Solving a linear programming model using simplex method

3/9/2017

Simplex Table

Coming up with the first simplex table

Interpreting the last simplex table

Steps: 1. Convert each inequality constraint into equation by adding a slack variable.

2. Set real variables equal to zero.

3. Build simplex table.

• Put variables with values into basic variable column and write their corresponding values in the quantity column.

• Write coefficients of all equations into the body of the table.

• Calculate Z and C – Z.

Things to interpret: 1. Optimal solution.

2. Resources.

• The utility • Shadow price • Change in the original amount 3. Special case: more than one optimal solution.

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Suppose that the following linear programming model represents the production of wedding clothes.

X1 = female clothes X2 = male clothes Maximize: P = 50X1 + 20X2 Subject to: 5X1 + 10X2 ≤ 300 (minutes of cutting time) 4X1 + 2X2 ≤ 80 (hours of sewing time) 90X1 + 15X2 ≤ 1500 (minutes of beading time) X1, X2 ≥ 0

1

Convert each inequality constraint into equations by adding a slack variable.

Maximize: Z = 50X1 + 20X2 Subject to: 5X1 + 10X2 + S1

= 300

4X1 + 2X2 + S2

= 80

90X1 + 15X2 + S3 = 1500 X1, X2, S1, S2, S3 ≥ 0 Maximize: Z = 50X1 + 20X2 + 0S1 + 0S2 + 0S3 Subject to: 5X1 + 10X2 + S1 + 0S2 + 0S3 = 300 4X1 + 2X2 + 0S1 + S2 + 0S3 = 80 90X1 + 15X2 + 0S1 + 0S2 + S3 = 1500 X1, X2, S1, S2, S3 ≥ 0

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2

Set real variables equal to zero X1 = 0 X2 = 0 ∴ S1 = 300, S2 = 80, S3 = 1500

3

Equations’ coefficients

Build simplex table

C= Objective function coefficients

C

→

50

20

0

0

0

↓

BASIC

X1

X2

S1

S2

S3

QTY

0

S1

5

10

1

0

0

300

0

S2

4

2

0

1

0

80

0

S3

90

15

0

0

1

1500 0

Z

0

0

0

0

0

C–Z

50

20

0

0

0

Z values for all variables are equal to zero in the first table

Now, suppose that the model has been solved by using simplex method and the following table is the last simplex table from the solution.

BASIC

X1

X2

S1

S2

S3

QTY

S1

0

0

1

‐6.88

0.25

125

X2

0

1

0

0.75

‐0.03

10

X1

1

0

0

‐0.13

0.02

15

Z

50

20

0

8.75

0.17

950

C–Z

0

0

0

‐8.75

‐0.17

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BASIC

X1

X2

S1

S2

S3

QTY

S1

0

0

1

‐6.88

0.25

125

X2

0

1

0

0.75

‐0.03

10

X1

1

0

0

‐0.13

0.02

15

Z

50

20

0

8.75

0.17

950

C–Z

0

0

0

‐8.75

‐0.17

X1 = 15 X2 = 10 S1 = 125 S2 = 0 S3 = 0 Z = 50X1 + 20X2 = 50(15) + 20(10) = RM950

1.

2.

3.

4.

List out all variables.

Look at the variables in the basic variable column and read their corresponding values in the quantity column.

The rest of variables that are not in the basic column automatically have zero values.

Read the Z value in the quantity column or calculate the Z value by substituting the values of all real variables into the objective function.

‐

Slack variables represent resources.

‐

The values of the constraints in the model represent the original amount of resources.

Maximize: P = 50X1 + 20X2 Subject to: 5X1 + 10X2 ≤ 300 (minutes of cutting time) 4X1 + 2X2 ≤ 80 (hours of sewing time)

Original amount of resources

90X1 + 15X2 ≤ 1500 (minutes of beading time) X1, X2 ≥ 0 ‐

The values of slack variables in the optimal solution represent the left amount of resources.

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Resource

Original amount

Left amount

Used amount

Utility

Cutting time

300 minutes

125 minutes

175 minutes

Not fully utilized

Sewing time

80 hours

0 hour

80 hours

Fully utilized

Beading time

1500 minutes

0 minutes

1500 minutes

Fully utilized

The values of constraints in the model

= Original amount – Left amount

The values of Slack variables n the optimal solution

BASIC

X1

X2

S1

S2

S3

QTY

S1

0

0

1

‐6.88

0.25

125

X2

0

1

0

0.75

‐0.03

10

X1

1

0

0

‐0.13

0.02

15

Z

50

20

0

8.75

0.17

950

C–Z

0

0

0

‐8.75

‐0.17

Shadow prices

Shadow price is the maximum amount that we could pay if we want to add one unit of resource.

Resource

Shadow price

Meaning

Cutting time

RM0

If we want to add 1 minute of cutting time, the maximum amount that we could pay is RM0.

Sewing time

RM8.75

If we want to add 1 hour of sewing time, the maximum amount that we could pay is RM8.75.

Beading time

RM0.17

If we want to add 1 minute of beading time, the maximum amount that we could pay is RM0.17.

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‐

Changing the original amount of resources will result in the change of optimal profit.

‐

To calculate the new optimal profit, use the following formula: New profit = Old profit + (Changed amount x Shadow price)

‐

Examples: • If sewing time is changed to 85 hours, New profit = RM950 + (5 x RM8.75) = RM993.75 • If sewing time is changed to 70 hours, New profit = RM950 + (‐10 x RM8.75) = RM862.50 • If beading time is changed to 1560 minutes, New profit = RM950 + (60 x RM0.17) = RM960.20

Steps in identifying the existence of more than one optimal solution: 1. Identify real variable that is not in the basic variable column.

2. Look at the C – Z value of that variable.

• If the C – Z value is zero, then there exists more than optimal solution.

• If the C – Z value is not zero, then there exists only one optimal solution.

BASIC

X1

X2

S1

S2

S3

QTY

S1

0

0

1

‐6.88

0.25

125

X2

0

1

0

0.75

‐0.03

10

X1

1

0

0

‐0.13

0.02

15

Z

50

20

0

8.75

0.17

950

C–Z

0

0

0

‐8.75

‐0.17

Both X1 and X2 are in the basic variable column.

Thus, no need to proceed to step 2.

This indicates that this problem has only one optimal solution.

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