Introduction to Optimum Design, 4th Edition Jasbir Arora complete guide with actual exam samples sol

SAMPLE OF EXAMS AND PROJECTS

SAMPLE PROJECTS

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Sample Projects

SAMPLE PROJECT 1: LINEAR PROGRAMMING Solid Waste Management Plan A residential solid waste management plan is required for the Iowa City-Cedar RapidsWaterloo region. Each of these cities generates quantities of solid waste that must be disposed of in a landfill or processed at an intermediate facility. At the intermediate facility, useful material, such as dry fuel, may be recovered from the solid waste and then sold, thereby reducing the overall disposal cost. Residue of the intermediate facility must be shipped to a landfill.

A proposed system is shown in the figure. In addition to the three sources S1, S2 and S3, representing Iowa City, Cedar Rapids and Waterloo, respectively, there is one intermediate facility, I, that converts 60% of the residential waste into a dry-fuel product and can process 5000 tons per week, an electric generating station, P, that can use up to 4500 tons of dry fuel per week, and two landfills, D1 and D2, that can accept up to 10,000 tons of waste per week each. There are nine transportation links in the system. Data on quantities and costs are listed in Tables 1, 2 and 3. The disposal/processing costs in Table 3 apply to all incoming material at a facility and include capital and operating costs less revenues from the sale of useful material.

Answers to the following questions are needed for the evaluation of the proposed system.

You must be able to answer all the questions without repeating the analysis performed to answer the first question. Do not repeat the analysis to answer the remaining questions. If you are unable to answer a question based on the results obtained in question 1, state that you are unable to answer the question and why.

1. What is the optimum operating plan?

2. What is the maximum operating cost at the intermediate processing facility that will allow its operation at capacity?

3. What is the consequence of enlarging the capacity of that facility to 9000 tons per week?

4. Which of the following increases in capacity will result in the most economical system and why or why not? a. The intermediate facility to 9000 tons per week. b. The landfill, D1, to 15,000 tons per week. c. The landfill, D2, to 18,000 tons per week. d. The power plant to 9000 tons per week.

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Table 1. Solid Waste Quantities

Table 2. Transportation Costs

Source

Link

Cost, $/ton

1 2 3 4 5 6 7 8 9

2.00 2.50 2.70 3.50 3.50 3.00 2.10 1.00 0.20

Amount, tons/week

S1 Iowa City S2 Cedar Rapids S3 Waterloo

4800 7900 9700

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Table 3. Disposal/Processing Costs Facility Cost, $/ton I Intermediate 1.90 P Power plant −0.80 D1 Dump I Unprocessed waste 3.40 Residue from I 1.10 D2 Dump 2 3.10

Sample Projects

SAMPLE PROJECT 2: LP Project No. 3 Air Pollution Control Model Electric power generating plants have been criticized for contributing excessively to air pollution problems. The emissions from plant smokestacks contain pollutants of various kinds that are considered harmful to the environment.

As a result, power plants are subject to regulations governing the amounts of specific pollutants that are allowed in the stack emissions.

A steam power plant has a choice of burning coal, oil and natural gas at the same time and in any combination to 8 generate power. The fuels must provide a heat input to the plant of 40×10 BTU/hr to generate the required power out.

Air pollution criteria must be met, and so choice of fuel is not based only on cost and efficiency. The following table gives the necessary data for cost in $per ton, heating value in BTU per pound (lb) and pollutants produced in lb per ton for each of the fuels. The maximum allowable limit for each of the pollutants in lb/hr is also given in the table.

Note that standard ton is used; i.e., 1 ton = 2000 lb Coal

Oil

Gas

Maximum Allowed

Hydrocarbons

1 lb/ton

2 lb/ton

1 lb/ton

100 lb/hr

NO, NO 2 , NO 3

1 lb/ton

1 lb/ton

1 lb/ton

100 lb/hr

SO 2

140 lb/ton

80 lb/ton

1 lb/ton

2,000 lb/hr

CO

100 lb/ton

90 lb/ton

90 lb/ton

20,000 lb/hr

Particulate Cost Heating Value

6 lb/ton $60/ton 14,000 BTU/lb

2 lb/ton $100/ton 19,000 BTU/lb

1 lb/ton $160/ton 22,000 BTU/lb

80 lb/hr

An electrostatic precipitator can also be installed to remove particulate matter for $5 per pound of particulate removed.

This could be a cost-effective way to remove particulate matter, and therefore needs to be considered in the formulation of the problem.

Using Excel Solver to generate your solution, address the following questions in your report: a. Formulate the problem and determine the fuel usage policy on per hour basis that minimizes the cost of fuels and meets the air pollution requirements. b. What is the consequence of an increase in the price of gas to $170/ton? c. What is the consequence of reducing the permissible SO 2 output to 1500 lb/hr to minimize the acid rain problem?

For parts b and c, predict the results using the sensitivity information for the original results obtained in part a, and then verify your results by re-solving the problem with the modified data using the program (Hint: Use the allowable ranges in the Sensitivity Sheet of the Solver output).

Note: You must define the design variables very precisely and watch the units of various quantities carefully.

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Project Reporting Requirements Submit your report using the provided Microsoft Word .doc file. Remember to download the file and before working on it to rename the file including your first and last name. For example, “project1.doc” is the name of the file that you download. Assuming that “Aye Ten” is a student name, the renamed file would be “project1_AyeTen.doc”.

Your report must be formatted properly and carefully, and include the following 4 parts: 1) Complete problem formulation process: problem description, data/information, clear definition of design variables, cost function, and constraints.

2) Solution for the parts a, b and c, and verification of the predictions in parts b and c.

3) A brief discussion and conclusions for the project.

4) Submit the Excel file for the problem which should be organized clearly and in a readable format.

Project Grading Rubric: Graded based on 50 points Report attributes Formulation of the problem (20 points)

Problem solution (10 points) Sensitivity analysis (10 points) Excel sheet for the problem (10 points)

Meets all expectations

Partially meets expectations

Below expectations

(20 points)

(10 points)

(5 points)

Complete and clear presentation that includes complete problem formulation process with proper formatting.

Complete, but sloppy presentation that either includes poorly defined design variables and/or cost function and/or constraints

Incomplete formulation that either lacks clear definition of design variables and/or cost function and/or constraints

(10 points) Complete solution, active constraints

(6 points) Partial solution

(2 points) Incorrect solution or no solution

(10 points)

(6 points)

(2 points)

Solution based on sensitivity analysis and its verification.

Incomplete sensitivity solution and/or missing verification.

Incorrect sensitivity solution. Missing verification

(10 points)

(5 points)

(0 points)

Nicely organized Excel Excel sheet incomplete Not included sheet in a clear readable and/or unreadable and not and logical format formatted well Overall Score: 1) Best effort: 45 or more; 2) Acceptable effort: 35 to 44; 3); Needs improvement: 20 to 34

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Sample Projects

Solution: Air Pollution Control Model Design Variables 𝑥𝑥1 = tons of coal used/hr 𝑥𝑥2 = tons of oil used/hr

𝑥𝑥3 = tons of gas used/hr

𝑥𝑥4 = pounds of particulate material removed/hr by the electrostatic precipitator

Cost function

Minimize the cost of fuel use/hr 𝑓𝑓 = 60𝑥𝑥1 + 100𝑥𝑥2 + 160𝑥𝑥3 + 5𝑥𝑥4

Constraints

Hydrocarbons: NO, NO 2 , NO 3 : SO 2 : CO: Particulate: Heat Input:

𝑥𝑥1 + 2𝑥𝑥2 + 𝑥𝑥3 ≤ 100 𝑥𝑥1 + 𝑥𝑥2 + 𝑥𝑥3 ≤ 100

140𝑥𝑥1 + 80𝑥𝑥2 + 𝑥𝑥3 ≤ 2000

100𝑥𝑥1 + 90𝑥𝑥2 + 90𝑥𝑥3 ≤ 20000

6𝑥𝑥1 + 2𝑥𝑥2 + 𝑥𝑥3 − 𝑥𝑥4 ≤ 80

(14000𝑥𝑥1 + 19000𝑥𝑥2 + 22000𝑥𝑥3 )2000 = 40 × 108

Nonnegative variables:

𝑥𝑥𝑖𝑖 ≥ 0; 𝑖𝑖 = 1, 2, 3, 4

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SAMPLE PROJECT 3: NONLINEAR PROGRAMMING Optimum Use of Sheet Metal A manufacturer is to produce circular steel end plates for cans for the food industry.

Three diameters of cans are involved. The plates will be punched using a set of three circular punches, one punch for each size can, in some optimal configuration on the manufacturer's sheet metal stock. The three plate diameters are 15 cm., 11 cm., and 7 cm. The sheet metal stock is 20 cm. wide and will be cut to the appropriate length prior to punching. The problem becomes one of locating the punches so as to use as little metal stock as possible for each set of three end plates. Some physical limitations exist in order to produce a quality end plate. The punches will not produce a clean punch if the edge of the punch is within 1 cm. of the edge of the stock nor will it produce a clean punch if the edge of the punches are within 0.25 cm. of each other.

Formulate this as an optimal design problem to determine the locations of the three punches and the length of stock required. The formulation must be normalized and in standard form. The program IDESIGN will be used to solve the resulting problem.

The manufacturer has the option of using a different metal stock which has a slightly greater resistance to damage from the punching. Because of this tolerance, the distance between the edges of the punches may be reduced to 0.12 cm. From the results of the problem solution above compute the savings in metal stock due to the additional tolerance.

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Sample Projects

SAMPLE PROJECT: PROBLEM FORMULATION Optimum Design of Compression Members to Satisfy AISC LRFD Manual Requirements Step 1. Project Statement: Columns made of wide flange sections are used in numerous applications. They need to be designed for minimum cost which is related directly to the mass of the column. Assume a length of the column as L ft, service dead load of P D kips and live load of P L kips. The optimum design must satisfy all the requirements of the AISC Load and Resistance Factor Design Manual. Formulate the design problem as an optimization problem using the five step procedure. State all the assumptions made in the formulation. Submit a word-processed report.

tf

h tw

b

Cross-section of wide flange sections or plate girders

Step 2. Data and Information Collection: Data: L, P D , P L , Steel grade (E, F y , etc) Expressions: Ag = 2bt f + ht w 2  bt 3 h tf   f    + 2 + bt f + Ix = 2 2    12 12    

twh3

Iy = rx =

ry =

b 3t f

t3 h + w 6 12 Ix Ag Iy Ag

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AISC LRFD Specifications: Design load Pu : Pu = max{1.2 PD + 1.6 PL , 1.4 PD } Column design: b E h E ≤ 0.56 ≤ 1.49 If and , local instability is prevented.

2t f Fy Fy tw

 K L λc = max  x x  rxπ

Fy K y L y , E ryπ

0.658λ2c F y  Fcr =  0.877  2 Fy  λc

Fy   E 

for λc < 1.5 for λc ≥ 1.5

Design strength: φc Pn = φc Fcr Ag = 0.85 Fcr Ag

Pu ≤ φc Pn or 1.0 −

φc Pn ≤0 Pu

For wide flange sections, torsional buckling and flexural-torsional buckling need not be considered.

Slenderness ratio (recommendation) (SPEC B7):

KL ≤ 200 r Step 3. Design Variables: b tf h tw

= flange width, in = flange thickness = web height, in = web thickness, in

Intermediate Variables: Ag , I x , I y , rx , ry , Pn , Fcr , λc Arora, Introduction to Optimum Design, 4e

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Step 4. Definition of Objective (Cost) Function The objective is to minimize the total cost of the wide flange column which can be directly related to the total volume of the column, as

f = Ag L , in 3 Or, just the area of the cross-section (since length of the column L is fixed), as

f = Ag Step 5. Design Constraints:

Design Strength: 1.0 −

φc Pn ≤0 Pu

Slenderness Ratio:

K x Lx rx K y Ly ry

≤ 200 ≤ 200

K x , Lx , K y , L y are determined from the practical situations.

Local Buckling:

h E ≤ 1.49 tw Fy b E ≤ 0.56 2t f Fy Design Variable Bounds: bmin ≤ b ≤ bmax

t f min ≤ t f ≤ t f max

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Sample Projects hmin ≤ h ≤ hmax t w min ≤ t w ≤ t w max

Note that it is permissible to use a cross-sectional shape that does not satisfy the widththickness ratio requirements (slender sections), but such a member may not be permitted to carry as large a load as the one that does satisfy the requirements. In other words, the design strength could be reduced because of local buckling. In the above formulation, local bucking constraints are imposed and so local bucking is not allowed.

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Sample Projects

SAMPLE PROJECT: NONLINEAR PROGRAMMING Optimum Design of Compression Members to Satisfy AISC LRFD Manual Requirements Consider the design optimization problem that you have formulated earlier. Solve the problem for the data given below. Use a numerical method to solve the problem, such as in Excel, Matlab, Mathematica, etc. Submit a brief word-processed report containing your problem formulation, solution, computer files and a discussion of the solution.

DATA FOR PROJECT: L = 30 ft ; PD = 2 × 10 5 lbs; PL = 2 × 10 5 lbs; E = 29 × 10 6 psi; Fy = 5 × 10 4 psi

Lower and upper bounds of the design variables:

1 ≤ h ≤ 20 in; 1 ≤ b ≤ 20 in; 0.1 ≤ t f ≤ 1 in; 0.1 ≤ t w ≤ 1 in; Solution for the problem is given in the below Excel sheets.

Project #1 (A) 1. Design variable name web height flange width flange thickness web thickness

Lower limit 1 1 0.1 0.1

Symbol h b tf tw

Value 8.5718583504302 16.1374523605781 0.598277890596105 0.238877211066465

Value

Units in psi psi lb lb

2. Parameter name Span length Modulus of elasticity Yield stress Dead load Live load

Symbol L E sigma_y PD PL

=360 =29000000 =50000 =200000 =200000

3. Dependent variable name Cross sectional area Moment of inertia Bending Moment Radius of gyration w.r.t x axis Radius of gyration w.r.t y axis K ramda_c Fcr Pu Pn

Symbol Ag Ix Iy rx ry K ramda_c Fcr Pu Pn

Equation =2*b*tf+h*tw =tw*h^3/12+2*(b*tf^3/12+b*tf*(0.5*h+0.5*tf)^2) =b^3*tf/6+tw^3*h/12 =SQRT(Ix/Ag) =SQRT(Iy/Ag) =1 =MAX(K*L/(rx*PI())*SQRT(sigma_y/E),K*L/(ry*PI())*SQRT(sigma_y/E)) =IF(ramda_c<1.5,(0.658^(ramda_c^2))*sigma_y,0.877*sigma_y/ramda_c^2) =MAX(1.2*PD+1.6*PL,1.4*PD) =Fcr*Ag

4. Objective fuction name Volume of material

Symbol Vol

=Ag*L

5. Constraints Design strength Slenderness ratio Slenderness ratio Local web buckling Local flange buckling Height to Flange width

Equation

Value/Eq. =1-0.85*Pn/Pu =K*L/rx =K*L/ry =h/tw =b/(2*tf) =b-h

Arora, Introduction to Optimum Design, 4e

</>/= < < < < < <

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20 20 1 1

Units in^3

Value/Eq. =0 Allowable bending s Allowable shear stre =200 Allowable deflection =200 =1.49*SQRT(E/sigma Local buckling =0.56*SQRT(E/sigma Local buckling =0

Sample Projects Microsoft Excel 15.0 Answer Report Worksheet: [Project-Optimization of W Shape.xlsx]Problem Report Created: 3/31/2016 4:31:20 PM Result: Solver has converged to the current solution. All Constraints are satisfied.

Solver Engine Engine: GRG Nonlinear Solution Time: 0.062 Seconds.

Iterations: 17 Subproblems: 0 Solver Options Max Time 100 sec, Iterations 10000, Precision 0.000001 Convergence 0.0001, Population Size 100, Random Seed 0, Derivatives Forward, Require Bounds Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 5%, Solve Without Integer Objective Cell (Min) Cell $C$28 Vol

Variable Cells Cell $D$3 h $D$4 b $D$5 tf $D$6 tw

Name

Original Value 21600

Final Value 7688.514072

Name

Original Value 20 20 1 1

Final Value Integer 8.57185835 Contin 16.13745236 Contin 0.598277891 Contin 0.238877211 Contin

Constraints Cell Name $B$31 Design strength Value/Eq. $B$32 Slenderness ratio Value/Eq. $B$33 Slenderness ratio Value/Eq. $B$34 Local web buckling Value/Eq. $B$35 Local flange buckling Value/Eq. $B$3 web height Lower limit $B$4 flange width Lower limit $B$5 flange thickness Lower limit $B$6 web thickness Lower limit $D$3 h $D$4 b $D$5 tf $D$6 tw

Arora, Introduction to Optimum Design, 4e

Cell Value Formula Status -5.26407E-07 $B$31<=$D$31 Binding 81.27159123 $B$32<=$D$32 Not Binding 81.27165432 $B$33<=$D$33 Not Binding 35.88395189 $B$34<=$D$34 Binding 13.48658593 $B$35<=$D$35 Binding 1 $B$3<=$D$3 Not Binding 1 $B$4<=$D$4 Not Binding 0.1 $B$5<=$D$5 Not Binding 0.1 $B$6<=$D$6 Not Binding 8.57185835 $D$3<=$E$3 Not Binding 16.13745236 $D$4<=$E$4 Not Binding 0.598277891 $D$5<=$E$5 Not Binding 0.238877211 $D$6<=$E$6 Not Binding

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Slack 0 118.7284088 118.7283457 0 0 7.57185835 15.13745236 0.498277891 0.138877211 11.42814165 3.862547639 0.401722109 0.761122789

Sample Projects

SAMPLE PROJECT: PROBLEM FORMULATION Optimum Design of Crane Girder Step 1. Problem Statement Plate girders are used as cranes in mechanical shops to transfer loads from one place to another. A crane girder needs to be designed with the objective of minimizing its total mass which is directly related to the total cost. The maximum wheel load is P and the span of girder is L. The hook center is moving between the ends of the girder. The design must satisfy the deflection, bending stress, shear stress, lateral-torsional buckling and local bucking constraints. Formulate the design problem as an optimization problem using the five step procedure. State all the assumptions made in your formulation. Submit a word-processed report.

tf

h tw

b

Step 2. Data and Information Data: L, P, δ a , σ a , τ a , Steel grade (E, F y , etc) Expressions: I=

2 1 1 1   1 t w h 3 + 2  bt 3f + bt f  h + t f   12 2   2 12

PL3 48 EI Mc 1 PL σ= , c =  t f + h  , M = 2  4 I  VQ 1 1 τ = , Q = twh 2 + bt f ( h + t f ), Itw 8 2

δ =

Arora, Introduction to Optimum Design, 4e

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ry =

Iy A

=

1 3  1 ht w + 2 t f b 3  12   12 ht w + 2t f b

Step 3. Design Variables b : width of flange t f : thickness of flange h : height of web t w : thickness of web Step 4. Cost Function Minimization total mass f =ρL(2bt f + ht w ) Step 5. Constraints Deflection:

δ ≤ δa

Bending stress:

σ ≤σa

Shear stress:

τ ≤τa

Buckling: E Fy

Lateral torsional buckling:

L ≤ 1.76ry

Local flange buckling:

b E ≤ 0.38 2t f Fy

Local web buckling:

h E ≤ 3.76 tw Fy

Design Variable bounds: bmin ≤ b ≤ bmax

t f min ≤ t f ≤ t f max hmin ≤ h ≤ hmax t w min ≤ t w ≤ t w max

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SAMPLE PROJECT: NONLINEAR PROGRAMMING Optimum Design of Crane Girder Consider the design optimization problem that you have formulated earlier. Solve the problem for the data given below. Use a numerical method to solve the problem, such as in Excel, Matlab, Mathematica, etc. Submit a brief word-processed report containing your problem formulation, solution, computer files and a discussion of the solution.

DATA FOR PROJECT: L = 30 ft ; P = 2 × 10 lbs; E = 29 × 10 6 psi; Fy = 5 × 10 4 psi; 5

L ; σ a = 3 × 10 4 psi; τ a = 18 × 10 3 psi δ a = 240

Lower and upper bounds of the design variables: 1 ≤ h ≤ 40 in; 1 ≤ b ≤ 40 in; 0.1 ≤ t f ≤ 3 in;

0.1 ≤ t w ≤ 3 in;

The following Excel Answer Report gives the optimal solution for the problem. It is seen that the flange width is much larger than the height of the web which may not be practical. Therefore a constraint 𝑏𝑏 ≤ ℎ is imposed. The Excel sheets on the following pages give the solution for this case.

Microsoft Excel 15.0 Answer Report Worksheet: [Project-Opt of Crane Girder.xlsx]Crane Girder Report Created: 4/1/2016 1:13:18 PM Result: Solver found a solution. All Constraints and optimality conditions are satisfied.

Solver Engine Engine: GRG Nonlinear Solution Time: 0.062 Seconds.

Iterations: 19 Subproblems: 0 Solver Options Max Time 100 sec, Iterations 10000, Precision 0.000001 Convergence 0.0001, Population Size 100, Random Seed 0, Derivatives Central, Require Bounds Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 3%, Solve Without Integer Constraints Objective Cell (Min) Cell $C$25 Vol

Variable Cells Cell $D$3 h $D$4 b $D$5 tf $D$6 tw

Name

Original Value 16200

Final Value 40785.27746

Name

Original Value 15 15 1 1

Final Value Integer 11.27013505 Contin 30.77866011 Contin 1.681597762 Contin 0.867583703 Contin

Constraints Cell Name $B$28 Bending stress Value/Eq. $B$29 Shear stress Value/Eq. $B$30 Deflection Value/Eq. $B$31 Lateral torsional buckling Value/Eq. $B$32 Local flange buckling Value/Eq. $B$33 Local web buckling Value/Eq. $B$3 web height Lower limit $B$4 flange width Lower limit $B$5 flange thickness Lower limit $B$6 web thickness Lower limit $D$3 h $D$4 b $D$5 tf $D$6 tw

Cell Value Formula Status 29469.90185 $B$28<=$D$28 Not Binding 18000 $B$29<=$D$29 Binding 1.5 $B$30<=$D$30 Binding 360 $B$31<=$D$31 Binding 9.15161188 $B$32<=$D$32 Binding 12.99025674 $B$33<=$D$33 Not Binding 1 $B$3<=$D$3 Not Binding 1 $B$4<=$D$4 Not Binding 0.1 $B$5<=$D$5 Not Binding 0.1 $B$6<=$D$6 Not Binding 11.27013505 $D$3<=$E$3 Not Binding 30.77866011 $D$4<=$E$4 Not Binding 1.681597762 $D$5<=$E$5 Not Binding 0.867583703 $D$6<=$E$6 Not Binding

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Slack 530.09815 0 0 0 0 77.56253449 10.27013505 29.77866011 1.581597762 0.767583703 28.72986495 9.221339893 1.318402238 2.132416297

Sample Projects The Excel sheet showing the additional constraint 𝑏𝑏 ≤ ℎ.

Project - Optimum Design of Crane 1. Design variable name Lower limit web height 1 flange width 1 flange thickness 0.1 web thickness 0.1

Symbol h b tf tw

2. Parameter name Span length Modulus of elasticity Yield stress Concentrated load

Symbol L E sigma_y P

360 29000000 50000 200000

3. Dependent variable name Cross sectional area Moment of inertia Bending Moment Bending stress Shear force Deflection Shear stress Radius of gyration w.r.t y axis

Symbol A I M sigma V D tau ry

Equation =2*b*tf+h*tw =1/12*tw*h^3+2*(1/12*b*tf^3+b*tf*(1/2*h+1/2*tf)^2) =P*L/4 =M*(tf+h/2)/I =P =P*L^3/(48*E*I) =V*(1/8*tw*h^2+1/2*b*tf*(h+tf))/(I*tw) =SQRT((1/12*h*tw^3+2*(1/12*tf*b^3))/A)

4. Objective fuction name Volume of material

Symbol Vol

=A*L

5. Constraints Bending stress Shear stress Deflection Lateral torsional buckling Local flange buckling Local web buckling b<=h

Value

Units in^3

</>/= < < < < < < <

Value/Eq. 30000 18000 =L/240 =1.76*ry*SQRT(E/sigma_y) =0.38*SQRT(E/sigma_y) =3.76*SQRT(E/sigma_y) 0

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Upper limit 40 40 3 3

Units in psi psi lb

Equation

Value/Eq. =sigma =tau =D =L =b/(2*tf) =h/tw =b-h

Arora, Introduction to Optimum Design, 4e

Value 30.8949952738024 30.8949952738024 1.6879537549947 0.346633586010688

Name Allowable bending stress Allowable shear stress Allowable deflection Lateral torsional buckling Local flange buckling Local web buckling

Units in in in in

Sample Projects

Microsoft Excel 15.0 Answer Report Worksheet: [Project-Opt of Crane Girder.xlsx]Crane Girder Report Created: 4/1/2016 1:24:04 PM Result: Solver converged in probability to a global solution.

Solver Engine Engine: GRG Nonlinear Solution Time: 0.343 Seconds.

Iterations: 0 Subproblems: 31 Solver Options Max Time 100 sec, Iterations 10000, Precision 0.000001 Convergence 0.00001, Population Size 100, Random Seed 0, Derivatives Central, Multistart, Require Bounds Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 3%, Solve Without Integer Constraints, Ass Objective Cell (Min) Cell $C$25 Vol

Variable Cells Cell $D$3 h $D$4 b $D$5 tf $D$6 tw

Name

Original Value 16200

Final Value 41402.84024

Name

Original Value 15 15 1 1

Final Value Integer 30.89499527 Contin 30.89499527 Contin 1.687953755 Contin 0.346633586 Contin

Constraints Cell Name $B$28 Bending stress Value/Eq. $B$29 Shear stress Value/Eq. $B$30 Deflection Value/Eq. $B$31 Lateral torsional buckling Value/Eq. $B$32 Local flange buckling Value/Eq. $B$33 Local web buckling Value/Eq. $B$34 b<=h Value/Eq. $B$3 web height Lower limit $B$4 flange width Lower limit $B$5 flange thickness Lower limit $B$6 web thickness Lower limit $D$3 h $D$4 b $D$5 tf $D$6 tw

Cell Value Formula Status 10800.13724 $B$28<=$D$28 Not Binding 18000 $B$29<=$D$29 Binding 0.234725073 $B$30<=$D$30 Not Binding 360 $B$31<=$D$31 Binding 9.15161188 $B$32<=$D$32 Binding 89.12868378 $B$33<=$D$33 Not Binding 0 $B$34<=$D$34 Binding 1 $B$3<=$D$3 Not Binding 1 $B$4<=$D$4 Not Binding 0.1 $B$5<=$D$5 Not Binding 0.1 $B$6<=$D$6 Not Binding 30.89499527 $D$3<=$E$3 Not Binding 30.89499527 $D$4<=$E$4 Not Binding 1.687953755 $D$5<=$E$5 Not Binding 0.346633586 $D$6<=$E$6 Not Binding

Slack 19199.86276 0 1.265274927 0 0 1.424107456 0 29.89499527 29.89499527 1.587953755 0.246633586 9.105004726 9.105004726 1.312046245 2.653366414

It is seen that the height of the web has increased substantially, and the optimum material has also increased accordingly.

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Sample Projects

SAMPLE PROJECT: GRAPHICAL OPTIMIZATION Optimum Design of Torsion Rod Design a hollow torsion rod shown in Fig.E3.34 to satisfy the following: 1. The calculated shear stress, 𝜏𝜏 , shall not exceed the allowable shear stress 𝜏𝜏𝑎𝑎 under the normal operation torque T o (N·m).

2. The calculated angle of twist, 𝜃𝜃, shall not exceed the allowable twist, 𝜃𝜃𝑎𝑎 (radians).

3. The member shall not buckle under a short duration torque of T max (N·m).

Requirements for the rod and material properties are given in Table E3.34(A) and E3.34(B) (use length = 0.75m and Titanium material). Use the following design variables: x 1 = outside diameter of the shaft; x 2 = ratio of inside/outside diameter, d i /d o.

Formulate the design optimization problem, and using graphical optimization, determine the inside and outside diameters for a minimum mass rod to meet the above design requirements.

Compare the hollow rod with an equivalent solid rod (d i /d o = 0). Use consistent set of units (e.g.

Newtons and millimeters) and let the minimum and maximum values for design variables be given as 𝑑𝑑𝑖𝑖 ≤ 0.999 0.02 ≤ 𝑑𝑑𝑜𝑜 ≤ 0.5 m, 0.60 ≤ 𝑑𝑑𝑜𝑜 Useful expressions for the rod are: 𝜋𝜋 Mass of rod: 𝑀𝑀 = 𝜌𝜌𝜌𝜌(𝑑𝑑𝑜𝑜2 − 𝑑𝑑𝑖𝑖2 ), 𝑘𝑘𝑘𝑘 4 𝑐𝑐 Calculated shear stress: 𝜏𝜏 = 𝑇𝑇𝑜𝑜 , 𝑃𝑃𝑃𝑃 𝐽𝐽 𝑙𝑙 Calculated angle of twist: 𝜃𝜃 = 𝑇𝑇𝑜𝑜 , 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝐺𝐺𝐺𝐺 Critical buckling torque: 𝜋𝜋𝑑𝑑𝑜𝑜3 𝐸𝐸 𝑑𝑑𝑖𝑖 2.5 𝑇𝑇𝑐𝑐𝑐𝑐 = �1 − � , N. m 𝑑𝑑𝑜𝑜 12√2(1 − 𝜈𝜈 2 )0.75 Notation M = mass of the rod (kg) 𝑑𝑑𝑜𝑜 = outside diameter of the rod (m) 𝑑𝑑𝑖𝑖 = inside diameter of the rod (m) 𝜌𝜌 = mass density of material (kg/m3) l = length of the rod (m) T 0 = Normal operation torque (N ⋅ m) c = Distance from rod axis to extreme fiber (m) J = Polar moment of inertia (m4) θ = Angle of twist (radians) G = Modulus of rigidity (Pa) T cr = Critical buckling torque (N ⋅ m) E = Modulus of elasticity (Pa) 𝜈𝜈 = Poisson’s ratio.

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FIGURE E3-34 Hollow torsion rod.

Sample Projects TABLE E3-34(A) Rod Requirements Torsion rod Length, number l (m) 1 0.50 2 0.75 3 1.00

Normal torque, T 0 (kN ⋅ m) 10.0 15.0 20.0

Max. torque, T max (kN ⋅ m) 20.0 25.0 30.0

TABLE E3-34(B) Materials and Properties for the Torsion Rod Material Density, Allowable 3 Shear stress, 𝜌𝜌 (kg/m ), 𝜏𝜏𝑎𝑎 (MPa) 1. 4140 alloy steel 2. Aluminum alloy 24 ST4 3. Magnesium alloy A261 4. Berylium 5. Titanium

7850 2750 1800 1850 4500

275 165 90 110 165

Allowable twist, 𝜃𝜃𝑎𝑎 (degrees) 2 2 2

Elastic modulus, E (GPa)

Shear modulus, G (GPa)

Poisson’s ratio (𝜈𝜈)

210 75 45 300 110

80 28 16 147 42

0.30 0.32 0.35 0.02 0.30

Project 1 Reporting Requirements Submit your report using the provided Microsoft Word .doc file. Remember to download the file and before working on it to rename the file including your first and last name. For example, “project1.doc” is the name of the file that you download. Assuming that “Aye Ten” is a student name, the renamed file would be “project1_AyeTen.doc”.

Your report should include the following 4 parts: 5) Formulation of the problem: clear definition of design variables, cost function, and constraints.

6) Graphical representation of the problem using MATLAB; copy the MATLAB code in the end of the Word report in an Appendix.

7) Final solution for the problem and discussion about the solution, which should include: a) Final design variables and cost function values b) Active/inactive constraints at optimum c) Discussion of the problem formulation and the final solution.

8) Submit the MATLAB “.m” file for the problem (in addition to including it in the body of the report); the code should be organized clearly and in a readable format.

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Sample Projects

Project 1 Grading Rubric: Graded based on 50 points Report attributes Formulation of the problem (15 points)

Graphical representation of the problem (15 points)

Final solution for the problem and its discussion (10 points)

MATLAB code for the problem (10 points) Overall Score

Meets all expectations

Partially meets expectations

Below expectations

(15 points)

(8 points)

(3 points)

Complete and clear presentation that includes clearly defined design variables, accurate cost function and constraints

Complete, but sloppy presentation that either includes poorly defined design variables and/or cost function and/or constraints

Incomplete formulation that either lacks clear definition of design variables and/or cost function and/or constraints

(15 points) Complete legible graph with correct labels and correct identification of the optimum point

(8 points) Partially complete graph with either missing or incorrect labels and/or missing or incorrect optimum point (5 points)

(10 points) Complete solution that includes correct: Design variables Cost function value Active/inactive constraints at optimum.

Complete discussion of the formulation and the solution

Partial solution that contains all of the components with either incorrect design variables and/or incorrect cost function and/or incorrect active/inactive constraints at optimum; incomplete or missing discussion of problem formulation or solution

(3 points) Graph illegible and/or labels missing and/or optimum point missing

(2 points) Incomplete solution with either missing or incorrect design variables; missing or incorrect cost function; incorrect active/inactive constraints at optimum; missing and or incomplete discussion of problem formulation and/or solution.

(10 points)

(5 points)

(0 points)

Nicely organized MATLAB code in a clear readable and logical format Best effort 45 or more

MATLAB code incomplete and/or unreadable and not formatted well Acceptable effort 35 to 44

Not included

Arora, Introduction to Optimum Design, 4e

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Needs improvement 20 to 34

Sample Projects Solution Design Variables: x 1 = outside diameter of the shaft; x 2 = ratio of inside/outside diameter, d i /d o Units of mass, force and length are kg, N and mm respectively.

Cost Function: minimize mass of hollow shaft

(

(

)

) ( π ρ l 4) x (1 − x )

2 d i2 l ρ ( π 4 ) d o2 1 − ( di d= f = ρ ( π 4 ) d o −= o) l 2

2 1

2 2

Constraints:

g1 =τ − τ a =To c J −τ a =To ( 0.5 x1 ) ( π x 14 (1 − x2 ) 32 ) − τ a =(16To π ) x 13 (1 − x 42 ) − τ a ≤ 0

( G π ( d − d ) 32 ) − θ =( 32T l G π ) x (1 − x ) − θ ≤ 0

g 2 =θ − θ a =T0 l GJ − θ a =T0 l

4 o

4 i

4 1

0

a

4 2

a

  πE   x 3 (1 − x ) 2.5 + T −Tcr + Tma x = − g3 = 2 ma x ≤ 0  12 2 1 − υ 2 0.75  1 ( )  

(

)

Transform the parameters used in the foregoing equations to have consistent units. Note the first case of requirements and first case of material properties in Tables E3.34(A) and E3.34(B) are used. l = 0.5 m = 500 mm; T o = 10.0 kN.m = 10.0(103)(103) = 107 N.mm; T max = 20.0 kN.m = 2.0 × 10 7 N.mm;

θ a = 2° = 2 ( π 180 ) = π 90 rad ; ρ = 7850 kg/m3 = 7850(10 −9 ) = (7.85 × 10 −6 ) kg/mm3;

τ a = 275 MPa = 275 N/mm2 E = 210 GPa = (2.1 × 105 ) N/mm2; G = 80 GPa = (8.0 × 10 4 ) N/mm2; υ = 0.3 2 f = ( 7.85 × 10− 6 ) ( π 4 )( 500 ) x 12 (1 − x= 2)

( 3.08269 × 10 ) x (1 − x ) g 1 = (16 × 10 π ) x (1 − x ) − 275 = 5.093 × 10 x (1 − x ) − 275 ≤ 0 32 (10 ) ( 500 ) g x (1 − x ) − π 90 = = 6.36619 × 10 x (1 − x ) − 3.49066 × 10 ≤ 0 (8.0 × 10 ) π 7

3 1

−3

4 2

7

3 1

2 1

2 2

4 1

4 2

4 2

7

2

= g3

4 1

4

−π ( 2.1×10 5 )

12 2 (1 − 0.3 )

2 0.75

4 2

5

( x (1 − x ) 2.5) + ( 2.0 ×10 ) 3 1

7

2

= 2.0 × 107 − ( 4.17246 × 104 )( x 13 ) (1 − x 2 )

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≤0

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−2

Sample Projects g 4 = 20 − x1 ≤ 0; g 5 =x1 − 500 ≤ 0; g 6 = 0.6 − x2 ≤ 0; x2 0.999 ≤ 0 g 7 =−

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Sample Projects

∗

∗

Optimum solution: x 1 = 103.0 mm, x 2 = 0.955, f ∗ = 2.9 kg; g 1 (shear stress constraint) and g 3 (buckling constraint) are active.

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Sample Projects MATLAB Code [x1,x2]=meshgrid(90:0.1:110, 0.9:0.001:1); %Enter functions for the minimization problem f=(3.08269*10^-3)*(x1.^2).*(1-x2.^2); g1=(5.093*10^7)-(275)*(x1.^3).*(1-x2.^4); g2=(6.36619*10^5)-(3.49066*10^-2)*(x1.^4).*(1-x2.^4); g3=(2*10^7)-(4.17246*10^4)*(x1.^3).*(1-x2).^2.5; g4=20-x1; g5=x1-500; g6=0.6-x2; g7=x2-0.999; cla reset axis auto %Minimum and maximum values for axes are determined automatically xlabel('x1'),ylabel('x2') %Specifies labels for x- and y-axes hold on %retains the current plot and axes properties for all subsequent plots cv1=[0 0]; const1=contour(x1,x2,g1,cv1,'k','LineWidth',3); text(100.5,0.948,'g1') cv11=[0.01:0.001:0.1]; const1=contour(x1,x2,g1,cv11,'c'); const2=contour(x1,x2,g2,cv1,'k','Linewidth',3); const2=contour(x1,x2,g2,cv11,'c'); text(92,0.925,'g2') const3=contour(x1,x2,g3,cv1,'k','Linewidth',3); const3=contour(x1,x2,g3,cv11,'c'); text(108,0.952,'g3') const4=contour(x1,x2,g4,cv1,'k','Linewidth',4); const4=contour(x1,x2,g4,cv11,'c'); text(0.1,0.06,'g4') const5=contour(x1,x2,g5,cv1,'k','Linewidth',3); const5=contour(x1,x2,g5,cv11,'c'); text(0.1,0.02,'g5') const6=contour(x1,x2,g6,cv1,'k','Linewidth',3); const6=contour(x1,x2,g6,cv11,'c'); text(2.5,0.005,'g6') const7=contour(x1,x2,g7,cv1,'k','Linewidth',3); const7=contour(x1,x2,g7,cv11,'c'); text(6,200,'g7') text(98,0.93,'Feasible Region') fv=[1.5 2.9 5]; %Defines contours for the minimization function fs=contour(x1,x2,f,fv,'k'); %'k' specifies black dashed lines for function contours clabel(fs) %Automatically puts the contour value on the graph hold off %Indicates end of this plotting sequence Arora, Introduction to Optimum Design, 4e

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Sample Projects

SAMPLE PROJECT: VERIFICATION OF KKT CONDITIONS Optimum Design of a Tripod Tripods are used in many military and commercial applications. The objective of this project is to design a minimum mass tripod of height H to support a vertical load W = 60 kN. The tripod base is an equilateral triangle with sides B = 1200 mm. The struts have a solid circular cross section of diameter D (Fig. E3.54).

The axial stress in the struts must not exceed the allowable stress in compression, and axial load in the strut P must not exceed the critical buckling load P cr divided by a safety factor FS = 2. Use consistent units of Newtons and centimeters. The minimum and maximum values for design variables are 0.5≤ H ≤1m and 1 ≤ D ≤10 cm. Material properties and other relationship are given below: Material: aluminum alloy 2014-T6 Allowable compressive stress,

𝜎𝜎𝑎𝑎 = 150MPa

Young’s modulus,

E = 75 GPa

𝜌𝜌 = 2800 kg/m3

Mass density,

1 𝑙𝑙 = (𝐻𝐻 2 + 𝐵𝐵 2 )0.5 3 2 𝜋𝜋 𝐸𝐸𝐸𝐸 𝑃𝑃𝑐𝑐𝑐𝑐 = 2 𝑙𝑙 𝜋𝜋 4 𝐼𝐼 = 𝐷𝐷 64 𝑊𝑊𝑊𝑊 𝑃𝑃 = 3𝐻𝐻

Strut length, Critical buckling load, Moment of inertia, Strut load,

FIGURE E3.54 A tripod.

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Sample Projects

Project 2 Reporting Requirements Submit your report using the provided Microsoft Word .doc file. Remember to download the file and before working on it to rename the file including your first and last name. For example, “project1.doc” is the name of the file that you download. Assuming that “Aye Ten” is a student name, the renamed file would be “project1_AyeTen.doc”.

Your report should include the following 4 parts: 9) Complete problem formulation process: problem description, data/information, clear definition of design variables, cost function, and constraints.

10) Graphical representation and optimum solution of the problem using MATLAB (design variable values, cost function, active constraints; zoomed-in figure for the optimum point); copy the MATLAB code in the end of the Word report in an Appendix.

11) Verify the KKT necessary conditions for the graphical solution: a) Write KKT conditions for the solution case. b) Calculate the Lagrange multipliers for the active constraints and verify their sign. c) A brief discussion of the solution process, final solution, KKT Conditions, and conclusions.

Note: This discussion should only contain information pertinent to the scope of this project.

12) Submit the Matlab/Mathematica “.m/.nb” file for the problem (in addition to including it in the body of the report); the code should be organized clearly and in a readable format.

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Sample Projects

Project 2 Grading Rubric: Graded based on 50 points Report attributes Formulation of the problem (10 points)

Meets all expectations

Partially meets expectations

Below expectations

(10 points)

(6 points)

(2 points)

Complete and clear presentation that includes complete problem formulation process with proper formatting.

Complete, but sloppy presentation that either includes poorly defined design variables and/or cost function and/or constraints

Incomplete formulation that either lacks clear definition of design variables and/or cost function and/or constraints

Graphical representation of the problem (10 points)

(10 points) Complete legible graph with correct labels and correct identification of the optimum point

Final solution, verification of KKT conditions, discussion (20 points)

(20 points)

(6 points) (2 points) Incomplete graph Graph illegible and/or labels missing and/or with either optimum point missing or missing incorrect labels and/or missing or incorrect optimum point (10 points) (4 points)

MATLAB code for the problem (10 points)

Complete solution; complete and correct KKT conditions.

Correct solution of Lagrange multipliers.

Proper discussion of results/KKT conditions.

Incomplete solution.

Incomplete or incorrect verification of KKT conditions.

Lack of discussion of results/KKT conditions.

Incorrect solution.

Incorrect verification of KKT conditions; no discussion of results/KKT conditions.

(10 points)

(5 points)

(0 points)

Nicely organized MATLAB code in a clear readable and logical format

MATLAB code incomplete and/or unreadable and not formatted well

Not included

Overall Score: 1) Best effort: 45 or more; 2) Acceptable effort: 35 to 44; 3); Needs improvement: 20 to 34

Arora, Introduction to Optimum Design, 4e

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Sample Projects Solution: Design of a tripod-

Design Variables: H = height of the tripod; D = diameter of cross-section of the struts Data and Information: Using units of Newtons and centimeters, the data are calculated as: Area, 𝐴𝐴 =

𝜋𝜋𝐷𝐷 2

; 4

𝜋𝜋𝐷𝐷 4

Moment of inertia, 𝐼𝐼 = 64

σa = 150 MPa = 1.5 × 104 N/cm 2 ; E = 75 GPa = 7.5 × 106 N/cm 2 ;

2800 kg/m3 = 2.8 × 10−3 kg/cm3 ; B == 1200 mm 120 cm; W = 60 kN = 6.0 × 104 N ρ= Various expressions are defined in the problem statement.

Cost Function: minimize mass; f = 3 ( ρ Al ) , kg Constraints: P g1 = − σ a ≤ 0 A

g2 = P −

Pcr ≤0 FS

g 3 = H − 100 ≤ 0; − H + 50 ≤ 0; g4 = g5 = D − 10 ≤ 0; g 6 =− D + 1 ≤ 0 The optimization problem is to find D and H to minimize mass f subject to inequality constraints g 1 to g 7 .

In the computer program, the above formulation may be used as it is. Or, it may be reduced to be only in terms of the design variables by substituting various constants and expressions into the cost and constraints:

f = 3 ( 2.8 × 10−3 )( πD 2 4 )( H 2 + 120 2 3) 2 = ( 6.59734 × 10−3 ) D 2( H 2 + 4800 ) 2 1

1

( 2.546475 × 10 )( H + 4800 ) D H − 1.5 × 10 ≤ 0; ( 2.0 × 10 )( H + 4800 ) − 1.816774 × 10 D H + 4800 ≤ 0; = g ( ) ( ) H = g1

4

4

2

1 2

2

2

4

1 2

6

4

2

2

g 3 = H − 500 ≤ 0; g 4 = 50 − H ≤ 0; g 5 = D − 50 ≤ 0; g 6 = 0.5 − D ≤ 0

Optimum solution from the graph: H ∗ = 50.0 cm, D ∗ = 3.42 cm, f ∗ = 6.6 kg; g 2 (buckling load constraint) and g 4 (maximum height constraint) are active.

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Sample Projects Verification of KKT conditions:

Arora, Introduction to Optimum Design, 4e

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Sample Projects

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Sample Projects

Optimum solution: H ∗ = 50.0 cm, D ∗ = 3.42 cm, f ∗ = 6.6 kg; g 2 (buckling load constraint) and g 4 (maximum height constraint) are active.

Arora, Introduction to Optimum Design, 4e

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Sample Projects

Zoomed-in graph Exercise 3.54 - Zoomed-in 6 5.5 5

g4

4.5

Feasible Region

9

4

D

6.6 Optimum X

3.5 3

g2

8

5

2.5 2 g1 1.5 1 10

20

30

40

50 H

Arora, Introduction to Optimum Design, 4e

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60

70

80

Sample Projects

MATLAB Code: 3.54 [H,D]=meshgrid(1.0:1.0:600.0, 0.1:0.1:60.0); f=6.59734e-3*(D.^2).*(sqrt(H.^2+4800)); g1=2.546479e4*sqrt(H.^2+4800)./((D.^2).*H)-1.5e4; g2=2.0e4*sqrt(H.^2+4800)./H-1.816774e6*(D.^4)./(H.^2+4800); g3=H-500; g4=50-H; g5=D-50; g6=0.5-D; cla reset axis([-20,600,-2,60]) xlabel('H'),ylabel('D') title('Exercise 3.54') hold on cv=[0 0]; const1=contour(H,D,g1,cv,'k'); text(550,3,'g1') cv1=[1000.0:1000.0:20000.0]; const11=contour(H,D,g1,cv1,'m'); const2=contour(H,D,g2,cv,'k'); text(400,4.5,'g2') cv2=[1000.0:1000.0:5000.0]; const21=contour(H,D,g2,cv2,'r'); const3=contour(H,D,g3,cv,'k'); cv3=[0.5:0.5:5.0]; text(515,30,'g3') const31=contour(H,D,g3,cv3,'r'); const4=contour(H,D,g4,cv,'k'); text(20,30,'g4') const41=contour(H,D,g4,cv3,'r'); const5=contour(H,D,g5,cv,'k'); cv5=[0.05:0.05:0.5]; text(250,53,'g5') const51=contour(H,D,g5,cv5,'r'); const6=contour(H,D,g6,cv,'k'); text(250,-1,'g6') const61=contour(H,D,g6,cv5,'m'); text(200,45,'Feasible Region') fv=[6.6, 100, 500, 1000]; fs=contour(H,D,f,fv,'k--'); clabel(fs) hold off

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Sample Projects Code for zoomed-in figure [H,D]=meshgrid(10.0:1.0:80.0, 1:0.1:6.0); f=6.59734e-3*(D.^2).*(sqrt(H.^2+4800)); g1=2.546479e4*sqrt(H.^2+4800)./((D.^2).*H)-1.5e4; g2=2.0e4*sqrt(H.^2+4800)./H-1.816774e6*(D.^4)./(H.^2+4800); g3=H-500; g4=50-H; g5=D-50; g6=0.5-D; cla reset axis([10,80,1,6]) xlabel('H'),ylabel('D') title('Exercise 3.54 - Zoomed-in') hold on cv=[0 0]; const1=contour(H,D,g1,cv,'k'); text(70,1.7,'g1') cv1=[200.0:100.0:5000.0]; const11=contour(H,D,g1,cv1,'m'); const2=contour(H,D,g2,cv,'k'); text(75,3.8,'g2') cv2=[300.0:100.0:6000.0]; const21=contour(H,D,g2,cv2,'r'); const3=contour(H,D,g3,cv,'k'); cv3=[0.04:0.02:2.0]; text(515,30,'g3') const31=contour(H,D,g3,cv3,'r'); const4=contour(H,D,g4,cv,'k'); text(52,5,'g4') const41=contour(H,D,g4,cv3,'r'); const5=contour(H,D,g5,cv,'k'); cv5=[0.02:0.01:0.4]; text(250,53,'g5') const51=contour(H,D,g5,cv5,'r'); const6=contour(H,D,g6,cv,'k'); text(250,-1,'g6') const61=contour(H,D,g6,cv5,'m'); text(55,4.5,'Feasible Region') text (50,3.42, 'X') text (51,3.6,'Optimum') fv=[5, 6.6, 8, 9]; fs=contour(H,D,f,fv,'k--'); clabel(fs) hold off

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Sample Projects

%Exercise 3.54 - Alternate Formulation [H,D]=meshgrid(1.0:1.0:120.0, 0.1:0.1:12.0); %Data for the problem W=60000; B=120; ro=2.8/1000; E=7.5e6; sigma_a=15000; D_min=1; D_max=10; H_min=50; H_max=100; FS=2; %Analysis Expressions L=(H.*H+B.*B./3).^0.5; I=pi.*D.^4./64; A=pi.*D.*D./4; P_cr=pi.*pi.*E.*I./(L.*L); P=W.*L./(3.*H); Mass=3.*ro.*A.*L; sigma=P./A; %Formulation f=Mass; g1=sigma - sigma_a; g2=P-P_cr./FS; %f=6.59734e-3*(D.^2).*(sqrt(H.^2+4800)); %g1=2.546479e4*sqrt(H.^2+4800)./((D.^2).*H)-1.5e4; %g2=2.0e4*sqrt(H.^2+4800)./H-1.816774e6*(D.^4)./(H.^2+4800); g3=H - H_max; g4=H_min - H; g5=D - D_max; g6=D_min - D; cla reset axis([1,120,.1,12]) xlabel('Height, H'),ylabel('Diameter, D') title('Exercise 3.54') hold on cv=[0 0]; const1=contour(H,D,g1,cv,'k'); text(80,2,'g1') cv1=[1000.0:100.0:20000.0]; const11=contour(H,D,g1,cv1,'m'); const2=contour(H,D,g2,cv,'k'); text(110,4.5,'g2') cv2=[1000.0:100.0:10000.0]; const21=contour(H,D,g2,cv2,'r'); const3=contour(H,D,g3,cv,'k'); cv3=[0.5:0.1:5.0]; text(95,9,'g3') const31=contour(H,D,g3,cv3,'r'); const4=contour(H,D,g4,cv,'k'); text(52,9,'g4') const41=contour(H,D,g4,cv3,'r'); const5=contour(H,D,g5,cv,'k'); cv5=[0.05:0.01:0.5]; text(75, 9.5,'g5') const51=contour(H,D,g5,cv5,'r'); const6=contour(H,D,g6,cv,'k'); text(15,1.3,'g6') Arora, Introduction to Optimum Design, 4e

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Sample Projects const61=contour(H,D,g6,cv5,'m'); text(60,7,'Feasible Region') fv=[4, 6.6, 8, 10, 12]; fs=contour(H,D,f,fv,'k--'); clabel(fs) hold off

Exercise 3.54 12 11 10 g5 9

g3

g4

Diameter, D

8 7

Feasible Region

6 5

10

4

6.6

12 8

g2

3 4 2 1

g1 g6 20

40

Arora, Introduction to Optimum Design, 4e

60 Height, H

P-39

80

100

120

SAMPLE EXAMS

Arora, Introduction to Optimum Design, 4e

E-1

Sample Exams

SAMPLE EXAM 1: MID –TERM EXAM for First Course on Optimization (Time allowed = 2 hrs) 1. A proposal to provide on-site power generation for a hospital is to be evaluated. The proposal is to purchase an initial equipment package now costing $750,000. This package has a life of 25 years and no salvage value. Maintenance costs are $18000 per year. Ten years from now, an additional equipment package costing $200,000 is needed for a planned expansion. The life of this package is 15 years and the salvage value is $15000.

Maintenance and operating costs are an additional $18000 per year. What is the present worth of this proposal? Assume 8% interest compounded annually.

(15 points) 2.

2

2

Minimize

f(x) = x1 + x2

Subject to

x1 − 9x2 ≥ 0 ;

2

x1 ≥ 4

x1 + x2 ≤ 10 ; x2 ≥ 0 a. Show the feasible region on the graph sheet. b. State whether this region is (circle your choice) (1) open or closed (2) bounded or unbounded (3) convex or not convex? c. Determine x* graphically and label this point.

(15 points) 3.

The cross section of a corn storage building is shown below. At least 1000 ft3 of corn is to be stored per foot of length of the building. Formulate the problem to provide the least cost of building per unit volume of corn stored. Neglect costs associated with the ends of the building.

(20 points) Cost of building per foot of length Roof 3S2 2 Walls 2H each Floor 5S Ends Neglect

Arora, Introduction to Optimum Design, 4e

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Sample Exams

4. (i)

Answer True or False _____ The number of independent equality constraints can be larger than the number of design variables for a design optimization problem. (ii) _____ The number of "≤ type" constraints must be less than the number of design variables for a design problem. (iii) _____ A function can have more than one global minimum point. (iv) _____ Gradient of a function at a point is normal to the level surface defined by the function. (v) _____ A point satisfying first order necessary conditions may not be a minimum point for the function. (vi) _____ If Hessian of the cost function for unconstrained problem is positive definite at a candidate point, the point is a local maximum for the function. (vii) _____ A point satisfying KKT necessary conditions for a general optimum design problem can be a maximum point for the cost function. (viii) _____ In optimum design problems, "≥ type" constraints cannot be treated. (ix) _____ A linear equality constraint always defines a nonconvex region. (x) _____ The minimum point for a constrained optimization problem gives a stationary value for the Lagrange function.

(10 points) 5.

The equilibrium state for an engineering system can be obtained by minimizing the following function: 2 2 f(x1, x2) = 132.6 x1 − 53.1 x1x2 + 207.6 x2 − 1.732 x1 − x2 (i) Determine equilibrium state for the system (ii) Check sufficient conditions and discuss global optimality of the solution.

(15 points)

6.

An optimum design problem is formulated as minimize f(x1, x2) = 2 2x1 + x2; subject to:

1 1 1 + − 2 ≤ 0; − 1 ≤ 0; x1 ≥ 0; x2 ≥ 0 x1 x1 + 2 x2 x1 + 2 x2 (i) Write KKT necessary conditions.

Arora, Introduction to Optimum Design, 4e

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Sample Exams (ii) Ignoring the constraints x1 ≥ 0, x2 ≥ 0, define the cases to be solved for the candidate minimum points (iii) Find solution for any two cases defined in (ii).

(25 points)

Arora, Introduction to Optimum Design, 4e

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Sample Exams

SAMPLE EXAM 2: MID-TERM EXAM for First Undergraduate Course (Time allowed = 90 minutes) 1.

A company is considering the purchase of a new piece of testing equipment which is expected to produce $8,000 additional profit at the end of the first year of operation; this amount will decrease by $500 per year for each additional year of ownership. The equipment costs $18,000 and will have a salvage value of $3,000 after 4 years of use. The same equipment could also be rented for $6,500 per year, payable in advance on the first day of each rental year. The interest rate is 25% compounded annually. Use a present worth analysis to compare the alternatives. What is your recommendation to the company - purchase, rent, neither?

(15 points)

2.

A rectangular storage structure is to be constructed for which the following costs apply: 2 Cost, $/ft Roof 3.00 ‰W (e.g., $30/ft2 if W = 10 ft) Walls 2.00 ‰H Floor 5.00 The structure is to be placed at the center of a plot of land measuring 40 ft ‰60 ft, as shown below, with 5 ft minimum clearance to the property line. Because of solar collectors on the adjacent property to the north, the tank must not cast a noon-day shadow outside the property line when the sun is at an angle of 70 ο with the horizontal. At least 15,000 ft3 of volume is required (neglect the thickness of the walls and roof). Formulate the problem to determine the dimensions for the least structure cost per unit volume. Express the formulation in the standard form. It is not necessary to normalize the constraints. Do not solve. (15 points)

Arora, Introduction to Optimum Design, 4e

E-5

Sample Exams 3.

An optimum design problem is formulated as minimize f(x) = (x1 − 3)2 + (x2 − 3)2 subject to:

(I) 2x1 − 4x2 ≥ − 4 2

(II) − x1 + x2 ≤ 0

x1 ≤ 2

(III)

x1 ≥ 0

(IV)

(V) x2 ≥ 0 a. Plot and label the constraints and identify the feasible region. b. Plot iso-cost lines and identify the optimum point. c. Identify active constraints at the optimum point; give values of design variables and the cost function at the optimum point.

(15 points) 4.

An unconstrained optimum design problem is formulated as 25 minimize f(a, b) = a + + 5b ab Check whether or not a = 5 and b = 1 is a minimum point for the problem.

(8 points)

5.

Check for convexity of the following design problem: 2 2 maximize f(x) = − x1 − x2 + 3x1 − 3x2 + x1x2 subject to the constraints 2x1 − 4x2 ≥ − 4 2

x1 − x2 ≥ 2

x1 ≤ 2

x1 ≥ 0 x2 ≥ 0

(7 points) 6. (i)

Answer True or False _____ The number of independent equality constraints can be greater than the number of design variables for a design optimization problem. (ii) _____ The number of "≤ type" constraints must be less than the number of design variables for a design problem. (iii) _____ A function can have more than one global minimum point. (iv) _____ Gradient of a function at a point is normal to the level surface defined by the function. (v) _____ A point satisfying first order necessary conditions is always a minimum point for the function. (vi) _____ If Hessian of the cost function for unconstrained problem is positive definite at a candidate point, the point is a local maximum for the function.

Arora, Introduction to Optimum Design, 4e

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Sample Exams (vii) _____ A point satisfying KKT necessary conditions for a general optimum design problem can be a maximum point for the cost function. (viii) _____ In optimum design problem formulation, "≥ type" constraints cannot be treated. (ix) _____ A nonlinear equality constraint always defines a nonconvex region. (x) _____ The minimum point for a constrained optimization problem gives a stationary value for the Lagrange function.

(15 points) 7.

A circular tank that is closed at both ends is to be fabricated to have a volume of 250π m3.

The fabrication cost is found to be proportional to the surface area of the sheet metal needed 2 for fabrication of the tank and is $400/m . The tank is to be housed in a shed with sloping roof which limits the height of the tank by the relation H ≤ 8D where H is the height and D is the diameter of the tank. The problem is formulated as minimize f = 400(2 π D2/4 + π DH) ; subject to the constraints: πD 2 H = 250 π; H ≤ 8D; H ≥ 0, D ≥ 0 4 a. Write KKT necessary conditions for the design problem b. Identify all the cases for solution of the KKT necessary conditions c. Check if the following case gives a solution that satisfies all the KKT necessary conditions: H ≤ 8D as active; Lagrange multipliers for the other two inequalities as zero.

(20 points)

8.

An optimum design problem is formulated as minimize f(x1, x2) = − x1 − 2x2

subject to g1(x1, x2) = 0.1x1 + 0.4x2 − 225 ≤ 0 g2(x1, x2) = x1 + x2 − 800 ≤0

x x g3(x1, x2) = 1 + 2 − 1.0 ≤ 0 600 1200 950 1450 The point x1 = 3 , x2 = 3 satisfies the KKT necessary conditions. The Lagrange multipliers for the constraints are: 10/3, 2/3, 0. a. What is the effect on the cost function if the right hand side of constraint one is changed to 5? b. What is the effect on the cost function if the right hand side of constraint two is changed to−50?

(5 points)

Arora, Introduction to Optimum Design, 4e

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Sample Exams

SAMPLE EXAM 3: MIDTERM EXAM for First Undergraduate Course (Time allowed = 70 minutes) Problem 1 (25 Points) An engineering design optimization problem is formulated as follows: Optimize f = 2 x1 + 3x 2 − x1 − 2 x 2 3

2

Subject to

g1 : x1 + 3x 2 ≤ 6 g 2 : 5 x1 + 2 x 2 ≤ 10 g 3 : − x1 ≤ 0 g 4 : − x2 ≤ 0 A graphical representation for the problem is shown in the figure. The dashed lines show the constraint boundaries and solid lines show the cost function contours. ANSWER THE FOLLOWING QUESTIONS.

(a) Label each constraint on the graph as g 1 to g 4 . (8)

(b) Hatch-out the infeasible side of each constraint. What is the feasible region for the problem? (5) FGIH (c) For each of the following points, identify whether the point is a local minimum, local maximum or neither of the two by circling your selection: (12) Point A: Min

Max

Neither

Point B: Min

Max

Neither

Point F: Min

Max

Neither

Point G: Min

Max

Neither

Point H: Min

Max

Neither

Point I:

Min

Arora, Introduction to Optimum Design, 4e

Max

Neither

E-8

Sample Exams

Arora, Introduction to Optimum Design, 4e

E-9

Sample Exams Problem 2 (25 Points) An engineering design problem is formulated as the following unconstrained optimization problem: minimize f ( x, y, z ) = x + y + 2 y + 2 z − 4 zx − 8 z 2

2

3

(i) (8 Points) Write gradient vector and Hessian matrix for the function.

 2x − 4z  ∇f =  2 y + 2   2  6 z − 4 x − 8  2 H= 0   − 4

0 − 4 2 0  0 12 z 

(ii) (7 Points) Check if (2, -1, 1) is a stationary point for the function?

The gradient vector should be zero.

 2 × 2 − 4  0  ∇f =  − 2 + 2  =  0  ≠ 0     − 6 − 8 − 8 10     The gradient is not zero, so it is not a stationary point.

Arora, Introduction to Optimum Design, 4e

E-10

Sample Exams (iii) (10 Points) Check if the point (4, -1, 2) is a local minimum point for the function.

 8 − 8  0  ∇f =  − 2 + 2  =  0      24 − 16 − 8 0 The gradient is zero, so it is a stationary point. Check the form of the Hessian.

 2 H= 0  − 4

0 − 4 2 0  0 24

M1 = 2 > 0 M2 = 4 > 0 M 3 = (2)(2)(24)+(-4)(8) = 64 > 0 Hessian is positive definite, so (4, -1, 2) is a local minimum point because it satisfies both second order necessary and sufficient conditions.

Arora, Introduction to Optimum Design, 4e

E-11

Sample Exams Problem 3 (25 Points) An engineering design problem is formulated as minimize

f ( x1 , x 2 ) = 2 x12 + x 22 − 11x1 − 4.5 x 2 + 2

subject to

g1 : 10 ≥ 2 x1 + 5 x 2

(1)

g 2 : 2 x1 + x 2 ≤ 8

(2)

The point (2.5, 1) has been identified as a local minimum point. Using constraint variation sensitivity result, determine the change in the minimum value for the cost function if the constant 10 in the first constraint (g 1 ) is changed to 9.5?

(5 Points) Write problem in the standard form:

f ( x1 , x 2 ) = 2 x12 + x 22 − 11x1 − 4.5 x 2 + 2

minimize subject to

g1 : 2 x1 + 5 x 2 − 10 ≤ 0 ;

g 2 : 2 x1 + x 2 − 8 ≤ 0

(5 Points) Calculate cost function and check constraints at (2.5, 1):

f = 2( 2.5 ) 2 + 1 − 11( 2.5 ) − 4.5( 1 ) + 2 = −16.5 g1 = 2( 2.5 ) + 5 − 10 = 0 ( active ) g 2 = 2( 2.5 ) + 1 − 8 = −2 < 0 ( inactive ) (8 Points) Calculate Lagrange multiplier for g 1 :

L = 2 x12 + x 22 − 11x1 − 4.5 x 2 + 2 + u1 (2 x1 + 5 x 2 − 10 )

∂L = 4 x1 − 11 + 2u1 = 0; ∂x1

u1 = 0.5

∂L = 2 x 2 − 4.5 + 5u1 = 0; u1 = 0.5 ∂x 2 (7 Points) Calculate change in cost

∆f = −u1 (∆e1 ); ∆e1 = e1new − e1old = 9.5 − 10 = −0.5 ∆f = −0.5( −0.5 ) = 0.25 The cost function will increase by 0.25.

Arora, Introduction to Optimum Design, 4e

E-12

Sample Exams Problem 4 (25 Points) A trucking company wants to purchase several new trucks. It has $2 million to spend. The investment should yield a maximum of trucking capacity for each day in tonnes×kilometers. Data for the three available truck models are given in the table below; i.e., truck load capacity, average speed, crew required/shift, hours of operations for three shifts, and the cost of each truck. There are some limitations on the operations that need to be considered. The labor market is such that the company can hire at the most 150 persons to operate the trucks. Garage and maintenance facilities can handle at the most 25 trucks. How many trucks of each type should the company purchase? Formulate the design optimization problem. DO NOT SOLVE THE PROBLEM. No. of hours of Truck load Average truck Crew required operations per Cost of each capacity speed (km/h) per shift day Truck model (tonnes) (3 shifts) truck ($) A 10 55 1 18 40,000 B 20 50 2 18 60,000 C 18 50 2 21 70,000

(a) Identify and clearly define the design variables for the problem. (6) A = Number of Type A to be purchased B = Number of Type B to be purchased C = Number of Type C to be purchased (b) Describe in words the objective function for the problem. (2) Objective function: maximize the trucking capacity, tonnes × kilometers (c) Write an expression for the objective function. (5) Capacity

= A(10×55×18) + B(20×50×18) + C(18×50×21) = 9900A + 18000B + 18900C

(d) Clearly describe each constraint in words. (4) The total cost of trucks should not exceed $2 mllion Number of drivers hired cannot exceed 150 Maintenance facility cannot handle more than 25 trucks Number of trucks purchased of each type must be positive or at the most zero

Arora, Introduction to Optimum Design, 4e

E-13

Sample Exams

(e) Develop an expression for each constraint. (8) Available capital constraint:

40000A + 60000B + 70000C ≤ 2,000,000

Limitation of drivers:

3(A + 2B +2C) ≤ 150

Limitation of maintenance facility:

A + B +C ≤ 25

Nonnegativity of variables:

A, B, C ≥ 0

Arora, Introduction to Optimum Design, 4e

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Sample Exams

SAMPLE EXAM 4: FINAL EXAM for First Undergraduate Course (Time allowed = 2 hrs) 1.

2.

An airlines company is considering the purchase of new jet passenger airplanes. Pertinent data on the planes under consideration are listed in the table below. A maximum of $150 million is available for the purchase. Pilots are available to fly up to 30 new airplanes and the available area of maintenance facilities is 900,000 ft2. For overall fleet balance, at least a third of the planes purchased should be short range airplanes. Formulate this problem to maximize the annual profit resulting from the purchase. Ignore the fact that the number of airplanes must be an integer.

Express the formulation in standard form, but do not normalize. DO NOT SOLVE. (10 points) _________________________________________________________________________ Net annual profit including Area needed Minimum Purchase Cost Capital recovery costs for maintenance Number 2 Required Type $million/plane $1000/plane ft /plane _______________________________________________________________________________ A Long Range 6.7 420 40,000 1 B Medium Range 5.0 300 30,000 1 C Short Range 3.5 230 24,000 3 _______________________________________________________________________________ A function f(x) is to be minimized subject to the constraints x1 − x2 ≤ 1; 6x1 + 5x2 ≤ 30 ; x1, x2 ≥ 0 A plot of several constant values of f(x) are shown. a) Identify on the plot all of the local minimum points for the constrained problem. b) Identify the global minimum point.

(10 points)

Arora, Introduction to Optimum Design, 4e

E-15

Sample Exams

3.

An optimum design problem is given as: minimize f(x1, x2) = 3(x1−1)2 + (x2−2)2 subject to 3x1 + 2x2 − 6 ≤ 0 2

2

(x1 + 1) + (x2 + 1) − 9.61 = 0 a) Determine whether or not the point (−1, 2.1) satisfies the necessary conditions for a minimum point. b) The point (1.846, 0.231) is a local minimum point. Determine the approximate minimum value of f(x) in this region if the equality constraint is changed to (x1 + 1)2 + (x2 + 1)2 − 9 = 0 (10 points) 4.

Maximize z = 6x1 + 4x2

subject to 3x1 + 5x2 ≤ 15 2x1 + x2 ≥ 2

x1 − 2x2 = −2 x1 ≥ 0

x2 is unrestricted in sign a) Set up the initial Simplex tableau for the given LP problem. Define any variables that you add. b) Complete one iteration and circle the pivot element for the second iteration, but do not proceed further. c) List the values of every variable and function (i) at the initial point (ii) after the first iteration (15 points)

Arora, Introduction to Optimum Design, 4e

E-16

Sample Exams

5.

Given the following LP problem (10 points) minimize f = 9x1 + 3x2 + 2x3 subject to g1: 2x1 − 3x2 + x3 ≥ 5

g2: −x1 − 2x2 + 2x3 ≤ 2 x1, x2 ≥ 0

The final tableau is: ______________________________________________________________________ x1 x2 x3 x4 x5 x6 ______________________________________________________________________ 1 − 0·8 0 − 0·4 −0·2 0·4 1.6 0 − 1·2 1 − 0·2 0·4 0·2 1.8 ______________________________________________________________________ 0 13 0 4 1 −4 f −18 where x4 = surplus variable for g1 x5 = slack variable for g2 x6 = artificial variable for g1 a) What are the optimal x and f(x). Identify the basic variables and the active constraints. b) List the Lagrange multipliers for the two constraints c) Is the solution unique? Explain d) The RHS of g2 is increased from 2 to 5. (i) Does the basis change? (ii) If not, what is the new f(x)? e) What are the upper and lower bounds on the cost coefficient of x1 such that the optimal point x does not change?

6.

The minimum of f(x) is desired. Starting from the point x(0) = (1,1), the first iteration of (1) the Steepest Descent method gives the point x = (2.111, 0.444).

2 2 f(x) = 2x1 + x2 − 8x1 a) Determine the search direction at point x(1) using: 1. Steepest Descent 2. Conjugate Gradient 3. Newton's Method b) Establish whether d = (1.5, −1) is a descent direction starting from x(1). c) Starting from x(0) = (1, 1), find x(1) using d = (1, −1).

(15 points)

7.

An optimum design problem is formulated as find y1 and y2 to minimize : f(y1, y2) = y1(y1 + 6y2)

Arora, Introduction to Optimum Design, 4e

E-17

Sample Exams

subject to the constraints : h1 = ≤0

y12 y2 − 1.0 = 0; g1 = 1 − y2/2y1 ≤ 0; g2 = y2/3y1 − 1.0 200

g3 = y2/20 − 1.0 ≤ 0 ;

g4 = − y1 ≤ 0 ; g5 = − y2 ≤ 0

At a given design point (4, 8), a) define the linear programming subproblem using 50% move limits b) define the quadratic programming subproblem. points) 8.

(15

An optimum design problem is formulated in a normalized form as minimize f = 200 x1 + 100 x2 x1 x2 x ≤ 0 ; g2: 1 − 1.0 ≤ 0 10000 100 x2 x x g3: − 1.0 ≤ 0 ; g4: − 1 + 2 ≤ 0 200 50 100 At the design point (90, 120), the solution of QP subproblem is given as d = (−30, 30); u = (15000, 0, 0, 0) Calculate an appropriate step size and the improved design in the given direction. Let initial R = 1.0 and γ = 0.5.

(15 points)

subject to g1: 1.0 −

Arora, Introduction to Optimum Design, 4e

E-18

Sample Exams

SAMPLE EXAM 5: FINAL EXAM for First Undergraduate Course (Time allowed = 2 hrs) 1.

The circular tube shown below is to be designed for two conditions: when P = 40 kN, axial stress σ ≤ σy when P = 0, deflection due to weight δmax ≤ 0.001l

Limits for dimensions are t = 2.0 to 15.0 mm, D = 15.0 to 125.0 mm, and design is to be optimized for minimum weight.

Material Properties: σy = 165 MPa ; ρ = 80 kN/m3 ;

D ≤ 20. The t

E = 210 GPA

Weight W = ρAl ; I = πD3t/8 5W l 3 Constraint Relations: σ = P/A; δmax = when P = 0 384 E I Formulate the optimal design problem. Express the results in standard normalized form.

(10 points) Member Properties: Area A = πDt;

2.

Given the following optimum design problem: 2 2 minimize f(x1, x2) = 2x1 − 6x1x2 + 9x2 − 18x1 + 9x2 subject to x1 + 2x2 ≤ 10

4x1 − 3x2 − 20 ≤ 0 x1 ≥ 0

x2 ≥ 0 Check if point x1 = 70/11 and x2 = 20/11 is a candidate local minimum point.

(10 points)

Arora, Introduction to Optimum Design, 4e

E-19

Sample Exams 2. Solve the following problem using the Simplex method up to the point that the first feasible solution is found. List the values of all the variables at each iteration.

(15 points) Minimize f = x1 − 2x2 subject to g1: 2x1 − x2 ≥ 2

g2: x1 + x2 ≥ 3 g3:

4.

x2 ≤ 2

x1, x2 ≥ 0

Given the following LP problem: minimize f = 2x1 − 4x2

subject to g1: 10x1 + 5x2 ≤ 15 g2: 4x1 + 10x2 ≤ 36 x1, x2≥ 0

Slack variables for g1 and g2 and x3 and x4 respectively. The final tableau is _______________________________________ x2 x3 x4 b x1 _______________________________________ 2 1 1/5 0 3 −16 0 −2 1 6 _______________________________________ 10 0 4/5 0 f +12 a. Determine optimal values of f and x. b. Determine Lagrange multipliers for g1 and g2. c. Determine the right-hand-side ranges for g1 and g2. d. What is the smallest value that f can have, with the current basis, if the right hand side of g1 is changed? What is the right hand side of g1 for this case?

(10 points) 5.

3

2

2

2

Consider the problem : minimize f(x1, x2) = x1 + 12x1x2 + 2x2 + 5x1 + 3x2 a. Find steepest descent direction at the point (2,1). b. Find Newton direction at the point (2,1). points)

6.

(15 2

2

Consider the problem : minimize f(x1, x2) = 5x1 − x1 x2/16 + x2/4x1 Determine if the direction (2, 5) is that of descent for f(x1, x2) at the point (1, 2). points)

Arora, Introduction to Optimum Design, 4e

E-20

(5

Sample Exams 7.

Consider the problem : minimize f(x1,x2) = 250 ‰103 x1x2 subject to the constraints x1x2 ≥ 6.4 ‰10-3 3 x1x2 ≤ 1.6 ‰10-4 x1 ≥ 0, x2 ≥ 0 a. Define the problem in the standard form normalizing the constraints. b. Linearize the problem at the point (0.4, 0.01) and define a linear subproblem with proper move limits. c. Define the quadratic programming subproblem at the point (0.4, 0.01).

(20 points)

8.

The minimum weight design of a closed and thin-walled cylindrical pressure vessel subjected to stress, strain and volume constraints can be formulated as 2 minimize x + 320 xy x subject to −1≤0 60 y x( x − y ) 1− ≤0 3600 x ≥ 0, y ≥ 0 where x is the radius and y is the thickness. At a design point of x = 40, y = 0.5, the search direction for the descent function is given as d = (25.6, 0.45). The Lagrange multipliers for the constraints are u = (4880, 19400). Choose γ = 0.5 and calculate the step size for the constrained steepest descent method at the given point.

(15 points)

Arora, Introduction to Optimum Design, 4e

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Sample Exams

SAMPLE EXAM 6: FINAL EXAM for First Undergraduate Course (Time allowed = 75 minutes) 1. (15 points) An engineering design problem is formulated as follows: Maximize z = −2 x + 3 x 1

2

subject to

6 − 3 x1 + 2 x 2 ≥ 0 − x1 + x 2 − 4 ≥ 0 x1 is free in sign; x 2 ≥ 0 (i) Transcribe the problem to the standard LP format. (7)

(2) Minimize

(

)

f = 2 x1+ − x1− − 3 x 2 ;

x1 = x1+ − x1− (2)

( ) x3 = slack var iable + − (3) − (x1 − x1 ) + x 2 − x 4 + x5 = 4; x 4 = surplus var iable 3 x1+ − x1− − 2 x 2 + x3 = 6; x1+ , x1− , x 2 ≥ 0;

x5 = artificia ; var iable

(ii) Set up the initial tableau for the problem so that the Simplex method can be used to solve the problem. DO NOT SOLVE THE PROBLEM. (6)

+

(3) Artificial cost, w = x 5 = 4 + x1

− x1− − x 2 + x 4

(3)

x1+

x1-

x2

x3

x4

x5

b

x3

3

-3

-2

1

0

0

6

x5

-1

1

1

0

-1

1

4

Cost

2

-2

-3

0

0

0

f-0

Artificial cost

1

-1

1

0

1

0

w-4

(iii) What are the initial values for the basic and nonbasic variables? (2)

Basic: x 3 = 6, x 5 = 4; all others are nonbasic – have zero value

Arora, Introduction to Optimum Design, 4e

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Sample Exams 2. (16 points) A linear design optimization problem is to be solved using the Simplex method. The initial tableau for the problem is given below, where x 3 is a slack variable, x 4 is a surplus variable, and x 5 is an artificial variable.

X1

x2

x3

x4

x5

b

x3

-2

1

1

0

0

4

x5

1 (3)

1

0

-1

1

2

Cost

-4

-2

0

0

0

f-0

Artificial cost

-1 (2)

-1

0

1

0

w-2

x3

0

3

1

-2

2

8

x1

1

1

0

-1

1

2

Cost

0

2

0

-4

4

f+8

Art. (5)

0

0

0

0

1

w-0

ANSWER THE FOLLOWING QUESTIONS. (i) Identify basic and nonbasic variables for the initial tableau. (2) Basic: x3, x5; (ii)

Nonbasic: x1, x2, x4

Complete (ONLY) one iteration of the Simplex method to obtain the next tableau. (10)

(iii) Does your tableau give an optimum solution for the linear design problem?

YES/NO (1)

(iv) Does your tableau indicate end of Phase I of the Simplex method?

YES/NO (1)

(v)

Does your tableau indicate the problem to be infeasible?

(vi) Does your tableau indicate the problem to be unbounded?

Arora, Introduction to Optimum Design, 4e

E-23

YES/NO (1) YES/NO (1)

Sample Exams 3. (15 points) A linear engineering design problem is formulated as follows: Minimize f = 10 x1 + 6 x 2 + 2 x3 − 6 x 4 subject to

− x1 + x 2 + x3 − x 4 ≥ 1 3x1 + x 2 − x3 − 3x 4 ≥ 2 x1 , x 2 , x3 , x 4 ≥ 0

The final Simplex tableau for the problem is given as follows: x1

x2

x3

x4

x5

x6

x7

x8

b

x2

0

1

0.5

-1.5

-0.75

-0.25

0.75

0.25

1.25

x1

1

0

-0.5

-0.5

0.25

-0.25

-0.25

0.25

0.25

Cost

0

0

4

8

2

4

-2

-4

f - 10

Artificial

0

0

0

0

0

0

1

1

w-0

x 5 and x 6 are surplus variables for constraint #1 and 2, respectively. x 7 and x 8 are artificial variables for constraint #1 and 2, respectively.

ANSWER THE FOLLOWING QUESTIONS. (1.5) each i

x 1 is a nonbasic variable

YES/NO

ii

Value of x 2 at the optimum point is 0.25

YES/NO

iii

Value of x 6 is zero at the optimum point

YES/NO

iv

Value of the cost function at the optimum is 10

YES/NO

v

Constraint #1 is inactive at the optimum point

YES/NO

vi

Constraint #2 is active at the optimum point

YES/NO

vii

Lagrange multiplier for constraint #1 is 2

YES/NO

viii

Lagrange multiplier for constraint #2 is -4

YES/NO

ix

The problem has multiple optimum points

YES/NO

x

Cost coefficient of x 3 can be increased to 4 without affecting the optimum cost function

YES/NO

Arora, Introduction to Optimum Design, 4e

E-24

Sample Exams 4. (20 points) An engineering design problem is formulated in terms of the design variables x and y as follows:

(2

Minimize f = x 1 x − y Subject to

2

)

h1 : y ( x + 4 y ) = 10 g1 : − 2 x 2 + 13 y ≤ 0 g2 : − y − 4 ≤ 0

(i) Evaluate cost and constraint functions at the point (2,3). (4)

(

)

f = 2 12 × 2 − 3 2 = −16

h1 := 3(2 + 4 × 3) − 10 = 32 g1 = −2 × 2 2 + 13 × 3 = −7 g 2 = − 3 − 4 = −7 (ii) Evaluate gradients of the cost and constraint functions at the point (2,3). (8)

x − y 2   − 7  ∇f =  = ;  − 2 xy  − 12

 y  3 ∇h1 =  =   x + 8 y  26

− 4 x   − 8  ∇g1 =  = ;  1 / 3  1 / 3

0 ∇g 2 =   − 1

(iii) Define the quadratic programming subproblem at the point (2,3). (8) Minimize

(

f = −16 − 7 d1 − 12d 2 + 12 d12 + d 22

)

subject to

32 + 3d1 + 26d 2 = 0

− 7 − 8d1 + 13 d 2 ≤ 0 − 7 − d2 ≤ 0

Arora, Introduction to Optimum Design, 4e

E-25

Sample Exams 5. (20 Points) An engineering design problem is formulated as the following unconstrained problem: Minimize f = 8 x

2 + 2 x1 x 2 + 5 x 22 − 2 x1 + 3x 2 1

Starting from a point (-1, -1), one iteration of the steepest descent method has been completed yielding a new point as (0.21, 0.45). (i)

Calculate the steepest descent direction at the point (-1, -1). 6

16 x + 2 x 2 − 2 − 20 c = ∇f =  1 =   2 x1 + 10 x 2 + 3  − 9  Steepest descent direction

20 d = −c =   9 (ii)

Calculate the steepest descent direction at the point (0.21, 0.45). 2

16 x + 2 x 2 − 2 2.26 c ( 1 ) = ∇f =  1 =   2 x1 + 10 x 2 + 3 7.92 (iii)

Calculate the conjugate gradient direction at the point (0.21, 0.45). 8

67.8 2.26 20  0.82  = 0.141 ; β = d ( 1 ) = −c( 1 ) + β 0 d ( 0 ) = −  + . = 0 141     1 481 7.92  9  − 6.65 (iv)

Show that the conjugate gradient direction calculated in part (iii) is a direction of descent for the cost function. 4

d ( 1 ) • −c( 1 ) = 1.8532 − 52.668 < 0 The descent condition is satisfied.

Arora, Introduction to Optimum Design, 4e

E-26

Sample Exams 6. (14 points) An engineering design problem is formulated in terms of the design variables x, y, and z as follows: Minimize f = 2xz subject to

2

g1 : (4 × 10 −4 )( 5 zy + 2 z 2 ) − 1 ≤ 0 g 2 : 1 − 2.5 × 10 −5 ( x + y ) z 2 ≤ 0 The quadratic programming subproblem has been defined and solved at the point (50, 10, 25). The Lagrange multipliers for the constraints are obtained as (20000, 70000). Using the initial value for the penalty parameter R 0 as 1, calculate the value of the descent function Φ at the given point.

SOLUTION:

f = 2 × 50 × 25 2 = 31250 (2)

g1 = (4 × 10 −4 )( 5 × 25 × 10 + 2 × 25 2 ) − 1 = 0 g 2 = 1 − 2.5 × 10 −5 (50 + 10) × 25 2 = 0.0625

(4)

r = 20000 + 70000 = 90000 (2)

R = max(1,90000 ) = 90000 (2) V = max{0; 0.0.0625} = 0.0625 (2)

Φ = f + RV = 62500 + 0.0625 × 90000 = 68125 (2)

Arora, Introduction to Optimum Design, 4e

E-27

Sample Exams

SAMPLE EXAM 7: PROBLEM FORMULATION for First Undergraduate Course (Time allowed = 20 minutes) A chemical manufacturer plans to construct an underground tank for liquid propane storage, as shown in the figure. The required volume of the cylindrical tank of diameter D and height H (ignore thickness of the tank wall) must be at least 400,000 ft3. A sketch of the tank is shown below along with the cost data for construction and operations. The depth of the cover C must be at least 8 ft. The depth of the excavation cannot exceed 150 ft because rock is encountered at that level and it is uneconomical to dig through the rock. Another requirement for the problem is that the diameter to height ratio cannot exceed 1.2. The objective is to minimize the total cost of construction and operations of the tank facility. Also the annual cost of operations of the tank should not exceed $400,000.

Soil

Cover, C Height, H

150 ft

Diameter, D

Rock Construction Costs: Excavation: Tank top and bottom: Tank wall:

Assume that the shape of excavation is cylindrical with diameter D and depth (C + H). Cost of excavation = $540 per cubic yard.

Cost = $100 per ft2 of surface area.

Cost = $[0.10 D(C+H)] per ft2 of wall area.

Annual Operating Costs: The annual operating cost depends on the total surface area of the tank and the depth of the cover, and is given as [200(total surface area of the tank) - 6000C] $.

The design variables for the problem are identified as: C = depth of the cover, ft D = diameter of the tank, ft H = height of the tank, ft

Arora, Introduction to Optimum Design, 4e

E-28

Sample Exams

CIRCLE YOUR CHOICE (each question carries 5 points).

1. The cost of excavation is given as A. $135𝜋𝜋𝐷𝐷 2 (𝐶𝐶 + 𝐻𝐻) B. $5𝜋𝜋𝐷𝐷 2 (𝐶𝐶 + 𝐻𝐻) C. $135𝜋𝜋𝐷𝐷 2 𝐻𝐻

D. $5𝜋𝜋𝐷𝐷 2 𝐻𝐻

2. Cost of the tank’s bottom is given as A. $25𝜋𝜋𝐷𝐷 2 B. $100𝜋𝜋𝐷𝐷 2 C. $50𝜋𝜋𝐷𝐷 2

D. $100𝜋𝜋𝜋𝜋

3. Cost of tank wall is given as A. $0.1𝜋𝜋𝜋𝜋(𝐶𝐶 + 𝐻𝐻)𝐻𝐻 B. $0.1𝜋𝜋𝜋𝜋(𝐶𝐶 + 𝐻𝐻)𝐷𝐷

C. $0.1𝜋𝜋𝐷𝐷 2 𝐻𝐻(𝐶𝐶 + 𝐻𝐻) D. $0.1𝐷𝐷𝐷𝐷(𝐶𝐶 + 𝐻𝐻)

4. The constraint for the volume is given as A. 𝜋𝜋𝐷𝐷 2 𝐻𝐻 ≥ 400,000 B. 𝜋𝜋𝐷𝐷 2 𝐻𝐻 ≤ 400,000

C. 𝜋𝜋𝐷𝐷 2 𝐻𝐻 = 1,600,000

D. 𝜋𝜋𝐷𝐷 2 𝐻𝐻 ≥ 1,600,000

5. The constraint on the depth of excavation is given as A. 𝐶𝐶 + 𝐻𝐻 = 150 B. 𝐻𝐻 ≤ 150

C. 𝐶𝐶 + 𝐻𝐻 ≤ 150

D. 𝐶𝐶 + 𝐻𝐻 ≥ 150

6. The annual operating cost constraint is given as A. 100𝜋𝜋𝜋𝜋[𝐻𝐻 + 𝐷𝐷] − 6000𝐶𝐶 ≤ 400,000 B. 100𝜋𝜋𝜋𝜋[2𝐻𝐻 + 𝐷𝐷] − 6000𝐶𝐶 = 400,000

C. 100𝜋𝜋𝜋𝜋[𝐻𝐻 + 𝐷𝐷] − 6000𝐶𝐶 = 400,000

D. 100𝜋𝜋𝜋𝜋[2𝐻𝐻 + 𝐷𝐷] − 6000𝐶𝐶 ≤ 400,000

Arora, Introduction to Optimum Design, 4e

E-29

Sample Exams

SAMPLE EXAM 8: GRAPHICAL OPTIMIZATION for First Undergraduate Course (Time allowed = 20 minutes) The graphical representation for a nonlinear design optimization problem with one equality constraint is given below. The equality constraint represents a circle, identified as h(x) = 0 in the figure. Four iso-cost curves (ellipses) with cost function values of 19, 41, 248 and 457 are shown in the figure.

Circle True or False (5 points each) (i)

Point C is a local maximum point.

True False

(ii)

Point A is a local minimum point.

True False

(iii)

Point F is a local maximum point.

True False

(iv)

Point B is a local minimum point

True False

(v)

Point E is a local maximum point when the equality constraint is changed to an inequality such that the region outside the circle is infeasible.

True False

(vi)

Point D is a local minimum point when the equality constraint is changed to an inequality such that the region outside the circle is infeasible.

True False

Arora, Introduction to Optimum Design, 4e

E-30

Sample Exams

SAMPLE EXAM 9: OPTIMALITY CONDITIONS for First Undergraduate Course (Time allowed = 30 minutes) 1. (10 Points) Hessian (2×2 matrix) of a function at its stationary point is given as 0 −2 𝐇𝐇 = � � −2 0

Circle your choice: A. H is positive definite B. H is positive semidefinite C. H is negative definite D. H is negative semidefinite E. H is indefinite

2. (10 Points) An engineering design problem is formulated as maximize

F ( x1 , x 2 ) = − x12 − 2 x 22 + 5 x1 + 12 x 2 − 2

subject to

5 ≥ 5 x1 + 2 x 2 x1 + 2 x 2 ≤ 8

(1)

(2) The Lagrangian function for the problem is defined as (Circle your choice):

(

) (

)

A.

L = x12 + 2 x 22 − 5 x1 − 12 x 2 + 2 + u1 5 x1 + 2 x 2 − 5 + s12 + u 2 x1 + 2 x 2 − 8 + s 22

B.

L = − x12 − 2 x 22 + 5 x1 + 12 x 2 − 2 + u1 5 x1 + 2 x 2 − 5 + s12 + u 2 x1 + 2 x 2 − 8 + s 22

C.

L = x12 + 2 x 22 − 5 x1 − 12 x 2 + 2 + u1 5 − 5 x1 − 2 x 2 + s12 + u 2 x1 + 2 x 2 − 8 + s 22

(

) (

(

) (

)

(

) (

)

D. L = x12 + 2 x 22 − 5 x1 − 12 x 2 + 2 − u1 5 x1 + 2 x 2 − 5 + s12 − u 2 x1 + 2 x 2 − 8 + s 22

Arora, Introduction to Optimum Design, 4e

E-31

)

Sample Exams

3. (30 Points) An engineering design problem is formulated as the following unconstrained optimization problem: minimize f ( x, y ) = 2 x 3 + y 2 − 4 xy − 8 x

Circle YES or NO for the following questions:

A. Point (1, -1/2) is a stationary point for the function f.

YES

NO

B. Point (2, 4) is a local minimum point for the function f.

YES

NO

C. Point (-2/3, -4/3) is a local maximum point for the function f.

YES

NO

Arora, Introduction to Optimum Design, 4e

E-32

Sample Exams

4. An Optimization problem is formulated as minimize f ( r , t ) = − ( r − 3) − ( t − 2 ) 2

2

subject to g1 = r + t − 10 ≤ 0

g2 = t − 5 ≤ 0 The Lagrangian for the problem is defined as 𝐿𝐿 = −(𝑟𝑟 − 3)2 − (𝑡𝑡 − 2)2 + 𝑢𝑢1 (𝑟𝑟 + 𝑡𝑡 − 10 + 𝑠𝑠12 ) + 𝑢𝑢2 (𝑡𝑡 − 5 + 𝑠𝑠22 )

Circle True or False (5 points each) (i)

This problem has 2 KKT solution cases

True False

(ii)

At the point (3, 2), constraint g 1 is active

True False

(iii)

At the point (3, 2), the Lagrange multiplier u 1 for constraint g 1 is zero

True False

(iv)

At the point (3, 2), constraint g 2 is active

True False

(v)

At the point (3, 2), the Lagrange multiplier u 2 for constraint g 2 is positive

True False

(vi)

At the point (5.5, 4.5), constraint g 1 is active

True False

(vii)

At the point (5.5, 4.5), constraint g 2 is active

True False

(viii)

At the point (5.5, 4.5), the Lagrange multiplier u 1 for constraint g 1 is 5

True False

(ix)

At the point (5.5, 4.5), the Lagrange multiplier u 2 for constraint g 2 is zero

True False

(x)

The point (5, 5) is determined to be a local minimum point. The Lagrange multiplier u 2 for constraint g 2 could be negative

True False

Note: Point (3, 2) implies r = 3 and t = 2

Arora, Introduction to Optimum Design, 4e

E-33

Sample Exams

Karush-Kuhn-Tucker (KKT) Optimality Conditions The Problem: Minimize Subject to

f(x)

h i (x) = 0; i = 1 to p g j (x) ≤ 0; j = 1 to m KKT Optimality Conditions: 1. Define the Lagrangian p

m

i =1

j=1

(

L = f + ∑ vihi + ∑ uj g j + sj2

)

2. The Gradient Conditions ∂L = 0; k = 1 to n ∂xk

∂L = 0 = hi; i = 1 to p ∂vi ∂L j = 1 to m = 0 = g j + sj2 ; ∂uj

(

)

3. The Switching Conditions ∂L j = 1 to m = 0 = 2ujsj; ∂sj 4. The Feasibility Check for Inequality Constraints

s 2j ≥ 0; (or g j ≤ 0; j = 1 to m) 5. Non-negativity of Lagrange Multipliers for Inequalities uj ≥ 0; j = 1 to m 6. Regularity Check Gradients of active constraints should be linearly independent In such a case the LM for the constraints are unique.

Arora, Introduction to Optimum Design, 4e

E-34

Sample Exams

SAMPLE EXAM 10: LINEAR PROGRAMMING for First Undergraduate Course (Time allowed = 40 minutes) 1. (10 points) An engineering design problem is formulated as a linear programming (LP) problem: Maximize z = x1 − 2x 2 subject to

10 ≥ −3 x1 + 2 x 2 2 x1 − x 2 ≥ 6 x2 ≥ 0

ANSWER THE FOLLOWING QUESTIONS FOR THE LP. i

Standard definition of the LP problem requires minimization of the function

YES/NO

ii

x1 must be decomposed as x1 = x1+ − x1− ; with x1+ , x1− ≥ 0 to transcribe the

YES/NO

problem to the standard form iii

x 2 must be decomposed as x 2 = x 2+ − x 2− ; with x 2+ , x 2− ≥ 0 to transcribe the

YES/NO

iv

problem to the standard form Right sides of the LP constraints in the standard form are free in sign

YES/NO

v

Constraint #1 requires introduction of a slack variable to transcribe it into the standard

YES/NO

form vi

The problem requires Two-Phase Simplex method for its solution

YES/NO

vii

Constraint #2 requires introduction of a surplus variable to transcribe it into the standard

YES/NO

form viii

Constraint #1 requires addition of an artificial variable

YES/NO

ix

Solution of the problem requires use of an artificial cost function in the Simplex procedure In Phase II of the Simplex iterations, the artificial cost function determines the pivot element

YES/NO

x

Arora, Introduction to Optimum Design, 4e

E-35

YES/NO

Sample Exams 2. (20 points) A linear design optimization problem is to be solved using the Two-Phase Simplex method. The initial tableau for Phase I of the problem is given below, where x 3 is a slack variable, x 4 is a surplus variable, and x 5 is an artificial variable. x1

x2

x3

x4

x5

b

x3

-2

1

1

0

0

4

x5

1

1

0

-1

1

2

Cost

-4

-2

0

0

0

f-0

Artificial cost

-1

-1

0

1

0

w-2

x3

0

3

1

-2

2

8

x1

1

1

0

-1

1

2

Cost

0

2

0

-4

4

f+8

Art.

0

0

0

0

1

w-0

(i) ANSWER THE FOLLOWING QUESTIONS. (10 points) i a21 can be the pivot element for this tableau

YES/NO

ii

a12 can be the pivot element for this tableau

YES/NO

iii

x5 is a basic variable having value of 2

YES/NO

iv

x1 is a basic variable having a value of 4

YES/NO

v

x4 is a nonbasic variable

YES/NO

vi

Artificial cost function value is 2 for the initial tableau

YES/NO

vii

If a12 is selected as the pivot element, x3 will become a nonbasic variable

YES/NO

viii

a11 can be selected as a pivot element

YES/NO

ix

x3 is a basic variable in the initial tableau

YES/NO

x

If a22 is selected as the pivot element, it will make x1 a basic variable

YES/NO

(ii) Choosing a21 as the pivot element, complete the Simplex iteration in the blank tableau shown above. (10 Points)

Arora, Introduction to Optimum Design, 4e

E-36

Sample Exams 3. (10 points) A linear design optimization problem is to be solved using the Simplex method. During the Simplex iteration, the following tableau is obtained, where x 3 is a slack variable, x 4 is a surplus variable, and x 5 is an artificial variable. x1

x2

x3

x4

x5

b

x2

0

1

3

-2

2

8

x1

1

0

1

-1

1

2

Cost

0

0

2

-4

4

f+8

Art. Cost

0

0

0

0

1

w-0

ANSWER THE FOLLOWING QUESTIONS. (1 Point each) (i)

x1 is a nonbasic variable

YES/NO

(ii)

x5 is a nonbasic variable

YES/NO

(iii)

Does the tableau give an optimum solution for the linear design problem?

YES/NO

(iv)

Does the tableau indicate end of Phase I of the Simplex method?

YES/NO

(v)

Does the tableau indicate the problem to be unbounded?

YES/NO

(vi)

x2 is a basic variable

(vii)

Value of x2 in the tableau is 2

YES/NO

(viii)

Value of the cost function at the current point is -8

YES/NO

(ix)

x3 is the pivot column for the next iteration

YES/NO

(x)

In Phase II, x5 column can become a pivot column

YES/NO

Arora, Introduction to Optimum Design, 4e

YES/NO

E-37

Sample Exams 4. (10 points) A linear engineering design problem is formulated as follows: Minimize f = 10 x1 + 6 x 2 + 2 x3 − 6 x 4 subject to

− x1 + x 2 + x3 − x 4 ≥ 1 3x1 + x 2 − x3 − 3x 4 ≥ 2 x1 , x 2 , x3 , x 4 ≥ 0

The final Simplex tableau for the problem is given as follows: x1

x2

x3

x4

x5

x6

x7

x8

b

x2

0

1

0.5

-1.5

-0.75

-0.25

0.75

0.25

1.25

x1

1

0

-0.5

-0.5

0.25

-0.25

-0.25

0.25

0.25

Cost

0

0

4

8

2

4

-2

-4

f - 10

Artificial

0

0

0

0

0

0

1

1

w-0

x 5 and x 6 are surplus variables for constraints #1 and #2, respectively. x 7 and x 8 are artificial variables for constraints #1 and #2, respectively.

ANSWER THE FOLLOWING QUESTIONS (2 Points each). i

Value of x 2 at the optimum point is 0.25

YES/NO

ii

Value of x 6 is zero at the optimum point

YES/NO

iii

Value of x 1 is 0.25 at the optimum point

YES/NO

iv

Value of x 5 is 2 at the optimum point

YES/NO

v

Value of the cost function at the optimum is 10

YES/NO

vi

Constraint #1 is inactive at the optimum point

YES/NO

vii

Constraint #2 is active at the optimum point

YES/NO

viii

Lagrange multiplier for constraint #1 is 2

YES/NO

ix

Lagrange multiplier for constraint #2 is -4

YES/NO

x

The problem has multiple optimum points

YES/NO

Arora, Introduction to Optimum Design, 4e

E-38

Sample Exams

SAMPLE EXAM 11: NONLINEAR PROGRAMMING for First Undergraduate Course (Time allowed = 45 minutes) INDICATE YOUR ANSWER TO EACH QUESTION IN THE FOLLOWING TABLE

Question #1: Part (i) (3 points)

A

B

C

Question #1: Part (ii) (3 points)

A

B

C

Question #1: Part (iii) (3 points)

A

B

C

Question #2 (8 points)

A

B

C

Question #3 (8 points)

A

B

Question #4 (9 points)

A

B

C

D

Question #5 (8 points)

A

B

C

D

Question #6 (8 points)

A

B

C

D

Total: 50 points

Arora, Introduction to Optimum Design, 4e

E-39

Sample Exams 1. (9 points) An engineering design problem, formulated in terms of the design variables x and y, has following two constraints:

h1 : y ( x + 4 y ) = 10 g1 : − 2 x 2 + 13 y ≤ 0 (iv) At the design point (2, 3), constraint g 1 is (Mark your selection): (A) Active

(B) Violated (C) Inactive SOLUTION: g1 = −2 × 2 2 + 13 × 3 = −7 < 0;

inactive

(v) At the design point (2, 1), constraint h 1 is (Mark your selection): (A) Active (B) Violated

(C) Inacive SOLUTION: h1 := 1(2 + 4 × 1) − 10 = −4 ≠ 0; violation (vi) Gradient of the constraint function h 1 at the point (2, 3) is given as (Mark your selection): (A) ∇ℎ1 = � 2 � 26

(B) ∇ℎ1 = �

3 � 26

𝑦𝑦 3 SOLUTION: ∇ℎ1 = �𝑥𝑥 + 8𝑦𝑦� = � � 26

Arora, Introduction to Optimum Design, 4e

E-40

(C) ∇ℎ1 = � 3 � 14

Sample Exams 2. (8 points) The cost function for an engineering design problem, formulated in terms of the design variables x and y, is given as:

(

Minimize f = x x − 2 y 2

)

subject to some constraints. At the point (2, 2), the cost function for the quadratic programming (QP) subproblem is given as (Mark your selection): (𝐴𝐴) 𝑓𝑓 ̅ = −12 − 4𝑑𝑑1 − 8𝑑𝑑2 + 0.5(𝑑𝑑12 + 𝑑𝑑22 )

(𝐵𝐵) 𝑓𝑓 ̅ = −12 − 4𝑑𝑑1 − 16𝑑𝑑2 − 0.5(𝑑𝑑12 + 𝑑𝑑22 ) (𝐶𝐶) 𝑓𝑓 ̅ = −12 − 4𝑑𝑑1 − 16𝑑𝑑2 + 0.5(𝑑𝑑12 + 𝑑𝑑22 )

SOLUTION: 𝑓𝑓(2,2) = 2(2 − 2 × 2 × 2) = −12

∇𝑓𝑓 = �

2𝑥𝑥 − 2𝑦𝑦 2 2×2−2×2×2 −4 �=� �=� � −4𝑥𝑥𝑥𝑥 −4 × 2 × 2 −16

𝑓𝑓 ̅ = −12 − 4𝑑𝑑1 − 16𝑑𝑑2 + 0.5(𝑑𝑑12 + 𝑑𝑑22 )

3. (8 Points) An unconstrained cost function for a design problem is formulated as: Minimize f = 3 x12 + x1 x 2 + 2 x 22 − x1 + x 2

At the design point (-1, 2), the direction d = (3, 5) is a direction of descent for the cost function (Mark your selection): (A) Yes (B)

No

SOLUTION: The gradient of the cost function is given as

6 x + x 2 − 1 − 5 c = ∇f =  1 =   x1 + 4 x 2 + 1  8  𝐜𝐜 ∙ 𝐝𝐝 = −5 × 3 + 8 × 5 = 25 > 0 𝑁𝑁𝑁𝑁𝑁𝑁 𝐴𝐴 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷

Arora, Introduction to Optimum Design, 4e

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Sample Exams 4. (9 points) An engineering design problem is formulated in terms of the design variables x and y as:

(

Minimize f = x x − 2 y 2

)

subject to the constraints.

1 (xy + 3 y 2 ) − 1 = 0 10 1 g1 : (−6 x 2 + y ) ≤ 0 100 1 g2 : − y −1≤ 0 4

h1 :

During the line search along a search direction, a design point is obtained as (1.345, 2.160). The descent function value (Φ) at this point using the penalty parameter value of 100 is most nearly (Mark your selection):

(A)

58.28

(B)

79.76

(C)

143.26

(D)

-164.74

SOLUTION: The problem functions at the point (1.345, 2.160) are evaluated as 𝑓𝑓(1.345,2.16) = 1.345(1.345 − 2 × 2.162 ) = −1.345 × 7.9962 = −10.741439

1 (1.345 × 2.16 + 3 × 2.162 ) − 1 ℎ1 (1.345, 2.16) = 10 1 (2.9052 + 13.9968) − 1 = = 10

16.902 10

− 1 = 0.6902

1 1 (−6 × 1.3452 + 2.16) = 100 (−10.85415 + 2.16) = −0.0869415 𝑔𝑔1 (1.345, 2.16) = 100

𝑔𝑔2 (1.345, 2.16) = −14(2.16) − 1 = −0.54 − 1 = −1.54

𝑉𝑉(1.345,2.16) = 𝑚𝑚𝑚𝑚𝑚𝑚(0.6902, −0.0869415, −1.54) = 0.6902 Φ(1.345,2.16) = −10.741439 + 100 × 0.6902 = 58.28

Arora, Introduction to Optimum Design, 4e

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Sample Exams

5. (8 Points) An engineering design problem is formulated as the following unconstrained problem: Minimize f = 8 x

2 + 2 x1 x 2 + 5 x 22 − 2 x1 + 3x 2 1

Starting from a design point (-1, -1), one iteration of the steepest descent method has been completed yielding a new design point as (0.21, 0.45).

The value of the scale parameter 𝛽𝛽1 for the conjugate gradient direction is most nearly (Mark your selection): (A) -0.141

(B) 0.317

(C) 0.141

(D) 7.094

SOLUTION: Gradient at the starting design point (-1, -1) is

16 x + 2 x 2 − 2 − 20 c ( 0 ) = ∇f =  1 =   2 x1 + 10 x 2 + 3  − 9  Gradient at the new design point (0.21, 0.45) is

16 x + 2 x 2 − 2 16 × 0.21 + 2 × 0.45 − 2 3.36 + 0.9 − 2 2.26 c (1) = ∇f =  1 = = =   2 x1 + 10 x 2 + 3  2 × 0.21 + 10 × 0.45 + 3 0.42 + 4.5 + 3 7.92 The value of the parameter 𝛽𝛽1 at the new design point (0.21, 0.45) is

 c (1)    β1 =  c ( 0)   

β1 =

2

2.26 2 + 7.92 2 5.1076 + 62.7264 67.8 = = 0.141 = 481 400 + 81 20 2 + 9 2

Arora, Introduction to Optimum Design, 4e

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Sample Exams 6. (8 Points) An engineering design problem is formulated as the following unconstrained problem: Minimize f = 7 x12 + 12 x 22 − x1 Starting from a design point (2, 1), one iteration of the steepest descent method has been completed yielding a new design point as (0.58, -0.31).

The conjugate direction at the new design point if the parameter 𝛽𝛽1 = 0.1 is used, is most nearly (Mark your selection):

(𝐴𝐴) �−4.42� 9.84

(𝐵𝐵) � 4.42 � −9.84

(𝐶𝐶) �−7.12� 7.44

SOLUTION: Gradient at the starting design point (2, 1) is

(𝐷𝐷) �−9.82� 5.04

14 x − 1 27 c = ∇f =  1  =    24 x 2  24 Gradient at the new design point (0.58, -0.31) is

14 x − 1  7.12  c (1) = ∇f =  1  =    24 x 2  − 7.44 Calculate the conjugate gradient direction at the new design point (0.58, -0.31) with 𝛽𝛽1 = 0.1.

 7.12  − 27 − 7.12 − 2.7  − 9.82 + 0.1 d (1) = −c (1) + β 0 d (0) = −  =  =  − 7.44 − 24  7.44 − 2.4   5.04 

Arora, Introduction to Optimum Design, 4e

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