DOS Inv:

6500 114 19 7279 779 1519 5

0 0 20

556 9 21

3250 47 23

500

500

750

5000 72 23 5649 649

(units) (000$)

1270 2223

1270 2223

1270 2223

1944 3402

500 875

750 1313

Actual (units) Days of Supply

2265 10

2730 15

3944 13

19

5

5

a) Target inventory levels for the three months based on 5 days of supply: January = 5 x 2000 / 20 = 500; February = 5 x 2000 / 20 = 500; March = 5 x 3000 / 20 = 750 Planned build for each month required to achieve the target accounting for the forecast demand and the inventory in the previous period: Build plan = forecast demand + target inventory – previous month inventory January planned build is zero since 3944 units remain in inventory at the end of December.

February planed build = 2000 + 500 – 1944 = 556 March planned build = 3000 +750 – 500 = 3250

4-6

Chapter 04 Solution Manual

b) Qualitative factors: The plan indicates no production in January and very light production in February.

This could be implemented as a plant shutdown that may be very disruptive to work force moral and cause an employee retention problem.

It can also have quality and productivity issues as more problems are likely at shutdown and start-up. Key skills are not practiced.

A better alternative might be to maintain some production below customer demand and gradually reduce inventory levels.

Consider going to a 4 day or other form of shorten workweek. Restrict the use of overtime. Consider the use of a planned shutdown during the summer vacation season or force the use of accrued vacation time to reduce the number of workers available.

4-7

3. TRAPPER LAWN EQUIPMENT COMPANY REVISITED a) The average forecast error is calculated as the difference between the total forecast and actual demand divided by the total forecast. In this case, since the 3 month cumulative error is given in the table: Forecast error % = -1425 / (5000 + 4000 + 6500) x 100 = -9.2% Reducing each of the forecast values by 9.2% for January to June yields the projected values units sales and resulting inventory levels and days of supply shown in the table below.

Trapper Lawn Equipment Company Sales and Operations Planning Spreadsheet Riding Mowers Product Group (Make-to-Stock) History Plan Sales Oct Nov Dec Jan Forecast (M$) 12.50 10.00 16.25 5.00 (units) 5000 4000 6500 2000 Actual / Projected (units) 1816 4384 3626 6065 Diff: Month -616 -374 -435 Cumulative -990 -1425 Avg % Error -9.2% Operations Plan (units) 5000 4000 6500 2000 (# employ) 72 70 114 33 # Work Days/Mo. 23 19 19 20 Actual (units) 5649 4091 7279 Diff: Month 649 91 779 Cumulative 740 1519 Inventory Plan / Projected

(units) (000$)

Actual (units) Days of Supply / Projected

Target DOS Inv: 1270 1270 2223 2223 2265 10

2730 15

Feb 5.00 2000 1816

Mar 7.50 3000 2724

Apr 10.00 4000 3632

May 12.50 5000 4540

Jun 17.50 7000 6356

2000 32 21

3000 43 23

4000 67 20

5000 76 22

7000 106 22

5 1270 2223

500 4128 7224

500 4312 7546

750 4588 8028

1000 4956 8672

1250 5416 9476

1750 6060 10602

3944 13

46

50

39

27

26

21

b) Options for consideration Change the forecast. This would require the marketing and production mangers coming to agreement on what the new forecast should be.

Adjust the production plan to compensate for the fact that the forecast seems to have a relatively consistent negative bias. This option has little risk in the near term since inventory levels are relatively high.

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Chapter 04 Solution Manual

4. SKI & SEA, INC. a. Level Plan Aggregating the forecast Quarter Jet Skis Snowmobiles Total

1 10,000 9,000 19,000

2 15,000 7,000 22,000

3 16,000 19,000 35,000

4 3,000 10,000 13,000

Total 44,000 45,000 89,000

Determining the production rate: (Total forecast - beginning inventory) / 4 quarters (89,000 - 1,000) / 4 = 22,000 units per quarter The Plan and its costs: Quarter 1 2 3 Demand 19,000 22,000 35,000 Production 22,000 22,000 22,000 Beginning Inventory 1,000 4,000 4,000 Ending Inventory 4,000 4,000 0 Average Inventory* 2,500 4,000 2,000 Backorders 0 0 9,000 *(beginning inventory + ending inventory) / 2 Costs Regular time Inventory Backorders Total

$15.00  $ 3.00  $24.00 

Consequences: Low levels of inventory Substantial back order in quarter 3 b. Cumulative Chart

4-9

88,000 8,500 9,000

4 13,000 22,000 0 0 0 0

Total = $ 1,320,000 =$ 25,500 = $ 216,000 $ 1,561,500

Total 89,000 88,000

8500 9000

100,000

Cum. production and demand in units

Cumulative Forecast 80,000

Cumulative Output

60,000

40,000

20,000

Quarter 1

Quarter 2

Quarter 3

Quarters

c. Inventory Space = 20 cubic feet x 4000 = 80,000 cubic feet d. Investment = $ 600.00 x 4,000 = $ 2,400,000

4-10

Quarter 4

Chapter 04 Solution Manual

5. IVAR JORGENSON a. Overtime Quarter 1 2 3 4 Total Jet Skis 10,000 15,000 16,000 3,000 44,000 Snowmobiles 11,000 7,000 19,000 10,000 47,000 Total 21,000 22,000 35,000 13,000 91,000 Production rate = (91,000 - 1,000) / 4 = 22,500 units per quarter Quarter

1 2 3 Demand 21,000 22,000 35,000 Overtime 500 500 500 Regular 22,000 22,000 22,000 Output 22,500 22,500 22,500 Beginning Inventory 1,000 2,500 3,000 Ending Inventory 2,500 3,000 0 Average Inventory* 1,750 2,750 1,500 Backorders 0 0 9,500 *(beginning inventory + ending inventory) / 2 Costs Regular time Overtime Inventory Backorders Total b.

c.

$15.00  $22.50  $ 3.00  $24.00 

Subcontracting Subcontracting Cost Overtime Cost Net Increase/(Decrease) New Total Cost

Hiring a New Worker Hiring Regular Overtime Cost Net Increase/Decrease New Total Cost

4 13,000 500 22,000 22,500 0 0 0 0

91,000 2,000 88,000 90,000

6000 9500

88,000 2,000 6,000 9,500

Total = $ 1,320,000 =$ 45,000 =$ 18,000 = $ 228,000 $ 1,611,000

$ 30.00  $ 22.50 

2,000 2,000

=$ 60,000 =$ 45,000 $ 15,000 $ 1,626,000

$ 300.00  $ 15.00  22.5 

1 2,000 2000

=$ 300 =$ 30,000 =$ 45,000 $ (14,700) $ 1,596,300

4-11

6. SUSIE SVENSON a. No back order plan Production rate = (66,000 - 1,000) / 3 = 25,000 units per quarter Since each full-time employee produces 500 units/quarter, 50 employees are required.

Quarter

1 2 3 Demand 19,000 22,000 35,000 Production 25,000 25,000 25,000 Beginning Inventory 1,000 7,000 10,000 Ending Inventory 7,000 10,000 0 Average Inventory* 4,000 8,500 5,000 Backorders 0 0 0 *(beginning inventory + ending inventory) / 2

4 13,000 25,000 0 12,000 6,000 0

Slope is equal to 25,000 per quarter

Costs Regular time Hiring Inventory

$ 15.00  $ 300.00  $ 3.00 

b. 12,000 units

4-12

100,000 6 23,500

Total = $ 1,500,000 =$ 1,800 =$ 70,500 $ 1,572,300

Total 89,000 100,000

23,500 0

Chapter 04 Solution Manual

7. THOR LEDGER a.

Quarter

1 2 3 4 Demand 19,000 22,000 35,000 13,000 Production 18,000 22,000 35,000 13,000 Beginning Inventory 1,000 0 0 0 Ending Inventory 0 0 0 0 Average Inventory* 500 0 0 0 Backorders 0 0 0 0 Workers 36 44 70 26 Hire 0 8 26 0 Layoff 0 0 0 44 *Average Inventory = (Beginning Inv. + Ending Inv.)/2

Costs Regular time Inventory Hiring Layoff

$ 15.00  $ 3.00  $ 300.00  $ 1,500.00 

4-13

88,000 500 34 44

= $ 1,320,000 =$ 1,500 =$ 10,200 =$ 66,000 $ 1,397,700

Total 89,000 88,000

500 0 176 34 44

b.

Quarter 1 2 3 4 Demand 19,000 22,000 35,000 13,000 Regular 18,000 18,000 18,000 13,000 Overtime 0 4,000 7,200 Subcontract 0 0 5,000 Part-time 0 0 4,800 Production 18,000 22,000 35,000 13,000 Beginning Inventory 1,000 0 0 0 Ending Inventory 0 0 0 0 Average Inventory* 500 0 0 0 Backorders 0 0 0 0 Workers 36 36 36 26 Layoff 0 0 0 10 *Average Inventory = (Beginning Inv. + Ending Inv.)/2

Costs Regular time Inventory Overtime Subcontracting Part-time Firing

$ 15.00  $ 3.00  $ 22.50  $ 30.00  $ 36.00  $ 1,500.00 

67,000 500 11,200 5,000 4,800 10

= $ 1,005,000 =$ 1,500 = $ 252,000 = $ 150,000 = $ 172,800 =$ 15,000 $ 1,596,300

c. Hiring and Layoff actions are shown on the charts in (a) and (b)

4-14

Total 89,000 67,000 11,200 5,000 4,800 88,000

500 0 134 10

Chapter 04 Solution Manual

8. LOVELL BRADLEY

Period: 1 2 3 Beginning Inventory 400 200 0 Production 8,000 8,000 8,000 Demand 8,200 10,000 7,000 Ending Inventory 200 0 0 Average Inventory* 300 100 0 Backorders 0 1,800 800 *Average Inventory = (Beginning Inv. + Ending Inv.)/2

4 0 7,600 6,600 200 100 0

9. THE N.L. HYER CO. a. Daily production = 210. Note: Need to build to third quarter peak. Cumulative forecast = 38,590 - 1,000 (inv.) = 37,590. 37,590/179 days = 210 units /day. Number of employees required = 21.

b. The number of units in inventory at the end of quarter IV will be 5,700.

Quarter: Beginning Inventory Production Demand Ending Inventory Backorders Employees Required *21 = (12,600/10) / 60

1 1,000 12,600 12,000 1,600 0 *21

4-15

2 1,600 12,180 13,000 780 0 21

3

4

780 12,810 13,590 0 0 21

0 14,700 9,000 5,700 0 21

10. JOAN'S JOYOUS NATURE FOOD a. Joan should produce 135 units each month. [(120 + 160 - 10)/2 = 135]

Cum. production and demand in units

600

500

Cumulative Output

400

300

Cumulative Demand

200

100

Month 1

Month 2

Month 3

Month 4

Months

b. The ending inventory for month 4 is 180 units. [(10 + (4  135) - 370) = 180] c. Joan should produce 90 units each month. [(120 + 160 + 20 + 70 - 10) / 4 = 90]

d.

Month: 1 2 Beginning Inventory 10 0 Production 90 90 Demand 120 160 Ending inventory 0 0 Average inventory 5 0 Carrying cost $25 $0 Backorders (cumulative) 20 90 Backorder cost $160 $720 Total Inventory Cost = $5  5 = $25 Total Backorder Cost = $8  130 = $1040

4-16

3 0 90 20 0 0 $0 20 $160

4 0 90 70 0 0 $0 0 $0

Chapter 04 Solution Manual

11. ORO DEL MAR CO. a.

Cum. demand and prod. in 1,000 pounds

400

Cumulative Output

300

200

Cumulative Demand

100

January

February

March

Months

b. A production rate of 100 units per month is required in order to avoid backorders and result in no ending inventory in March. [(100 + 300 - 100) / 3]

4-17

12. CRAZY TUBB THUMPERS a.

Cum. demand and prod. in 1,000 units

40

Cumulative Output

30

Ending Inventory (5,000 units)

20

10

Cumulative Demand January

February

March

Months

b.

Total monthly production Regular time production (9 Emps. x 1,000 units) Overtime production (10,000 units - 9,000 units)

10,000 units 9,000 units 1,000 units

Regular time labor cost (9,000 units x $1.20*) Overtime labor cost (1,000 units x $1.80**) Total monthly labor cost *Regular time labor rate / unit = $1,200 / 1,000 units = $1.20 **Overtime labor rate / unit = $1.20 x 1.5 = $1.80

4-18

$10,800 1,800 $12,600

Chapter 04 Solution Manual

13. THE BI-PRODUCT CO. a.

Quarter: Beginning Inventory (Prod. A) Demand (Prod. A) Beginning Inventory (Prod. B) Demand (Prod. B)

1 2,400 9,800 900 14,500

2

3

4

12,000

14,000

31,000

30,000

19,500

25,000

 Demand = (9,800 + 14,500) + (12,000 + 30,000) + (14,000 + 19,500) + (31,000 + 25,000) = 155,800 - Beginning Inventory (2,400 + 900) = 152,500  days = 68 + 56 + 62 + 58 = 244 Production / day = 152,500/244 = 625. b. 625 units/day => 25 employees Quarter Total Beginning Inventory* Total Production Total Demand Total Ending Inventory Total Backorders Employees Required *Total of product A and product B.

1 3,300 42,500 24,300 21,500 0 25

4-19

2 21,500 35,000 42,000 14,500 0 25

3 14,500 38,750 33,500 19,750 0 25

4 19,750 36,250 56,000 0 0 25

14 CUBBY COMPRESSORS COMPANY a) Target inventory levels based on 5 days of supply: 5 x 900 / 20 = 225 Planned build for each month required to achieve the target accounting for the forecast demand and the inventory in the previous period: Build plan = forecast demand + target inventory – previous month inventory January planed build = 900 + 225 – (-52) = 1,177 February planed build = 900 + 225 – 225 = 900 March planned build = 900 +225 – 225 = 900, etc.

Revised plan: Cubby Sales and Operations Planning Spreadsheet Sales Forecast Actual Diff: Month Cumulative

History Oct 0.56 800 826 26

(in Million $) (in units) (in units)

Operations Plan (in units) (in # employees) Number Working Days/Mo. Actual (in units) Diff: Month Cumulative

800 34 23 758 -42

Nov 0.60 850 851 1 27

Plan Dec Jan 0.63 0.63 900 900 949 49 76

800 42 19 842 42 0

800 42 19 824 24 24 5

Target DOS Inv: Inventory Plan

(in units) (in 000 $)

Actual (in units) Days of Supply /Projected Line Fill / Projected

150 74

100 49

82 2.0 100%

73 1.7 100%

Feb 0.63 900

Mar 0.63 900

Apr 0.63 900

May 0.63 900

Jun 0.63 900

1177 59 20

900 43 21

900 39 23

900 45 20

900 41 22

900 41 22

225

225

225

225

225

225

225 110

225 110

225 110

225 110

225 110

225 110

-52 5.0 -1.1 95% 100%

5.0 100%

5.0 100%

5.0 100%

5.0 100%

5.0 100%

0 0

4-20

Chapter 04 Solution Manual

b) The 45-employee “hiring freeze” creates a constraint on how quickly the inventory can be built up to the targeted level. The 5-day target is not achieved until May. Since each employee can only build the equivalent of one unit per workday*, the capacity for 45 employees in a 20day month is exactly the 900-units/month constant demand forecast. Inventory above demand is accumulated on months with more than 20 workdays. The build plan is determined as follows: a. The required build including existing inventory or backlog and the target days of buffer inventory: January build requirement = 900 +225 – (-52) = 1,177 b. The available capacity for the month: January max available capacity = 20 x 45 = 900 c. If the max available capacity < build requirement, then build to max available capacity.

If not then build to requirement. d. Repeat the process for each month. * Note: The output per worker used for planning purposes is not explicitly stated in the problem, but can be calculated from any of the given table monthly columns: Output/worker day = planned output / (# employees x # work days) The results for this problem are shown in the following table:

4-21

Revised plan: Cubby Sales and Operations Planning Spreadsheet (45 employee max) Sales Forecast

(M$) (units) (units)

Actual Diff: Month Cumulative Operations Plan (units) (# employ) # Work Days/Mo. Actual (units) Diff: Month Cumulative Inventory Plan

(units) (000$)

Actual (units) Days of Supply / Projected

History Oct 0.56 800 826 26

800 34 23 758 -42

Feb 0.63 900

Mar 0.63 900

Apr 0.63 900

May 0.63 900

Jun 0.63 900

800 42 19 824 24 24

900 45 20

945 45 21

1035 45 23

900 45 20

997 45 22

900 40 22

5 0 0

225 -52 0

225 -7 0

225 128 63

225 128 63

225 225 110

225 225 110

-52 -1.1

-1.2

-0.2

2.8

2.8

5.0

5.0

Dec 0.63 900 949 49 76

800 42 19 842 42 0

Target DOS Inv: 150 100 74 49 82 2.0

Plan Jan 0.63 900

Nov 0.60 850 851 1 27

73 1.7

c) Options for improving customer service with the hiring freeze: •

Use overtime.

•

Increase the workweek by adding additional hours, days (i.e. weekends) or shifts (i.e. split, 2nd, or 3rd).

•

Use temporary workers or subcontract work.

•

Improve forecast accuracy.

•

Reduce published availability or advertising in selected regions or markets (demand management).

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Chapter 04 Solution Manual

15. ABC CONSUMER ELECTRONIC COMPANY In this case, the level production schedule is simply recalculated using the 10 day of supply target in month 3: 1. Total build required for forecast = (2.1+2.3+2.5) x 1,000,000 = 6,900,000 2. Inv target for Sept at 10 DOS = 10 x 2.5/30 x 1,000,000 = 833,333 3. Added inventory required from current = 833,333 – 250,001 = 583,333 4. Total 3 month build = 6,900,000 + 583,333 = 7,483,333 5. Build per work day = 7,483,333 / (10 + 21 +23) = 138,580 6. Implied constant work force = 138,580/1000 = 138.58 →139 7. Planned monthly builds based on 139 employees and given work days: July build plan = 139 x 10 x 1000 = 1,390,000 August = 139 x 21 x 1000 = 2,919,000 September = 139 x 23 x 1000 = 3,197,000 Completed spreadsheet: ABC Sales and Operations Planning Spreadsheet Sales Forecast

History Apr (M$) $22.50 (units) 1.50 (units) 0.78 -0.72

Plan May $25.50 1.70 1.95 0.25 -0.47

Jun $28.50 1.90 1.63 -0.27 -0.74

Jul $31.50 2.10

Aug $34.50 2.30

Sep $37.50 2.50

1.50 75 20 1.51 0.01

1.70 77 22 1.70 0.00 0.01

1.90 86 22 1.88 -0.02 -0.01

1.39 139 10

2.92 139 21

3.20 139 23

(units) (000$)

250000 $2,625

250000 $2,625

250000 $2,625

Actual (units) Days of Supply

250001 9.6

250000 3.9

250001 4.6

Actual Diff: Month Cumulative Operations Plan (units) (# employ) # Work Days/Mo. Actual (units) Diff: Month Cumulative Inventory Plan

-459999 159001 856001 $0 $1,669,508 $8,988,008

-6.6

4-23

2.1

10.3

Sum 6900000

7506000 54 250001 Beg inv 833333 Tgt inv for DOS in Sept 583333 Additional inv required 7483333 3 month total build 138580 Build per day 139 Implied constant labor

16. GENERAL AVIONICS

Quarter 1 2 3 4

Sales 5,000 10,000 8,000 2,000

Production 4,000 10,000 8,000 2,000

Workforce 40 100 80 20

Cost Item Inventory Carrying Cost Overtime Cost Firing Cost (100 x $400) Hiring Cost (60 x $200) Regular Payroll Cost (240 x $1,200) Total Cost

Ending Inventory 0 0 0 0 Cost $

0 0 40,000 12,000 288,000 $340,000

17. GENERAL AVIONICS REVISITED

Quarter 1 2 3 4

Sales 5,000 10,000 8,000 2,000 25,000

Production 6,000 6,000 6,000 6,000 24,000

Workforce 60 60 60 60 240

Cost Item Inventory Carrying Cost Overtime Cost .

Firing Cost ...

Hiring Cost Regular Payroll Cost (240 x $1,200) Stockout Cost ($5 x $6,000) .

Total Cost .

4-24

Stockouts 0 2,000 4,000 0 6,000

Ending Inventory 2,000 0 0 0 2,000

Cost $ 4,000 0 0 0 288,000 30,000 $322,000

Overtime Production 0 0 0 0 0

Chapter 04 Solution Manual

18. GENERAL AVIONICS AGAIN Chase Sales Plan Quarter 2 3 4

Ending Overtime Sales Production Workforce Inventory Production 8,000 7,000 70 1,000 6,400 6,400 64 1,000 1,600 1,600 16 1,000 16,000 15,000 150 3,000

Cost Item Inventory Carrying Cost (3000 x $2) Overtime Cost Firing Cost (54 x $400) Hiring Cost (20 x $200) Regular Payroll Cost (150 x $1,200) Total Cost

0 0 0 0

Cost $ 6,000 0 21,600 4,000 180,000 $211,600

Level Production Plan Quarter 2 3 4

Sales 8,000 6,400 1,600 16,000

Production 7,000 6,400 5, 000 18,400

Workforce 50 50 50 150

Ending Inventory 1,000 1,000 3,400 5,400

Cost Item Inventory Carrying Cost ($2 x 5,400) Overtime Cost ($14* x 3,400) Firing Cost Hiring Cost Regular Payroll Cost(150 x $1,200) Total Cost *$14 = $12 for regular + $2 overtime premium

4-25

Cost $ 10,800 47,600 0 0 180,000 $238,400

Overtime Production 2,000 1,400 0 3,400

Chapter 04 A Solution Manual

CHAPTER 4A Solution Manual Discussion Questions 1. Some students may be tempted to suggest the forecasts are the least certain of the data required, but our vote is for the shortage costs. There may be some idea of the direct costs of a back order, in terms of the extra handling, paperwork, and so forth, but the opportunity costs of service failures are very tough to measure. The use of what-if analysis can help to overcome this uncertainty.

2. Technically the mixed integer formulation adds constraints, computer time and "elegance." In terms of utility it allows one to take into account whole persons, whole units of product and fixed costs like setups costs. In some cases it may not be necessary to use integer formulations. When the coefficients in the linear programming formulation provide integer multiples, the LP solution will provide integer answers.

3. There are three areas where managers should be concerned. The first of these relates to providing the appropriate data, the second the solution procedure itself, and the third in the disaggregating of the results. The data problem is self evident (see also discussion question 1).

The procedure may provide a false sense of accuracy to people and/or allow them to get out of "owning" the solution. Finally, the formulation of the model may require aggregations that are difficult to disaggregate in the shop 4. The argument that we make here is that data accuracy, implementation of the formal system dictates and appropriate loop closing feedback all depend upon an effective MPC system.

Many firms have tried some of the techniques described in this chapter and have abandoned them after a short period of time. They were never incorporated into an effective formal system where they could become a part of the fabric of the daily life on the factory floor.

Until there is an effective formal MPC system, it is hard to imagine any of these techniques living beyond the interest of the champion.

4A-1

Chapter 04 A Solution Manual

Chapter 4A - Problem Summaries Problem

Difficulty

Problem Type

1

Moderate

Production planning using linear programming. The student must formulate the problem as an LP then solve it.

2

Easy

Calculating the total cost of a production plan using spread sheets.

3

Moderate

Calculating the total cost of a production plan and then formulating it as a linear programming problem.

4

Moderate

Calculating the total cost of two production plans using spread sheets. Examines the trade-off between level production and a fixed employment plan with overtime.

5

Moderate

Calculating the total cost of a production plan with limited capacity and subcontracting using spreadsheets.

6

Moderate

Developing and incremental cost expression for a production planning problem and the formulating as a linear program. Then calculating the cost of constant production plan and a chase plan.

7

Hard

Formulate a production plan using linear programming and then formulating using the linear decision rule.

8

Easy

Assessing inventory carrying cost and the labor budget of a level production plan.

4A-2

Chapter 04 A Solution Manual

1. SEYMORE BICYCLE LP Formulation: Let: Wi = Employment level in Quarter i Hi = Number of hires in Quarter i Fi = Number of fires in Quarter i Xi = Regular time production in Quarter i Oi = Overtime production in Quarter i Ii = Inventory at the end of Quarter i Si = Stockouts during Quarter i

i = 0 to 4 i = 1 to 4 i = 1 to 4 i = 1 to 4 i = 1 to 4 i = 0 to 4 i = 1 to 4

4

Minimize  ( 2  Ii + 200  Hi + 400  Fi + 12  O i + 1,200  Wi + 5  Si ) i =1

Subject to: X1 + O1 + I0 - I1 + S1 = 5000 X2 + O2 + I1 -I2 + S2 = 10,000 X3 + O3 + I2 - I3 + S3 = 8,000 X4 + O4 + I3 - I4 + S4 = 2,000 W0 + H1 - F1 - W1 = 0 Wl + H2 - F2 - W2 = 0 W2 + H3 - F3 - W3 = 0 W3 + H4 - F4 - W4 = 0 100*W1 - X1 = 0 100*W2 - X2 = 0 100*W3 - X3 = 0 100*W4 - X4 = 0 40*W1 - O1  0 40*W2 - O2  0 40*W3 - O3  0 40*W4 - O4  0 I0 = 1,000 W0 = 60 Note: All variables are greater than or equal to zero and Wi is integer.

We hope your students can see very quickly that it is much better to incur stockouts than to produce with the problem as stated. The entire plan consists of firing workers and taking the stockouts. The answers are: a. No production. b. $144,000 (fire 60 and 24,000 short) If stockouts aren't allowed, a very different picture emerges as, seen on the next page.

4A-3

Chapter 04 A Solution Manual

Quarter 0 1 2 3 4

No Stockouts Allowed: Production Closing Regular Overtime Inventory 1,000 6,000 0 2,000 6,000 2,000 0 6,000 2,000 0 2,000 0 0 20,000 4,000

Demand 5,000 10,000 8,000 2,000 25,000

Budget Regular Payroll: 200  $1,200/quarter Overtime: 4,000  $14 Inventory carrying cost: 2,000  $2 Stockout Cost: Firing Cost: 40  $400 Total Budget

= = = = =

Number Employees 60 60 60 60 20 200

$240,000 56,000 4,000 0 16,000 $316,000

2. KOSAR MANUFACTURING a. Level Production Plan

Quarter 0 1 2 3 4 Average

Demand

Production

3,000 2,250 2,000 2,700 2,487.5

2,400 2,400 2,400 2,400

Ending Inventory 350 -250 -100 300 0

b. Total Cost: = (4) (300) ($15) + ($5) (300) + ($10) (350) = 18,000 + 1,500 + 3,500 = $23,000

4A-4

Regular Production

Extra Production

2,100 2,100 2,100 2,100

300 300 300 300

Chapter 04 A Solution Manual

3. KEW TOY COMPANY a.

Quarter 0 1 2 3 4 Average

Sales Forecast -50 80 120 150 100

Level Production Plan Ending Production Inventory -0 100 50 100 70 100 50 100 0

Total Cost Hiring Cost: Firing Cost: Inventory Cost ($1)(50 + 70 + 50):

Quarter 0 1 2 3 4 Average

= = =

Chase Sales Production Plan Sales Ending Forecast Production Inventory --0 50 50 0 80 80 0 120 120 0 150 150 0 100 Total Cost Hiring Cost: ($10)(3 + 4 + 3) Firing Cost: ($5)(5) Inventory Cost:

= = =

Employment 10 10 10 10 10

$0 0 170 $170

Employment 10 5 8 12 15

$100 25 0 $125

b. Total annual cost for matching sales plan = $125. c. Of the two policies, level production and matching sales, the latter plan is the lowest cost.

The production rate for each quarter for this plan is: Quarter 1 = 50, Quarter 2 = 80, Quarter 3 = 120, and Quarter 4 = 150.

4A-5

Chapter 04 A Solution Manual

d. LP formulation Let: L1 = Employment level increase in Quarter 1 L2 = Employment level increase in Quarter 2 L3 = Employment level increase in Quarter 3 L4 = Employment level increase in Quarter 4 D1 = Employment level decrease in Quarter 1 D2 = Employment level decrease in Quarter 2 D3 = Employment level decrease in Quarter 3 D4 = Employment level decrease in Quarter 4 X1 = Amount of production in Quarter 1 X2 = Amount of production in Quarter 2 X3 = Amount of production in Quarter 3 X4 = Amount of production in Quarter 4 Y1 = Ending inventory in Quarter 1 Y2 = Ending inventory in Quarter 2 Y3 = Ending inventory in Quarter 3 Y4 = Ending inventory in Quarter 4 E0 = Employment level in Quarter 0 Minimize: 1  (Y1 + Y2 + Y3) + 10  (L1 + L2 + L3 + L4) + 5  (D1 + D2 + D3 + D4) Subject to: X1 + Y0 - Y1 = 50 X2 + Y1 - Y2 = 80 X3 + Y2 - Y3 = 120 X4 + Y3 - Y4 = 150 Y0 = 0 Y4 = 0 E0 = 10 E0 - 1/10X1 + L1 - D1 = 0 1/10 X1 - 1/10 X2 + L2 - D2 = 0 1/10 X2 - 1/10 X3 + L3 - D3 = 0 1/10 X3 - 1/10 X4 + L4 - D4 = 0 Xi, Yi, Ei, Di > 0

4A-6

Chapter 04 A Solution Manual

4 . MISCHIEF COMPANY Fixed Employment Plan Demand December January February March April

200 220 200 220

Total Cost Regular Time: Overtime: Hiring / Firing: Inventory Cost:

Employ.

20 20 20 20 20

Prod.

Overtime

200 200 200 200

0 20 0 20

($2,000)(20)(4): = ($3,000)(20/10)(2): =

Ending Inv. 0 0 0 0 0

$160,000 12,000 0 0 $172,000

=

Level Production Plan Demand December January February March April Total Cost Regular Time: Overtime: Hiring / Firing: Inventory Cost:

200 220 200 220

Employ.

20 21 21 21 21

Prod.

Overtime

210 210 210 210

0 0 0 0

($2,000)(21)(4): = = ($1,200)(1): ($150)([10 + 0 + 10 + 0]/2 ): =

4A-7

Ending Inv. 0 10 0 10 0

$168,000 0 1,200 1,500 $170,700

Chapter 04 A Solution Manual

5. FAIRMOUNT COMPANY a. Level Production Plan Period 0 1 2 3 4 Average

Demand

Production

Subcontract

6,000 4,500 6,000 5,100 5,400

4,600 4,600 4,600 4,600

200 300 1000 500

Costs Production Change Cost Inventory Cost Subcontracting Total

Ending Inventory 1,200 0 400 0 0

600  $20  4 = 400  $4 = 2000  $18 =

$48,000 1,600 36,000 $85,600

b. Chase Production Plan Period 0 1 2 3 4

Demand

Production

Subcontract

6,000 4,500 6,000 5,100

4,600 4,500 4,600 4,600

200 0 1000 500

Shortage

400

Ending Inventory 1,200 0 0 0 0

No! Since Maximum Production = 4,600 you can not cover the demand in period 3 6. ABC COMPANY a. Total Cost Expression TC/yr. = $2 (X1 + X2 + X3 + X4) + $3 (Y1 + Y2 + Y3 + Y4) + $2 (I1 + I2 + I3 + I4) + $1,600 (H1 + H2 + H3 + H4 + F1 + F2 + F3 + F4) + $2 (S1 + S2 + S3 + S4) Where: X1 = Regular production in Quarter 1 Y1 = Overtime production in Quarter 1 I1 = Ending inventory in Quarter 1 H1 = No. hired in Quarter 1 F1 = No. fired in Quarter 1 S1 = Slack regular time production in Quarter 1 E1 = Number of employees in Quarter 1 (Etc.) 4A-8

Chapter 04 A Solution Manual

b. Minimize: 2  (X1 + X2 + X3 + X4) + 3  (Y1 + Y2 + Y3 + Y4) + 2  (I1 + I2 + I3 + I4) + 2  (S1 + S2 + S3 + S4) + 1,600  (H1 + H2 + H3 + H4 + F1 + F2 + F3 + F4) Subject To: I0 + X1 + Y1 - I1 = 9,000 I1 + X2 + Y2 -I2 = 12,000 I2 + X3 + Y3 -I3 = 16,000 I3 + X4 + Y4 -I4 = 12,000 E0 = 12 1,000 E1 - X1 - S1 = 0 1,000 E2 - X2 - S2 = 0 1,000 E3 - X3 - S3 = 0 1,000 E4 - X4 - S4 = 0 E0 - E1 + H1 - F1 = 0 E1 - E2 + H2 - F2 = 0 E2 - E3 + H3 - F3 = E3 - E4 + H4 - F4 = 0 I4  1,000 I0 = 1,000 Y1 - 250 E1  0 Y2 - 250 E2  0 Y3 - 250 E3  0 Y4 - 250 E4  0 c. Constant production rate: 49,000/4 = 12,250 units / quarter Quarter Sales Regular Production O. T. Production Beginning Inventory Ending Inventory

1 9,000 12,000 250 1,000 4,250

2 12,000 12,000 250 4,250 4,500

4A-9

3 16,000 12,000 250 4,500 750

4 12,000 12,000 250 750 1,000

Totals 49,000 48,000 1,000 10,500

Chapter 04 A Solution Manual

Costs: Regular Production O.T. Production Inventory Total Cost

48,000  $2 1,000  $3 10,500  $2

$96,000 $3,000 $21,000 $120,000

d Production rate equal to sales Quarter Sales Regular Production Unused Production Employees O. T. Production Beginning Inventory Ending Inventory

1 9,000 9,000 3,000 12 0 1,000 1,000

Costs: Regular Production Unused Production O.T. Production Hire / Fire Inventory Total Cost

2 12,000 12,000 0 12 0 1,000 1,000

3 16,000 14,000 0 13 2,000 1,000 1,000

47,000  $2 3,000  $2 2,000  $3 2  $1,600 4,000  $2

4 12,000 12,000 0 12 0 1,000 1,000

$94,000 $6,000 $6,000 $3,200 $8,000 $117,200

A production rate equal to sales lowers costs by $2,800.

7. BOSTON PAINT COMPANY a. Linear Programming Model Assume 2,400/600 = 4 units / person / quarter Let: X1 = Regular production during Quarter 1 Y1 = Overtime production during Quarter 1 I1 = Ending inventory Quarter 1 H1 = Number hired in Quarter 1 F1 = Number fired in Quarter 1 E1 = Number of employees in Quarter 1 S1 = Slack regular production during Quarter 1 (Etc.) Minimize: 20*(I1 + I2 + I3 + I4) + 200*(H1 + H2 + H3 + H4 + F1 + F2 + F3 + F4) + 320*(X1 + X2 + X3 + X4) + 380(Y1 + Y2 + Y3 + Y4) 4A-10

Totals 49,000 47,000 3,000 2 2,000 4,000

Chapter 04 A Solution Manual

Subject To: I0 + X1 + Y1 - I1 = 3,000 I1 + X2 + Y2 -I2 = 1,800 I2 + X3 + Y3 -I3 = 2,400 I3 + X4 + Y4 -I4 = 3,500 I4  100 I0 = 300 E0 = 600 4 E1 - X1 - S1 = 0 4 E2 - X2 - S2 = 0 4 E3 - X3 - S3 = 0 4 E4 - X4 - S4 = 0 E0 - E1 + H1 - F1 = 0 E1 - E2 + H2 - F2 = 0 E2 - E3 + H3 - F3 = 0 E3 - E4 + H4 - F4 = 0 b. Linear Decision Rule Model Formulation Minimize: 200(E1-600)2 + 200(E2-E1)2 + 200(E3-E2) 2+ 200(E4-E3)2 + 60(X1-4E1)2 + 60(X2-4E2)2 + 60(X3-4E3)2 + 60(X4-4E4)2 + 320(X1 + X2 + X3 + X4) + 20(I1 + I2 + I3 + I4) Subject To: I0 = 300 I0 + X1 - I1 = 3,000 I1 + X2 -I2 = 1,800 I2 + X3 -I3 = 2,400 I3 + X4 -I4 = 3,500 I4  100 X1 - 4E1 = 0 X2 -4 E2 = 0 X3 -4 E3 = 0 X4 -4 E4 = 0 (all Xi, Ei, Ii, > 0)

4A-11

Chapter 04 A Solution Manual

The linear programming model in Part a will not solve the revised production planning problem with the new employment and over/under time costs. The revised objective function and constraints shown above must be considered in the formulation of a linear decision rule model.

4A-12

Chapter 04 A Solution Manual

8. PICKWICK COMPANY a. 950 + 1,200 + 1,420 + 1,630 / 4 = 1,300 units b. 200 + (1,300 -950) + (1,300-1,200) = 650 units 650  $500 = $325,000 c. Quarterly labor budget in dollars = 1,300 units X $200 = $260,000.

Quarterly labor budget in hours = 260,00  5 = 52,000.

4A-13

Chapter 05 Solution Manual

CHAPTER 5 Solution Manual Discussion Questions 1. It's important here that the students recognize that the master production schedule really tells the factory what to build. This may be in anticipation of later sales, it may be to replace inventory of already sold items or it could be contracted requirements for another firm. The trade-offs for these decisions have already been made by the time the MPS is put into effect.

2. Clearly the key to expanding output is to expand capacity. To increase the MPS without simultaneous adjustments in the capacity to meet the production plan, and therefore the company game plan is to severely unbalance the company's activities.

The result is likely to be a return to the informal system.

3. This is a difficult question designed to make a student think about the differences in the three environments. For the make-to-stock environment, the MPS needs to resupply the inventory to meet customer service levels. This means balancing the production capacity, MPS stability and uncertainty in demand. In the make-to-order environment, the MPS must provide the ability to track individual customer order progress and to promise delivery for new orders. This might also involve providing the flexibility to adapt to the changes that take place as the product is designed. In the assemble-to-order environment, the MPS provides the basic materials from which the final product is assembled to the customer's specifications. This means the MPS must provide the flexibility to accommodate variations in the specific product mix.

4. The students must assess the utility for getting bad news honestly. Is it better for them to be told that their laundry will be ready on Thursday when it actually won't be ready until Saturday, or to be told ahead of time that, in fact, it won't be ready until Saturday? Which is the appropriate stance for a company to take?

5. The master production scheduler uses the information in the available row to indicate when he or she ought to consider scheduling more production to meet a future need (as determined from the forecast). On the other hand, once the master production quantities have been determined they provide the basis for actually promising deliveries to customers through the available to promise row. Thus the available to promise row provides the future product that is promised to customers as their orders come in.

6. If the forecast is an unbiased forecast of future demand, anytime a larger quantity than the forecast is used to reduce the availability in the future, an earlier or greater MPS quantity could be scheduled. This would lead to an overstatement which would not be reduced until the quantity is netted off against actual customer orders. This convention leads to a form of additional buffering in the system.

5-1

Chapter 05 Solution Manual

7. The role of the master production scheduler in a good MPS system is indeed a critical one. The master production scheduler does make a good number of the trade-offs that are essential in the marketing / manufacturing interface and can affect both the service levels and economics dramatically. The qualities to look for in a person, then, are perhaps more on the managerial and judgmental side (for making the trade-offs) than on the technical side.

8. Firm planned orders do not get changed by the computer, so changes are in the hands of the master production scheduler. The master production scheduler can, therefore, look at other ways of solving problems than moving orders around. This can provide a great deal of stability and predictability on the shop floor, thereby building increased confidence that, when changes are made, they are needed.

Chapter 5 Problem Summaries Problem Difficulty

Problem Type

1

Easy

Using time phased MPS records and revising the requirements.

Need to revise record once actual demand is given.

2

Moderate

Time phased MPS, EOQ, and capacity analysis. This problem integrates many aspects of the MPC system.

3

Moderate

Processing MPS time phased records. The problem also covers making order promising decisions and then updating the MPS time phased record.

4

Easy

Linking MPS and MRP record processing together for a product.

Note that this requires knowledge of MRP explosion, which is covered later in the book.

5

Easy

MPS record processing with order promising. Students have to update the record after accepting an order.

6

Hard

Processing an MPS and rolling through time after actions occur during the first week.

7

Moderate

8

Easy

This problem requires building the MPS time phased record and then examines cumulative lead times and time fences.

9

Moderate

This problem requires building the MPS time phased record and

Creating a superbill and discussing the advantages and disadvantages of using them.

5-2

Chapter 05 Solution Manual

Chapter 5 Problem Summaries Problem Difficulty

Problem Type then examines time fences. It also looks at order promising.

10

Hard

Looks at using planning bills for forecasting. It also examines order promising using the MPS records for each option.

1. EXCELSIOR SPRINGS, LTD. a. MPS record

Forecast Projected Available Balance MPS On Hand = 30

Quarter 2 3 70 70 20 30 80 80

1 20 10

4 20 10

b. Inventory balances

Quarter 1 2 3 4

Closing Inventory Balances 10 20 30 10

c. Updated record

2 30 80

Quarter 3 50 30

4 70 40 80

Forecast Projected Available Balance MPS On Hand = 110 *Assuming that the 80 MPS in quarter 2 came in.

The order for 80 MPS units in period 3 was shifted to period 4.

5-3

Chapter 05 Solution Manual

2. NEPTUNE MANUFACTURING COMPANY a.

Item: FunRay Series Forecast Projected Available Balance MPS

Item: SunRay Series Forecast Projected Available Balance MPS Item: StingRay Series Forecast Projected Available Balance MPS

15,000

3,000

5,000

Q1 8,000 7,000

Q2 9,000 3,000 5,000

Q3 6,000 2,000 5,000

Q4 6,000 1,000 5,000

Q1 4,000 2,000 3,000

Q2 5,000 0 3,000

Q3 2,000 1,000 3,000

Q4 2,000 2,000 3,000

Q1 9,000 0 4,000

Q2 10,000 0 10,000

Q3 6,000 0 6,000

Q4 7,000 0 7,000

Maximum Capacity 15,000 15,000 15,000 15,000 Capacity Required 7,000 18,000 14,000 15,000 Capacity Available 8,000 -3,000 1,000 0 Note: problem with capacity in Q2 could be resolved by moving 3000 units of StingRay to Q1. b.

Product Line FunRay Series SunRay Series StingRay Series

Carrying Cost $10 $15 $20

Forecast 29,000 13,000 32,000

5-4

Set up Cost $5,000 $5,000 $5,000

EOQ 5,385 2,944 4,000

Chapter 05 Solution Manual

3. MURPHY MOTORS a.

Item: P24 Week

1

2

3

4

5

6

7

8

Forecast Orders Projected Available Balance 20 Available to Promise MPS MPS lot size = 70

30 13

30 8

30 4

40

40

40

45

45

60 69 70

30

70 66 70

30

60 70 70

20

45 70 70

70 70 70

b.

Order 1 2 3 4

Amount 40 30 30 25

Desired Week 4 6 2 3

Accept (Y or N) Y Y Y Y

Updated Record Item: P24 Week

1

2

3

4

5

6

7

8

Forecast Orders Projected Available Balance 20 Available to Promise MPS MPS lot size = 70

30 13

30 38

30 29

40 40

40

40 30

45

45

60 39 70

30

70 1 70

30

60 40 70

20

45 70 70

70 70 70

5-5

Chapter 05 Solution Manual

4. THE SPENCER OPTICS COMPANY a.

Week

1

2

3

4

5

6

7

8

9

10

Forecast Orders Projected Available Balance 140 Available to Promise MPS MPS lot size = 300

100 110

100 80

100 50

100 20

100

100

100

100

100

100

30 30

230 150 300

130

30

230 300 300

130

30

230 300 300

130

30

5. NINO SPIRELLI a.

Week

1

2

3

4

5

6

Forecast Orders Projected Available Balance Available to Promise MPS MPS lot size = 50

20 14 35 33 50

30 8 5

20 6 35 44 50

30

20

30

5

35 50 50

55 50 50

5

b. No Problems. c. He can deliver 33 immediately and 11 in week 3 or all 44 in week 3. d.

Week

1

2

3

4

5

6

Forecast Orders Projected Available Balance Available to Promise MPS MPS lot size = 50

20 14 35 33 50

30 8 5

20 6 35 29 50

30 15 5

20

30

35 50 50

55 50 50

5

5-6

Chapter 05 Solution Manual

6. GEORGIA CLAY AND GRAVEL a.

Smell Fresh Week

1

2

3

4

5

6

7

8

Forecast Orders Projected Available Balance Available to Promise MPS MPS lot size = 50

20 5 50 60 50

20 3 30

20 2 10

30

30

30

30

30

30 50 50

50 50 50

20

40 50 50

10

2

3

4

5

6

7

8

9

20

b.

Smell Fresh Week

Forecast 20 20 30 30 30 30 30 40 Orders 13 8 20 Projected Available Balance 45 25 5 25 45 15 35 5 15 Available to Promise 24 30 50 50 50 MPS 50 50 50 50 On-hand = 45 = 20 period 1 on-hand + 50 MPS - 25 period 1 demand, MPS lot size = 50

5-7

Chapter 05 Solution Manual

7. FALCON SPORTS INC. a.

Common Parts 600 400

Horsepower

10 602 410

Seating Style

12 601 610

Single 315 360

Starter

Dual 320 350

Manual 235 250

Auto 230 254

b.

Super Jet Ski

10 HP 602, 410 Usage = .5

12 HP 601, 610 Usage = .5

Single Seat 315, 360 Usage=.40

Dual Seat 320, 350 Usage=.60

Manual Starter 235, 250 Usage=.25

Auto Starter 230, 254 Usage=.75

Common 600, 400 Usage=1.0

c. Super Bill Approach Advantages 1. Easy to incorporate safety stock.

2. Easy to compute available to promise.

Disadvantages 1. Need separate bills of material for each option.

5-8

Chapter 05 Solution Manual

8. THE ACE ELECTRONICS COMPANY a. MPS Record Item - #2400 Forecast Orders Projected Available Balance Available To Promise MPS MPS lot size = 200

0

1 100 95 100 0 200

2 100 105 -5

3 100 70 95 130 200

4 100

5 100 10 95 190 200

-5

6 100

7 100

8 100

-5

95 200 200

-5

b. If the cumulative lead time for item #2400 was 8 weeks, the master scheduler should look at the available in week 9. Since it is negative, he/she should release a new MPS quantity. That is, there is an implied time fence at the end of week 8.

9. BRANDY BOARDS a.

Sound Xapper Week

1

2

3

4

5

6

7

Forecast Orders Projected Available Balance Available to Promise MPS MPS lot size = 150

50 60 10 0

50 70 90 0 150

50 20 40

50 70 -30

50 40 70 80 150

50 30 20

50

70

-30

b. Issue an MPS order for 150 units in week 7. c. Can not accept the first two orders but can accept the order for 20 units in week 6.

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Chapter 05 Solution Manual

10. THE PARKER CORPORATION a. Forecasting Demand Using the Current Approach Catalog Number Body H.P. Drive T 3 C T 3 G T 4 C T 4 G T 5 C T 5 G O 3 C O 3 G O 4 C O 4 G O 5 C O 5 C B 3 C B 3 G B 4 C B 4 G B 5 C B 5 G

Sales Forecast (Current) (20% Increase)* -* -* 1 1.1 4 4.4 18 19.8 17 18.7 -* -* 13 14.3 9 9.9 10 11.0 8 8.8 -* -* 6 6.6 7 7.7 2 2.2 5 5.5 100 110 + 3 HP *10% of the sales increase (10 units) is to be made up of the 3 H.P. machines.

Using The Planning Bill Approach Projected Planning Option Sales Incr. (%) Forecast Bill (%) T-Body 40 20 48 40 O-Body..

40 20 48 40 G-Body 20 20 24 20 3 H.P. --10* 8 4 H P. 40 10 44 37 5H P. 60 10 66 55 120 Chain 50 20 60 50 Gear 50 20 60 50 Common Parts 100 20 120 100 *10% of growth is to be accounted for by the new 3 H.P. machines thus resulting in only a 10% increase for the 4 H.P. and 5. H.P. machines.

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Chapter 05 Solution Manual

b.

Common "Kit" Week

1

2

3

4

5

6

7

8

9

10

11

Forecast Orders Projected Available Balance ATP MPS MPS lot size = 5

3 3

2 3

3 2

2 1

3 1

2 1

3

2

3

2

3

2 -1 5

-1

1 2 5

-1

1 3 5

-1

1 5 5

-1

1 5 5

-1

-4

Week

1

2

3

4

5

6

7

8

9

10

11

Forecast Orders Projected Available Balance ATP MPS MPS lot size = 5

2 1

1

2 1

1 1

2 1

1

1

1

2

1

2

8 7

7

5

4

7 4 5

6

5

4

2

1

-1

Week

1

2

3

4

5

6

7

8

9

10

11

Forecast Orders Projected Available Balance ATP MPS MPS lot size = 5

2 2

0 2

1 1

1

1

1 1

1

1

1

1

1

1 -1

-1

3 3 5

2

1

0

-1

3 5 5

2

1

0

0

5 H.P. Motor

10

4 H.P. Motor

3

*The convention used in rounding is to round up, but carry forward the cumulative round-up amount so that the cumulative round-up is never greater than 1. E.g., the 5 H.P. option in week 1 is (3 x .55) = 1.65 and 2 units are shown in the forecast. In week 2, (2 x .55) = 1.1, but .35 round-up units have been carried forward. The forecast is shown as 1 and the cumulative round-up is reduced to .25.

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Chapter 05 Solution Manual

3 H.P. Motor Week

1

Forecast Orders Projected Available Balance ATP MPS MPS lot size = 0

1

2

3

4

5

6

7

8

9

10

1

11 1

1 0

4 4 5

3

3

3

2

2

2

2

2

2

1

Week

1

2

3

4

5

6

7

8

9

10

11

Forecast Orders Projected Available Balance ATP MPS MPS lot size = 10

2 1

1 1

1 2

1

2

1

1

1

2

1

1

12 10

11

9

9

6

5

4

3

1

0

-1

Week

1

2

3

4

5

6

7

8

9

10

11

Forecast Orders Projected Available Balance ATP MPS MPS lot size = 10

2 2

1 2

1

1 1

2 1

1 1

1

1

2

1

1

4 2

2

11 7 10

10

8

7

6

5

3

2

1

Chain Drive

14

Gear Drive

6

Delivery Promises: T3G-- Promise for delivery in week 3 since a common parts set is unavailable until then.

Note that one of the two available will be needed to cover the "over-promise" of one 4 H.P. model in the first two weeks. That exhausts the available-to-promise for week 3.

04C--Promise to deliver in week 5 since a common parts set is unavailable until then after T3G is promised and the over promise of 4 H.P. model is taken care of.

5-12

Chapter 06 Solution Manual

CHAPTER 6 Solution Manual Discussion Questions 1. The analogy here is somewhat like a rear-wheel drive car. The MRP system takes its direction from the "front end" and provides the detailed plans (power) to the "back end." This is somewhat similar to the engine applying the power to the rear wheels in order to reach the direction set by the front wheels.

2. Many students like to have information on planned receipts and net requirements shown in the records until they get familiar with them. These two lines do add to ease of understanding of the records.

3. The bill of material is most often developed for making a single unit of something (in later chapters we will see variations on the bill of material theme). A recipe, on the other hand, often provides the ingredients for making a quantity of something, e.g., 4 servings. If the unit is considered a "dish" then the two are very similar. The bill of material shows level by level product buildup where the instructions in a recipe provide this information, e.g., mix flour and milk in one bowl and beat egg whites stiff in another; when completed fold egg whites into the flour and milk mixture.

4. The student should be able to express the linkages between records, like that in Figure 2.8, which communicate changes in one record to other records. This communication provides the coordination between individual items. The back schedule results from the explosion process itself. The end item due dates are used with the lead times to indicate start times which become due dates for subsequent items. This is equivalent to specifying the latest start as indicated in Figure 2.7.

5. Several examples should come to the student's mind. In the classroom, for example, the number of seats that the registrar assumes may not correspond to the number of seats in the room due to "borrowing" for other classes. The consequence is clear -- there may be excess seats in one room and insufficient seats in another which could lead to additional "borrowing." Students move and don't report their new addresses or they sit in on a different section of a course than that for which they are registered. The consequences are they may not be easily found in the event of an emergency.

6. Several examples of information decay are provided in the chapter, but the students should be pressed to think of others and their consequences. For example, customers may cancel orders which, if not communicated to the factory floor, would result in using manufacturing capacity to produce parts that may not be needed. If a raw material delivery is late, speedy processing of the information may provide opportunities to find ways of meeting the original customer due date. The generalization that the students should be able to come to is that the later the processing of information, the less likely it is that the factory is working on what is really required at the moment. This is what gives rise to the informal system.

6-1

Chapter 06 Solution Manual

7. It may be hard for students to think of other examples, but the shopping list and paid for bag of groceries is a common one. The tentative plans for the classes to be taken to graduate versus enrollment in a course is another, and the idea of holding options open on a major until the last possible moment (i.e., commitment to specific classes required for one and not for another) may, be helpful examples. In a shop, the separation is important because the scheduled receipts have already had resources committed to them while the plans remain just that. In the academic environment, the process of dropping courses (sometimes at a cost to the student) should help the student see that a scheduled receipt is different and more costly to change than a planned one (planned courses).

8. The obvious implication of this is that the material will be used for something else, and when it comes time to build the shop order, the material won't be available. The inventory records will overstate the availability of inventory to make additional units, which means that the informal system on the shop floor will have to decide how to allocate the insufficient inventory to the competing products.

9. There are several examples of student transactions: registration for classes, posting fees and tuition, posting grades, processing drop/adds, etc. It is fairly obvious for each of these what happens if it is not done well. Your students should be able to generate many examples (and horror stories) once they get started.

Chapter 6 - Problem Summaries Problem

Difficulty

Problem Type

1

Easy

Level by level processing and gross to netting process using the BOM without using MRP time phased records.

2

Moderate

Drawing product structure trees from assembly instructions, applying low level codes and level by level processing to determine gross requirements

3

Hard

Drawing a product structures tree including low level coding and then creating Gantt Charts to show front and back scheduling logic.

4

Easy

Gross to netting process using a word description of the BOM. This problem can be a little tricky because the BOM is for six servings. This problem also first to require the use of time phased records.

5

Moderate

Using a spreadsheet and time phased records to do the gross to netting process and analyze the impact of safety stock and safety lead time on the timing of orders and average inventory.

6-2

Chapter 06 Solution Manual

Chapter 6 - Problem Summaries Problem

Difficulty

Problem Type

6

Moderate

Multiple usage of a common component in a single end item. This problem requires the student to understand that gross requirements can come from more that one parent in the same product.

7

Moderate

This problem extends the concepts from problem 6 to include gross requirements from parents in two different end items

8

Hard

This problem builds on the concepts in problem 7 of two end items with common components and includes the MPS links to the end items

9

Easy

This problem analyzes the impact of inventory availability on product delivery.

10

Easy

Impact of fixed lot sizes versus lot sizes that take into account the information collected by the MRP system.

11

Moderate

This product examines the impact of firm planned orders on inventory planning. It also touches on the concept of action messages.

12

Easy

This problem examines the impact of inaccurate inventory information for a product on factory operations.

13

Moderate

Updating a single product’s MRP record after several transactions have occurred and a week has transpired.

14

Hard

Updating a product’s MRP record after several transactions have occurred and a week has transpired.

15

Hard

Replanning the MRP records for two items after there were errors in the inventory counts. It also rolls the MRP records forward one period after actual demand was greater than planned.

6-3

Chapter 06 Solution Manual

1. JOE’S BURGERS a. Product mix - hamburger meat Percent Number of burgers Meat per burger (lbs.)

Total meat (lbs.)

Big Burger 20 40 1/4 10

Giant Burger 45 90 1/2 45

Football Burger Total 35 100 70 200 2 140

Giant Burger 90 2 180

Football Burger Total 70 200 4 280 500

Giant Burger 90

Football Burger 70

90

70

Total meat per day = 195 pounds b. Pickle slices Number of Burgers Slices per Burger Total Slices

Big Burger 40 1 40

Total pickle slices per day = 500 c. Buns - no finished goods Number of Burgers Regular Buns Sesame Buns

Big Burger 40 40

Regular Buns Sesame Buns

Required

On-Hand

Net Order

80 320

25 20

55 300

Bun Inventory 25 20

Big 10

Total 200 40 160

d. Buns - finished goods

Regular Buns Sesame Buns

2-day Requirement 80 320

6-4

Finished Good Giant Football 5

15

Net Order 45 280

Chapter 06 Solution Manual

2. P301 COMPUTER a.

P301 Computer

P Processor Unit

V Video Unit

M Memory Board (3 Req.)

C Ram Chip (4 Req.)

S Switch

A Arithmetic Board (2 Req.)

B Box Casing

X Board

K Keyboard Unit

Z MicroProcessor

S Switch (4 Req.)

R Rom Chip (2 Req.)

S Switch

b., c.

Item Low-Level Code A - Arithmetic Board 2 B - Box Casing 2 C - Ram Chip 3 K - Keyboard Unit 1 M - Memory Board 2 P - Processor Unit 1 R - ROM Chip 3 S - Switch 3 V - Video Unit 1 X - Board Type X 3 Y - Board Type Y 3 Z - Microprocessor 3 *Assuming no inventory for any item

6-5

Quantity Required for One Unit* 2 1 12 1 3 1 4 9 1 3 2 2

Y Board

Chapter 06 Solution Manual

3. PETER’S POWER TOOLS (PPT) a.

Saw

A (2 required)

C (3 required)

B

E (3 required)

D (2 required)

D

F (3 required)

E (2 required)

D (2 required)

b.

FRONT LOAD SCHEDULE E D F A C B SAW

1

2

3

4

5 Week Number

6-6

6

7

8

9

Chapter 06 Solution Manual

c.

BACK LOAD SCHEDULE F D E B C A SAW

1

2

3

4

5

6

7

8

9

Week Number

The front schedule logic causes all the work to be done ahead of schedule which creates additional inventory and occupies resources that could be used on other orders. MRP applies back schedule logic so that parts and components are completed when they are needed to fill customer orders.

6-7

Chapter 06 Solution Manual

4. MARTHA'S TRIPLE CHOCOLATE SMOOTHIES To make 6 Triple Chocolate Smoothies, we need 2 dashes of cinnamon, 1.5 liters of vanilla ice cream, 2 liters of chocolate ice cream, ½ liter of chocolate milk and 1 bag of chocolate chips.

Triple Chocolate Smoothies (6 Drinks)

Cinnamon (4 dashes)

Vanilla Ice Cream (1 1/2 liters)

Chocolate Ice Cream (2 liters)

Chocolate Chips (1 Bag)

Chocolate Milk (1/2 liters)

a. Chocolate Ice Cream Chocolate Ice Cream required: (2 @ 2 liters) Less Chocolate Ice Cream on hand:

4 liters - 0 liters

Chocolate Ice Cream to purchase:

4 liters

b. Vanilla Ice Cream Vanilla Ice Cream required: (2 @ 1 1/2 liters) Less Vanilla Ice Cream on hand:

3 liters 1 liters

Vanilla Ice Cream to purchase:

2 liters

c. Chocolate Milk November Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 1, LT = 1, SS = 0

1/2

6-8

11

12

13

14

15 1

1/2

1/2

1/2

1/2 1

1/2

Chapter 06 Solution Manual

5. MRP SPREADSHEET a. No safety stock Period Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 50, LT = 1, SS = 0

1 20 50 32

2

2 20

3 40

4 30

5 30

6 30

12 50

22 50

42

12 50

32

There are three planned orders and they are in weeks 2, 3, and 5. Average projected inventory is 25.33 units. b. Safety Stock of 15 units Period Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 50, LT = 1, SS = 15

1 20 50 32 50

2

2 20

3 40

4 30

5 30

6 30

62

22 50

42 50

62

32

There are three planned orders and they are in weeks 1, 3, and 4. Two of planned orders were moved up because of the safety stock requirement. Average projected inventory is 42 units. c. Safety Lead Time of 1 week Period Gross Requirements Scheduled Receipts Projected Available Balance 2 Planned Order Release Q = 50, LT = 1, SS = 0, SLT = 1 week

1 20 50 32 50

2 20

3 40

4 30

5 30

6 30

62 50

72

42 50

62

32

There are three planned orders and they are in weeks 1, 2, and 4. All of the planned orders were moved up by 1 week to account for the safety lead time. Average projected inventory is 50.33 units.

Using either safety stocks or safety lead time causes the system to hold more inventory and order earlier. The safety lead time causes all orders to be held an extra week in inventory, which increases inventory more than the safety stock.

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Chapter 06 Solution Manual

6. MULTI-LEVEL ITEM EXAMPLE A

B (2 Req.)

C

C

Part: A Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 20, LT = 1, SS = 0

21

1 5

2 15

3 18

4 8

5 12

6 22

16

1 20

3 20

15

3 20

1

1

2 40

3 40

4

5 40

6

32 52 40

12

12 40

12

12

12

1 40

2 20

3 60

4

5 20

6

30

10 60

10

10 20

10

10

Part: B Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 40, LT = 2, SS = 0

20

Part: C Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = LFL, LT = 1, SS =10

70

6-10

Chapter 06 Solution Manual

7. TRACI’S TOMAHAWK Product Structures A

G (2 required)

C

F

D

G

H

Part A Week Gross Requirement Scheduled Receipt Projected Available Balance Planned Order Release Lead Time = 1, Q = L4L, SS = 0

10

1 5 5 10

2 3

3 10

4 15

5 15

7 3

0 15

0 15

0

1 10

2 20 15 0 15

3 25

4 15

5 5

0 5

0

0

2 21

3 35

4 30

5

24 40

29 20

19

19

Part F Week Gross Requirement Scheduled Receipt Projected Available Balance Planned Order Release Lead Time = 2, Q = L4L, SS = 0

15

5 25

Part G Week Gross Requirement Scheduled Receipt Projected Available Balance 30 Planned Order Release Lead Time = 1, Q = multiples of 20, SS = 10

1 25 20 25 20

An order for 20 units must be planned for Part G in period 1 so that safety stock is maintained in period 3.

6-11

Chapter 06 Solution Manual

8. Ajax Manufacturing A

X

B

Y

G

H

Y (2 Req.)

Z

a.

Part: G Period Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = LFL, LT = 1, SS = 0

0

1 8 10 2 6

2 8

3 8

4 8

5 8

6 8

7 8

0 8

0 8

0 8

0 8

0 8

0

1 32 10 45 27

2 36

3 36

4 36

5 36

6 36

7 36

9 36

0 36

0 36

0 36

0

0

1 32 10 45 36

2 36

3 36

4 36

5 36

6 36

7 36

10 36

10 36

10 36

10 36

10

10

Part: Y Period Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = LFL, LT = 2, SS = 0

67

b.

Part: Y Period Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = LFL, LT = 2, SS = 10

67 1

There will be a past due order of 1 that must be expedited and then the first order would increase to 36. The other option would be to order 37 in period 1 which would resolve the safety stock problem in period 3 instead of period 2.

6-12

Chapter 06 Solution Manual

9. PRODUCT A

Item A B C D

Inventory 10 40 60 60

Limit of A's by Item a b c 10 10 10 20 40 40 30 30 25 60 40 60

Start 10 40 30 60

The table above provides the basis for answering the three parts of the question. There are two difficulties that students are likely to have with this problem. The first is that you must add the restriction caused by A itself plus the minimum from B, C, or D. The second is they will forget that it takes 2 Cs to make an A. The answers are: a. 30 units of A (10 A’s in inventory + 20 A’s can be built) b. 40 units of A (10 A’s in inventory + 30 A’s can be built) c. 37 units of A (10 A’s in inventory + 27 A’s can be built) 10. XYZ Manufacturing Item A Lot Size = 150

Average Inventory = 79.1

Period

1

2

3

4

5

6

7

8

9

10

Gross Requirements

71

46

49

55

52

47

51

48

56

51

79

33

134

79

27

130

79

31

125

74

0

150

0

0

150

0

0

150

0

0

Scheduled Receipts Proj. Available Bal.

150

Planned Order Rel. Q = 150, LT = 1, SS = 0 Item B

Lot Size = 150 Average Inventory = 69. 5

Period

1

2

3

4

5

6

7

8

9

10

Gross Requirements

77

83

90

22

10

10

16

19

27

79

73

140

50

28

18

8

142

123

96

17

150

0

0

0

0

150

0

0

0

0

Scheduled Receipts Proj. Available Bal. Planned Order Rel. Q = 150, LT = 1, SS = 0

150

6-13

Chapter 06 Solution Manual

Item A

Lot Size = 3 week supply, Average Inventory = 46.9

Period

1

2

3

4

5

6

7

8

9

10

Gross Requirements

71

46

49

55

52

47

51

48

56

51

79

33

107

52

0

99

48

0

51

0

0

123

0

0

146

0

0

107

0

0

Scheduled Receipts Proj. Available Bal.

150

Planned Order Rel. Q = 3 weeks, LT = 1, SS = 0

Barf Lot Size = 3 week supply, Average Inventory = 51.4 Period

1

2

3

4

5

6

7

8

9

10

Gross Requirements

77

83

90

22

10

10

16

19

27

79

73

112

22

0

26

16

0

106

79

0

122

0

0

36

0

0

125

0

0

176

Scheduled Receipts Proj. Available Bal.

150

Planned Order Rel. Q = 3 weeks, LT = 1, SS = 0

The requirements for item A have substantially less variation than those for item B. Still, there is little difference between the average inventories compared to the difference between the fixed lot sizes of 150 units and the 3-week-supply lot size. However, the elimination of the "residual" inventories (having a zero balance every three periods) with the 3 week supply, means that the average inventory is reduced to about 60% of that for the fixed lot size.

11. a.) Any transaction that causes projected available balance at the end of period 2 to go below 5; e.g. any demand over 15 in period one, any scrap loss over 5 in the scheduled receipt, or any error in inventory reducing it below 5. Any transaction that causes the projected available balance at end of period 2 to go above 24; e.g. any discovered extra inventory, reduced demand or extra scheduled receipts combination that exceeds 14 units. b.) Any transaction that causes projected available balance at the end of period 4 to go below 5; e.g. any demand over 10 in period one, any scrap loss on the scheduled receipt of 40, or any error in inventory reducing it below 10 Any transaction that causes the projected available balance at end of period 4 to go above 24; e.g. any discovered extra inventory, reduced demand or extra scheduled receipts combination that exceeds 19 units.

The action message when the inventory drops below 5 would be to move the order earlier or else there will be a stockout. If the inventory is greater than 24 then the message would tell them to release the order later.

6-14

Chapter 06 Solution Manual

12.

Without recording the scrap, the MRP record will appear as it does in the problem. If the scrap had been reported, it would be: Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = LFL, LT = 5, SS = 0

1 25 35

10

2 30 35 15

3 5

4 15

10

-5

5 5 15 5

6 10 -5

The 5 unit shortage may not be discovered until week 4. At that time, it will be necessary to expedite the scheduled receipt of 15 from week 5 into week 4, and cover the deficiency in week 6.

13.

Toaster Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 50, LT = 2, SS = 5

10

1 20 50 40 50

2 20

3 35

4 25

5 35

6 35

7 35

8 35

20

35 50

10 50

25

40 50

5

20

Toaster Week 2 3 4 5 6 7 8 9 Gross Requirements 20 35 25 35 35 35 35 40 Scheduled Receipts 50 Projected Available Balance 57* 37 52 27 42 7 22 37 47 Planned Order Release 50 50 50 50 Q = 50, LT = 2, SS = 5 *57 = 10 (Previous) + 45 (Receipt) + 7 (Adjustment) - 5 (Disbursement) **To get the required on-hand inventory at the beginning of week 2 need to release an order for 50 at the beginning of week 1.

6-15

Chapter 06 Solution Manual

14. ABC MANUFACTURING COMPANY a.

Component C Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 40, LT = 2, SS = 0

5

1 15 40 30

2 30

3 25

4 25

5 10

6 20

0 40

15 40

30

20

0

2 55

3 0 40 5

4 15

5 10

6 20

7 15

b.

Component C Gross Requirements Scheduled Receipts Projected Available Balance 20* -35 30 20 0 Planned Order Release 40 40 Q = 40, LT = 2, SS = 0 *20 = 5 (Previous) + 30 (Receipt) - 5 (Adjustment) - 10 (Disbursement)

25

The inventory planner should "expedite" the scheduled receipt from week 3 to be delivered in week 2.

6-16

Chapter 06 Solution Manual

l5. ITEMS A AND B Note: Tell your students that the lead time for part B is 1 period a. There is one planned order for Item B in period 2 Item A Period Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = Lot-for-lot, LT = 1, SS = 10

68

1 50

2 50

3 50

4 50

5 50

6 50

18 42

10 50

10 50

10 50

10 50

10

1 84 250 174 0

2 100

3 100

4 100

5 100

6

74 250

224 0

124 0

24 0

24

Item B Period Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 250, LT = 1, SS = 0

8

b. A new planned order for Item B is generated in period 4.

Item A Period Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = Lot-for-lot, LT = 1, SS = 10

55

1 50

2 50

3 50

4 50

5 50

6 50

5 55

10 50

10 50

10 50

10 50

10

1 110 250 148 0

2 100

3 100

4 100

5 100

6

48 250

198 0

98 250

248 0

248

Item B Period Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 250, LT = 1, SS = 0

8

6-17

Chapter 06 Solution Manual

c. Items A & B With Extra Sales. The planned order of 50 Item A’s in period 2 is now 60.

Item A Period

2

3

4

5

6

7

8

9

10

11

Gross Requirements

50

50

50

50

50

50

50

50

50

50

Scheduled Receipts

42 0

10

10

10

10

10

10

10

10

10

60

50

50

50

50

50

50

50

50

Period

2

3

4

5

6

7

8

9

10

11

Gross Requirements

120

100

100

100

100

100

100

100

100

0

54

204

104

4

154

54

204

104

4

4

Proj. Available Bal.

8

Planned Order Rel. Q= LFL, LT = 1, SS = 10 Item B

Scheduled Receipts Proj. Available Bal. Planned Order Rel. Q = 250, LT = 1, SS = 10

174

250

250

6-18

250

Chapter 06A Solution Manual

CHAPTER 6A Solution Manual Discussion Questions 1. Since the discrete procedures order a number of periods of requirements, the order size can change if the requirements change. This means that an order that may have been started for 100 units last week may want to be changed to 110 units this week. This, of course, is sometimes impossible. On the other hand, fixed order quantities have only their priorities changed, and this is easier to manage on the shop floor.

2. Experiments have shown that as uncertainty increases the difference between the procedures decreases. This can be seen by comparing experiment 2 with experiments 1 and 3 in Figure 6A.6. The absolute value of the cost goes up as uncertainty increases. This is because the amount of safety stock and/or service penalties are increasing.

3. Each of the procedures illustrated assumes that an effective time-phased requirement schedule is available. This can only be true if the purchase requirements are derived from the MRP system and not from some additional or external forecast of purchased requirements. Thus the techniques can only be used after an MPC system is in place.

4. If the due date is not changed, the result is simply inflated lead-time. Any due date oriented priority rule will simply focus on the unchanged due date and will not complete the order with any safety lead-time. The result of not changing the due date will be excess work in process inventories and no additional buffering.

5. This clearly runs counter to the whole thrust of the MPC system whose job it is to reduce organizational slack. Some of the difficulties with using the slack as a buffer revolve around management and control. It is difficult to measure the cost and quantity of some forms of organizational slack and so it is more difficult to manage and control. In contrast, safety lead time and safety stock can be explicitly measured and managed. A second difficulty is the continual one that reintroduction of the organizational slack tends to support the informal system to solve problems.

6. The primary merit of the suggestion is that the foreman must use the formal system to make his or her decision. The priority information communicated at the time of need would be the only information available on which to base the decision. On the other hand, by not providing at least some look ahead and planning information to the foreman, some opportunities for foreman discretion in making economic combinations might be overlooked.

6A-1

Chapter 06A Solution Manual

Chapter 6A - Problem Summaries Problem

Difficulty

Problem Type

1

Easy

Comparing EOQ and POQ on a single level single item problem using time phased requirements.

2

Easy

Comparing L4L and EOQ on a single level single item problem using time phased requirements.

3

Easy

Comparing EOQ and PPB on a single level single item problem using time phased requirements.

4

Moderate

Comparing EOQ, PPB, POQ based on carrying cost, average inventory, and setup cost.

5

Moderate

Using part period balancing on a single level single item problem using time phased requirements.

6

Moderate

Comparing part period balancing to LUC on an item with a purchase discount.

7

Hard

8

Moderate

Using POQ and PPB in multilevel MRP records and examining the impact of changes in the MPS.

9

Moderate

Spreadsheet comparison of EOQ and PPB on a single level single item problem using time phased requirements.

10

Moderate

Analyzing the impact of alternative ordering plans on two levels of MRP records and then has the student devise their own ordering plan.

11

Easy

Comparing LUC, LPC on an item that has a purchase discount available.

Using a spreadsheet to determine the ordering pattern using LUC with and without a purchase discount available. Does not allow requirements to be split.

6A-2

Chapter 06A Solution Manual

Chapter 6A - Problem Summaries Problem

Difficulty

Problem Type

12

Moderate

Impact of rolling through time on lower level requirements This problem demonstrates how the orders change with Wagner-Whitin depending on the amount of demand information available. It also show the impact on lower level requirements.

13

Easy

Using safety lead time in the ordering policy to adjust for the delivery reliability of the supplier.

14

Moderate

Adjusting lot sizes for scrap rates. It requires the student to make two adjustments: (1) for minimum scrap by increasing the lot size and (2) for the variation in scrap by adding a safety stock.

15

Moderate

This problem demonstrates the nervousness impact of changing requirements on lot sizes. It show how a lower level item can become past due by reduced requirements at a higher level.

1. COMPARISON OF EOQ AND POQ a. Annual usage = 52 x 12 = 624 units/year

EOQ =

2  12  45 = 33 units 1

Orders/year = 624  33 = 19 Orders b.

EOQ (33 units): Week Number Requirements Order Quantity Ending Inventory

1 15 33 18

2 2

3 10

16

6

4 12 33 27

POQ (33  12 = 2.75 weeks)  3 weeks Week Number 1 2 3 Requirements 15 2 10 Order Quantity 27 Ending Inventory 12 10 0

4 12 18 6

6A-3

5 6

6 0

21

21

5 6

6 0

0

0

7 14 7 7

8 9 33 31

7 14 23 9

8 9 0

Chapter 06A Solution Manual

EOQ $180.00 147.00 $327.00

Ordering Cost Inventory Carrying Cost Total Cost

6A-4

POQ $135.00 37.00 $172.00

Chapter 06A Solution Manual

2. COMPARISON OF LOT-FOR-LOT AND POQ LOT SIZING a. Lot-for-Lot Item: A Week Gross Requirements Scheduled Receipts Projected Avail. Bal. Planned Order Release LT = 1 SS = 0

3

1 1

2 4

3 2

4 8

5 1

6 0

7 6

8 2

9 1

10 3

2 2

0 2

0 8

0 1

0

0 6

0 2

0 1

0 3

0

1 2

2 2

3 8

4 1

5

6 6

7 2

8 1

9 3

10

2 8

0 1

0

0 6

0 2

0 1

0 3

0

0

0

1 1

2 4

3 2

4 8

5 1

6 0

7 6

8 2

9 1

10 3

2 4

2

0 9

1

0

0 8

2

0 4

3

0

1 4

2

3 9

4

5

6 8

7

8 4

9

10

0 9

0

0

0 8

0

0 4

0

0

0

0

Item: B Week Gross Requirements Scheduled Receipts Projected Avail. Bal. Planned Order Release LT = 2 SS = 0

4

b. POQ Item: A Week Gross Requirements Scheduled Receipts Projected Avail. Bal. Planned Order Release LT = 1, SS = 0

3

Item: B Week Gross Requirements Scheduled Receipts Projected Avail. Bal. Planned Order Release LT = 2, SS = 0

4

6A-5

Chapter 06A Solution Manual

3. COMPARISON OF EOQ AND PART PERIOD BALANCING

2  113  300 = 184 units 2

a. Total demand = 1355, Average demand = 113 EOQ = Economic Order Quantity Month Number 1 2 Requirements 100 10 Order Quantity 184 Beg. Inventory 184 84 End. Inventory 84 74

Part Period Balancing Month Number 1 Requirements 100 Order Quantity 145 Beg. Inventory 145 End. Inventory 45

3 15

4 20

74 59

59 39

2 10

3 15

4 20

45 35

35 20

20 0

5 70 184 223 153

6 250 184 337 87

7 250 184 271 21

8 250 229 250 0

9 250 250 250 0

10 40 184 184 144

5 70 70 70 0

6 250 250 250 0

7 250 250 250 0

8 250 250 250 0

9 250 250 250 0

10 40 140 140 100

Part Period Balancing Calculations: Ordering Cost = $300 1 2 3 4

($2) x [(1/2) x (100)] = $100 ($2) x [(1/2) x (100) + (3/2) x (10)] = $130 ($2) x [(1/2) x (100) + (3/2) x (10) + (5/2) x (15)] = $205 ($2) x [(1/2) x (100) + (3/2) x (10) + (5/2) x (15) + (7/2) x (20)] = $345

5 ($2) x [(1/2) x (700)] = $70 6 ($2) x [(1/2) x (70) + (3/2) x (250)] = $820 6 ($2) x [(1/2) x (250)] = $250 7 ($2) x [(1/2) x (250) + (3/2) x (250)] = $1,000 7 ($2) x [(1/2) x (250)] = $250 8 ($2) x [(1/2) x (250) + (3/2) x (250)] = $1,000 8 ($2) x [(1/2) x (250)] = $250 9 ($2) x [(1/2) x (250) + (3/2) x (250)] = $1,000 9 ($2) x [(1/2) x (250)] = $250 10 ($2) x [(1/2) x (250) + (3/2) x (40)] = $370 10 ($2) x [(1/2) x (40)] = $40 11 ($2) x [(1/2) x (40) + (3/2) x (0)] = $40 12 ($2) x [(1/2) x (40) + (3/2) x (0) + (5/2) x (100)] = $540

6A-6

11 0

12 100

144 144

144 44

11 0

12 100

100 100

100 0

Chapter 06A Solution Manual

12 ($2) x [(1/2) x (40) + (3/2) x (0) + (5/2) x (100)] = $540 b. The Economic Order Quantity procedure assumes constant uniform demand. The Part Period Balancing procedure assumes that equating setup and inventory costs will lead to minimum total cost.

4. THE THRIFTY COMPUTER COMPANY a. Total demand = 400, Average demand = 50 1. EOQ =

Month Number Requirements Order Quantity Beg. Inventory End. Inventory Avg. Inventory

1 65 82 82 17 49.5

2 45 82 99 54 76.5

3 35

4 10

54 19 36.5

19 9 14

2  50  100 = 82 units 150 .

5 115 164 173 58 115.5

6 25 58 33 45.5

7 85 82 115 30 72.5

8 20

8 20

8 20

30 10 20

2. POQ = 82/50 = 1.64 weeks  2 weeks Month Number Requirements Order Quantity Beg. Inventory End. Inventory Avg. Inventory

1 65 110 110 45 77.5

2 45 45 0 22.5

3 35 45 45 10 27.5

4 10

3 35 45 45 10 27.5

4 10

10 1 5

5 115 140 140 25 82.5

6 25 25 0 12.5

7 85 105 105 20 62.5

5 115 115 115 0 57.5

6 25 25 25 0 12.5

7 85 105 105 20 62.5

20 0 10

3. Part Period Balancing Month Number Requirements Order Quantity Beg. Inventory End. Inventory Avg. Inventory

1 65 110 110 45 77.5

2 45 45 0 22.5

6A-7

10 0 5

20 0 10

Chapter 06A Solution Manual

Calculations (Part Period Balancing): Setup cost = $100 1 2.

($1.50) x [(1/2) x (65)] = $48.75 ($1.50) x [(1/2) x (65) +(3/2) x (45)] = $150

3.

4 5.

($1.50) x [(1/2) x (35)] = $26.25 ($1.50) x [(1/2) x (35) +(3/2) x (10)] = $48.75 ($1.50) x [(1/2) x (35) +(3/2) x (10) + (5/2) x (115)] = $480

5.

6

($1.50) x [(1/2) x (115)] = $86.25 ($1.50) x [ (1/2) x (115) + (3/2) x (25) ] = $142.5

6.

7.

($1.50) x [(1/2) x (25)] = $18.75 ($1.50) x [(1/2) x (25) +(3/2) x (85)] = $210

7.

8.

($1.50) x [(1/2) x (85)] = $63.75 ($1.50) x [(1/2) x (85) +(3/2) x (20)] = $108.80

EOQ $400 645 $1045

Setup Cost Inventory Carrying Cost ..

Total Cost

EOQ 1. Simplicity 2. Limited information required 3. Time between orders can vary

EOQ 1. Assumes constant demand 2. Doesn't utilize all information 3. Fixed lot size

Advantages POQ 1. Simplicity 2. Limited information required 3. Lot size can vary

Disadvantages POQ 1. Doesn't utilize all information 2. Fixed time between orders 3. Does not evaluate all material ordering possibilities.

4. Does not evaluate all material ordering possibilities.

6A-8

POQ $400 450 $850

PPB $500 412 $912

PPB 1. Utilizes all information 2. Lot size can vary 3. Time between orders can vary

PPB 1. Complexity 2. Does not evaluate all material ordering possibilities

Chapter 06A Solution Manual

5. PART PERIOD BALANCING

Month Number Requirements Order Quantity Beg. Inventory End. Inventory Calculations: 1.

Q = 100 2.

Q = 135 3.

Q = 120 4.

Q = 145 5.

Q = 60 6.

Q = 130 7.

Q = 115 8.

Q = 185 9.

Q = 150 10.

Q = 205

1 70 135 135 65

2 30

3 35

65 35

35 0

4 60 120 120 60

5 60 60 0

6 25 60 60 35

7 35 35 0

8 70 115 115 45

9 45 45 0

10 70 150 115 80

($1)(1)(30) = $ 30 < $100 30 + ($1)(2)(35) = $ 100 = $100 ($1)(1)(60) = $60 < $100 60 + ($1)(2)(25) = $110 > $100 ($1)(1)(35) = $ 35 < $100 35 + ($1)(2)(60) = $155 > $100 ($1)(1)(45) = $45 < $100 45 + ($1)(2)(70) = 185 > $100 ($1)(1)(80) = $80 < $100 80 + ($1)(2)(55) = 190 > $100

Cost of the solution = 6 ($100) + 1 (65 + 35 + 60 + 35 + 45 + 80) = 600 + 320 = $920

6A-9

11 80 150 0

12 55 55 55 0

Chapter 06A Solution Manual

6. MOM WITH PURCHASE DISCOUNTS a. Target Level =

Period 1 2 3 4 5 6 7 8 9 10 11 12

(50 − 40)(100) = 500 1

Order 135

145

150

150 55

Reqs.

70 30 35 60 60 25 35 70 45 70 80 55

Cum. Reqs.

70 100 135 60 120 -35 105 -70 150 --

Part-Periods 70 X 0 = 0 30 X 1 = 30 35 X 2 = 70 60 X 0 = 0 60 X 1 = 60 -35 X 0 = 0 70 X 1 = 70 -70 X 0 = 0 80 X 1 = 80 --

Cumulative Part-Periods 0 30 100 0 60 -0 70 -0 80 --

1.

Look ahead (Period 3): Plus(Period 4):

35 x 2 Periods x $1 = $ 70  $100 60 x 3 Periods x $1 = $180 > $100

2.

Look ahead (Period 5): Plus (Period 6): Plus(Period 7):

20 x 1 Period x $1 = $ 20  $100 25 x 2 Periods x $1 = $ 50  $100 35 x 3 Periods x $1 = $105 > $100

3.

Look ahead (Period 8): Plus (Period 9): Plus(Period 10):

5 x 1 Period x $1 = $ 5  $100 45 x 2 Periods x $1 = $ 90  $100 70 x 3 Periods x $1 = $210 > $100

4.

Look ahead (Period 11): Plus(Period 12):

50 x 1 Period x $1 = $ 50  $100 55 x 2 Periods x $1 = $110 > $100

6A-10

Chapter 06A Solution Manual

b. Least Unit Cost Procedure

Period 1 2* 3

Reqs.

70 30 35

Cum. Reqs.

70 100 135

Setup Cost $100 $100 $100

Inv. Cost $0 $ 30 $100

Unit Price $50 $45 $45

Cum. Cost $3,600 $4,630 $6,275

Cost/ Unit $51.43 $46.30 $46.48

3 4 4* 5

35 60 5 60

35 95 100 155

$100 $100 $100 $100

$0 $ 60 $ 70 $180

$50 $50 $45 $45

$1,850 $4,910 $4,670 $7,255

$52.86 $51.68 $46.70 $46.80

5 6 6* 7

55 25 20 35

55 80 100 115

$100 $100 $100 $100

$0 $ 25 $ 65 $ 95

$50 $50 $45 $45

$2,850 $4,125 $4,665 $5,370

$51.82 $51.56 $46.60 $46.70

7 8 8* 9

15 70 15 45

15 85 100 130

$100 $100 $100 $100

$0 $ 70 $100 $160

$50 $50 $45 $45

$ 850 $4,420 $4,700 $6,110

$56.67 $52.00 $47.00 $47.00

10 10* 11 12

70 30 80 55

70 100 150 205

$100 $100 $100 $100

$0 $ 30 $ 80 $190

$50 $45 $45 $45

$3,600 $4,630 $6,930 $9,515

$51.43 $46.30 $46.20 $46.41

Period 1 3 5 7 10 12

Order Quantity 100 100 100 130 150 55

6A-11

Chapter 06A Solution Manual

7. THE FARR MACHINE CORPORATION MOM Procedure

T arg et Level =

(100 − 95)( 80) = 80 part periods 5

Week 1 2 3 4

Order 46

4 5 5

10

4 5

90

Cum. Reqs.

Reqs.

Part-Periods 44 44 44 x 0 = 0 2 46 2x1=2 10 56 10 x 2 = 20 24 80 24 x 3 = 72 No quantity discount should be taken

Cumulative Part-Periods 0 2 22 94

10 10 10 x 0 = 0 42 52 42 x 1 = 42 28 80 28 x 2 = 56 No quantity discount should be taken

0 42 98

42 46

42 88

42 x 0 = 0 46 x 1 = 46

0 46

Look ahead (Period 6): 2 x 2 Periods x $5 = $ 20  $50 Plus (Period 7): 30 x 3 Periods x $5 = $450 > $50

1.

7 8 9

40

30 10 4 Week 1 3 4 7 9

30 40 44

30 x 0 = 0 10 x 1 = 10 4x2=8

Order Quantity 46 10 90 40 4

Average Demand = 21.1 EOQ = 20.5 TBO = .97 OMT = 21.1 * (0) = 0

6A-12

0 10 18

Chapter 06A Solution Manual

Least Period Cost Procedure

Week 1 2 3 3*

Reqs.

44 2 10 24

Cum. Reqs.

44 46 56 80

Cum. Setup Cost $50 $50 $50 $50

4 4*

42 38

42 80

$50 $50

$0 $190

$100 $ 95

$4,250 $7,840

$4,250.00 $4,283.33

5 6 7*

46 2 30

46 48 78

$50 $50 $50

$0 $ 10 $310

$100 $100 $100

$4,650 $4,860 $8,160

$4,650.00 $2,430.00 $2,720.00

7 8 9

30 10 4

30 40 44

$50 $50 $50

$0 $ 50 $ 90

$100 $100 $100

$3,050 $4,100 $4,540

$3,050.00 $2,050.00 $1,513.33

Week 1 4 5 7

Cum. Inv. Cost $0 $ 10 $110 $470

Unit Price $100 $100 $100 $ 95

Cum. Cost $4,450 $4,660 $5,760 $8,120

Cost/ Period $4,450.00 $2,330.00 $1,920.00 $2,273.60

Order Quantity 56 42 48 44

6A-13

Chapter 06A Solution Manual

Least Unit Cost Procedure

Period 1 2

Reqs.

44 2

Cum. Reqs.

44 46

Setup Cost $50 $50

Inv. Cost $0 $ 10

Unit Price $100 $100

Cum. Cost $4,450 $4,660

Cost/ Unit $101.14 $101.30

2 3 4

2 10 42

2 12 54

$50 $50 $50

$0 50 $470

$100 $100 $100

$ 250 $1,300 $5,920

$125.00 $108.33 $109.63

4 4* 5

42 38 46

42 80 88

$50 $50 $50

$0 $190 $230

$100 $ 95 $ 95

$4,250 $7,840 $8,640

$101.19 $ 98.00 $ 98.18

5 6 7

8 2 30

8 10 40

$50 $50 $50

$0 $ 10 $310

$100 $100 $100

$ 850 $1,060 $4,360

$106.25 $106.00 $109.00

7 8

30 10

30 40

$50 $50

$0 $ 50

$100 $100

$3,050 $4,100

$101.67 $102.50

8 9

10 4

10 14

$50 $50

$0 $ 20

$100 $100

$1,050 $1,470

$105.00 $105.00

Week 1 2 4 5 7 8

Order Quantity 44 12 80 10 30 14

6A-14

Chapter 06A Solution Manual

MOM: Month Number Requirements Order Quantity Beg. Inventory End. Inventory

1 44 46 46 2

2 2

Least Unit Cost: Month Number Requirements Order Quantity Beg. Inventory End. Inventory

1 44 44 44 0

Least Period Cost: Month Number Requirements Order Quantity Beg. Inventory End. Inventory

1 44 56 56 12

MOM LUC LPC

Purchase Cost $18,550 $18,600 $19,000

3 10 10 10 0

4 42 90 90 48

5 46

6 2

48 2

2 0

2 2 12 12 10

3 10

4 42 80 80 38

5 46 10 48 2

6 2

2 2

3 10 10 0

5 46 48 48 2

6 2

12 10

4 42 42 42 0

2 0

10 0

Ordering Cost $250 $300 $200

2 0

2 0

Inventory Carrying $ 785 $ 745 $ 685

7 30 44 44 10

8 10

9 4

10 0

4 0

7 30 30 30 0

8 10 14 14 4

9 4

7 30 44 44 14

8 10

9 4

14 4

4 0

Total Cost $19,585 $19,645 $19,885

Recommendation: Use the ordering scheme resulting from the MOM procedure.

6A-15

4 0

Chapter 06A Solution Manual

8. THE FISHER PRODUCTS COMPANY a. D = 23.13 units S = $9/ order C = $10/ unit/ week

2  23.13  9 = 64.52 units 2 64.52 POQ = = 2.79  3 weeks 23.13

EOQ =

b. D = 35.25 ounces S = $5/ order C = $.04/ ounce / week

2  32.25  5 = 93.87 units .04 93.87 POQ = = 2.66  3 weeks 32.25

EOQ =

EXHIBIT A Material Requirements Planning Worksheet End products Week Game A master schedule Game B master schedule Toy Cup Week Gross Requirements Scheduled Receipts* Projected Available Balance** Planned Order Release

44

Plastic Molding Material Week Gross Requirements Scheduled Receipts* Projected Available Balance** 20 Planned Order Release *Received at the beginning of each week. **Measured at the end of each week.

1 21 2

2 0 2

3 0 9

4 21 0

5 20 6

6 0 3

7 15 0

8 0 9

1 44

2 2

3 9

4 42

5 46

6 3

7 30

8 9

0 53

51

42

0 79

33

30

0 9

0

1 106 90 4

2

3

4 158

5

6

7 18

8

4

4 154

0

0

0 18

0

0

6A-16

Chapter 06A Solution Manual

c. Toy Cup: D = 19.5 units S = $9/ order C = $10/ unit/ week PPB Calculations: 1. ($.10) x [(1/2) x (6)] = $.30 2. ($.10) x [(1/2) x (6) + (3/2) x (49)] = $7.65 3. ($.10) x [(1/2) x (6) + (3/2) x (49) + (5/2) x (33)] = $15.90 Order 55 in Week 1 1.

2.

3.

4.

($.10) x [(1/2) x (33)] = $1.65 ($.10) x [ (1/2) x (33) + (3/2) x (10)] = $3.15 ($.10) x [(1/2) x (33) + (3/2) x (10) + (5/2)] = $3.40 ($10) x [(1/2) x (33) + (3/2) x (10) + (5/2) x (1) + (7/2) x (2)] = $4.10 Order 46 in Week 3

Plastic Molding Material: D = 25.25 ounces S = $5/ order C = $.04/ ounce / week PPB Calculations: Not applicable since a single order for 26 should be placed in Week 1.

EXHIBIT B Material Requirements Planning Worksheet End products Week Game A master schedule Game B master schedule

1 9

6A-17

2 22 1

3 7

4 21 7

5 15 3

6

7

8

10

1

2

Chapter 06A Solution Manual

Toy Cup Week Gross Requirements (2 units/game A) (1 unit/game B) Scheduled Receipts* Projected Available Balance Planned Order Release

1

2

3

4

5

6

7

8

9

45

7

49

33

10

1

2

46 55

1

49 46

0

13

3

2

0

Week

1

2

3

4

5

6

7

8

Gross Requirements (2 oz./toy cup)

110

92

0

0

0

0

0

0

0

55

Scheduled Receipts* Projected Available Balance Planned Order Release

166 10

-100

66

26

d. There will be a shortage of 100 ounces of the plastic molding material in the first week that will prevent the company from meeting the revised master schedule. The alternative courses of action available to the planner are: 1) to expedite the order of the plastic molding material scheduled to be received in week 2 in order to receive 100 ounces one week early, or to revise the master schedule to reflect the shortage and re-acknowledge the revised shipping date(s) to the customer(s). If the scheduled receipt cannot be expedited, the master schedule should be revised to maintain the integrity of the current order dates.

6A-18

Chapter 06A Solution Manual

9. CHAMBERS INC. Total demand = 400, Average demand = 50 1. EOQ =

2  50  27 = 104 units .25

Month Number Requirements Order Quantity Beg. Inventory End. Inventory Avg. Inventory

1 51 104 104 53 78.5

2. Part Period Balancing Month Number 1 Requirements 51 Order Quantity 65 Beg. Inventory 65 End. Inventory 14 Avg. Inventory 39.5

2 14 53 39 46 2 14 14 0 7

3 59 104 143 84 113.5

4 93 104 188 95 141.5

5 44

6 12

95 51 73

51 39 45

3 59 152 152 93 122.5

4 93

5 44 56 56 12 34

6 12

Setup Cost Inventory Carrying Cost Total Cost

93 0 46.5

EOQ $108 163 $271

12 0 6

7 67 104 143 76 109.5

8 60

7 67 127 127 60 93.5

8 60

76 16 46

60 0 30

PPB $108 95 $203

Calculations (Part Period Balancing): Setup cost = $27 1.

2.

3.

($0.25) x [(1/2) x (51)] = $6.375 ($0.25) x [(1/2) x (51) + (3/2) x (14)] = $11.63 ($0.25) x [(1/2) x (51) + (3/2) x (14) + (5/2) x(59)] = $48.5

3.

4.

($0.25) x [(1/2) x (59)] = $7.375 ($0.25) x [(1/2) x (59) +(3/2) x (93)] = $42.25

5.

6 7.

($0.25) x [(1/2) x (44)] = $5.50 ($0.25) x [ (1/2) x (44) + (3/2) x (12) ] = $10.00 ($0.25) x [(1/2) x (44) + (3/2) x (12) + (5/2) x(67)] = $51.88

7.

8.

($0.25) x [(1/2) x (67)] = $8.375 ($0.25) x [(1/2) x (85) +(3/2) x (60] = $30.88 Chambers should use the PPB lot sizing method since it provides a lower cost than their old method of EOQ.

6A-19

Chapter 06A Solution Manual

10. THE SILVER BROTHERS a. Quick's Part Q = 40 LT = 1 SS = 5 Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Total Planned Inventory = 186

7

1 23 40 24 0

2 13

3 36

4 12

5 21

6 8

7 34

8 23

11 40

15 40

43 0

22 0

14 40

20 40

37 0

1 0

2 40

3 40

4 0

5 0

6 40

7 40

8 0

12 100

72 0

32 0

32 0

32 100

92 0

52 0

52 0

1 23 40 24 0

2 13

3 36

4 12

5 21

6 8

7 34

8 23

11 30

5 41

34 0

13 0

5 34

5 55

37 0

1 0

2 30

3 41

4 0

5 0

6 34

7 55

8 0

12 59

41 0

0 0

0 0

0 141

107 0

52 0

52 0

Hiho's Part Q = 100 LT = 1 SS = O Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Total Planned Inventory = 376

112

b. Here is one solution to the problem Quick's Part Q = varies LT = 1 SS = 5 Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Total Planned Inventory = 134

7

Hiho's Part Q = varies LT = 1 SS = O Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Total Planned Inventory = 264

12

This problem shows the need to take into account the multiple levels of the BOM. The discreteness of the demands means that inventory is less at both levels with the example shown.

6A-20

Chapter 06A Solution Manual

11. LUC a. Demand = 20 Price = 1.00 Order = 5 Carry = .2 Period Requirements Order Q Order $ Carry $ Price $ Total $ Per Unit

1

2

3

4

5

6

10 10 5.00 0.00 10.00 15.00 1.50

18 28 5.00 3.60 28.00 36.60 1.31

30 58 5.00 15.60 58.00 78.60 1.36

35 93 5.00 36.60 93.00 134.60 1.45

10 103 5.00 44.60 103.00 152.60 1.48

16 119 5.00 60.60 119.00 184.60 1.55

Q=28 Periods 1-6 Q= 30 Periods 3-6 Q= 35 Periods 4-6 Q= 26 Periods 5-6

1.50 1.17 1.14 1.50

1.31 1.18 1.16 1.32

1.36 1.21 1.22

1.45 1.28

1.48

1.55

This spreadsheet is used to decide each order independently. The unit costs are determined for periods 1 to 6 first. Once the order of 28 is determined, the demands are changed and the order for period 3 is found by computing the unit costs for periods 3 - 6. The order and unit costs for each iteration are shown at the bottom of the spreadsheet.

The inventory carrying costs are calculated by finding the cost of carrying the requirements for a period into that period (requirements  carrying cost  (period number - 1)) and adding that to the carrying cost of the previous periods. b. With discount (note, for simplicity, no split periods) Demand = 119 Order = 5 Carry = .2 Price = 1.00 Price = .95 for quantities > 50 units

Period Requirements Order Q Order $ Carry $ Price $

1 10 10 5.00 0.00 10.00

2 18 28 5.00 3.60 28.00

6A-21

3 30 58 5.00 15.60 55.10

4 35 93 5.00 36.60 88.35

5 10 103 5.00 44.60 97.85

6 16 119 5.00 60.60 113.05

Chapter 06A Solution Manual

Total $ Per Unit

15.00 1.50

36.60 1.31

75.70 1.31

129.95 1.40

147.45 1.43

178.65 1.50

Q = 58 Periods 1 - 6 Q = 35 Periods 4 - 6 Q = 26 Periods 5 - 6

1.50 1.17 1.50

1.31 1.16 1.32

1.31 1.14

1.40

1.43

1.50

12. XYZ COMPANY a.

Item B Beginning Week 1 2 3 4 5 6 1 438 0 516 0 2 0 580 0 0 3 207 555 0 0 4 555 0 0 Changes of +12% and -65% observed in week 3 for example

7

0

b. The changes in the gross requirements for item B are induced simply by the addition of one more period of gross requirements for item A each week that is known with certainty, and the subsequent re-lot sizing of the net requirements for item A. Recommend either the use of firm planned orders, an extended planning horizon, freezing a portion of the MPS planning horizon, or a change in the lot sizing procedure for item A. Such changes can cause excess inventories or expediting costs for item B, and similar problems in the supplier's factory.

13. K. C. JONES Week 1 2 3 Gross Requirements 33 12 23 Scheduled Receipts 45 Projected Available Balance 60 72 60 37 Planned Order Release 45 Q = 45 units, LT = 3 weeks, SS = 0 units, SLT = 1 week* 

4 37

5 26

6 3

7 8

8 31

0 45

19

16

8

22

Use a safety a safety lead time of one week because of timing variation.

6A-22

Chapter 06A Solution Manual

14. CANNONBALL ADDERLY Week Gross Requirements Scheduled Receipts Projected Available Balance 21 Planned Order Release Q = 35 units LT = 2 weeks SS = 3 units*

1 12

2 0

3 27

4 8

5 34

6 3

7 23

8 29

6 35**

6

14 35

6

7 35

4 35

16

22

*Use a safety stock of ((.27 - .17)  30) = 3 units because of quantity variation. **Use an order quantity of 30 x 1.17 = 35 to provide for an average scrap allowance of 5 units.

The projected available balance is based on this 17% scrap loss (i.e. order 35 and get 30 on the average).

15. ATLAS WRENCH COMPANY a.

Z Wrench Week Gross Requirements Scheduled Receipts Projected Available Balance 26 Planned Order Release POQ = 3 weeks, LT = 2 weeks, SS = 0

1 2

2 19

3 5

4 1

5 3

6 45

24

5 48

0

48

45

0

1

2 48

3

4

5

6

-40

0

0

0

0

Z Shaft Week Gross Requirements Scheduled Receipts Projected Available Balance 3 Planned Order Release 40 POQ = 3 weeks, LT = 3 weeks, SS = 0

5 8

This change in the gross requirements of the Z Wrench causes another period to be pulled into the POQ order quantity. The increased POQ causes the Z Shaft to have a past due order which is the 40 shown in the record above. This does create a problem on the shop floor because priorities are changed and schedules must be revised.

6A-23

Chapter 06A Solution Manual

b. Atlas Wrench can deal with these situations by using firm planned orders, freezing the MPS, or using a different lot sizing method to reduce the nervousness.

6A-24

Chapter 07 Solution Manual

CHAPTER 7 Solution Manual Discussion Questions 1. The primary consideration here is one of time. The decisions made in resource planning are those which provide the capacities that will be utilized in the future. Bricks and mortar, machine tools, and trained manpower are all constraints on the amount of capacity that can be used in master production scheduling later on. If insufficient capacity exists at the time the master production schedule is being devised, some of the product might have to be offloaded to subcontractors or additional expensive short term capacity such as overtime must be provided. The primary consideration here is one of time. The decisions made in resource planning are those which provide the capacities that will be utilized in the future.

Bricks and mortar, machine tools, and trained manpower are all constraints on the amount of capacity that can be used in master production scheduling later on. If insufficient capacity exists at the time the master production schedule is being devised, some of the product might have to be off-loaded to subcontractors or additional expensive short term capacity such as overtime must be provided.

2. JIT implies capacity flexibility. That is, the ability to absorb variations in volume and product mix. Accommodating this flexibility requires cross-trained workers and flexible assignment possibilities as well as process capability. If these are in place, the planning of flexible capacity requires, perhaps, even more capability to produce the current master schedule in order to have it available on short notice. Most of the detailed execution of the changes in people, machine, materials, etc. would be handled on the shop floor, however.

3. About the only advantage of rough-cut planning procedures are that they are quick. CRP uses significantly more information allowing a more precise calculation of capacity requirements. The key reason is that the exact timing of order quantities and timing are used in the calculations. Examples where it might be important to get the detail of CRP for capacity planning is in complex, highly engineered, long lead time situations. When several different products/parts are competing for limited resources, it is important to control capacity very carefully and to be able to know if customer delivery dates can be met or if changes are needed. In such cases the detail from CRP may be warranted.

4. Horizontal loading will complete whole jobs faster than vertical loading. In the case of horizontal loading the highest-priority shop order is scheduled in all work centers, then the job with the next highest priority. This is often in conflict with using the work centers to their highest capacity, since it will have more “holes” in the schedule then the vertical loading approach.

5. Input/Output control confirms that we're doing what we thought we could do under a given plan. In Input/Output control planned work input and planned output are compared to the work actual input and output.

6. The reason is that an hour of capacity lost at a bottleneck will actually decrease the total amount of work that can be completed in the shop. Once that hour of work has been lost, it can never be recovered. At non-bottlenecks this is not true since we have additional capacity.

7-1

Chapter 07 Solution Manual

7. Assuming that the “PLAN” columns represent actual demand forecasts and tie well into the sales forecasts, then the “FINL” can be compared directly to see how different the plans are from actual. These numbers simply need to be converted to monetary terms (dollars, for example) for the direct input/output comparison.

8. Differences between “load” and the three capacities represent Applicon’s ability to take on additional work in the next month. Large orders could be included into a trial run of the MPS to examine the orders’ impact in term so existing capacity availabilities.

9. If the Manugistics APS showed a negative days of supply value this would indicate a shortage or stockout condition. 5D indicates a projected 5 days available in inventory.

10. Certainly there are examples of capacity planning at the university. The brick and mortar, dormitory, and class room capacity (resource planning) all must be accomplished for future classes. The specific balance between majors, departments, and rooms for the current year must be met (rough cut), and the individual classroom sizes must accommodate the classes. The implications of poor capacity planning are clear; overcrowded classes, idle facilities, and frustration all around. The capacity must be available in faculty, teaching space, and laboratories to support such things as increases in the number of majors, increased enrollments, and new fields within the university.

Chapter 7 Problem Summaries Problem Difficulty

Problem Type

1

Easy

Simple capacity requirements calculations.

2

Easy

Extension of #1 to consider staffing requirements.

3

Easy

Extension of #1 to consider some additional staffing requirements.

4

Easy

Another extension of #1 to consider the impact of inspection time on capacity requirements.

5

Moderate

Capacity requirements calculations with setup and runtime parameters in multiple work centers.

6

Moderate

Capacity requirements calculations using master scheduling data.

7

Moderate

Work center capacity calculations from MRP data.

8

Moderate

Input/output analysis.

9

Moderate

Worker/Machine requirements and comparison to one and two shift operation.

10

Moderate

What if analysis from an example given in the chapter.

7-2

Chapter 07 Solution Manual

Chapter 7 Problem Summaries Problem Difficulty 11

Moderate

Problem Type Analyzes a situation where a new order puts the work center over capacity. Requires adjustment to the MPS or an addition of capacity. Refers to an actual example in the text.

1.

Week Leather Fabric

1 15 65

2 20 60

3 18 62

1.2 1

Sewing 0.55 0.66 0.55

Assy 0.45 0.54 0.45

Work Center Sewing 45.65 Assembly 37.35

46.2 37.8

45.98 37.62

Leather Fabric

4 21 59

5 15 65

6 13 67

46.31 37.89

45.65 37.35

45.43 37.17

5 75.00 1.25

6 65.00 1.08

2.

Average labor hours: Sewing = 45.865 -- largest variance = .445 hours Assembly= 37.53 -- largest variance = .36 hours It just does not look like a problem!

3.

Week Minutes Hours

1 75.00 1.25

2 100.00 1.67

3 90.00 1.50

4 105.00 1.75

Combining with Problem 1, inspect at the seat manufacturer: Combine times with those in assembly. Or, match leather colors in sets before sewing. Or, perhaps – get more uniform quality from leather supplier

7-3

Chapter 07 Solution Manual

4.

Sewing Assy Leather 0.66 0.62 Fabric 0.55 0.45 Note assembly has gone up by 5 minutes (or 5/60 hours) Work Center Sewing Assembly

45.65 38.60

46.20 39.47

45.98 39.12

46.31 39.64

45.65 38.60

5.

Week MPS

1 20

2 30

3 35

4 20

WC 200 Frame

34

49

56.5

34

WC 300 Part A Part B Total

23 19 42

33 27 60

38 31 69

23 19 42

WC 400 Part C

26

38

44

26

WC 500 Part D

22

32

37

22

Week MPS (bookcases) MPS (panels) MPS (shelves)

1 35 70 140

2 50 100 200

3 40 80 160

End Panel Operation 1 Operation 2 Operation 3

Machine Center Saw Planer Router

Setup Time 36 min.

18 min.

48 min.

Run Time Per Unit 5 min.

3 min.

4 min.

Shelf Operation 1 Operation 2 Operation 3 Operation 4

Saw Molder Router Sander

30 min.

78 min.

42 min.

6 min.

4 min.

5 min.

5 min.

1 min.

6. a.

7-4

45.43 38.25

Chapter 07 Solution Manual

Week Machine center Saw Minutes Hours

1 36+(35x5x2)=386 30+(35x4x4)=590 976 16.3

2 3 36+(50x5x2)=536 36+(40x5x2)=436 30+(50x4x4)=830 30+(40x4x4)=670 1366 1106 22.77 18.43

Minutes Hours

18+(35x3x2)=228 3.8

18+(50x3x2)=318 18+(40x3x2)=258 5.3 4.3

Minutes Hours

48+(35x4x2)=328 48+(50x4x2)=448 48+(40x4x2)=368 42+(35x5x4)=742 42+(50x5x4)=1042 42+(40x5x4)=842 1070 1490 1210 17.83 24.83 20.17

Minutes Hours

78+(35x5x4)=778 78+(50x5x4)=1078 78+(40x5x4)=878 12.97 17.97 14.63

Minutes Hours

6+(35x1x4)=136 2.27

Planer

Router

Molder

Sander 6+(50x1x4)=206 3.43

6+(40x1x4)=166 2.77

b.

Week Past Due 1 2 3

Setup Hours 1.5 1.5 1.5

Weekly Router Machine Load Run Total Hours Hours 16.83 17.83 23.33 24.83 18.67 20.17

MPS 35 50 40

c. Since the shelves must be processed at the sanding operation after routing they should be processed first at the router if the two parts arrived at the router on the same day. This way both parts would be in their final operation at the same time and the complete component sets would be available for assembly sooner.

7-5

Chapter 07 Solution Manual

7.

Work Center 100 200 300

1 5.33 6.12 13.25

2 6.17 5.5

3 6.17 5.5

Week 4 6.17 5.5 9.5

5 6.17

6

Examples: WC 100, week 1 = 2 + (20x10/60)=5.33.

WC 200, week 1 = 3+(30x5/60)+(15x5/2/60)= 6.12 8.

Week

1

2

3

4

5

6

7

8

Planned input

60

60

60

60

60

60

60

60

Actual input

68

70

75

70

68

60

55

55

Cumulative deviation

8

18

33

43

51

51

46

41

Planned output

65

65

65

65

65

65

65

65

Actual output

60

62

63

63

64

65

63

60

Cumulative deviation

-5

-8

-10

-12

-13

-13

-15

-20

58

66

78

85

89

84

76

71

Actual backlog

50

9.

Work Center 100 200 300

A 35 25 46

Product Line B 28 37 35

Total C 12 56 33

a.

Work Center Workers Machines 100 2 2 200 3 3 300 3 3 b.

Work Center Workers Machines 100 2 1 200 3 2 300 3 2

7-6

75 118 114

Chapter 07 Solution Manual

10.

MPS = PLAN12 + PLAN01 + PLAN02 = 1685 + 2598 +2530 = 6813 Sold = FINL12 + FINL01 + FINL02 = 2327+ 2610 +2758 = 7695 This indicates that the company is already oversold in this product line to the tune of 882 units during the next three months. The only hope is to increase capacity on the B1 line or to offload some of its production to other production lines.

11.

New Capacity Status Report Work Center

Load Std. Hrs.

Capacity (Standard)

ALF-A ALF-T HLT-A HLT-T MIS-A MIS-T MVX-A MVX-T OLD-P PCB-A PCB-H PCB-I PCB-M PCB-P PCB-T PCB-V PCB-W

170 45 488 155 340 64 469 179 81 92 44 124 441 608 918 123 216

480 80 800 160 800 80 1040 160 0 160 160 320 480 960 1680 160 180

Totals

4557

7680

The order moves the company up from a total load of 3634 hours to 4557 hours (from 47 % of standard capacity to 59%). There are now potential problems in work centers HLT-T, MVX-T, OLD-P, AND PCB-W, but the total capacity is more than adequate.

7-7

Chapter 08 Solution Manual

CHAPTER 8 Solution Manual Discussion Questions 1. This question is designed to get the students to think about priorities. The obvious priority is to assign seniors first to the required class, so they will graduate, but fill the class with juniors so that the capacity is utilized on the average. This is roughly equivalent to an earliest due date rule. Clearly if the capacity is 40 per year, no prioritizing is necessary and all students will graduate on time. If, on the other hand, there is only capacity for 10 students per year the number of majors graduating a year cannot exceed 10 and the prioritizing rule will not create capacity.

2. In addition to the status information discussed in the chapter, the students ought to be able to suggest things like late completions, over-optimistic estimates of capacity, machine breakdowns, changes in the labor situation, etc.

3. The queue time aspect of the lead time is the "rubbery" part of lead time. Since the queue time can vary considerably and it is a major part of the lead time, actual lead times to complete an item can vary from a minimum of their physical processing time to substantially greater than that, as items sit in queue on the shop floor.

4. The shop floor is concerned with when a part should be completed, not when it arrived. The examples given in the question are concerned with people so equitability is important. The FCFS rule is viewed as equitable. That is, it serves people as they arrive and is not concerned with when they should be finished. On the shop floor we are concerned with getting the parts completed when they are needed, not processing them in some "equitable" fashion.

5 . Bottleneck work centers are finite forward loaded using computer scheduling logic.

However, non-bottleneck work centers are not forward finite loaded by computer scheduling logic. Instead, due dates are set for these operations by considering the computer schedule for the bottleneck work centers, and fixed lead time off-sets for the non-bottleneck operations.

This difference in the scheduling approach for bottleneck and non-bottleneck work centers is desirable in order to focus major attention on the bottleneck work centers that constrain plant work flow.

6. Buffers are placed in front of bottleneck work centers to allow for unforeseen variations in production at earlier non-bottleneck work centers. This is done to protect the bottleneck work centers from having idle capacity because of the late arrival of work. The idea is not to disrupt the flow of material from a bottleneck work center. The size of these buffers is set in proportion to the level of delay expected in the arrival of orders from earlier work centers.

8- 1

Chapter 08 Solution Manual

7. The notion of overriding the formal system to gain efficiencies on the factory floor must be very carefully thought through. Clearly, one of the primary considerations in this determination is the objectives of the production activity control system and the company. If the shop floor economies are not great in relation to the benefits of sticking to the schedule, then the schedule ought to remain king. On the other hand, if equipment utilization is important, then discretion could be given to the foreman. A second level of concern is who has the best information to make the trade-off decision, presuming that all people have valid information.

8. The wands used by clerks in department and grocery stores, are used to collect sales and process inventory information. The tags usually contain detail data on the item for inventory control purposes and may have current price information for the cash register. The primary benefits are improved accuracy in entering data at the point of sale (as compared to a clerk manually putting data into the cash register) and improved inventory movement data.

9. The idea behind this question is to get students to think through some of the differences in the activities and time-frames of the two groups of people. Also, the buyers interact with the salespeople and, perhaps, the engineers at the vendors' locations while the schedulers are more often interacting with the order entry people. Having the groups separated in the organization helps maintain these distinctions in activities and focus. It is useful to help the students see the differences and understand the difficulty of trying to "simultaneously finish a term paper due tomorrow and decide on the elective courses to take in their senior year."

Chapter 8 - Problem Summaries Problem

Difficulty

Problem Type

1

Moderate

Machine center scheduling using Gantt charts. Need to consider job priorities and arrival dates.

2

Easy

Use priority scheduling rules to determine the processing sequence and create a Gantt chart for the sequence using shortest operation next rule.

3

Hard

Priority scheduling with jobs produced at two work centers. Includes order completion performance and move time.

4

Moderate

5

Easy

Splitting jobs to meet due dates, this uses the same approach as problem # 3 but allows one job to be split in two parts.

Updates critical ratio as a job is partially completed.

8- 2

Chapter 08 Solution Manual

Chapter 8 - Problem Summaries Problem

Difficulty

Problem Type

6

Moderate

7

Easy

Identifying the bottleneck department in a plant.

8

Hard

Similar to problem 6 (priority scheduling) but requires the students to sort through more information and to give the formulas for each rule.

9

Moderate

10

Hard

Uses MRP records and open orders tied to routing. You must determine if the timing of orders in the shop will allow the schedules to be delivered when they are required by the MRP system.

11

Moderate

Using the repetitive lots concept with transfer batches to minimize flow time.

12

Easy

Looks at the impact of the buyer’s MRP system on the supplier’s scheduling. Examines adjusting the schedule so the supplier can deliver the necessary goods.

13

Hard

Determining the minimum size of transfer batch to minimize flow when the process batch is split into two transfer batches.

14

Moderate

Looks at using critical ratio to schedule jobs and then updating the ratio’s after time has passed.

15

Moderate

Linking shop floor schedules to MRP records. It shows the impact of changes in on-hand inventory on and shop floor priorities.

16

Hard

Comparing repetitive lots to the FCFS dispatching rule to determine the impact on flow time.

17

Hard

Determining the OPT production schedule and the resultant profit.

18

Moderate

Using Pareto analysis to select suppliers for relationship building.

Priority scheduling problem with multiple jobs and multiple rules.

Students must determine the different sequences under each rule.

Minimizing setups by running the same types of jobs together.

8- 3

Chapter 08 Solution Manual

1. TOM’S SAILBOARD a.

WC1

B

WC2

G

G

WC3 IDLE

Order Arrival

IDLE

11/10

IDLE

B

B

G

11/11

B, G

K

11/12

H

H

11/13

11/14

K

H

H

IDLE

IDLE

11/15

11/16

11/17

K

11/18

11/19

IDLE

11/20

11/21

11/22

K

Which ever job arrives at each work center first starts being processed first because of the FCFS rule in each work center. The work centers do not wait for an order if there is already an order available. Use alphabetical order to break ties (e.g. both G and H arrive at work center 3 on 11/12 but G gets priority based on alphabetical order). b. The determination of how long jobs wait at work center 2 is as follows: B and G are processed immediately at work center 2, order K must wait 1 day (11/14) and order H waits 2 days (11/16 and 11/17).

2. THE KNOX MACHINE COMPANY a. 1. The shortest operation next rule: B-D-A-C 2. The first-come, first-served rule: D-A-B-C 3. The earliest due date rule: C-A-D-B b.

P&W Grinder

B 7/10

D 7/11

A 7/12

7/13

7/14

8- 4

C 7/15

7/16

7/17

7/18

7/19

7/20

7/21

Chapter 08 Solution Manual

3. ABC JOBS a. EARLIEST DUE DATE -- Jobs B and C are late.

Machine Center X Machine Center Y Job Start Days* Done Move Start Days Done A 0.0 4.5 4.5 2.0 6.5 2.5 9.0 B 4.5 12.0 16.5 2.0 18.5 3.0 21.5 C 16.5 7.5 24.0 2.0 26.0 3.5 29.5 *All times in days, i.e. 40/5 = 8 hours per day, 36/8 = 4.5 days for job A.

Due 10.0 17.0 25.0

b. CRITICAL RATIO -- Jobs A and C are late.

Critical Machine Center X Ratio Job Start Days* Done 1.00 B 0.0 12.0 12.0 *1.11 A 12.0 4.5 16.5 1.92 C 16.5 7.5 24.0 *(10 - 0)/(4.5 + 2 + 2.5) = 1.11

Move 2.0 2.0 2.0

Machine Center Y Start Days Done 14.0 3.0 17.0 18.5 2.5 21.0 26.0 3.5 29.5

Due 17.0 10.0 25.0

c. OVERTIME: EARLIEST DUE DATE -- All jobs can be completed.

Machine Center X Machine Center Y Job Start Days* Done Move Start Days Done A 0.0 3.0 3.0 2.0 5.0 2.5 7.5 B 3.0 8.0 11.0 2.0 13.0 3.0 16.0 C 11.0 5.0 16.0 2.0 18.0 3.5 21.5 *60/5 = 12 hours per day, 36/12 = 3.0 days for job A. CRITICAL RATIO -- Job A is late.

Critical Machine Center X Ratio Job Start Days* Done 1.31 B 0.0 8.0 8.0 1.33 A 8.0 3.0 11.0 2.38 C 11.0 5.0 16.0

8- 5

Move 2.0 2.0 2.0

Due 10.0 17.0 25.0

Machine Center Y Start Days Done 10.0 3.0 13.0 13.0 2.5 15.5 18.0 3.5 21.5

Due 17.0 10.0 25.0

Chapter 08 Solution Manual

d. 1 DAY MOVE TIME EARLIEST DUE DATE -- Jobs B and C are late.

Job A B C

Machine Center X Start Days* Done 0.0 4.5 4.5 4.5 12.0 16.5 16.5 7.5 24.0

Move 1.0 1.0 1.0

Machine Center Y Start Days Done 5.5 2.5 8.0 17.5 3.0 20.5 25.0 3.5 28.5

Due 10.0 17.0 25.0

CRITICAL RATIO -- Jobs A and C are late.

Critical Ratio 1.25 1.06 2.08

Job A B C

Machine Center X Start Days* Done 12.0 4.5 16.5 0.0 12.0 12.0 16.5 7.5 24

Move 1.0 1.0 1.0

Machine Center Y Start Days Done 17.5 2.5 20.0 13.0 3.0 16.0 25.0 3.5 28.5

4. ABC JOBS - SPLIT ORDER Machine Center X Machine Center Y Start Days Done Move Start Days Done 0.0 4.5 4.5 2.0 6.5 2.0 8.5 4.5 6.0* 10.5 2.0 12.5 1.5* 14.0 10.5 7.5 18.0 2.0 20.0 3.5 23.5 18.0 6.0* 24.0 2.0 26.0 1.5* 27.5 *Half of the time for the full order.

All the jobs will be completed on time by splitting job B into two parts.

Job A B1 C B2

5. YACHT PROJECT a. Critical Ratio=

b. Critical Ratio=

TimeRemaining 40− 11 = =1.04 Work Remaining 28

TimeRemaining 40−11 = =.91 Work Remaining 32

8- 6

Due 10.0 17.0 25.0 28.0

Due 10.0 17.0 25.0

Chapter 08 Solution Manual

6. BIG DAN’S MACHINE SHOP a.

Rule Critical Ratio Earliest Due Date Order Slack Shortest Operation Next

1 0.80 12 -3.0 8

2 1.25 15 3.0 5

Order 3 1.40 14 4.0 4

4 1.20 18 3.0 3

5 0.95 19 -1.0 6

b.

Rule Critical Ratio Earliest Due Date Order Slack Shortest Operation Next

Order To Run Next 1 1 1 4

7. MARUCHECK'S MAKESHIFT MANUFACTURING If the three shaping machines are run in parallel, here is the total time in each department to run a batch of 200 pieces: Department Shaping Pickling Packing

Setup 60 0 0

Run Time 1  (200/3) 4  60 (60  200)  (4  25)

Therefore, Pickling is the bottleneck operation.

8- 7

Total Time 127 minutes 240 minutes 120 minutes

Chapter 08 Solution Manual

8. THE ACE MACHINE COMPANY a.

1. Critical Ratio=

duedate−now lead timeremaining

2. Order slack = (due date - now) - (total processing time remaining) 3. Shortest operation next = smallest operation processing time 4. Slack per operation=

(due date−now)−(total processing time remaining) number of operations remaining

b. 1. Priority Index Calculations:

Rule Critical Ratio Order Slack Shortest Operation Slack per Operation

1 .467 -1.000 2.000 -.333

2 .733 3.000 3.000 .750

Order 3 -.100 -10.000 5.000 -5.000

4 -.333 -13.000 4.000 -4.333

Resulting Priority Sequences: Rule Critical Ratio Order Slack Shortest Operation Slack per Operation

Order To Run Next 4 4 5 3

8- 8

5 .200 -5.000 1.000 -.417

Chapter 08 Solution Manual

9. OPTIMA SHOP a. The Gantt Chart shows a schedule to minimize setup costs. Note that customer order #1 could be shipped sooner if Part A of order #2 were not run on the same setup.

BIG MESS

Run A Customer #1

Setup A

Run A Cust. #2

Setup B

Run B Customer #1

Run B Customer #3

Setup C

NO PROBLEM

5

Run C Cust. #1

10

Run C Cust. #3

Setup D

15

Run D Cust. #2

20

25

30

The No Problem work center is scheduled to complete Part D, the second part of customer order #2, at the same time as the other part on this order is completed. Part C for customer order #1 is completed 4 days earlier than necessary (the alternative is to make customer order #2 wait for part D or to re-setup C for customer order #3). b. Yes -- no problem at all. c. There is no change at all. The setups do not overlap between departments, so the set up person can do them all as scheduled in a. d. OPT is a finite loading approach. It would produce a result similar to the Gantt Chart in 1 a.

OPT would concentrate on Big Mess, since it is the bottleneck work center. It would combine the setups as indicated to maximize the capacity utilization.

10. THE XYZ COMPANY a. Shortest Operation Next Rule:

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Chapter 08 Solution Manual

Processing Time* Item (Machine 2) A 1 B 2 C 1.5 *In days Item A should be scheduled first, item C scheduled second, and item B scheduled third.

The implications of this schedule are that orders for items A and C will be finished on time, but the order for item B will be late. b. Critical Ratio Rule: Item A B C

Critical Ratio 5/5 = 1.0 10/14.5 = .690 5/3 = 1.667

Item B should be scheduled first, item A scheduled second, and item C scheduled third.

The implications of this schedule are that the orders for all items will be finished when needed according to the MRP system. c. The inventory planner should reschedule the order for 30 units of item B to be received at the start of Week 7 instead of Week 3. The critical ratio priority would change to 30/14.5 = 2.07 and would be scheduled last.

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Chapter 08 Solution Manual

11. THE ACE TOOL COMPANY a.

NO TRANSFER BATCHES

Hour 40.67

WORK CENTER 1 Hour 65.00 WORK CENTER 2

TRANSFER BATCHES

Hour 40.67

WORK CENTER 1 Hour 43.40 WC 2

1

WC 2

WC 2

2

WC 2

WC 2

3

WC 2

WC 2

4

WC 2

WC 2

5

8-11

WC 2

6

7

8

9

Chapter 08 Solution Manual

b. Under No Transfer Option: Work center 1:

2.4 min./unit X 1,000 units + 40 min. setup time = 2,440 min. or 40.67 hours.

Work center 2:

1.44 min./unit X 1,000 units + 20 min. setup time = 24.33 hours.

(The last operation is completed at hour 65.0 at work center 2.)

Under Transfer Option: Work center 1:

2.4 min./unit X 1,000 units + 40 min. setup time =2,440 min. or 40.67 hours.

Work center 2:

•

First batch starts as early as 40 min. + 100 X 2.4 = 280 minutes or 4.67 hours.

•

Each batch of 100 takes 4 hours to process at work center 1 and 20 min. + 1.44 min./unit X 100 = 164 min. or 2.73 hours on work center 2.

•

The last batch can clear work center 2 as early as hour 43.40.

c. Benefits of Using Transfer Batches: Shorter production lead time Less work-in-process inventory Improved customer service Costs Using Transfer Batches: There could be as many as 10 setups required at work center 2 if none of the batches were run under the common setup. Such a schedule is shown below.

Work center 2

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Chapter 08 Solution Manual

Under Transfer Batches Batch # Early Start Time 1 4.67 2 8.67 3 12.67 4 16.67 5 20.67 6 24.67 7 28.67 8 32.67 9 36.67 10 40.67

8-13

Time 7.40 11 .40 15.40 19.40 23.40 27.40 31.40 35.40 39.40 43.40

Chapter 08 Solution Manual

12. FLATBUSH PRODUCTS, INC. a.

Circuit #101 Week

1

Gross Reqs.

50

2

3

4

5

6

7

8

9

10

11

12

50

50

50

15

15

15

Scheduled Rec. Proj. Avail. Bal.

65

15

15

15

15

15

15

Plan. Order Rel. Q = 50; LT = 3; SS = 2

15

15

15

50

50

50

b. Pete should level out the requirements for circuit #101 in order to meet the limited capacity restrictions of the vendor. This can be accomplished by planning to release orders earlier in the planning horizon as follows:

Circuit #101 Week

1

Gross Reqs.

50

2

3

4

5

6

7

8

9

10

11

12

50

50

50

65

15

65

Scheduled Rec. Proj. Avail. Bal.

65

Plnd. Order Rel. Q = 50; LT = 3; SS = 2 * Firm planned order

15

15

15

65

50*

65 50*

8-14

65

65

115

115 50*

Chapter 08 Solution Manual

13. ED’S SHEET METAL a. General Gantt chart for a transfer batch of Qt: Operation 1

S1

Run Qt

Run (800-Qt)

Transfer

T

Operation 2

T

S2

A

Run Qt

Run (800-Qt)

B

C

In order for there to be no idle time at operation 2 and to have the minimum transfer lot, the run time on operation 1 for the remainder of the order plus the transfer time must equal the transfer time plus setup and run time for the transfer batch at operation 2 (see B above). In mathematical terms:

20 + 90 + 8Q t = 6  (800 − Q t ) + 20 90 + 8Q t = 4800 − 6Q t 14Q t = 4710 Q t = 337 Total manufacturing lead time for transfer batch approach: For A For B For C

60 + (6  337) 20 + 90 + (8  337) 8  463 Total

= 2082 = 2806 = 3704 8592

b. No transfer batches: Operation 1: Operation 2:

60 + (6  800) 20 + 90 + (8  800) Total

Savings: 11,370 - 8,592 = 2,778 minutes.

8-15

= 4860 = 6510 11370

Chapter 08 Solution Manual

14. FRAMKRANTZ FACTORY The spreadsheet program calculates the priorities as a function of the times in the table plus move and queue time. a. Job D is the most critical job and is behind time as is job C. Job D C E A B

Time Remaining 15 21 27 19 40

Work Remaining by Activity Machine Move Queue Total 11 4 8 23 13 3 6 22 16 4 8 28 5 2 4 11 13 3 6 22

Priority .65 .95 .96 1.73 1.82

b. Since all jobs are waiting, the time saved at machine 1 made no difference in relative priorities. The absolute priorities did change, though. Four of the jobs are now behind schedule.

Shop Day = 96 Job D C E A B

Time Remaining 2 8 14 6 27

Work Remaining by Activity Machine Move Queue Total 5 3 6 14 10 2 4 16 12 3 6 21 4 1 2 7 10 2 4 16

Priority 0.14 0.50 0.67 0.86 1.69

c. As shown below, the priority for job A drops to zero. It is now most critical.

Job A

Time Remaining 0

Work Remaining by Activity Machine Move Queue 4 1 2

8-16

Total 7

Priority .00

Chapter 08 Solution Manual

15. PART NUMBER 483 The MRP records for each of the situations are: a.

Part No. 483 Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 40, LT = 2, SS = 5 *Shop Order Number 32

1 14 20

6

2 4 40* 42

3 10

4 20

5 3

6 10

32

12 40

9

39

4 20

5 3

6 10

15 40

12

42

There is nothing for the planner to do. The part is right on schedule. b.

Part No. 483 Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 40, LT = 2, SS = 5 *Shop Order Number 32

23

1 14

2 4

9

5

3 10 40* 35

There is nothing the planner should do here either, even though the part is ahead of schedule. The priority on the shop floor should reflect the fact that the part is ahead of schedule and move it through only if there is nothing of higher priority. c.

Part No. 483 Week

1

2

3

4

5

6

Gross Requirements

14

4

10

20

3

10

Scheduled Receipts

40* 39

29

9

6

36

Projected Available Balance

17

Planned Order Release Q = 40, LT = 2, SS = 5 *Shop Order Number 32

43

40

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Chapter 08 Solution Manual

Now there is a problem, but only a minor one! There will be a need to use some safety stock to cover the immediate order, and the planner should indicate that the part won't be finished until next week in case some subassembly is expecting to use it.

Situation: a. b. c.

105 110 100 100 100 100 5 10 0

Part No. 483 Due date Current date Time remaining Work remaining: Move time Machine time Queue time Total The critical ratio

2 2 1 5 1

2 2 1 5 2

2 2 1 5 0

16. BORING MACHINE CENTER a.

Job A B C A B B C C A A B C A B B

Start Setup Time Time 8:24 AM 15 min.

8:49 AM 10 9:14 AM 20 9:54 AM 15 10:19 AM 10 10:44 AM 10:59 AM 20 11:39 AM 11:59 AM 15 12:24 PM 12:34 PM 10 12:59 PM 20 1:39 PM 15 2:04 PM 10 2:29 PM

Run Time 10 min.

15 20 10 15 15 20 20 10 10 15 20 10 15 15

Completion Time 8:49 AM 9:14 AM 9:54 AM 10:19 AM 10:44 AM 10:59 AM 11:39 AM 11:59 AM 12:24 PM 12:34 PM 12:59 PM 1:39 PM 2:04 PM 2:29 PM 2:44 PM

First-come, first-served finish time is 2:44 P.M. Total flow time is 6 hours, 20 minutes.

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Chapter 08 Solution Manual

b.

Job A A A A A B B B B B B C C C C

Arrival Time 8:24 AM 8:34 AM 8:56 AM 9:03 AM 9:18 AM 8:28 AM 8:39 AM 8:40 AM 9:12 AM 9:21 AM 9:31 AM 8:31 AM 8:42 AM 8:44 AM 9:14 AM

Start Setup Time Time 8:24 AM 15 8:49 AM 8:59 AM 9:09 AM 9:19 AM 9:29 AM 10 9:54 AM 10:09 AM 10:24 AM 10:39 AM 10:54 AM 11:09 AM 20 11:49 AM 12:09 PM 12:29 PM

Run Time 10 10 10 10 10 15 15 15 15 15 15 20 20 20 20

Completion Time 8:49 AM 8:59 AM 9:09 AM 9:19 AM 9:29 AM 9:54 AM 10:09 AM 10:24 AM 10:39 AM 10:54 AM 11:09 AM 11:49 AM 12:09 PM 12:29 PM 12:49 PM

Finish time is 12:49 P.M. Total flow time is 4 hours, 25 minutes.

17. HOWIE’S HANDICRAFT a.

Step 1: Determine the bottleneck Product Witches Goblins Ghosts Ghouls Total

Weekly Potential 60 50 40 30

1 600 500 400 450 1950

Station 2 600 750 400 600 2350

3 2400 2000 800 600 5800

Station 3 is the bottleneck since it exceeds 2400 mins. when using full weekly potential for all the products.

Step 2: Determine the most profitable product based on contribution per bottleneck minute.

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Chapter 08 Solution Manual

Product Witches Goblins Ghosts Ghouls

Price $35 $40 $45 $50

Material $15 $15 $20 $20

Contribution $20 $25 $25 $30

Contribution per minute $0.500 $0.625 $1.250 $1.500

Priority 4 3 2 1

Step 3: Determine the production schedule and contribution Product Ghouls Ghosts Goblins

Production Schedule 30 40 25

Contribution $900 $1000 $625 $2525

Capacity Used 600 800 1000

Capacity Remaining 1800 1000 0

b. Profit = $2525 - $2000 = $525

18. MARLBOROUGH MANUFACTURING COMPANY a. The companies with the greatest New Zealand dollars were spent that also have multiple orders which (e.g. Whist, Gentry) would be good candidates for building relationships.

Exceptions might be firms with low dollar value but critical items or those with single orders (these might be capital equipment; e.g. Locket) which might be used in the future.

These aren’t good candidates for starting the program, however.

Company Whist Gentry Kolst Jacobs Locket Axel Booker Zydec Farmic Hooker Hume Kume

NZD Cumulative NZD Percent Percent (NZD) 2,171,886 27.02 27.02 1,850,120 23.02 50.04 1,367,484 17.01 67.06 965,283 12.01 79.07 563,087 7.01 86.07 321,760 4.00 90.08 186,242 2.32 92.39 121,555 1.51 93.91 80,440 1.00 94.91 64,532 0.80 95.71 63,253 0.79 96.50 57,305 0.71 97.21

8-20

Cumulative Percent of Cos. 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 50.00 55.00 60.00

Chapter 08 Solution Manual

Backer First Grist Wolf Young Century Kelvin Neive

55,122 48,262 40,220 31,523 17,053 16,088 8,124 7,944 8,037,283

0.69 0.60 0.50 0.39 0.21 0.20 0.10 0.10

97.89 98.50 99.00 99.39 99.60 99.80 99.90 100.00

65.00 70.00 75.00 80.00 85.00 90.00 95.00 100.00

Cumulative Perccent of Purchases

100

80

60

40

20

100

90

80

70

60

50

40

30

20

10

0

0

Cumulative Pe rce nt of Supplie rs

b. Those firms with low dollar amounts would be candidates for deletion. c. Marlborough might want to look at their order policies. Many of the companies in the sample only get a small number of orders per year. Does this mean that Marlborough is not managing their inventory well? If they can’t do that, the company, probably isn’t prepared to enter into closer supply chain relationships.

8-21

Chapter 08A Solution Manual

CHAPTER 8A Solution Manual Discussion Questions l. Murphy's Law! The fact that conditions change continually on the factory floor means that rescheduling and / or providing dynamic scheduling tools is very important. Note that rescheduling does not imply "killing" a job. It simply means that the priorities for something else are greater now.

2. Certainly the students ought to come up with lateness as an important criterion. If there is a substantial degree of average lateness then perhaps the sequencing, capacity, or difficulty of the courses should be evaluated. The same might hold true if there were a large proportion of early graduates. In some colleges, facility utilization is an extremely important criterion as well.

3. Static scheduling problems could involve a standard tune-up on an automobile, harvesting several fields at harvest time, redoing all of the billboards for a company, and painting the rooms in an office building. Dynamic scheduling problems could involve the day-to-day scheduling of automobiles in a repair facility, cutting hair at a barber shop, bagging groceries at a grocery store and processing duplicating orders at a copy center.

4. Almost all students will use a combination rule of some kind. A common one is SPT with an urgency for proximity to due date. This is very similar to what many firms actually do use.

Chapter 8A - Problem Summaries Problem Difficulty

Problem Type

1

Easy

Priority scheduling problem using SPT and evaluating the sequence with average completion time and average job lateness.

2

Easy

Priority scheduling problem using slack per operation and evaluating the sequence with average number of jobs in the system and average job lateness.

3

Moderate Priority scheduling problem using FCFS, EDD, SPT, and slack per operation. The schedules are evaluated using average number of jobs in the system and average job lateness.

8A-1

Chapter 08A Solution Manual

4

Easy

Two machine sequencing problem with and without an intermediate storage buffer.

5

Easy

Two machine sequencing problem with an intermediate storage buffer.

It also requires an interpreting the results of the schedule.

6

Easy

Two machine sequencing problem with an intermediate storage buffer.

7

Moderate Determining customer delivery due dates using TWK and SLK procedures.

8

Moderate Two resource assignment problem; you must determine what work center and what job to assign a worker.

9

Moderate Using the repetitive lots concept with transfer batches to minimize flow time.

10 11 12 13 14

Hard

Determining the minimum size of transfer batch to minimize flow when the process batch is split into two transfer batches.

Moderate Determining the flow time when using repetitive lots on three machine centers.

Hard

Comparing repetitive lots to the FCFS dispatching rule to determine the impact on flow time.

Moderate Three machine sequencing problem using the Campbell, Dudek, and Smith heuristic.

Hard

Four machine sequencing problem using the Campbell, Dudek, and Smith heuristic.

8A-2

Chapter 08A Solution Manual

1. THE POHL POOL COMPANY a. Shortest processing time rule: D-A-C-G-F-E-B b. Average completion time (in days) of the above sequence: (3 + 7 + 13 + 21 + 30 + 41 + 54)  7 = 24.14 days c. Average job lateness (in days) of the above sequence: Completion Time 3 7 13 21 30 41 54

Job D A C G F E B

Job Due Date 7 8 8 16 21 39 37

Job Lateness -4 -1 5 5 9 2 17

(-4 + -1 + 5 + 5 + 9 + 2 + 17)  7 = 4.71 days

2. THE THOMPSON TOILET COMPANY a. Slack Per Operation Rule: Priority Index Calculations Job B: (8 - 4)/4 = 1.0 Job O: (7 - 8)/3 = -.33 Job W: (24 - 12)/4 = 3.0 Job L: (7 - 7)/2 = 0.0 Job S: (4 - 9)/2 = - 2.5

b. Average job lateness (in days) of the above sequence:

Job S O L B W

Completion Time 9 17 24 28 40

Job Due Date 4 7 7 8 24

Job Lateness 5 10 17 20 16

(5 + 10 + 17 + 20 + 16)  5 = 13.6 days

8A-3

Chapter 08A Solution Manual

c. Average number of jobs in the system: Days Jobs Days X Jobs 9 5 45 8 4 32 7 3 21 4 2 8 12 1 12 (45 + 32 +21 + 8 + 12)  40 = 2.95 jobs 3. THE HYER-THAN-EVER KITE MANUFACTURING EMPORIUM a. First-come, first-served rule: 1. A-B-C-D-E Job A B C D E

Processing Time 10 12 7 5 8

Flow Time 10 22 29 34 42

2. Average completion time (frame-making): (10 + 22 + 29 + 34 + 42)  5 = 27.4 days 3. Average number of jobs (frame-making): (10 + 22 + 29 + 34 + 42) (10 + 12 + 7 + 5 + 8 ) = 3.26 jobs b. Earliest due date rule: 1. E-B-C-D-A Job E B C D A

Processing time 8 12 7 5 10

Flow time 8 20 27 32 42

Job Due Date 12 15 16 17 25

2. Jobs late in leaving frame-making: B, C, D, A 3. Average job lateness in frame-making: (-4 + 5 + 11 + 15 + 17)  5 = 8.8 days

8A-4

Job Lateness -4 5 11 15 17

Chapter 08A Solution Manual

c. Shortest processing time rule: 1. D-C-E-A-B Job D C E A B

Processing time 5 7 8 10 12

Flow time 5 12 20 30 42

Job Due Date 17 16 12 25 15

Job Lateness -12 -4 8 5 27

2. Order E in process: Days 13-20 3. Average job lateness in frame-making: (-12 - 4 + 8 + 5 + 27)  5 = 4.8 days d.

Job A B C D E

Due Date 25 15 16 17 12

Total Processing 20 18 12 12 10

Job Slack 5 -3 4 5 2

Job Opers.

4 2 2 3 2

Ratio 1.25 -1.50 2.00 1.67 1.00

Rank 3 1 5 4 2

1. Slack per operation rule*: B-E-A-D-C Job B E A D C

Processing Time 12 8 10 5 7

Flow Time 12 20 30 35 42

2. Order D will be completed: day 35 3. Average completion time (frame-making): (12 + 20 + 30 + 35 + 42)  5= 27.8 days Note: original calculation results. This rule must be recalculated after each job completion to get the appropriate schedule.

8A-5

Chapter 08A Solution Manual

4. JOHNSON PROCEDURE a. Job sequence: B-D-A-C b. Gantt Chart of above sequence 0

Machine 1

1

9

B

D

Machine 2

13 A

B 0

21 C

D

1

8

23

A

9

14

C 17

21

23

c. Gantt Chart of above sequence (assuming no buffer): 0

Machine 1

1

9

B

D

Machine 2 1

22

A

B 0

13 14 C

D 8

9

24

A 14

C 17

22

Note: Depending upon assumptions, the idle time after job A could appear before job A on machine 1.

8A-6

24

Chapter 08A Solution Manual

5. GRUBBS AUTO BODY a. Sequence for the shortest time span to complete the orders:.

Sequence 1 3 4 5 2

Car (F)ord (J)aguar (L)exus (C)adillac (H)onda

Gantt Chart of the Above Sequence: 0

Washing

1

4

F

8

J

Waxing

L

F 0

1

13 C

J 3

4

b. 17 hours are required to complete the orders. c. Waxing is idle 2 hours. d. The Cadillac will be completed in 16 hours. e. Average completion time (In hours): (3 + 9 + 13 + 16 + 17)  5 = 11.6 hours

8A-7

17

H

L 9

15

C 13

H 16 17

Chapter 08A Solution Manual

6. WOMBAT UNIVERSITY a. Sequence of Crew Tours: Building No. 1 2 3 4 5 6

Building Name Biology Foreign Language Drama English Chemistry Astronomy

Gantt Chart of the Above Sequence: 0

Sweeping

8

20

1

2

Waxing

3

1 0

8

35

78

4

2 17 20

52

96

5

3 37

6

4 53

106

5 73

78

6 91

96

106

b. The waxing crew will be working in the chemistry building during hours 79-91.

Alternatively, they can be idle from 74-83, and work hours 84-96. c. Total available time for both crews: 31 hours (10 hours during sweeping, 21 hours during waxing).

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Chapter 08A Solution Manual

7. BAXTER BOARDS, INC. a. Delivery time using the TWK Procedure Given an average delivery time for all orders of 8 weeks:

 (3K ) + (5K ) + (8K )   =8 3   K = 1.5

Order 1 2 3

Processing time 3X 5X 8X

K 1.5 1.5 1.5

Delivery time = 4.5 = 7.5 = 12.0

b. Delivery time using the SLK Procedure Given an average delivery time for all orders of 6 weeks:

 (3 + K ) + (5 + K ) + (8 + K )   =8 3   K = 2.67 Order 1 2 3

Processing Time 3+ 5+ 8+

K 2.67 2.67 2.67

Delivery Time = 5.67 = 7.67 = 10.67

c. Delivery time = 10 + k ( = 1) (using a k proportional to uncertainty) Large k values represent increased probability of meeting due date promises (which are longer).

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Chapter 08A Solution Manual

8 . THE SCM CORPORATION a. Shortest processing time dispatching rule: Labor Assignment Rule 1. Shortest Job 2. Oldest Job 3. Longest Queue

Machine Assigned A C B

Job 1 4 2

Machine Assigned A C B

Job 1 4 6

b. Slack time dispatching rule: Labor Assignment Rule 1. Shortest Job 2. Oldest Job 3. Longest Queue

c. A due date oriented labor assignment rule, such as critical ratio, might improve due date performance over the rules used in a and b.

9. THE ACE TOOL COMPANY a.

NO TRANSFER BATCHES

Hour 40.67

WORK CENTER 1 Hour 65.00 WORK CENTER 2

TRANSFER BATCHES

Hour 40.67

WORK CENTER 1 Hour 43.40 WC 2

1

WC 2

WC 2

2

WC 2

WC 2

3

WC 2

WC 2

4

WC 2

WC 2

5

WC 2

6

7

b. Under No Transfer Option: Work center 1:

2.4 min./unit X 1,000 units + 40 min. setup time = 2,440 min. or 40.67 hours.

Work center 2:

1.44 min./unit X 1,000 units + 20 min. setup time = 24.33 hours.

(The last operation is completed at hour 65.0 at work center 2.)

8A-10

8

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Chapter 08A Solution Manual

Under Transfer Option: Work center 1:

2.4 min./unit X 1,000 units + 40 min. setup time =2,440 min. or 40.67 hours.

Work center 2:

•

First batch starts as early as 40 min. + 100 X 2.4 = 280 minutes or 4.67 hours.

•

Each batch of 100 takes 4 hours to process at work center 1 and 20 min. + 1.44 min./unit X 100 = 164 min. or 2.73 hours on work center 2.

•

The last batch can clear work center 2 as early as hour 43.40.

c. Benefits of Using Transfer Batches: Shorter production lead time Less work-in-process inventory Improved customer service Costs Using Transfer Batches: There could be as many as 10 setups required at work center 2 if none of the batches were run under the common setup. Such a schedule is shown below.

Work center 2 Under Transfer Batches Batch # Early Start Time 1 4.67 2 8.67 3 12.67 4 16.67 5 20.67 6 24.67 7 28.67 8 32.67 9 36.67 10 40.67

8A-11

Time 7.40 11 .40 15.40 19.40 23.40 27.40 31.40 35.40 39.40 43.40

Chapter 08A Solution Manual

10. ED’S SHEET METAL a. General Gantt chart for a transfer batch of Qt: Operation 1

S1

Run Qt

Run (800-Qt)

Transfer

T

Operation 2

T

S2

A

Run Qt

Run (800-Qt)

B

C

In order for there to be no idle time at operation 2 and to have the minimum transfer lot, the run time on operation 1 for the remainder of the order plus the transfer time must equal the transfer time plus setup and run time for the transfer batch at operation 2 (see B above). In mathematical terms:

20 + 90 + 8Q t = 6  (800 − Q t ) + 20 90 + 8Q t = 4800 − 6Q t 14Q t = 4710 Q t = 337 Total manufacturing lead time for transfer batch approach: For A For B For C

60 + (6  337) 20 + 90 + (8  337) 8  463 Total

= 2082 = 2806 = 3704 8592

b. No transfer batches: Operation 1: Operation 2:

60 + (6  800) 20 + 90 + (8  800) Total

Savings: 11,370 - 8,592 = 2,778 minutes.

8A-12

= 4860 = 6510 11370

Chapter 08A Solution Manual

11. TRANSFER BATCHES

SCHEDULE OPERATION

1 1000

2 150

250

350

450

550

650

750

625

750

850

950 1050

3 250 200

400

600

800 TIM E

12. BORING MACHINE CENTER a.

Job A B C A B B C C A A B C A B B

Start Setup Time Time 8:24 AM 15 min.

8:49 AM 10 9:14 AM 20 9:54 AM 15 10:19 AM 10 10:44 AM 10:59 AM 20 11:39 AM 11:59 AM 15 12:24 PM 12:34 PM 10 12:59 PM 20 1:39 PM 15 2:04 PM 10 2:29 PM

Run Time 10 min.

15 20 10 15 15 20 20 10 10 15 20 10 15 15

Completion Time 8:49 AM 9:14 AM 9:54 AM 10:19 AM 10:44 AM 10:59 AM 11:39 AM 11:59 AM 12:24 PM 12:34 PM 12:59 PM 1:39 PM 2:04 PM 2:29 PM 2:44 PM

First-come, first-served finish time is 2:44 P.M. Total flow time is 6 hours, 20 minutes.

8A-13

1125 1000

1200

Chapter 08A Solution Manual

b.

Job A A A A A B B B B B B C C C C

Arrival Time 8:24 AM 8:34 AM 8:56 AM 9:03 AM 9:18 AM 8:28 AM 8:39 AM 8:40 AM 9:12 AM 9:21 AM 9:31 AM 8:31 AM 8:42 AM 8:44 AM 9:14 AM

Start Setup Time Time 8:24 AM 15 8:49 AM 8:59 AM 9:09 AM 9:19 AM 9:29 AM 10 9:54 AM 10:09 AM 10:24 AM 10:39 AM 10:54 AM 11:09 AM 20 11:49 AM 12:09 PM 12:29 PM

Run Time 10 10 10 10 10 15 15 15 15 15 15 20 20 20 20

Completion Time 8:49 AM 8:59 AM 9:09 AM 9:19 AM 9:29 AM 9:54 AM 10:09 AM 10:24 AM 10:39 AM 10:54 AM 11:09 AM 11:49 AM 12:09 PM 12:29 PM 12:49 PM

Finish time is 12:49 P.M. Total flow time is 4 hours, 25 minutes.

13. GRUBBS AUTO BODY AGAIN a. Two scheduling problems need to be solved: the M problem (problem #1) and the M-1 problem (problem #2) In problem #1 the (M problem) the application of CDS heuristic produces the following processing sequence: F - J - L - C - H In problem #2 the (M-1 problem) the application of CDS heuristic produces the following processing sequence: J - F - L - H - C. Problem #1, Sequence: F - J - L - C - H Job (F)ord (J)aguar (L)exus

Wash 1 3 4

8A-14

Wax 2 5 4

Chapter 08A Solution Manual

(C)adillac (H)onda

Washing

F

J

5 2

L

0 1

C

4

Buffing

H

8

13

F

J

1

15 L

9

Waxing

3 1

C

13 F

9

18 J

11

H 21

27

L

13

18

C 22

H 25

27 28

Make-

span = 28 hours Problem #2, Sequence: J - F - L - H - C Job (J)aguar (F)ord (L)exus (H)onda (C)adillac

Washing

J 0

F

L

H

3 4

Buffing

Wash & Buff 7 9 9 8 8

8

C 10

J 3

15 F

L

7

Waxing

15 J

7

Buff & Wax 9 10 9 7 6

H 20

F 12

15

C 26

L 17

20

29

H 24

26 27

C 29

32

Make-span = 32 hours b. The best sequence is F - J - L - C - H c. See a.

14. PHILIP a. Three scheduling problems need to be solved: the M problem (problem #1), the M- 1 problem (problem #2), and the M-2 problem (problem #3).

In problem #1 (the M problem) the application of CDS heuristic produces the following processing sequence: B - D - C - A.

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Chapter 08A Solution Manual

In problem #2 (the M-1 problem) the application of CDS heuristic produces the following processing sequence: B - D - A - C. In problem #3 (the M-2 problem) the application of CDS heuristic produces the following processing sequence: D - B - C - A.

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Chapter 08A Solution Manual

Problem #1 Sequence: B - D - C - A Job A B C D

M1

B

D

0 1

M2

Machine 1 6 1 5 4

C

Machine 3 2 5 3 6

A

5

10

B

16 D

1

9

M3

C

A

13

20

B

D

9

C

16 17

M4

22

20

A 22

B 16

D 21

Make-span = 32 Problem #2 (the M-1 problem) Sequence B - D - A - C Job A B C

Machines 1 & 2 8 9 12

8A-17

26

Machines 2 & 3 6 12 5

C 27

A 30

32

Chapter 08A Solution Manual

D

M1

B

8

D

7

A

0 1

M2

C

5

11

B

D

1

16 A

9

13

M3

C 15 16

B

23

D

9

A

16 17

M4

C 21

23

B

25 D

16

A

21

27

C 29

32

Make-span = 32 hours Problem #3 Sequence: D - B - C - A Job A B C D

M1

D 0

B

Machines 1, 2, & 3 12 16 14 9 C

A

4 5

M2

10 D

4

M3

16 B

C

8

16 D

A 23

B

8 9

M4

Machines 2, 3,& 4 8 20 12 11

16

C 23

D 15

Make-span = 33 hours b. The best sequence is either B - D - C - A or B - D - A - C c. See part A.

8A-18

25

23

A 25

29

B

C 28

A 31

33

Chapter 09 Solution Manual

CHAPTER 9 Solution Manual Discussion Questions 1. JIT techniques largely apply to the back end of the MPC system. Also much of the early publicity for JIT came out of the Toyota experience with Kanban. The focus was on the "pull" aspects of the Kanban system and the reductions in inventories Toyota was achieving. Since the two were being publicized at the same time, many people associated the inventory reductions with the Kanbans, not all the other activities that were involved. Some of that history has stuck.

2. This can be a fun question for the students to answer (you might invite the registrar to the session). As starters, the students will think of the wasted time (theirs) in the system. But you can push them to think of what the product is, what is quality, where there are inventories, etc.

This will lead directly to ways to get at improvement. They must start by knowing the objectives, the current process, whose waste it is (what administrator cares about student time?) and the resources available. From there they can make suggestions.

3. Each of these institutions has some capacity for handling surges. The football stadium is the most limited, having fixed (almost) seating. It takes on additional people by having them at the sidelines, in the aisles and in temporary seats. The clothing store must rely on inventory and/or very fast delivery. In the accounting firm (as with students) the surge capacity comes from the people working faster and more hours.

4. It is necessary to distinguish between closely located and distant facilities for some of the signals. Yelling, for example, requires proximity, as does a light, a hand signal, and some noise systems (restaurant). For some of the physical things, like golf balls in tubes, model rail cars, and wire guided vehicles, distance is also limited. For card-like signals (empty containers, paper and computer messages) the distance is not so critical.

5. Clearly such firms need to think differently about their inbound transportation. Can they run "bus routes" to pick up smaller amounts from several suppliers at a time? Some trucking companies are designing “JIT” trailers just for this purpose. Would the savings in inventory and other expenses offset the use of commercial package delivery systems for small quantities? Can they combine with other firms in the area to have suppliers deliver more frequently to all firms in the group? Can they locate some of their activities nearer to their suppliers?

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6. Unfortunately, this is very close to some of the early complaints about U.S. automotive company JIT. There are two clues as to what might be going on. The first is that the company got rid of some of its inventory. This probably happened by telling the suppliers that they were going to have to make frequent small deliveries, thereby shifting the inventory to the suppliers (hence the mountain in their shops). This is not an uncommon interpretation of what JIT means. The second clue is the weekly specification of what is needed with no evidence of any other information being shared. The week is unlikely to be long enough to cover the suppliers' lead times (another reason for big inventories). There is no evidence in the quote that there is any level loading or forward planning for the suppliers.

7. This just shows how far we have to go. It is clear that the manager just hired brawn and hours.

He could not think of such things as quality circles, cleaning up the plant, working on product or process improvements, learning another skill, studying quality control, etc. This attitude is very prevalent in Western industry today.

8. Wow! Many managers are wondering about this very same question right at this moment.

Some principles do come from the chapter, however. Clearly one starts with the people. It is a person oriented approach as opposed to a material, equipment or technology oriented approach. This implies training and persuasion. It also implies moving slowly and deliberately, with achieved successes at each step (i.e., cherry pick the first projects). Since the approach requires getting people to think differently, this is the first priority for management. The specific project is of less importance, as long as it meets the criteria above.

Chapter 9 Problem Summaries Problem Difficulty

Problem Type

1

Easy

Requires calculating the number of Kanbans and demand conversion.

One part requires the conversion of demand to weeks from days.

2

Moderate

Requires calculating the number of Kanbans and requires the lead time to be converted from hours to days.

3

Hard

This problem looks at the reduction of inventory investment by moving to a Kanban environment from a traditional batch environment. It is also examines the savings in WIP investment from reducing lead time.

4

Moderate

5

Easy

Cost analysis of the value of moving to JIT The problem is a simple problem to create a level schedule for a mixed model assembly line and calculating the number of Kanbans for a component.

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Chapter 09 Solution Manual

Chapter 9 Problem Summaries Problem Difficulty

Problem Type

6

Moderate

Leveling the production schedule for a mixed model line when capacity is not given.

7

Hard

Mixed model schedule and minimum batch size for a mixed model line.

The difficulty of this problem is that some items are only produced once in an four day cycle.

8-11

Moderate

Demonstrates master scheduling, MRP explosion, the use of phantom bills of materials, and rough capacity planning all the in context of a companying using JIT concepts.

12-15

Moderate

Demonstrates demand pull logic and does a comparison to MRP logic with level schedules.

1. ABC Company Product 1

Product 2

Kanbans 300(1)(1.15)/20 = 17.25 or 18 150(5)(2)(1)/30 = 50 Note that Product 2’s demand is per day so it must be converted to weekly demand by multiplying it by 5.

2. ABC Company

D L (in days) S (in decimal) C (D *L * (1+S))/C = Kanbans

W

X

Y

Z

900 0.25 0.25 25 11.25 12

250 0.625 0.2 40 4.69 5

1200 0.125 0.15 50 3.45 4

350 0.375 0.1 20 7.22 8

3. BCD Company a.

Q/2 x $150 = 6000/2 x (150)= $450,000 b.

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Chapter 09 Solution Manual

Kanbans = (D x L x (1+S))/ C = 250 x (10) x (1.1)/100 = 2750/100 = 27.5 ≈ 28 Average inventory = 28 kanbans (100 units) = 2800 Change investment = 3000-2800 = 200 x $150 = $30,000 c.

Kanbans = (D x L x (1+S))/ C = 250 x (5) x (1.1)/100 = 1375/100 = 13.75 ≈ 14 Average inventory =14 kanbans(100 units) = 1400 Change in investment = 2800 - 1400 = 1400 x $150 = $210,000 $210,000/$100,000 = 210% return 4. CDE Company

Savings Purchase Orders Obsolescence Vendor Maintenance Shipping Expediting Total

Cost $60,000 20,000 10,000 5,000 5,000 $100,000

$25,000

Cost Savings

$75,000

5. DEF Company a.

Product Line

1

2

Forecast for year 20,000 10,000 Daily Batch 80 40 Hourly Batch 5 2.5 Mixed Model Schedule 4 2

3 5,000 20 1.25 1

b. Daily schedule: 1,1,1,1,2,2,3 – repeated 20 times c. Kanbans = 15 Kanbans = DL(1 + S)/C = 80 (2) (2) (1.15)/25 = 14.72 ≈ 15 containers 6. EFG Company a.

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Chapter 09 Solution Manual

Product Line

I

II

III

IV

Forecast for year Daily Batch Hourly Batch Mixed Model Schedule

500 2 .25 1

1500 6 .75 3

4500 18 2.25 9

6000 24 3 12

b. I, 3IIs, 9IIIs, 12IVs – repeated once

7. FGH Company Number Of Batches Produced Of Each Product Each Day Product

Daily Requirement

1 2 3 4 5 6 7 8 9

720 600 480 480 360 120 60 30 30

Batches/day

1 6 5 4 4 3 1 1

Day of cycle 2 3

4

6 5 4 4 3 1

6 5 4 4 3 1

6 5 4 4 3 1 1

1 1 24

24

24

8. GHI Company a.

Model

A

B

C

D

Forecast for year 1000 3000 7000 9000 Daily Batch 4 12 28 36 Hourly Batch .5 1.5 3.5 4.5 Mixed Model 1 3 7 9 b. 9Ds, 7Cs, 3Bs, 1A – repeated four times

9-5

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Chapter 09 Solution Manual

9.

GHI Company

Week

1

2

3

4

5

6

7

Gross Requirements 150 150 Scheduled Receipts Proj. Avail. Bal 200 50 700 Planned Ord. Rel 800 Lot size =800, safety stock = 0, lead time =1

150

150

150

150

150

550

400

250

100 800

750

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Chapter 09 Solution Manual

10. GHI Company a.

Air Conditioner Part no. 101 (1 required) Part no. 102 (2 required) Part no. 103 (1 required) Coupling (1/4 required) Bracket (1/4 required) Alternator (1/4 required) b.

Week

1

2

3

4

5

6

7

150

150

150

150

150

0 150

0 150

0 150

0 150

0

2

3

4

5

6

7

Gross Requirements 100 150 Scheduled Receipts Proj. Avail. Bal 250 150 200 Planned Ord. Rel 200 200 Lot size =200, safety stock = 20, lead time =2

150

150

150

150

50 200

100 200

150

200

Gross Requirements 150 150 Scheduled Receipts Proj. Avail. Bal 200 50 0 Planned Ord. Rel 100 150 Lot size = lot-for-lot, safety stock = 0, lead time =0 c. MRP record for coupling Week

1

11. GHI Company a. (Beginning time + ending time)/2 = (10.5 + 8)/2 = 9.25 b. 7.2 x90% = 6.5. Therefore 14000-15000 is a good estimate (another doubling from 7000) c. (10 hours – 5.5 hours)/5.5 hours (125) = daily surge capacity = 103.

Total capacity per week =(125+103)x5= 1136 This problem lends itself to some nice in-class questions as to the costs and benefits of the extra time available for the assembly team. One can ask what his costs, whether the firm should try to reduce the team size, what else the team might do, and if they could work on mass customization

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Chapter 09 Solution Manual

12. ZMW Motorwerks a. 500/2 (10 models) + 5 safety stock x 10 models= 2550 units b. Slippery maintains a finished goods inventory of one batch for each type. FG inventory is only held while waiting for the next order to come in. After an order is shipped from Slippery, the cycle looks like this: Days 1-3 inventory is replenished when producing the batch of 500 to replace the batch just shipped. Average inventory over those three days is 0 (to avoid double-counting day 3 as both WIP and FG, you would not say 500/3=166.66; see the answer for 12c). Days 4-10 wait for the next order to come in to ship out the batch of 500. Average inventory for the 7 days is 500. Thus, the average inventory for each item is (7*500)/10 = 350 x 10 = 3500 c. Demand = 500/week x 3 week lead time = 1500 d. Transaction costs: ordering, shipping, receiving invoicing, picking, inventory monitoring, quality checking, returns, repairs, selecting units at line, production scheduling at Slippery 13. Slippery a. The schedule does not make every model every day, nor does it schedule the amounts of production in accordance with the demand mix. All batches are 500 b. The average batch size would be 10, since all ten models would be made every day and the daily production at ZMW is 100 units. In fact, the actual quantities would vary greatly from this with some products not made every day. c. ZMW would need to keep smaller inventories, and not the same amount of each model.

They should be able to reduce ordering as well, since they would not like to order the smaller batches with the same approach. d. Slippery needs to have visibility into ZMWs inventory, adjusting its build schedule to reflect actual demand. Slippery also needs to change its building approach so that it can operate easily with smaller batches and a daily mixed model schedule 14. Slippery a. $3 (250x100) = $75,000 b. Negative – since they probably will incur more damage costs c. $3 (250x100) = $75,000 d. No clear benefit indicated unless cost savings passed on. e. Need to create a win-win approach so that perhaps the $150,000 is split three ways

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Chapter 09 Solution Manual

15. ZMW Motorwerks a. 2500 units versus new average inventory = 25/2 = 12.5 units + 10 units of safety stock = 22.5 units b. Virtually no finished goods inventory, WIP goes from 3 weeks to less than 1 day, raw materials inventory = ?? c. Hidden factory is much reduced d. Build schedule actual sequence must be passed to Slippery e. Flawless execution!

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Chapter 10 Solution Manual

CHAPTER 10 Solution Manual Discussion Questions 1. There are a couple of analogies here that are helpful. One is that of a duck hunter, the idea being that a duck hunter is trying to hit a moving target. He must be mobile, quick and responsive. The same is true of a kick returner in football. At the same time he is thinking of the other team (market) trying to get him, he must think of deploying his blockers to intercept them. Both of the analogies deal with the issue of anticipation and directing resources where they are needed. The master scheduler must direct the material resources (that have been in process for awhile) to the market need and start new production to meet future needs. He/she must be concerned with where as well as when. It doesn't count to get the shipment to New York exactly when the customer wanted it, say at 8:27, if the customer is in Toronto.

2. We view the DRP records as a "bridge" carrying information into and out of the market. On the basis of this information, the physical product is moved from the factory to the customer, again a form of bridge.

3. In a word, planning! The data that can be provided by DRP to the warehouse manager in the field provides the basis for planning. Knowing when shipments will arrive allows the manager to plan manning levels at the loading/unloading dock and to schedule vehicles to make outbound deliveries. The savings from such planning can be enormous. One trucking firm, using forward planning information, was able to cut the labor costs of the warehouses and drivers nearly in half. This was from a combination of savings in overtime and fewer people on the payroll.

4. The benefits are those mentioned in the chapter, having a closer match between market need and product availability. The risks, however, are more subtle. Apart from the obvious, of someone being malicious, there is always the chance that the field will start to outguess the marketing people. Then there is nobody responsible for the forecast and making the plan work.

5. There is one big difference. The TPOP record provides management with planning information. The system used by the consumer goods manufacturer does not base orders on the anticipated future need, but on the current inventory. Thus no information is available to do future planning. There is only the order (or no order).

6. This question is no different from the monitoring of any forecast. How good is it? In this case, there is a basis of comparison. Is it as good as the forecast from a computer? The way to monitor is quite clear. The data on the original forecast and subsequent changes must be kept and later compared to actual demand.

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Chapter 10 Solution Manual

7. There are three key differences. First, the economics of the order size are quite different. The shipping quantity depends on packing alternatives and transport costs, not machine setups.

Second, the lead times are related to moving times only. There is no processing and little queue time associated with transportation (some people argue that rail may be the exception).

Finally, there is not much that can be done to speed up or slow down the progress of a shipment once on its way to the destination.

8. Both the supply chain and materials management concepts have the same integrated view of materials flows. They both view the process from purchase of raw material to delivery of finished product as a continuous activity. On the other hand, a supply chain is much more related to the physical process involved in transforming the raw material into final product (perhaps that is why it was first popular among chemical and food companies). The materials management view, on the other hand, is much more organizational. It looks at the functions performed and argues that they should be organized under one department.

7

Chapter 10 - Problem Summaries Problem

Difficulty

Problem Type

1

Moderate

Rolling through time using DRP records and allowing orders to move based on actual demand.

2

Moderate

Rolling through time using DRP records. This problem compares updating records with and without error add-back.

3

Easy

Using DRP records with EOQ.

4

Easy

Processing requirements with DRP records.

5

Moderate

Processing DRP records using time phased order point with a central warehouse configuration.

6

Hard

Comparing EOQ versus DRP in a distribution system. This problem requires a cost calculation for setup and inventory.

7

Hard

Using DRP at a central warehouse and linking it to plant purchasing.

Part (c) requires updating the records after several transactions.

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Chapter 10 Solution Manual

8

Moderate

Using a spread sheet for processing DRP records and then replanning the records because of limited capacity.

9

Moderate

Using a spread sheet for processing DRP records and assessing the impact of changing the order quantity at the central warehouse.

10

Moderate

Extends #9 by adjusting for reduced demand at one warehouse.

11

Moderate

Allocating a monthly sales forecast to a weekly forecast at a specific warehouse.

12

Moderate

Using safety lead time in a DRP system.

13

Moderate

Opening a new warehouse and closing an old one in a DRP system.

14

Moderate

Using DRP records to deal with perishable items by creating exception notices based on projected inventory levels.

15

Hard

Rolling through time and updating the records for actual demand in a multilevel DRP system.

1. BOB STAHL a.

Forecast Requirements In Transit Projected Available Balance Planned Shipments Actual Demand (1) = 24

Forecast Requirements In Transit Projected Available Balance Planned Shipments Actual Demand (2) = 38

10

46

10-3

1 30 60 40

2 30 10 60

2 30

3 30

16 60

46

Week 3 30 40

Week 4 30 16 60

4 30

5 30

10 60

40

5 30

6 30

46

16

Chapter 10 Solution Manual

Forecast Requirements In Transit Projected Available Balance Planned Shipments Actual Demand (3) = 27

Forecast Requirements In Transit Projected Available Balance Planned Shipments Q = 60; lead time = 1; SS = 10

8

41

3 30 60 38 60

4 30 60 71 60

4 30 68

5 30 41

Week 5 30 38 60

Week 6 30 11 60

6 30

7 30

68

38

7 30

8 30

41

11 60

4 30

5 30

10 60

40

5 30

6 30

40

10

b.

Forecast Requirements In Transit Projected Available Balance Planned Shipments Actual Demand (1) = 24

Forecast Requirements In Transit Projected Available Balance Planned Shipments Actual Demand (2) = 38

10

1 30 60 40

2 30

Week 3 30

10 40 60 cumulative error = +6

2 36 46

3 30

Week 4 30

10 40 10 60 60 cumulative error = -2

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Chapter 10 Solution Manual

Forecast Requirements In Transit Projected Available Balance Planned Shipments Actual Demand (3) = 27

Forecast Requirements In Transit Projected Available Balance Planned Shipments Q = 60; lead time = 1; SS = 10

8

3 28 60 40

4 30

Week 5 30

10 40 60 cumulative error = +1

41

4 31

5 30

10 60

40

Week 6 30 10 60

6 30

7 30

10 60

40

7 30

8 30

40

10 60

The shipment in period 3 is delayed to period 4 when err add-back is used. This conforms to the plans made in period 1

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Chapter 10 Solution Manual

2. DRP RECORDS a.

Forecast Requirements In Transit Projected Available Balance Planned Shipments Actual Demand (1) = 24

Forecast Requirements In Transit Projected Available Balance Planned Shipments Actual Demand (2) = 16

Forecast Requirements In Transit Projected Available Balance Planned Shipments Actual Demand (3) = 18

Forecast Requirements In Transit Projected Available Balance Planned Shipments

6

22

6

28

1 20 40 26

2 20 6 40

2 20

3 20

2 40

22 40

3 20 40 26

4 20 6 40

4 20

5 20

8 40

28

Week 3 20 26

Week 4 20 42

Week 5 20 26

Week 6 20 8 40

4 20

5 20

6 40

26

5 20

6 20

22 40

42

6 20

7 20

6 40

26

7 20

8 20

28

8

The forecast error in period 1 causes the PAB to drop below safety stock in week 2. In general, there is little point in doing anything. That is what safety stock is for -- to be used to buffer forecast errors. As seen in the third record, the problem self corrects after week 2. Note the shift of the planned order for week 4 to week 3 and back again.

10-6

Chapter 10 Solution Manual

b. The error add back method leads to the following cumulative errors: Week 1 2 3

Cumulative Error -4 0 +2

Since the errors are within the safety stock, this method appears to work.

3. HAZY COMPANY a. ROP = demand during lead time + safety stock ROP = 5 + 4 = 9 units (the probability of demand > 9 units = .09  .1 )

EOQ =

2  5 10 = 10units 1

b.

Forecast Requirements In Transit Projected Available Balance Planned Shipments Q = 10; LT = 1; SS = 2 c.

Forecast Requirements In Transit Projected Available Balance Planned Shipments Q = 10; LT = 1; SS = 2

9

3

1 5

2 5

3 5

4 5

4 10

9

4 10

9

2 5 10 8

3 5

4 5

5 5

3 10

8

3 10

10-7

Week 5 6 5 5

7 5

8 5

9 5

10 5

4 10

4 10

9

4 10

9

Week 6 7 5 5

8 5

9 5

10 5

11 5

8

8

3 10

8

3 10

9

3 10

Chapter 10 Solution Manual

4. TASMANIAN TIRE COMPANY

Forecast Requirements In Transit Projected Available Balance Planned Shipments Q = 70; LT = 1; SS = 0

10

1 35 70 45

2 35

3 35

10 70

45

Week 4 5 35 35

6 35

7 35

8 35

10 70

10 70

45

10

Week 3 4 20 20

5 20

6 20

5 40

2 40

25

Week 3 4 10 10

5 10

6 10

5 20

5 20

15

Week 3 4 60

5 60

6

0

0

0

45

5. XYZ COMPANY

Warehouse A Forecast Requirements Scheduled Receipts Projected Available Balance Planned Shipments Q = 40; LT = 1; SS = 0

Warehouse B Forecast Requirements Scheduled Receipts Projected Available Balance Planned Shipments Q = 20; LT = 1; SS = 0

Central Warehouse Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = lot-for-lot; LT = 1; SS = 0

10

15

60

10-8

1 20 15 5 40

2 20 25

1 10

2 10

5 20

15

1 60

2

0

0 60

25

15

0 60

Chapter 10 Solution Manual

$3000

INVENTORY INVESTMENT $

$2500 $2000 $1500 $1000 $500 0 1

2

3

4

5

6

WEEK

6. BLUES BROTHERS’ SUNGLASSES a. Using EOQ Warehouse 1 or 2

EOQ =

2  40  15 = 69.2870units .25

Forecast Requirements In Transit Projected Available Balance Planned Shipments Q = 70, LT = 1, SS = 0

40

1 40

2 40

3 40

0 70

30 70

60

20 70

5 40

6 40

7 40

50

10 70

40

(210  $.25)  7 = (4  $15)  7 =

Inventory cost per week Setup Cost per week Cost per week for each warehouse Joliet Plant

EOQ =

Week 4 40

2  80 15 = 97.97units 100units .25

10-9

$7.50 $8.57 $16.07

Chapter 10 Solution Manual

Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 100, LT = 1, SS = 0

140

1 80

2 80

3 80

Week 4 80

60 100

80

0 100

20 100

5 80

6 80

7 80

40 100

60 100

80

Note: These are planned costs, assuming independent management (340  $.25)  7 = (5  $15) 7 =

$12.14 $10.71 $22.85

$16.07  2 + $22.85 =

$54.99

Inventory cost per week Setup Cost per week Cost per week for plant Total Cost per week (all locations) b. Using DRP Warehouse 1 or 2

Week Forecast Requirements In Transit Projected Available Balance Planned Shipments Q = 70, LT = 1, SS = 0

40

1 1 40

2 2 40

3 3 40

0 70

30 70

60

Week 4 4 40 20 70

5 5 40

6 6 40

7 7 40

50

10 70

40

(210  $.25)  7 = (4  $15) 7 =

Inventory cost per week Setup Cost per week Cost per week for each warehouse

$7.50 $8.57 $16.07

Joliet Plant

Gross Requirements Scheduled Receipts Projected Available Balance 140 Planned Order Release S = Lot-for-Lot, LT = 1, SS = 0

1 140

2 140

3

0 140

0

0 140

10-10

Week 4 140 0

5

6 140

0 140

0

7

Chapter 10 Solution Manual

(0  $.25)  7 = (3  $15)  7 =

$0.00 $6.42 $6.42

$16.07  2 + $6.42 =

$38.56

Inventory cost per week Setup Cost per week Cost per week for plant Total Cost per week (all locations)

With coordinated shipments and lot-for-lot order policy at the plant, the costs at the plant are reduced substantially while warehouse costs remain the same as in question a.

7. MVA PET FOOD COMPANY a.

Gro-Pup Seattle Warehouse Forecast Requirements In Transit Projected Available Balance Planned Shipments Q = 60; LT = 1; SS = 0.

Gro-Pup Portland Warehouse Forecast Requirements In Transit Projected Available Balance Planned Shipments Q = 15; LT = 1; SS = 0

Gro-Pup Plant Warehouse Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Releases Q = 50; LT = 2; SS = 0

43

2

1 20

2 20

3 20

Week 4 5 20 20

23

3 60

43

23

2 10

3 10

Week 4 5 10 10

6 10

7 10

8 10

12

2 15

7 15

2 15

7 15

12

3 15

Week 4 5 15 60

6 15

7 15

8

25

0

35

1 10 15 7 15

1 15 30

15 50

10-11

2 60 50 5

40 50

3 60

12

15 50

6 20

7 20

8 20

43

23

3

Chapter 10 Solution Manual

b.

Gro-Pup Packaging Material Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Releases Q = 100; LT = 4; SS = 0

25

1 50 100 75 100

2

3 50

Week 4 5 50

75

25

25

Week 5 6 20 20

6

7

8

7 20

8 20

9 20

41

21

1 60

41

7 10

8 10

9 10

0 15

5 15

10

75

c.

Gro-Pup Seattle Warehouse Forecast Requirements In Transit Projected Available Balance 21 Planned Shipments Q = 60; LT = 1; SS = 0.

Beginning Inventory 43 - 3 - 19 = 21

Gro-Pup Portland Warehouse Forecast Requirements In Transit Projected Available Balance 0 Planned Shipments Q = 15; LT = 1; SS = 0 Beginning Inventory = 2 + 3 + 15 - 20 = 0

2 20

3 20

4 20

1 60

41

21

2 10 15 5 15

10-12

1 60

3 10

4 10

Week 5 6 10 10

10

0 15

5 15

10

Chapter 10 Solution Manual

Gro-Pup Plant Warehouse Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Releases Q = 50; LT = 2; SS = 0 Beginning Inventory = 30 - 15 = 15

15

2 75 50 -10

Gro-Pup Packaging Material 2 Gross Requirements Scheduled Receipts Projected Available Balance 69 69 Planned Order Releases Q = 100; LT = 4; SS = 0 Beginning Inventory = 25 + 4 + 90 - 50 = 69

3 50 40 50

4 15 25

Week 5 6 75 0 50

0 50

7 15

8 75

35

10

7

8

3 50

4

19

19

Week 5 6 50 50 100 69 19

2 30

Period 3 4 30 60

5 60

6 60

32 40

42 40

22 48

10 60

10 0

2 40

Period 3 4 40 48

5 60

6 0

0 40

0 48

0

0

8. THE DRIP PRODUCERS

Warehouse Gross Requirements In Transit Projected Available Balance Planned Order Releases Q = 40; LT = 1; SS = 10

Central Warehouse Forecast Requirements Scheduled Receipts Projected Available Balance Planned Order Releases Q = Lot-for-Lot; LT = 1; SS = 0

12

0

b.

10-13

1 30 40 22 40

1 40 40 0 40

0 60

9

9

Chapter 10 Solution Manual

Central Warehouse 1 2 Gross Requirements 40 40 Scheduled Receipts 40 Projected Available Balance 0 0 0 Planned Order Releases 40 48 Q = Lot-for-Lot; LT = 1; SS = 0; Cap = 50 / period

Period 3 4 40 48 8 50

10 50

5 60

6 0

0

0

There are requirements for 47 units per period at central. Producing 47 per period will work, as does the schedule above.

9. CRANSTABLE COMPANY This problem has the three records together with both of the warehouses creating demands on the central facility. The planned order releases for quantities of 100 and 200 are shown for the central warehouse.

Warehouse 1 Forecast Requirements In Transit Projected Available Balance Planned Shipments Q = 48; LT = 1; SS = 10

Warehouse 2 Forecast Requirements In Transit Projected Available Balance Planned Shipments Q = 60; LT = 1; SS = 10

23

12

10-14

1 20 48 51 0

1 30 60 42 0

2 20

Period 3 4 20 20

5 20

6 20

31 0

11 48

39 0

19 48

47 0

2 30

Period 3 4 30 30

5 30

6 30

12 60

42 0

42 0

12 0

12 60

Chapter 10 Solution Manual

Central Warehouse Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Releases Q = 100; LT = 1; SS = 0

0

1 0

2 60

Period 3 4 48 60

0 100

40 100

92 0

32 100

84 0

84 0

200

0

0

200

0

0

5 48

6 0

Planned order releases for Q = 200 Planned Order Releases

10. CRANSTABLE COMPANY -- DRP This problem illustrates the "dying" product attributes of DRP. At warehouse 1, the planned shipments disappear, indicating a potential age / date problem. At Central, the planned order releases are reduced and delayed. If there is work in process, there may be a need to sharply change priorities or, if not possible, to run a promotion to clear the inventories.

Warehouse 1 Forecast Requirements In Transit Projected Available Balance Planned Shipments Q = 48; LT = 1; SS = 10

Warehouse 2 Forecast Requirements In Transit Projected Available Balance Planned Shipments Q = 60; LT = 1; SS = 10

23

12

10-15

1 10 48 61 0

1 30 60 42 0

2 10

Period 3 4 10 10

5 10

6 10

51 0

41 0

31 0

21 0

11 0

2 30

Period 3 4 30 30

5 30

6 30

12 60

42 0

42 0

12 0

12 60

Chapter 10 Solution Manual

Central Warehouse Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Releases Q = 100; LT = 1; SS = 0

2 60

0 100

40 0

40 100

40 0

80 0

80 0

200

0

0

0

0

0

0

1 0

Period 3 4 0 60

5 0

6 0

Planned order releases for Q = 200 Planned Order Releases

11. CRANKY TEA COMPANY Month

Constant Torment / Warehouse C Forecast Requirements* In Transit Projected Available Balance Planned Shipments Q = lot-for-lot; SS = 0; LT = 3 weeks

1 100 105

5 190

1 Week 2 3 250 450 260 445 15 10 100 250

4 200

1 100

0 450

0 200

2 Week 2 3 250 450

4 200

0

0

0

*National Sales X Warehouse Split = 4,000 units / mo. X .25 = 1000 units / mo. 1000 units / mo.

X weekly split

12. EXCELLO CORPORATION

Product X 1 Forecast Requirements 200 In Transit 410 Projected Available Balance 15 225 Planned Shipments 600 Q = 600; SLT = 1 week; LT = 1 week

10-16

2 200

3 200

25

425

Week 4 5 200 200 225 600

25

6 200

7 200

8 200

425

225

25

Chapter 10 Solution Manual

13. ALLIED PRODUCTS COMPANY Product B / San Diego Warehouse 1 2 Forecast Requirements 100 100 In Transit Projected Available Balance 205 10 5 Planned Shipments 105 100 Q = lot-for-lot; LT = 2 weeks; SS = 10 units Product B / Los Angeles Warehouse 1 Forecast Requirements In Transit Projected Available Balance 0 0 Planned Shipments Q = lot-or-lot; LT = 1 week; SS = 10 units

3 100

Week 4 5 100

6

7

8

10

10

10

10

10

10

2

3

Week 4 5 100

6 100

7 100

8 100

0

0

0 110

10 100

10 100

10 100

10

Week 4 5 110 100

6 100

7 100

8

185

185

85

85

Product B / Plant 1 2 Gross Requirements 105 100 Scheduled Receipts Projected Available Balance 300 195 95 Planned Order Releases 200 Q = 200 units; LT = 3 weeks; SS = 15 units

10-17

3

95 200

85

Chapter 10 Solution Manual

14. STASIK PHARMACEUTICAL COMPANY

Product W Forecast Requirements In Transit Projected Available Balance Firm Planned Order Planned Order Release Weeks of supply:

1 300 350

50

Week 4 5 300 300

6 300

7 300

8 300

2 300 900 650 900

3 300 350

950 900

650

1250 900

950

1550

2.2

1.2

3.2

2.2

4.2

3.2

5.2

0.2

Q = 900; SS = 50 units; LT = 2 weeks * Excess inventory in weeks 6 and 8. * Need to consider rescheduling 900 unit orders in weeks 4 and 6 or reducing the firm planned order quantities.

15. AMC CHEMICAL COMPANY a.

Product D/Dallas Warehouse Forecast Requirements In Transit Projected Available Balance 85 Planned Shipments Q = LFL: SS = 5 units; LT = 2 weeks.

1 40

2 40

3 40

Week 4 5 40 40

45 40

5 40

5 40

5 40

5 40

5 40

5

5

6

7

8

6 40

7 40

8 40

Product D/Plant Warehouse

1

2

3

Week 4 5

Gross Requirements

40

40

40

40

40

40

Scheduled Receipts

46 8

8

8

8

8

8

40

40

40

40

40

Projected Available Balance Planned Order Releases Q = LFL; SS = 8 units; LT = 1 week.

2

10-18

Chapter 10 Solution Manual

Ingredient X

1

2

3

Week 4 5

Gross Requirements

80

80

80

80

80

Scheduled Receipts

320 164

84

4

244

Projected Available Balance

4

244

Planned Order Releases Q = 320 units; SS = 2 units; LT = 4 weeks.

320

6

7

8

7

8

9

b.

Product D/Dallas Warehouse

2

3

4

Week 5 6

Forecast Requirements

40

40

40

40

40

40

40

40

5

5

8

9

8

9

In Transit Projected Available Balance Planned Shipments Q = LFL: SS = 5 units; LT = 2 weeks.

Old On Hand - Sales = New On Hand

40 37

-3

-3

5

5

5

5

48

40

40

40

40

40

7

85 - 48 = 37

Product D/ Plant Warehouse

2

3

4

Week 5 6

Gross Requirements

48

40

40

40

40

40

Scheduled Receipts

40 8

8

8

8

8

Projected Available Balance

8

0

Planned Order Releases 48 40 40 40 40 Q = LFL: SS = 8 units; LT = 1 week.

Old On Hand - Disbursement + Receipts = New On Hand 2 + 46 - 40 = 8

Ingredient X

2

3

4

Week 5 6

Gross Requirements

96

80

80

80

Scheduled Receipts Projected Available Balance

80

320 244

148

68

Planned Order Releases Q = 320 units; SS = 2 units; LT = 4 weeks.

Old On Hand - Disbursements + Receipts = New On Hand

10-19

-12

228

148

4 - 80 + 320 = 244

7

Chapter 10 Solution Manual

•

If there is a real concern that demand is increasing, then try to expedite the shipment of 320 units due in week 5 into week 4. c.

Product D/Dallas Warehouse

3

4

5

Week 6 7

Forecast Requirements

40

40

40

40

40

40

40

40

In Transit

40

48

6

14

5

5

5

5

5

5

Planned Shipments 31 40 40 Q = LFL: SS = 5 units; LT = 2 weeks.

Old On Hand - Sales + Receipts = New On Hand 37 - 31 = 6

40

40

40

8

9

10

9

10

Projected Available Balance

6

8

9

10

Product D/Plant Warehouse

3

4

5

Week 6 7

Gross Requirements

31

40

40

40

40

40

Scheduled Receipts

48 8

8

8

8

8

Projected Available Balance

0

17

Planned Order Releases 31 40 40 40 40 Q = LFL: SS = 8 units; LT = 1 week.

Old On Hand - Disbursement + Receipts = New On Hand 8 + 40 - 48 = 0

Ingredient X

3

4

5

Week 6 7

Gross Requirements

62

80

80

80

80

326

166

86

Scheduled Receipts Projected Available Balance

8

320 148

86

326

Planned Order Releases Q = 320 units; SS = 2 units; LT = 4 weeks.

Old On Hand - Disbursements = New On Hand 244 - 96 = 148 •

Try to de-expedite the order for 320 units now due in week 4 to week 5.

•

Suggest using firm planned orders in the MPS of, say, 40 units / week to avoid passing sales uncertainty to the suppliers.

10-20

Chapter 10A Solution Manual

CHAPTER 10A Solution Manual Discussion Questions 1. The strength of supply chain logistics is being able to look at the whole logistics process at once. This allows a firm to actually try to reduce the total cost to the final customer. The weakness of this approach is that some of the players in the supply chain actually have increased their cost by following actions that reduce their overall cost to the customer.

2. The key barrier to implementing the total cost approach is that some supply-chain members may be forced to incur extra cost and reduced profit, so that the total the delivered cost to the final customer is less. These firms will have a less than optimal solution for their own business in order to optimize the solution for the entire supply-chain 3. Over the course of their college careers, many students become better at studying. This is a form of evolution that occurs by learning what works best and what does not work. Students can review their performance in different courses to see what studying techniques worked and which didn't. Through time, they should improve by only keeping those techniques that are successful.

4. Customers have to be shown that not voting safety stock is not a detriment to their business.

The company can also offer financial incentives to make it worthwhile for the customer.

Another technique that can be used is to create a small demonstration that will show the customers the benefit of centralized safety stock.

5. Items that have more erratic demand may be a stronger candidate for an ROP/EOQ system.

These items may be harder to predict into the future and may have low-flying usage that will not justify a shipment from the base stock system. Theoretically nature of these items may also not lend themselves to a DRP system sends for prediction of demand is difficult. They stock systems would be useful on items that are delivered from the same suppliers. In those systems are able to couple orders together to take advantage of decrease shipping cost. The DRP system would be best for items that are more predictable and therefore more able to project ahead to usage in the future and use the detailed planning record accordingly.

6. Way to reduce the data requirements for sites that are not going to be used is to do a first level cut of potential sites. A firm can start by only looking at the most important criteria, and any sites that don't meet that criteria will be eliminated from further analysis. Their elimination from further analysis will reduce the amount of data that a firm must collect.

7. Them the way they should choose a customer performance measure is you should find out what customers value first. The performance measure could be different for different value segments in your business. But the most important step is to talk to the customer first and find out what is important to them. Many firms made that mistake of not talking to the customer first.

10A-1

Chapter 10A Solution Manual

Chapter 10 - Problem Summaries Problem

Difficulty

Problem Type

1

Easy

A pair in a quantity discounts for shipping versus a regular overnight rate.

2

Moderate

A warehouse location problem with five warehouse, two that cannot serve all customers.

3

Moderate

A problem using the time saved heuristic that has a depot and three customer locations. There is a time and added a second part of the problem.

4

Moderate

Another problem using the time saved to your heuristic with a warehouse, and four customer locations. There is also capacity constraint on the vehicle.

5

Moderate

This problem determines the best shipping lanes shipping lanes to use to supply two cities.

10A-2

Chapter 10A Solution Manual

1. CORD COMPANY a. Comparing Methods ship by month versus 1,000 Ship by Month January February March April May June Total Demand 50 100 100 50 100 50 Shipment 50 100 100 50 100 50 Inventory 0 0 0 0 0 0 Inventory Cost $0 $0 $0 $0 $0 $0 $0 Landed Cost $1,055 $ 2,110 $2,110 $1,055 $2,110 $1,055 $9,495

Ship 1,000

Shipment Inventory Inventory Cost LandedCost

January February March April May June Total 50 100 100 50 100 50 1,000 0 0 0 0 0 950 850 750 700 600 550 $20,000 $20,000 $950 $850 $750 $700 $600 $550 $4,400 $24,000

There is no reason to try and ship 1,000 units since the inventory cost makes it way too expensive. b. Cost when the restriction is 200 units 1,000 Ship to take advantage of discount

Shipment Inventory InventoryCost LandedCost

January February March April May June Total 50 100 100 50 100 50 50 200 0 200 0 0 0 100 0 150 50 0 $$100 $$150 $50 $$300 $1,055 $4,000 $$4,000 $$$9,055 $9,355

Use the competing firm and ship 200 in February and April. If you ship 250 in January this method becomes more expensive that using the regular overnight.

10A-3

Chapter 10A Solution Manual

2. WAREHOUSE LOCATION Step 1 Warehouse

A C E

35 * 38 40

Add Warehouse A

2 Warehouse A

D C E B

33 * 36 36 37

Add Warehouse D

2 Warehouse A D

C B E

37 38 40

Don’t add any more warehouse since cost are increasing

Only consider warehouses A, C, and E in the first step since E and B do not cover all customers 3. USING THE TIME SAVED HEURISTIC a. Best route with no time restrictions is D-B-C-A-D Time Saved Matrix B C A 1 2 B 7 Rank of Pairing by time saved B-C 7 A-C 2 A-B 1

Unconstrained Route B-C A-C D-B-C-A-D

Constrained Route D-B-C-D D-A-D

b. Best route with 15 hour restriction is D–B-C-D and D-A-D

10A-4

13 hours 6 hours

Chapter 10A Solution Manual

4. TIME SAVED HEURISTIC REVISITED a. The best route is A-B-C-A and A-D-E-A b. It will take 10 total hours to make the trips. There will be two trips because vehicle capacity is exceeded c. Detail work is show below:

B C D Rank Pairing by time saved B-C 2 B-D 1 C-E 1 C-D 0 D-E 0

Time Saved Matrix C D 2 1 0

E 0 1 0

Constrained Route A-B-C-A

4 hours

60 units

A-D-E-A

6 hours

80 units

10 hours 4. SUPERIOR STEEL

10A-5

Chapter 10A Solution Manual

Direct Route

< 5000 $10 cwt >5000 $7 cwt

Via Route

<10000 $6 cwt >10000 $4 cwt

Albany

Wayword

$4 cwt <8000 $5 cwt >8000 $4 cwt

< 5000 $4 cwt >5000 $2 cwt

Leftover

$8 cwt

Wayword Albany-Wayword Direct Albany-Wayword Direct Wayword via Leftover Albany-Wayword Direct Leftover-Wayword Direct

Leftover Albany-Leftover Direct Leftover via Waysword Albany-Leftover Direct Wayword-Leftover Direct Albany-Leftover Direct

Wayword Leftover Weight rate total Weight rate total

sum

5000

$6 $300

6000

$5 $300 $600

5000

$6 $300

6000

$7 $420 $720

5000

$8 $400

6000

$5 $300 $700

11000

$4 $440

6000

$2 $120 $560

5000

$4 $200

11000

$4 $440 $640

The lowest cost alternative is to ship everything direct to from Albany to Wayword and then to ship direct to Leftover at a cost of $560

10A-6

Chapter 11 Solution Manual

CHAPTER 11 Solution Manual Discussion Problems 1. For the snacks at a football game the dependency relationship is with the game schedule and the independent demand occurs in the specific selection of products by customers during the game. The company producing bumpers for automobiles has a dependent demand relationship on the automobile assembly schedule and an independent demand relationship with garages that repair the automobiles. The greeting card situation is similar to the football snacks in that the season defines the nature of the cards (dependent demand) while the customer defines a specific product mix (independent demand).

2. The ski manufacturer is likely to be holding a great portion of his or her inventory in anticipation stock. The tugboat manufacturer is likely to have a high level of movement (work in process) inventory, and the printer is likely to have a great deal of cycle stock since the books would be run on the printing press one at a time.

3. Most students and inventory experts would suggest that the service or shortage costs are the most difficult to measure. Some students may argue that the others are difficult as well, and that leads to the second question. Given that it will be difficult to measure any of the inventory costs, what should be done about lack of precision? The use of sensitivity analysis to determine the importance of errors in the estimates on the inventory policy is important in answering this question. Other students may say it's a matter of cost and benefit, implying that if the errors did not lead to a very high cost the benefit of an improved estimate may not be worthwhile.

4. The key to this lies in the notion of "lumpiness." One of the assumptions for the EOQ is that the demand be relatively constant and we have seen in the dependent demand situations that it could be quite different from relatively constant.

5. The first part of this question is clearly to review the ideas of having a sound data base transaction system and understanding of the basic trade-offs and theory before embarking on more advanced concepts. Those arguments are pretty well raised in the chapter. The second question, however, might require the student to think a little bit about the arguments for going directly, for example, to a joint ordering system. Such arguments might be where the potential user could demonstrate a very high payoff from going directly to that sort of strategy and where the fundamentals were very well understood and the data issues could be unambiguously solved in implementing the more advanced systems.

6. Most students can tell you how may suits they own but probably not how many pairs of socks.

Clearly they're implicitly rating suits A and socks B or C. You might ask them to think of other examples of the ABC concept as well. For example, how many $100 bills do they have in their billfold, how many pennies in their coin purse, etc.?

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Chapter 11 Solution Manual

7. The analogy is fairly straightforward. The demand corresponds to the daily amount spent, which could be variable, and the lead time is the period between posting the letter and when the funds are received, which also could be variable. The amount spent during the number of days waiting for the money to come could be quite variable. This, of course, is equivalent to the demand during lead time. The average during lead time would be the average amount spent between the time the letter was posted and the money was received (observed for a large number of requests for funds from home).

Chapter 11 - Problem Summaries Problem

Difficulty

Problem Type

1

Easy

2

Moderate

Determining the probability of stocking out and the shortage cost for a specific reorder point using a discrete demand distribution.

3

Hard

Finding the best ordering quantities using EOQ and calculating the safety stock and reorder point with a discrete demand distribution to meet a specific service level

4

Moderate

Determining the reorder point using the customer service criterion with a continuous demand distribution.

5

Moderate

Determining the reorder point using the customer service criterion with a continuous demand distribution.

6

Moderate

Using ABC analysis to determine cycle counting procedures for a set of products.

7

Moderate

Multiple criteria ABC analysis.

Finding the best ordering quantities using EOQ.

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Chapter 11 Solution Manual

1. MICHAEL TRACI a. Optimal order quantity using the EOQ model: EOQ =

2C p A CH

Where: CP = $30 (preparation cost per order) A = 30 units demand period CH = $2.00 carrying cost per period EOQ = optimal order lot size EOQ =

(2)($30)(30units) = 30units ($2.00)

b. Average cycle stock = 30 units /2 = 15 units c. 12 periods x 30 units = 360 units (annual unit demand) 360 units /30 units per order = 12 orders 12 orders x $30 = $360 ordering cost per year 15 units x $2.00 x 12 periods = $360 carrying cost per year Total = $720 per year 2. LOUIE’S LOBSTER POTS a. Here is one spreadsheet layout. The formulae for the expected number of units short differ for each value of R. Probs: .05 .15 Demand During LT 20 21 Cum. Prob. .05 .20 Pr D > DDLT .95 .80 Expected No. Short 2.90 1.95 Shortage cost for R = 20 is $72.50

.20 22 .40 .60 1.15

.30 23 .70 .30 .55

.10 24 .80 .20 .25

.15 25 .95 .05 .05

.05 26 1 0 0

Probs: .2 .4 Demand During LT 20 21 Cum. Prob. .20 .60 Pr D > DDLT .80 .40 Expected No. Short 1.50 .70 Shortage cost for R = 20 is $37.50

.2 22 .80 .20 .30

.1 23 .90 .10 .10

.1 24 1 0 0

.0 25 1 0 0

.0 26 1 0 0

b.

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Chapter 11 Solution Manual

3. THE ICU OPTICAL CLINIC a.

Demand / Month Probability 12 .10 13 .15 14 .15 15 .20 16 .20 17 .10 18 .05 19 .05 Average Monthly Demand: EOQ =

Expected Demand / Month 1.20 1.95 2.10 3.00 3.20 1.70 .90 .95 15.00

(2)($25)(12)(15)  35units ($.25)($30)

b.

ROP 15 *16 17 18 19 Buffer Stock = ROP - 15 = 1

Expected Numbers of Stockouts .75 .35 .15 .05 .00

Service Level .979 .990 .996 .999 1.000

*Selected ROP

c. The annual cost of carrying buffer stock is $7.50 (one unit).

4. WALLY’S WORLD a. E(Z) = [(100-99)(200)] / [(100)(9)] = .22 (Z  .37) Reorder point = (20) + (.37)(9) = 23.33 gallons Safety stock = 3.33 gallons b 23.33 > 18; therefore, an order for 200 gallons of ice cream should be made.

The grocery store orders a 10-day supply with each order. Since daily sales average 20 gallons, the lot size is 200 gallons.

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Chapter 11 Solution Manual

5. THE SELDOM SEEN RANCH a. The Z value for a .10 probability of stocking out is 1.282: Safety Stock = (1.282)(1.25)(40) = 64 bales b. Order Point = 1,000 + 64 = 1,064 bales c. E(Z ) Value for + 1.282 = .05 E(Z) = [(100 - SL)(Q)] /[(100)(1.25)(MAD)] .05 = [(100 - SL)(2,500)]/[(100)(1.25)(40)] SL = 99.90 d. Horace should place an order for 2,500 bales whenever the number of bales on hand goes below 1,064.

6. THE ALLIED BREAKWATER COMPANY

Unit Annual Part # Cost ($) Usage 1 .20 1,000 2 1.00 10 3 .25 12 4 3.00 100 5 10.00 300 6 7.00 2 7 .50 10 8 5.00 400 9 20.00 2 10 2.00 200 Total Annual Cost Volume Usage:

Annual Cost Volume Usage ($) 200 10 3 300 3,000 14 5 2,000 40 400 5,972

% of Annual 3.35 .17 .05 5.02 50.23 .23 .08 33.49 .67 6.70 100.00

ABC Category B C C B A C C A C B

Framkranz should schedule the reviews of the parts as follows: 1. Allot one full day to review each A item (parts #5 and #8 account for almost 84% of annual cost volume usage).

2. Allot one-half day to review each B item (Parts #1, #4, and #10 account for over 15% of annual cost volume usage).

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Chapter 11 Solution Manual

3. Allot one-quarter day to review each C item (parts #2, #3, #6, #7, and #9 account for only about 1% of annual cost volume usage).

Total Time in a Week A Items 2 days B Items* 2 days C Items 1 day *#9 assigned to the B category to fill out Framkranz's day.

7. SOARING EAGLE HANG GLIDER CO. a.

Item g i b c j a d e h f Total

Usage 56 48 47 105 81 14 24 75 5 43

Cost 94 87 68 23 24 83 45 10 51 2

$-Use 5,264 4,176 3,196 2,415 1,944 1,162 1,080 750 255 86 20,328

% .26 .21 .16 .12 .10 .06 .05 .04 .01 .00

Cum. % .26 .46 .62 .74 .84 .89 .95 .98 1

These are the sorted values. Two items account for nearly 50% of the total dollar-usage and the four items at the end account for about 10%.

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Chapter 11 Solution Manual

b. The classification matrix is presented below: Class A B C

I g c e

II b h

III i j f, a, d

The sorted values below indicate that two AA items account for 38% of the dollar- usage, the CC items about 22% and the last four items about 11% of the total.

Item g c e b i h j f a d Total

Class AI BI CI BII AIII CII BIII CIII CIII CIII

Multi-Class AA AA BB BB BB CC CC CC CC CC

Usage 56 105 75 47 48 5 81 43 14 24

11-7

Cost 94 23 10 68 87 51 24 2 83 45

$-Use 5,264 2,415 750 3,196 4,176 255 1,944 86 1,162 1,080 20,328

% .26 .12 .04 .16 .21 .01 .10 .00 .06 .05

Cum. % .26 .38 .42 .58 .79 .80 .90 .90 .96 1

Chapter 12 Solution Manual

CHAPTER 12 Solution Manual Discussion Questions 1. It is useful for the student to see that the order backlog does work as a buffer for the company, but the "costs" are borne by the customer that must wait. Order backlog has also been called negative inventory for these reasons.

2. Prepare for a lively discussion. Your students will really stumble around with this, which is one of the reasons that we don't feel the words are terribly descriptive. The terms arose because under MRP, a machine center would work as long as there is a queue “pushing” the work along. In a kanban controlled JIT approach, the using work center authorized production at the feed centers, “pulling” the work through the shop.

3. We're interested in having your students think through the implications of key requirements like small lots, frequent delivery; rapid or specific date of delivery needs; and/or very precise delivery promises on the design of the system. They will have trouble with this unless you prompt them a little.

4. Here again is a question that we have heard many times. We hope that your students are astute enough to see that the question itself doesn't make sense. What should really be asked is, "What are you trying to achieve with your MPC system?" From there one can suggest what might be most appropriate. Or favorite answer is: “Is a sock better than a shoe?”

5 . Your students should be able to see that a lot of the discussion of the assemble-to-order system gets at this need. Both capacity and inventory at other than finished goods points help to do this.

6. In general, this question raises the need to move from make-to-stock to assemble-to-order.

This means that the option items would need to be managed under the MPS. Forecasting, customer order promising and capacity management would all need to be managed at the option MPS level. One of your better students may point out that the automotive industry tends to reduce options (by specifying option packages) and still make-to-stock.

7. Applicon faced three market requirements that necessitated a new production process design and MPC approach. They were: pressure for lower prices, increased delivery speed and faster response to product technology changes.

8. For Moog to move from make-to-order and assemble-to-order to make-to-stock, they would need to move to a more narrow line of standardized products produced in high volume. This is quite different from their current product line of customized products produced in low volume.

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Chapter 12 Solution Manual

9. This is one of the arguments for doing MRP first, to get the discipline required. In both JIT and MRP first you want people to do what is required for the business, not just for proxy measures like utilization. Thus keeping busy (producing what is not needed now) may just be the wrong thing to do.

Chapter 12 - Problem Summaries Problem

Difficulty

Problem Type

1

Moderate

Requires the specification of MPC system design requirements for several product lines within a company.

2

Moderate

Requires the specification of MPC system design requirements for several product lines within a company. Similar to problem #1.

3

Easy

Impact of safety stock reduction caused by MPC design changes.

Requires calculation of safety stock under various manufacturing scenarios.

4

Moderate

Extends problem #3 with detailed cell data. It shows that if you have safety capacity you do not need safety stock. The safety capacity allows the company to respond to changes in demand.

5

Easy

Has the student assess the impact of manufacturing lead time on MPC design options.

6

Moderate

Requires the student to determine what transaction must take place to keep the shop on schedule. It also looks at the reduction of transactions by using phantom items.

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Chapter 12 Solution Manual

7

Moderate

Similar to problem #6. Looks at the total number of transactions required by a set of MRP records and then ask how much this would be reduced by phantom bills.

8

Moderate

MRP transactions and their reduction through switching to a JIT ratebased approach.

9

Moderate

Looks at the impact on safety stock by adding more end item options.

It also assess the reduction by limiting option combinations.

10

Moderate

Using the information from problem #9 it looks at the reduction of safety stock if they move to an assemble to order environment and hold options in inventory.

11

Easy

Looks at the impact of assemble to order on safety stock. It shows how safety stock is reduced by moving to assemble to order.

12

Moderate

Impact of time phased planning versus rate based planning on average inventory. It also analyzes adjusting the master schedule to meet safety stock goals.

13

Easy

14

Moderate

Extends problem #13 and switches the process to assemble to order.

It illustrates the change in safety stock by using assemble to order.

15

Moderate

Extends problem #13 and changes the production method again. It has the student determine the changes in safety stock under the new situation.

Impact of product variety on safety stock.

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Chapter 12 Solution Manual

1. WORLDWIDE BATTERIES

Mkt. Characteristics Replacement Batts. for Material Handling Equipment: Mkt. Requirements: Low Prices Delivery Speed

Manufacturing Task Features

MPS

Manufacturing Strategy Mat. Planning PAC

5-day Final Assembly LT

Low Cost Mfg. Short LTs

Purchased Components:

High Volume JIT Delivery

JIT-based

% Vol. Req. LT Low Volume 4 week lead time

60% @ 7 days 30% @ 14 days

Manufactured Components: JIT Mfg.

10% @ 3 days

MTS

Rate-based

JIT-based

ATO

Rate-based

JIT-based

MTO

Vendor MRP Shop-rate Based Material Planning

Replacement Batts. for Transport. Equip.: Mkt. Requirements: Low Prices

Vendor:

Delivery Reliability

Low Cost Mfg. Rel. Delivery

Low Volume Prods.

2 Week LT

Shop: -Rate-based Material Planning

-MRP Records

MTO

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JIT-based

Chapter 12 Solution Manual

2. CAMBRIDGE PLASTICS

Mkt. Characteristics

Manufacturing Task Features

Manufacturing Strategy MPS Mat. Planning

PAC

Injection Molded Plastic Bottles For Soft Drink Bottles: High Vol. Products High Volume Dedicated Equipment

Seasonal Variation

MTS

Rate-based

JITbased

MTO Cust. Order Promising Support Capacity Planning Support Pre-Production Activity Scheduling; e.g., Vendor Moldmaking

MRP

MRPbased

6 Mo. Blanket Orders Mkt. Requirements: Cap. Ramp-up Delivery Speed

Vol. Flexibility Quick Resp.

On Customer Schedule Changes

Custom Molded Plastic Bottles: Wide Prod. Range Low Vol. Products

Low Volume Batch Mfg

Custom Orders Mkt. Requirements Low Prices Del. Reliability

low cost Mfg. Rel. Delivery

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Chapter 12 Solution Manual

3. OAKMONT INC. a. Safety stock for a 8 week lead time: S.S. = 2.056 (8) 8 = 46.52

Safety stock for a 2 week lead time: S.S. = 2.056  (8) 2 = 23.26

The reduction in safety stock is 23.26 units or 50%. b. New safety stock investment: Safety stock for non-cell parts

= .60  200,000

Safety stock for cell parts

=

23.26  80,000 46.52

= $120,000

Total investment with cell

= 40,000 $160,000

Total investment without cell Net Investment

$200,000 $ 40,000

c. Redesign safety stock investment Safety stock for non-cell parts

= .40  200,000

Safety stock for cell parts

=

23.26  120,000 46.52

= $ 80,000

Total investment with resign

= 60,000 $140,000

Total investment without redesign Net Investment

$160,000 $ 20,000

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Chapter 12 Solution Manual

4. OAKMONT AGAIN a. Regular time capacity is 800 units, or 4 standard deviations beyond the mean demand. This corresponds to a service level of 99.997%. b. Since demand is being met beyond the 95% service level desired, no safety stock is required.

5. DIXON PLASTICS a. Dixon would not require different MPC design options. b. To service Amdur, Dixon might consider purchasing the raw material against the plans and finishing to order. It would depend on where in the process the differentiation in Amdur products occurred. If there was little differentiation from week to week, Dixon might make-to-stock Amdur's products. Further in this line of thinking, a separate manufacturing area may be set aside for Amdur. If the week-to-week production levels of Amdur products does not change greatly, a rate-based system may be employed, and a JIT shop floor control system used.

6. PRODUCT A a. Three transactions are required: 1. Release a shop floor order for 30 units of Part A. 2. Release a change of due date from week 2 to 3 for the open order for 100 units of Part B. 3. Release a shop order for 100 units of Part C. b. 6 weeks--1 for Part C, 2 for Part B, 1 for A and 2 for MRP processing.

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Chapter 12 Solution Manual

c. If everything went according to plan and all transactions were made, the records would look as follows: Product A Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 30, SS = 5, LT = 1

5

2 20

3 0

4 40

5 20

6 10

7 0

8 20

9 20

15

15 30

5 30

15

5

5 30

15 30

25

2

3 60 100 55

4 60

5

6

7 60

8 60

9

95

95

95 100

35

75

3

4

5

6 100

7

8

0

0

0 100

0

Part B Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 100, SS = 0, LT = 2.

15

15 100

Part C Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = LFL, SS = 0, LT = 1.

0

2 100 100 0

The transactions required, in addition to those made in the answer to Part A, would be: 1. Dispersal of 10 Part A's from inventory.

2. Close out of shop order for 30 Part A's due in week 2 and posting to inventory.

3. Dispersal of 60 Part B's from inventory.

4. Release a shop order for 100 Part B's due in week 4.

5. Close out shop order for 100 Part C's and post to inventory. d. Need MRP records for Item A only.

Need 3 transactions instead of 8.

Total lead-time is reduced from 4 weeks to 1.

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Chapter 12 Solution Manual

7. PRODUCT X a. Transactions: 40 X--Finished Product 4 orders opened.

5 orders closed.

5 inventory postings of 50 units received.

5 inventory releases of 50 units each.

Part Y 2 shop orders opened.

3 shop orders closed.

3 inventory postings (one of 50, 2 of 100).

4 inventory releases of 50 units each.

Part Z 1 order opened.

1 order closed.

1 inventory posting of 100 received.

2 inventory releases of 100 each.

Raw Material 1 order opened.

1 order closed.

1 inventory posting of 100 received.

1 inventory release of 100. b.

Product X Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = LFL, SS = 10, LT = l.

10

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1 50 50 10 50

2 50

3 50

4 50

5 50

10 50

10 50

10 50

10

Chapter 12 Solution Manual

Product Y Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = 100, SS = 0, LT = 0.

52

1 50

2 50

3 50

4 50

2

0 48

0 50

0 50

1

2 48

3 50

4 50

0

0

0 48

50

3 48

4 50

5

Product Z Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q = LFL, SS = 0, LT = 0.

0

5

Raw Material R Week Gross Requirements Scheduled Receipts Projected Available Balance Planned Order Release Q= LFL, SS = 0, LT = 1.

0

1

2

0

0 48

5

50

After conversion, the MRP records would now consist only of the records for Product X and Raw Material R. In terms of transactions, there might be a need to disperse 50 units of finished product per week and post its receipt into inventory. Raw material could be received right into the line if vendor relations and performance are adequate. If not there would be 2 order releases for X and order closings. Similarly for R there would be two receipts and disbursements plus two order releases and closings. This would total 16 transactions at most.

8. MELNICK MINES a. To create the record shown as of week 2, the following transactions must have been made: 1. Released an order for 80 units due in week 4.

2. Changed the due date on 80 units in week 2 to week 1.

3. Posted 75 units into inventory during week 1 and closed shop order.

4. Dispersed 30 units from inventory.

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Chapter 12 Solution Manual

b. All inventory, due date changes, and order release transactions could be cut out. The order size would now be 25 (or even 5 per day) with zero lead time.

9. FALCON SPORTS, INC. a. Current offering: = (2 horsepower) x (2 seating options) x (2 starter options) x 25 units/item = 200 units of safety stock New offerings = (2 horsepower) x (2 seating) x (2 starter) x (3 colors) x 25 units/item = 600 units of safety stock Safety stock increases by 400 units. b. Restricted Offering 10 horsepower models = (2 starters) x (2 seating) x (1 color) x (1 horsepower) = 4 end items 12 horsepower models = (2 starters) x (2 seating) x (2 colors) x (1 horsepower) = 8 end items Total End Items = 12 Safety stock per item = 600/12 = 50 units of safety stock per item

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Chapter 12 Solution Manual

10. FALCON SPORTS, INC. AGAIN a.

Total Options

= (2 starters) + (2 seating) + (3 colors) + (2 horsepower) = 9 options (9 options) x (40 units per item) = 360 units of safety stock

b. Safety stock per option = 600/9 = 66.67 11. RONSI RIST WATCH No! Don't return to the old system. The marketing manager has confused safety stock per unit and overall safety stock. There is a saving in safety stock inventory so that is not a reason to go back to the old system.

Make-to-stock safety stock required: 50 items  100 units per item = 5,000 units of safety stock Assemble-to-order safety stock required: Total number of items = 12 + 2 + 4 + 4 + 2 = 24 24 items  100 units per item = 2,400 units of safety stock A net savings of 5,000 - 2,400 = 2,600 units of safety stock.

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Chapter 12 Solution Manual

12. ISANDAR WEDGE The following simulations show the master production schedule that provides the desired inventory levels: a.

Week Opening Inventory MPS Demand Closing Inventory

1 0 50 25 25

2 25 50 25 50

3 50 50 25 75

4 75 50 25 100

5 100 25 25 100

6 100 25 25 100

Week Opening Inventory MPS Demand Closing Inventory

1 100 11 25 86

2 86 11 25 72

3 72 11 25 58

4 58 11 25 44

5 44 11 25 30

6 30 25 25 30

b.

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Chapter 12 Solution Manual

13. LEONE CO. Old offerings: New offerings:

(3 colors) X (10 units/item) = 30 units safety stock (3 colors) X (4 decal options) X (5 sequin options) X (10 units/item) = 600 units of safety stock.

14. LEONE CO. AGAIN Total Options:

(3 colors) + (3 decals) + (4 sequins) = 10 (10 option items) X (20 units / item) = 200 units safety of stock

15. LEONE ONCE MORE Safety Stock:

(1 color + 4 sequins) X (20 units / item) = 100 units of safety stock

Safety stock is reduced from Problem 14, but assembly lead time has increased. If assembly lead time becomes problematic, finished goods inventory may have to be held.

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