chapter 2

Quantitative Analysis BA 452 Homework 1 Questions

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Homework 1 covers the theory and applications in Lessons I-1 to I-3. This document has four parts: Objectives of doing your homework. Assignment of homework questions, with suggestions about which other questions may help you understand the homework questions. Homework 1 Supplemented Questions listing 58 questions: 4 of them are your homework, others may help you understand the homework questions, and the rest are there just in case you find them helpful. Some supplemental questions refer to a section of a chapter in the textbook (for example, Section 2.1 means Chapter 2 Section 1 of your textbook). Homework 1 Supplemented Answers listing answers to all 58 questions including your 4 homework questions.

Homework 1 covers the theory and applications in Lessons I-1 to I-3. This document has four parts: Objectives of doing your homework. Assignment of homework questions, with suggestions about which other questions may help you understand the homework questions. Homework 1 Supplemented Questions listing 58 questions: 4 of them are your homework, others may help you understand the homework questions, and the rest may help you understand fine points. Some supplemental questions refer to a section of a chapter in the textbook (for example, Section 2.1 means Chapter 2 Section 1 of your textbook). Homework 1 Supplemented Answers listing answers to all 58 questions excluding your 4 homework questions.


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Objectives By working through the homework questions and the supplemental questions, you will:

1. Obtain an overview of the kinds of problems linear programming has been used to solve.

2. Learn how to develop linear programming models for simple problems. 3. Be able to identify the special features of a model that make it a linear

programming model. 4. Learn how to solve two variable linear programming models by the graphical

solution procedure. 5. Understand the importance of extreme points in obtaining the optimal

solution. 6. Know the use and interpretation of slack and surplus variables.

7. Be able to interpret the computer solution of a linear programming problem. 8. Understand how alternative optimal solutions, infeasibility and

unboundedness can occur in linear programming problems. 9. Understand the following terms:

problem formulation feasible region constraint function slack variable objective function standard form solution redundant constraint optimal solution extreme point nonnegativity constraints surplus variable mathematical model alternative optimal solutions linear program infeasibility linear functions unbounded feasible solution


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Assignment Questions 11, 13, 27, and 53 is your homework assignment. Questions 11, 13, and 27 should be answered without referring to notes or using computers (Hint: Define decision variables I = Internet fund investment in thousands, B = Blue Chip fund investment in thousands. Then, the objective is to maximize the projected annual return 0.12I + 0.09B.) Question 53 can be answered with notes and computers. To supplement those homework questions, you should consider (but not turn in) the following questions. Questions answered without notes or computers: 1-13, 17-19, 21-26, 30-31, 34-36, 38, 42-46. Questions answered with notes and computers: 14-15, 20, 27-29, 33, 37, 39-41, 49-54. Tip: Those homework questions and supplementary questions are grouped into sets of similar type. Once you have mastered the questions in a set, you can skip the rest of the questions in that set. Tip: Some of your Exam 1 questions will be variations of some of those homework questions.


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Homework 1 Supplemented Questions 1. Which of the following mathematical relationships could be found in the linear programming model, and

which could not? For the relationships that are unacceptable for linear programs, state why. a. −1𝐴𝐴 + 2𝐵𝐵 ≤ 70 b. 2𝐴𝐴 − 2𝐵𝐵 = 50 c. 1𝐴𝐴 − 2𝐵𝐵2 ≤ 10 d. 3√𝐴𝐴 + 2𝐵𝐵 ≥ 15 e. 1𝐴𝐴 + 1𝐵𝐵 = 6 f. 2𝐴𝐴 + 5𝐵𝐵 + 1𝐴𝐴𝐵𝐵 ≤ 25

2. Find the solution that satisfy the following constraints: a. 4𝐴𝐴 + 2𝐵𝐵 ≤ 16 b. 4𝐴𝐴 + 2𝐵𝐵 ≥ 16 c. 4𝐴𝐴 + 2𝐵𝐵 = 16

3. Show a separate graph of the constraint lines and the solutions that satisfy each of the following

constraints: a. 3𝐴𝐴 + 2𝐵𝐵 ≤ 8 b. 12𝐴𝐴 + 8𝐵𝐵 ≥ 480 c. 5𝐴𝐴 + 10𝐵𝐵 = 200


constraints: a. 3𝐴𝐴 − 4𝐵𝐵 ≥ 60 b. −6𝐴𝐴 + 5𝐵𝐵 ≤ 60 c. 5𝐴𝐴 − 2𝐵𝐵 ≤ 0


constraints: a. 𝐴𝐴 ≥ 0.25(𝐴𝐴 + 𝐵𝐵) b. 𝐵𝐵 ≤ 0.10(𝐴𝐴 + 𝐵𝐵) c. 𝐴𝐴 ≤ 0.50(𝐴𝐴 + 𝐵𝐵)

6. Three objective functions for linear programming problems are 7𝐴𝐴 + 10𝐵𝐵, 6𝐴𝐴 + 4𝐵𝐵, and −4𝐴𝐴 + 7𝐵𝐵.

Show the graph of each for objective function values equal to 420.

7. Identify the feasible region for the following set of constraints: 0.5𝐴𝐴 + 0.25𝐵𝐵 ≥ 30 1𝐴𝐴 + 𝐵𝐵 ≥ 250 0.25𝐴𝐴 + 0.5𝐵𝐵 ≤ 50

𝐴𝐴,𝐵𝐵 ≥ 0


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8. Identify the feasible region for the following set of constraints: 2𝐴𝐴 − 1𝐵𝐵 ≤ 0

−1𝐴𝐴 + 1.5𝐵𝐵 ≤ 200 𝐴𝐴,𝐵𝐵 ≥ 0

9. Identify the feasible region for the following set of constraints:

3𝐴𝐴 − 2𝐵𝐵 ≥ 0 2𝐴𝐴 − 1𝐵𝐵 ≤ 200 1𝐴𝐴 ≤ 150


10. For the linear program Max 2𝐴𝐴 + 3𝐵𝐵 s.t.

1𝐴𝐴 + 2𝐵𝐵 ≤ 6 5𝐴𝐴 + 3𝐵𝐵 ≤ 15 𝐴𝐴,𝐵𝐵 ≥ 0

find the optimal solution using the graphical solution procedure. What is the value of the objective function at the optimal solution?

11. Solve the following linear program using the graphical solution procedure: Max 5𝐴𝐴 + 5𝐵𝐵 s.t.

1𝐴𝐴 ≤ 100 1𝐵𝐵 ≤ 80

2𝐴𝐴 + 4𝐵𝐵 ≤ 400 𝐴𝐴,𝐵𝐵 ≥ 0

12. Consider the following linear programming problem:

Max 3𝐴𝐴 + 3𝐵𝐵 s.t.

2𝐴𝐴 + 4𝐵𝐵 ≤ 12 6𝐴𝐴 + 4𝐵𝐵 ≤ 24 𝐴𝐴,𝐵𝐵 ≥ 0

a. Find the optimal solution using the graphical solution procedure. b. If the objective function is changes to 2A + 6B, what will the optimal solution be? c. How many extreme points are there? What are the values of A and B at each extreme point?


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13. Consider the following linear program: Max 1𝐴𝐴 + 2𝐵𝐵 s.t.

1𝐴𝐴 ≤ 5 1𝐵𝐵 ≤ 4

2𝐴𝐴 + 2𝐵𝐵 = 12 𝐴𝐴,𝐵𝐵 ≥ 0

a. Show the feasible region. b. What are the extreme points of the feasible region? c. Find the optimal solution using the graphical procedure.

14. RMC, Inc., is a small firm that produces a variety of chemical products. In a particular production

process, three raw materials are blended (mixed together) to produce two products: a fuel additive and a solvent base. Each ton of fuel additive is a mixture of 2/5 ton of material 1 and 3/5 of material 3. A ton of solvent base is a mixture is a mixture of ½ ton of material 1, 1/5 ton of material 2 and 3/10 ton of material 3. After deducting relevant costs, the profit contribution is $40 for every ton of fuel additive and $30 for every ton of solvent base. RMC’s production is constrained by the limited availability of the three raw materials. For the current production periods, RMC has available the following quantities of each new material:

Raw Material Amount Available for Production Material 1 20 tons Material 2 5 tons Material 3 21 tons

Assuming the RMC is interested in maximizing the total profit contribution, answer the following:

a. What is the linear programming model for the problem? b. Find the optimal solution using the graphical solution procedure. How many tons of each product should

be produced, and what is the projected total profit contribution? c. Is there any unused material? If so, how much? d. Are any of the constraints redundant? Is do, which ones?

15. Refer to the Par, Inc., problem described in Section 2.1 (Chapter 2 Section 1 of your text). Suppose that

Par’s management encounters the following situations: a. The accounting department revises its estimate of the profit contribution for the deluxe bag to $18 per

bag. b. A new low-cost material is available for the standard bag, and the profit contribution per standard bag

can be increased to $20 per bag. (Assume that the profit contribution of the deluxe bag is the original $9 value).

c. New sewing equipment is available that whole increase the sewing operations capacity to 750 hours. (Assume the 10A +9B is the appropriate objective function.) If each of these situations is encountered separately, what is the optimal solution and the total profit contribution?


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16. Refer to the feasible region for Par, Inc., problem in Figure 2.13. a. Develop an objective function that will make extreme point 5 the optimal extreme point. b. What is the optimal solution for the objective function you selected in part (a)? c. What are the values of the slack variables associated with this solution?

17. Write the following linear program in standard form:

Max 5𝐴𝐴 + 2𝐵𝐵 s.t.

1𝐴𝐴 − 2𝐵𝐵 ≤ 420 2𝐴𝐴 + 1𝐵𝐵 ≤ 610 6𝐴𝐴 − 1𝐵𝐵 ≤ 125


18. For the linear program Max 4𝐴𝐴 + 1𝐵𝐵 s.t.

10𝐴𝐴 + 2𝐵𝐵 ≤ 30 3𝐴𝐴 + 2𝐵𝐵 ≤ 12 2𝐴𝐴 + 2𝐵𝐵 ≤ 10 𝐴𝐴,𝐵𝐵 ≥ 0

a. Write the problem in standard form. b. Solve the problem using the graphical solution procedure. c. What are the values of the three slack variables at the optimal solution?

19. Given the linear program

Max 3A + 4B s.t. -1A + 2B ≤ 8 1A + 2B ≤ 12 2A + 1B ≤ 16 A, B, ≥ 0



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20. For the linear program Max 3A + 2B s.t. A + B ≥ 4 3A + 4B ≤ 24 A ≥ 2 A – B ≤ 0 𝐴𝐴,𝐵𝐵 ≥ 0


21. Consider the following linear program:

Max 2A + 3B s.t. 5A + 5B ≤ 400 Constraint 1 -1A + 1B ≤ 10 Constraint 2 1A + 3B ≥ 90 Constraint 3 𝐴𝐴,𝐵𝐵 ≥ 0

a. Graph the constraints, and place a number (1, 2, or 3) next to each constraint line to identify which constraint it represents.

b. Shade in the feasible region on the graph. c. Identify the optimal extreme point. What is the optimal solution? d. Which constraints are binding? Explain. e. How much clack or surplus is associated with the nonbinding constraint?


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22. Reiser Sports Products wants to determine the number of All-Pro (A) and College (C) footballs to produce in order to maximize profit over the next four-week planning horizon. Constraints affecting the production quantities are the production capacities in three departments: cutting and dyeing; sewing; and inspection and packaging. For the four-week planning period, 340 hours of cutting and dyeing time, 420 hours of sewing time, and 200 hours of inspection and packaging time are available. All-Pro footballs provide a profit of $5 per unit and College footballs provide a profit of $4 per unit. The liner programming model with production times expressed in minutes is as follows:

Max 5A + 4C

s.t.

12A + 6C ≤ 20,400 Cutting and dyeing

9A + 15C ≤ 25, 200 Sewing

6A + 6C ≤ 12,000 Inspection

A, C ≥ 0

a) Graph all constraints, then shade the feasible region for this problem. b) Determine the coordinates of each extreme point and the corresponding profit. Which extreme point

generates the highest profit. c) Draw the profit line corresponding to a profit of $4000. Move the profit line as far from the origin as you

can in order to determine which extreme point will provide the optimal solution. Compare your answer with the approach you used in part (b).

d) Which constraints are binding? Explain. e) Suppose that the values of the objective function coefficients are $4 for each All-Pro model produced

and $5 for each college model. Use the graphical solutions procedure to determine the new optimal solution and the corresponding value of profit.

23. Embassy motorcycles (EM)manufactures two lightweight motorcycles designed for easy handling and safety. The EZ-Rider model has a new engine and a low profile that make it easy to balance. The Lady-Sport model is slightly larger, uses a more traditional engine, and is specifically designed to appeal to women riders. Embassy produces the engines for both models at its Des Moines, Iowa plant. Each EZ-Rider requires 6 hours of manufacturing time and each Lady-Sport engine requires 3 hours of manufacturing time. The Des Moines plant has 2100 hours of engine manufacturing time available for the next production period. Embassy’s motorcycle frame supplier can supply as many EZ-Rider frames as needed. However, the Lady-Sport frame is more complex and the supplier can only provide up to 280 Lady-Sport frames for the next production period. Final assembly and testing requires 2 hours for each EZ-Rider model and 2.5 hours for each Lady-Sport model. A maximum of 1000 hours of assembly and testing time are available for the next production period. The company’s accounting department projects a profit contribution of $2400 for each EZ-Rider produced and $1800 for each Lady-Sport produced.

a. Formulate a linear programming model that can be used to determine the number of units of each

model that should be produced in order to maximize the total contribution to profit. b. Solve the problem graphically. What is the optimal solution? c. Which constraints are binding?


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24. Kelson Sporting Equipment, Inc., makes two different types of baseball gloves: A regular model and a catcher’s model. The firm has 900 hours of production time available in its cutting and sewing department, 300 hours available in its finishing department, and 100 hours available in its packaging and shipping department. The production time requirements and the profit contribution per glove are given in the following table:

Production Time (hours) Model Cutting and

Sewing Finishing Packaging and

Shipping Profit/Glove

Regular model 1 ½ 1/8 $5 Catcher’s model 3/2 1/3 ¼ $8

Assuming that the company is interesting in maximizing the total profit contribution, answer the following

a. What is the linear programming model for this problem? b. Find the optimal solution using the graphical solution procedure. How many gloves of each model

should Kelson manufacture? c. What is the total profit contribution Kelson can earn with the given production quantities? d. How many hours of production time will be scheduled in each department? e. What is the slack time in each department?

25. George Johnson recently inherited a large sum of money; he want to use a portion of this money to set

up a trust fund for his two children. The trust fund has two investment opportunities: (1) a bond fund and (2) a stock fund. The projected returns over the life of the investments are 6% for the bond fund and 10% for the stock fund. Whatever portion of the inheritance he finally decides to commit to the trust fund, he wants to invest at least 30% of that amount in the bond fund. In addition, he want to select a mix that will enable him to obtain a total return of at least 7.5%

a. Formulate a linear programming model that can be used to determine the percentage that should be allocated to each of the possible investment alternatives.

b. Solve the problem using the graphical solution procedure

26. The Sea Warf Restaurant would like to determine the best way to allocate a monthly advertising budget of $1000 between newspaper advertising and radio advertising. Management decided that at least 25% of the budget must be spent on each type of media, and that the amount of money spent on local newspaper advertising must be at least twice the amount of money spent on radio advertising. A marketing consultant developed an index that measure the audience exposure per dollar of advertising on a scale from 0 to 100, with higher values implying greater audience exposure. If the value of the index for local newspaper advertising is 50 and the value of the index for spot radio adverting is 80, how should the restaurant allocate its advertising budget in order to maximize the value of total audience exposure?

a. Formulate a linear programming model that can be used to determine how the restaurant should allocate its advertising budget in order to maximize the value of the total audience exposure.

b. Solve the problem using the graphical solution procedure.


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27. Blair &Rosen, Inc. (B&R, is brokerage firm that specializes investment portfolios designed to meet the specific risk tolerance of its clients. A client who contacted B&R this past week has a maximum of $50,000 to invest. B&R’s investment advisor decides to recommend a portfolio consisting of two investment funds: an Internet fund and a Blue Chip fund. The Internet fund has a projected annual return of 12%, whereas the Blue Chip fund has a projected annual return of 9%. The investment advisor requires that at most $35,000 of the client’s money should be invested in the Internet fund. B&R services include a risk rating for each investment alternative. The internet fund, which is the more risky of the two investment alternatives, has a risk rating of 6 per thousand dollars invested. The Blue Chip fund has a risk of 4 per thousand dollars invested. For example, if $10,000 is invested in each of the two investment funds, B&R’s risk rating for the portfolio would be 6(10)+4(10)=100. Finally, B&R developed s questionnaire to measure each client’s risk tolerance. Based on the responses, each client is classified as a conservative, moderate, or aggressive investor. Suppose that the questionnaire results classified current client as a moderate investor. B&R recommends that a client who is a moderate investor limit his or her portfolio to a maximum risk rating of 240.

a. What is the recommended investment portfolio for this client? What is the annual return for the portfolio?

b. Suppose that a second client with $50,000 to invest has been classified as an aggressive investor. B&R recommends that the maximum portfolio risk rating for an aggressive investor is 320. What is the recommended investment portfolio for this aggressive investor? Discuss what happens to the portfolio under the aggressive investor strategy.

c. Suppose that a third client with $50,000 to invest has been classified as a conservative investor. B&R recommends that the maximum portfolio risk rating for a conservative investor is 160. Develop the recommended investment portfolio for the conservative investor. Discuss the interpretation of the slack variable for the total investment fund constraint.

28. Tom’s Inc., produces various Mexican food products and sells them to Western Foods, a chain of grocery stores located in Teas and New Mexico. Tom’s, Inc., mails two salsa products: Western Foods Salsa and Mexico City Salsa. Essentially, the two products have different blends of whole tomatoes, tomato sauce, and tomato paste. The Western Foods Salsa is a blend of 50% whole tomatoes, 30% tomato sauce, and 20% tomato paste. The Mexico City Salsa, which has a thicker and chunkier consistency, consist of 70% whole tomatoes, 10% tomato sauce, and 20% tomato paste. Each jar of salsa produced weighs 10 ounces. For the current production period, Tom’s Inc., can purchase up to 280 pounds of whole tomatoes, 130 pounds of tomato sauce, and 100 pounds of tomato paste; the price per pound for these ingredients is $0.96, $0.64, and $0.56 respectfully. The cost of the spices and other ingredients as approximately $0.10 per jar. Tom’s Inc., buys empty glass jars for $0.02 each, and labeling and filling cost are estimated to be $0.03 for each jar of salsa produced. Tom’s contract with Western Foods resulting in sales revenue of $1.64 for each jar of Western Foods Salsa and $1.93 for each jar of Mexico City Salsa.

a. Develop a linear programming model that will enable Tom’s to determine the mix of salsa products that will maximize the total profit contribution.

b. Find the optimal solution.


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29. Auto Ignite produces electronic ignition systems for automobiles at a plant in Cleveland, Ohio. Each ignition system is assembled from two components produced at AutoIgnite’s plants in Buffalo, New York, and Dayton, Ohio. The Buffalo plant can produce 2000 units of component 1, 1000 units of component 2, or any combination of the two components each day. For instance, 60% of Buffalo’s production time could be used to produce component 1 and 40% of Buffalo’s production time could be used to produce component 2; in this case the Buffalo plant would be able to produce 0.6 (2000)=1200 units of component 1 every day and 0.4(1000)=400 units of component 2 each day. The Dayton plant can produced 600 units of component 1, 1400 units of component 2, or any combination of the two components each day. At the end of each day, the component production at Buffalo and Dayton is sent to Cleveland for assembly of the ignition systems on the following workday.

a. Formulate a linear programming model that can be used to develop a daily production schedule for the Buffalo and Dayton plants that will maximize daily production of ignition systems at Cleveland.


30. A financial advisor at Diehl Investments identified two companies that are likely candidates for a takeover in the near future. Eastern Cable is a leading manufacturer of flexible cable systems used in the construction industry, and ComSwitch is a new firm specializing in digital switching systems. Eastern cable is currently trading for $40 per share, and ComSwitch is currently trading for $25 per share. If the takeovers occur, the financial advisor estimates that the price of Eastern Cable will go to $55 per share and ComSwitch will go to $43 per share. At this point in time, the financial advisor has identified ComSwitch as the higher risk alternative. Assume that a client indicates a willingness to invest a maximum of $50,000 in the two companies. The client wasn’t to invest at least $15,000 in Eastern Cable and at least $10,000 in ComSwitch. Because of the higher risk associated with ComSwitch, the financial advisor has recommended that at most $25,000 should be invested in ComSwitch.

a. Formulate a linear programming model that can be used to determine the number of shares of Eastern Cable and the number of shares of ComSwitch that will meet the investment constraints and maximize the total return for the investment.

b. Graph the feasible region. c. Determine the coordinates of each extreme point. d. Find the optimal solution.


Min 3A+4B s.t. 1A+3B≥6 1A+1B≥4 A,B≥0 Identify the feasible region and find the optimal solution using the graphical solution procedure. What is the value of the objective function?

32. Identify the three extreme-point solutions for the M&D Chemicals problem (see section 2.5). Identify

the value of the objective function and the values of the slack and surplus variables at each extreme point.


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33. Consider the following linear programming problem: Min A+2B s.t. A+ 4B≤21 2A+ B≥7 3A+1.5B≤21 -2A+ 6B≥0 A,B≥0

a. Find the optimal solution using the graphical solution procedure and the value of the objective function. b. Determine the amount of slack or surplus for each constraint. c. Suppose the objective function is changed to max 5A+2B. Find the optimal solution and the value of the

objective function.

34. Consider the following linear program: Min 2A+2B s.t. 1A+3B≤12 3A+1B≥13 1A-1B=3 A,B≥0

a. Show the feasible region. b. What are the extreme points of the feasible region? c. Find the optimal solution using the graphical solution procedure.

35. For the linear program

Min 6A+4B s.t. 2A+1B≥12 1A+1B≥10 1B≤4 A,B≥0

a. Write the problem in standard form. b. Solve the problem using the graphical solution procedure. c. What are the values of the slack and surplus variables?


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36. As a part of quality improvement initiative, Consolidated Electronics employees complete a three-day training program on teaming and a two-day training program on problem solving. The manager of quality improvement has request that at least 8 training programs on teaming and at least 10 training programs on problem solving be offered during the next six months. In addition, senior-level management has been specified that at least 25 training programs must be offered during this period. Consolidated Electronics uses a consultant to teach the training programs. During this next quarter, the consultant has 84 days of training time available. Each training program on teaming cost $10,000 and each training programs and problem solving cost $8000.

a. Formulate a linear programming model that can be used to determine the number of training programs on teaming and the number of training programs on problem solving that should be offered in order to minimize total cost.

b. Graph the feasible region. c. Determine the coordinates of each extreme point. d. Solve the minimum cost solution.

37. The New England Cheese Company produces two cheese spreads by blending mild cheddar cheese with

extra sharp cheddar cheese. The Cheese spreads are packaged in 12-ounce containers, which are then sold to distributors through the Northeast. The Regular blend contains 80% mild cheddar and 20% extra sharp, and the Zesty blend contains 60% mild cheddar and 40% extra sharp. This year, a local dairy cooperative offered to provide up to 8100 pounds of mild cheddar cheese for $1.20 per pound and up to 3000 pounds of extra sharp cheddar cheese for $1.40 per pound. The cost to blend and package the cheese spreads, excluding the cost of the cheese, is $0.20 per container. If each container of Regular is sold for $1.95 and each container of Zesty is sold for $2.20, how many containers of Regular and Zesty should New England produce?


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38. Applied technology, Inc. (ATI), produces bicycle frames using two fiberglass materials that improve the strength to weight ratio for the frames. The cost of the standard grade material is $7.50 per yard and the cost of the professional grade material is$9.00 per yard. The standard and professional grade materials contain different amounts of fiberglass, carbon fiber, and Kevlar as shown in the following table: ATIATI signed a contract with a bicycle manufacturer to produce a new frame with a carbon fiber content of at least 20% and a Kevlar content of not greater than 10%. To meet the required weight specification, a total of 30 yards of material must be used for each frame.

a. Formulate a linear program to determine the number of yards of each grade of fiberglass material that ATI should use in each frame in order to minimize total cost. Define the decision variables and indicate the purpose of each constraint.

b. Use the graphical solution procedure to determine the feasible region. What are the coordinates of the extreme points?

c. Compute the total cost at each extreme point. What is the optimal solution? d. The distributor of the fiberglass material is currently overstocked with the professional grade material.

To reduce inventory, the distributor offered ATI the opportunity to purchase the professional grade for $8 per yard. Will the optimal solution change?

e. Suppose that the distributor further lowers the price of the professional grade material to $7.40 per yard. Will the optimal solution change? What effect would an even lower price for the professional grade material have on the optimal solution? Explain.

39. Innis Investments manages funds for a number of companies and wealthy client. The investment strategy is tailored to each client’s needs. For a new client, Innis has been authorized to invest up to $1.2 million in two investment funds: a stock fund and a money market fund. Each unit of the stock fund costs $50 and provides an annual rate of return of 10%; each unit of the money market fund cost $100 and provides an annual rate of return of 4%. The client wants to minimize risk subject to the requirement that the annual income from the investment be at least $60,000. According to Innis’s risk measurement system, each unit invested in the stock fund has a risk index of 8, and each unit invested in the money market fund has a risk index of 3; the higher risk index associated with the stock fund simply indicates that it is the riskier investment, Innis’s client also specified that at least $300,000 be invested in the money market fund.

a. Determine how many units of each fund Innis should purchase for the client to minimize the total risk index for the portfolio.

b. How much annual income will this investment strategy generate? c. Suppose the client desires to maximize annual return. How should the funds be invested?

Standard Grade Professional Grade Fiberglass 84% 58% Carbon Fiber 10% 30% Kevlar 6% 12%


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40. Photo Chemicals produces two types of photographic developing fluids. Both products cost Photo Chemicals $1 per gallon to produce. Based on an analysis of current inventory levels and outstanding orders for the next month, Photo Chemicals’ management specified that at least 30 gallons of product 1 and at least 20 gallons of product 2 must b produced during the next two weeks. Management also stated that an existing inventory of highly perishable raw material required in the production of both fluids must be used within the next two weeks. The current inventory of the perishable raw material is 80 pounds. Although more of this raw material can be ordered if necessary, any of the current inventory that is not used within the next two weeks will spoil – hence, the management requirement that at least 80 pounds be used in the next two weeks. Furthermore, it is known that product 1 requires 1 pound of this perishable raw material per gallon and product 2 requires 2 pounds of the raw material per gallon. Because Photo Chemicals’ objective is to keep its production costs at the minimum possible level, the firm’s management is looking for a minimum cost production plan that uses all the 80 pounds of perishable raw material and provides at least 30 gallons of product 1 and at least 20 gallons of product 2. What is the minimum cost solution?

41. Southern Oil Company produces two grades of gasoline: regular and premium. The profit contributions are $0.30 per gallon for regular gasoline and $0.50 per gallon for premium gasoline. Each gallon of regular gasoline contains 0.3 gallons of grade A crude oil and each gallon of premium gasoline contains 0.7 gallons of grade A crude oil. For the next production period, Southern has 18,000 gallons of grade A crude oil available. The refinery used to produce the gasoline has a production capacity of 50,000 gallons for the next production period. Southern Oil’s distributors have indicated that demand for the premium gasoline for the next production period will be at most 20,000 gallons.

a. Formulate a linear programming model that can be used to determine the number of gallons of regular gasoline and the number of gallons of premium gasoline that should be produced in order to maximize total profit contribution.

b. What is the optimal solution? c. What are the values and interpretations of the slack variables? d. What are the binding constraints?

42. Does the following linear program involve infeasibility, unbounded, and/or alternative optimal

solutions? Explain.

𝑀𝑀𝑀𝑀𝑀𝑀 4𝐴𝐴 + 8𝐵𝐵 𝑠𝑠. 𝑡𝑡. 2𝐴𝐴 + 2𝐵𝐵 ≤ 10 −1𝐴𝐴 + 1𝐵𝐵 ≥ 8 𝐴𝐴,𝐵𝐵 ≥ 0

43. Does the following linear program involve infeasibility, unbounded, and/or alternative optimal solutions? Explain. 𝑀𝑀𝑀𝑀𝑀𝑀 1𝐴𝐴 + 1𝐵𝐵 𝑠𝑠. 𝑡𝑡. 8𝐴𝐴 + 6𝐵𝐵 ≥ 24 2𝐵𝐵 ≥ 4 𝐴𝐴,𝐵𝐵 ≥ 0


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44. Consider the following linear program: 𝑀𝑀𝑀𝑀𝑀𝑀 1𝐴𝐴 + 1𝐵𝐵 𝑠𝑠. 𝑡𝑡 5𝐴𝐴 + 3𝐵𝐵 ≤ 15 3𝐴𝐴 + 5𝐵𝐵 ≤ 15 𝐴𝐴,𝐵𝐵 ≥ 0

a. What is the optimal solution for this problem? b. Suppose that the objective function is changed to 1𝐴𝐴 + 1𝐵𝐵. Find the new optimal solution.


𝑀𝑀𝑀𝑀𝑀𝑀 1𝐴𝐴 − 2𝐵𝐵 𝑠𝑠. 𝑡𝑡. −4𝐴𝐴 + 3𝐵𝐵 ≤ 3 1𝐴𝐴 − 1𝐵𝐵 ≤ 3 𝐴𝐴,𝐵𝐵 ≥ 0

a. Graph the feasible region for the problem. b. Is the feasible region unbounded? Explain. c. Find the optimal solution. d. Does an unbounded feasible region imply that he optimal solution to the linear program will be

unbounded?

46. The manager of a small independent grocery store is trying to determine the best use of her shelf space for soft drinks. The store carries national and generic brands and currently has 200 square feet of shelf space available. The manager wants to allocate at least 60% of the space to the national brands and, regardless of the profitability, allocate at least 10% of the space to the generic brands. How many square feet of space should the manager allocate to the national brands and the generic brands under the following circumstances?

a. The national brands are more profitable than the generic brands. b. Both brands are equally profitable. c. The generic brand is more profitable than the national brand.

47. Discuss what happens to the M&D Chemicals problem (see Section 2.5) if the cost per gallon for product

A is increased to $3.00 per gallon. What would you recommend? Explain.

48. For the M&D chemicals problem in Section 2.5, discuss the effect of management’s requiring total production of 500 gallons for the two products. List two or three actions M&D should consider to correct the situation you encounter.


18

49. PharmaPlus operates a chain of 30 pharmacies. The pharmacies are staffed by licensed pharmacists and pharmacy technicians. The company currently employs 85 full-time equivalent pharmacists (combination of full time and part time) and 175 full-time equivalent technicians. Each spring management reviews current staffing levels and makes hiring plans for the year. A recent forecast of the prescription load for the nest year shows that at least 250 full-time equivalent employees (pharmacists and technicians) will be required to staff the pharmacies. The personnel department expects 10 pharmacists and 30 technicians to leave over the next year. To accommodate the expected attrition and prepare for future growth, management stated that at least 15 new pharmacists must be hired. In addition, PharmaPlus’s new service quality guidelines specify no more than two technicians per licensed pharmacist. The average salary for licensed pharmacists is $40 per hour and the average salary for technicians is $10 per hour.

a. Determine a minimum-cost staffing plan for PharmaPlus. How many pharmacists and technicians are needed?

b. Given current staffing levels and expected attrition, how many new hires (if any) must be made to reach the level recommended in part (a)? What will be the impact on the payroll?

50. Expedition Outfitters manufactures a variety of specialty clothing for hiking, skiing, and mountain climbing. Its management decided to begin production on two new parkas designed for use in extremely cold weather: the Mount Everest Parka and the Rocky Mountain Parka. The manufacturing plant has 120 hours of cutting time and 120 hours of sewing time available for producing these two parkas. Each Mount Everest Parka requires 20 minutes of cutting time and 15 minutes of sewing time. The labor and material cost is $150 for each Mount Everest Parka and $50 for each rocky Mountain Parka, and the retail price through the firm’s mail order catalog are $250 for the Mount Everest Parka and $200 for the Rocky Mountain Parka. Because management believes that the Mount Everest Parka is a unique coat that will enhance the image of the firm, they specified that at least 20% of the total production must consist of this model. Assuming that Expedition Outfitters can sell as many coats of each type as it can produce, how many units of each model should it manufacture to maximize the total profit contribution?

51. English Motors, Ltd. (EML), developed a new all-wheel-drive sports utility vehicle. As part of the marketing campaign, EML produced a video tape sales presentation to send to both owners of current EML four-wheel-drive vehicles as well as to owners of four-wheel-drive sports utility vehicles offered by competitors; EML refers to these two target markets as the current customer market and the new customer market. Individuals who receive the new promotion video will also receive a coupon for a test drive of the new EML model for one weekend. A key factor in the success of the new promotion is the response rate, the percentage of individuals who receive the new promotion and test drive the new model. EML estimates that the response rate for the current customer market is 25% and the response rate for the new customer market is 20%. For the customers who test drive the new model, the sales rate is the percentage of individuals that make a purchase. Marketing research studies indicate that the sales rate is 12% for the current customer market and 20% for the new customer market. The cost for each promotion, excluding the test drive costs, is $4 for each promotion sent to the current customer market and $6 for each promotion sent to the new customer market. Management also specified that a minimum of 30,000 current customers should test drive the new model and a minimum of 10,000 new customers should text drive the new model. In addition, the number of current customers who test drive the new vehicle must be at least twice the number of new customers who test drive the new vehicle. If the marketing budget, excluding test drive costs, is $1.2 million, how many promotions should be sent to each group of customers in order to maximize total sales?


19

52. Creative Sports Design (CSD) manufactures a standard-size racket and an oversize racket. The firm’s rackets are extremely light due to the use of a magnesium-graphite alloy that was invented by the firm’s founder. Each standard-size racket uses 0.125 kilograms of the alloy and each oversize racket uses0.4 kilograms; over the next two-week production period only 80 kilograms of the alloy are available. Each standard-size racket uses 10 minutes of manufacturing time an each oversize racket uses 12 minutes. The profit contribution are $10 for each standard-size racket and $15 for each oversize racket, and 40 hours of manufacturing time are available each week. Management specified that at least 20% of the total production must be the standard-size racket. How many rackets of each type should CSD manufacture over the next two weeks to maximize the total profit contribution? Assume that because of the unique nature of their products, CSD can sell as many rackets as they can produce.

53. Management of High Tech Services (HTS) would like to develop a model that will help allocate their technicians’ time between service calls to regular contract customers and new customers. A maximum of 80 hours of technician time is available over the two-week planning period. To satisfy cash flow requirements, at least $800 in revenue (per technician) must be generated during the two-week period. Technician time for regular customers generates $25 per hour. However, technician time for new customers only generates an average of $8 per hour because in many cases a new customer contact does not provide billable service. To ensure that new customer contacts are being maintained, the technician time spend on new customer contacts must be at least 60% of the time spent on regular customer contacts. Given these revenue and policy requirements., HTS would like to determine how to allocate technician time between regular customer and new customers so that the total number of customers contacted during the two-week period will be maximized. Technicians require an average of 50 minutes for each regular customer contact and 1 hour for each new customer contact.

a. Develop a linear programming model that will enable HTS to allocate technician time between regular customers and new customers.


54. Jackson Hole Manufacturing is a small manufacturer of plastic products used in the automotive and computer industries. One of its major contracts is with a large computer company and involves the production of plastic printer cases for the computer company’s portable printers. The printer cases are produced on two injection molding machines. The M-100 machine has a production capacity of 25 printer cases per hour, and the M-200 machine has a production capacity of 40 cases per hour. Both machines use the same chemical material to produce the printer cases; the M-100 uses 40 pounds of the raw material per hour and the M-200 uses 50 pounds per hour. The computer company asked Jackson Hole to produce as many of the cases during the upcoming week as possible; it will pay $18 for each case Jackson Hole can deliver. However, next week is a regularly scheduled vacation period for most of Jackson Holes’ production employees; during this time, annual maintenance is performed for all equipment in the plant. Because of the downtime for maintenance, the M-100 will be available for no more than 15 hours, and the M-200 will be available for no more than 10 hours. However, because of the high setup cost involved with both machines, management requires that, each machine must be operated for at least 5 hours. The supplier of the chemical material used in the production process informed Jackson Hole that a maximum of 1000 pounds of the chemical material will be available for next week’s production’ the cost of this raw material is $6 per pound. In addition to the raw material cost, Jackson Hole estimates that the hourly cost of operation the M-100 and the M-200 are $50 and $75 respectively.


20

55. The Kartick Company is trying to determine how much of each of two products to produce over the coming planning period. There are three departments, A,B and C, with limited labor hours available in each department. Each product must be processed by each department and the per-unit requirements for each product, labor hours available , and per-unit profits are as shown below.

Labor required in each department Product (hrs./unit) Labor Hours Department Product 1 Product 2 Available

A 1.00 0.30 100 B 0.30 0.12 36 C 0.15 0.56 50

Profit Contribution $33.00 $24.00 A linear program for this situation is as follows: Let x₁ = the amount of product 1 to produce x₂ = the amount of product 2 to produce Maximize 33𝑀𝑀₁ + 24𝑀𝑀₂ 𝑠𝑠. 𝑡𝑡. 1.0𝑀𝑀₂₁ + .30𝑀𝑀₂ ≤ 100 . 30𝑀𝑀₁ + .12𝑀𝑀₂ ≤ 36 . 15𝑀𝑀₁ + .56𝑀𝑀₂ ≤ 50 𝑀𝑀₁, 𝑀𝑀₂ ≥ 0 Mr. Kartick (the owner) used trial and error with a spreadsheet model to arrive at a solution. His proposed solution is x₁=75 and x₂=60. He said he felt his proposed solution is optimal. Is his solution optimal? Without solving the problem, explain why you believe this solution is optimal or not optimal.

56. Assume you are given a minimization linear program that has an optimal solution. The problem is then modified by changing an equality constraint in the problem to a less-than-or-equal-to constraint. Is it possible that the modified problem is infeasible? Answer yes or no and justify.

57. Assume you are given a minimization linear program that has an optimal solution. The problem is then modified by changing a greater-than-or-equal-to constraint. Is it possible that the modified problem is infeasible? Answer yes or no and justify.

58. A consultant was hired to build an optimization model for a large marketing research company. The model is based on a consumer survey that was taken in which each person was asked to rank 30 new products in descending order based on their likelihood of purchasing the product. The consultant was assigned the task of building a model that selects the minimum number of products (which would then be introduced into the marketplace) such that the first, second, and third choice of every subject in the survey is included in the list of selected products. While building a model to figure out which products to introduce, the consultant’s boss walked up to her and said: “Look, if the model tells us we need to introduce more than 15 products, then add a constraint which limits the number of new products to 15 or less. Its’ too expensive to introduce more than 15 new products.” Evaluate this statement in terms of what you have learned so far about constrained optimization models.


21

Homework 1 Supplemented Answers 1. a, b, and e, are acceptable linear programming relationships. c is not acceptable because of 22B− d is not acceptable because of 3 A f is not acceptable because of 1AB

c, d, and f could not be found in a linear programming model because they have the above nonlinear terms. 2. a.

8

4

4 8 0

B

A

b.

B

A

8

4

4 8 0 c.

B

A

8

4

4 8 0

Points on line are only feasible points


22

3. a.

B

A 0

(0,9)

(6,0) b.

B

A 0

(0,60)

(40,0) c.

B

A 0

(0,20)

(40,0)

Points on line are only feasible solutions


23

4. a.

B

A (20,0)

(0,-15)

b.

B

A

(0,12)

(-10,0)

c.

B

A 0

(10,25)

Note: Point shown was used to locate position of the constraint line


24

5.

B

A 0 100 200 300

100

200

300

a

b

c

6. 7A + 10B = 420 is labeled (a) 6A + 4B = 420 is labeled (b) -4A + 7B = 420 is labeled (c)

80

40

40 80

B

A

20

60

100

20 60 0 100 -20 -40 -60 -80 -100

(c)

(a)

(b)


25

7.

0

B

A 150 50 100 200 250

50

100

8.

A

0

200

-200 -100 100 200

133 1 / 3 (100,200)

B


26

9.

B

A 0

200

100

-100

-200

200 300

(150,225)

(150,100)

100


27

10.

A + 2B = 6 (1) 5A + 3B = 15 (2)

(1) × 5 5A + 10B = 30 (3) (2) - (3) - 7B = -15 B = 15/7

From (1), A = 6 - 2(15/7) = 6 - 30/7 = 12/7

0

1

2

3

4

5

B

1 2 3 4 5 6

Optimal SolutionA = 12/7, B = 15/7

Value of Objective Function = 2(12/7) + 3(15/7) = 69/7

A


28

11.

0

100

B

100 200

Optimal SolutionA = 100, B = 50

Value of Objective Function = 750

A

A = 100

B = 80

My answer to this homework question will go here. To find your answer, you may want to study the answers to some of the similar questions.


29

12. a.

b.

c. There are four extreme points: (0,0), (4,0), (3,1,5), and (0,3).

(0,0)

1

2

3

4

5

B

1 2 3 4 5 6

Optimal SolutionA = 3, B = 1.5Value of Objective Function = 13.5

6

(4,0)

(3,1.5)

A

(0,0)

1

2

3

B

1 2 3 4 5 6

Optimal SolutionA = 0, B = 3Value of Objective Function = 18

A7 8 9 10


30

13. a.

b. The extreme points are (5, 1) and (2, 4). c.

0

2

4

6

8

B

2 4 6 8

Feasible Regionconsists of this linesegment only

A

0

2

4

6

8

B

2 4 6 8

Optimal Solution

A

A = 2, B = 4



31

14. a. Let F = number of tons of fuel additive S = number of tons of solvent base

Max 40F + 30S s.t. 2/5F + 1/2S ≤ 20 Material 1

1/5S ≤ 5 Material 2

3/5F + 3/10S ≤ 21 Material 3

F, S ≥ 0

b. Include isovalue lines to determine Material 1 and 3 constraints bind at the optimum.

c. Material 2: 4 tons are used, 1 ton is unused. d. No redundant constraints.

F


32

15. a.

b. Similar to part (a): the same feasible region with a different objective function. The optimal solution occurs at (708, 0) with a profit of z = 20(708) + 9(0) = 14,160.

c. The sewing constraint is redundant. Such a change would not change the optimal solution to the original

problem.

16. a. A variety of objective functions with a slope greater than -4/10 (slope of I & P line) will make extreme point (0, 540) the optimal solution. For example, one possibility is 3S + 9D. b. Optimal Solution is S = 0 and D = 540. c.

Department Hours Used Max. Available Slack Cutting and Dyeing 1(540) = 540 630 90 Sewing

5/6(540) = 450 600 150 Finishing

2/3(540) = 360 708 348 Inspection and Packaging

1/4(540) = 135 135 0

17. Max 5A + 2B + 0S1 + 0S2 + 0S3

s.t. 1A - 2B + 1S1 = 420

2A + 3B + 1S2 = 610

6A - 1B + 1S3 = 125

A, B, S1, S2, S3 ≥ 0


33

18. a. Max 4A + 1B + 0S1 + 0S2 + 0S3

s.t. 10A + 2B + 1S1 = 30

3A + 2B + 1S2 = 12

2A + 2B + 1S3 = 10

A, B, S1, S2, S3 ≥ 0 b.

c. S1 = 0, S2 = 0, S3 = 4/7

0

2

4

6

8

B

2 4 6 8

Optimal Solution

A

A = 18/7, B = 15/7, Value = 87/7

10

12

14

10


34

19. a. Max 3A + 4B + 0S

1 + 0S2 + 0S3

s.t. -

1A + 2B + 1S

1 = 8

(1) 1A + 2B + 1S2 = 12

(2) 2A + 1B + 1S3 = 16

(3) A, B, S1, S2, S3 ≥ 0

b.

c. S1 = 8 + A – 2B = 8 + 20/3 - 16/3 = 28/3 S2 = 12 - A – 2B = 12 - 20/3 - 16/3 = 0 S3 = 16 – 2A - B = 16 - 40/3 - 8/3 = 0

0

2

4

6

8

B

2 4 6 8

Optimal Solution

A

A = 20/3, B = 8/3Value = 30 2/3

10

12

14

10 12

(2)

(1)

(3)


35

20. a. Max 3A + 2B s.t. A + B - S1 = 4

3A + 4B + S2 = 24

A - S3 = 2

A - B - S4 = 0

A, B, S1, S2, S3, S4 ≥ 0 b.

c. S1 = (3.43 + 3.43) - 4 = 2.86 S2 = 24 - [3(3.43) + 4(3.43)] = 0 S3 = 3.43 - 2 = 1.43 S4 = 0 - (3.43 - 3.43) = 0


36

21. a. and b.

c. Optimal solution occurs at the intersection of constraints 1 and 2. For constraint 2, B = 10 + A Substituting for B in constraint 1 we obtain

5A + 5(10 + A)

= 400

5A + 50 + 5A = 400

10A = 350

A = 35 B = 10 + A = 10 + 35 = 45 Optimal solution is A = 35, B = 45

0

10

20

30

40

B

10 20 30 40

Optimal Solution

A

2A + 3B = 60

50

60

70

50 60

80

70 80

90

90 100

Constraint 3

Constraint 2

Constraint 1

Feasible Region


37

d. Because the optimal solution occurs at the intersection of constraints 1 and 2, these are binding constraints.

e. Constraint 3 is the nonbinding constraint. At the optimal solution 1A + 3B = 1(35) + 3(45) = 170. Because 170 exceeds the right-hand side value of 90 by 80 units, there is a surplus of 80 associated with this constraint.


38

22. a.

b.

Extreme Point

Coordinates Profit

1 (0, 0) 5(0) + 4(0) = 0 2 (1700, 0) 5(1700) + 4(0) = 8500 3 (1400, 600) 5(1400) + 4(600) =

9400 4 (800, 1200) 5(800) + 4(1200) =

8800 5 (0, 1680) 5(0) + 4(1680) = 6720

Extreme point 3 generates the highest profit. c. Optimal solution is A = 1400, C = 600 d. The optimal solution occurs at the intersection of the cutting and dyeing constraint and the inspection and packaging constraint. Therefore these two constraints are the binding constraints. e. New optimal solution is A = 800, C = 1200 Profit = 4(800) + 5(1200) = 9200

0

500

1000

1500

2000

C

500 1000 1500 2000

Feasible Region

A

5A + 4C = 4000

2500

3000

3500

2500 3000

Inspection andPackaging

Cutting andDyeing

Sewing

Number of All-Pro Footballs

5

4

3

21


39

23. a. Let E = number of units of the EZ-Rider produced L = number of units of the Lady-Sport produced

Max 2400E + 1800L s.t. 6E + 3L ≤ 210

0 Engine time

L ≤ 280 Lady-Sport maximum 2E + 2.5L ≤ 100

0 Assembly and testing

E, L ≥ 0

b. c. The binding constraints are the manufacturing time and the assembly and testing time. 24. a. Let R = number of units of regular model. C = number of units of catcher’s model.

Max 5R + 8C s.t.

1R + 3/2 C ≤ 900 Cutting and sewing

1/2 R + 1/3 C ≤ 300 Finishing


40

1/8 R + 1/4 C ≤ 100 Packing and Shipping

R, C ≥ 0 b.

c. 5(500) + 8(150) = $3,700 d. C & S 1(500) + 3/2(150) = 725 F 1/2(500) + 1/3(150) = 300 P & S 1/8(500) + 1/4(150) = 100 e.

Department Capacity Usage Slack C & S 900 725 175 hours

F 300 300 0 hours P & S 100 100 0 hours

800

400

400 800

C

R

200

600

1000

200 6000 1000

Optimal Solution(500,150)

F

C & S

P & S

Regular Model

Cat

cher

's M

odel


41

25. a. Let B = percentage of funds invested in the bond fund S = percentage of funds invested in the stock fund

Max 0.06 B + 0.10 S

s.t. B ≥ 0.3 Bond fund minimum 0.06 B + 0.10 S ≥ 0.075 Minimum return B + S = 1 Percentage requirement

b. Optimal solution: B = 0.3, S = 0.7 Value of optimal solution is 0.088 or 8.8% 26. a. a. Let N = amount spent on newspaper advertising R = amount spent on radio advertising

Max 50N + 80R s.t. N + R = 1000 Budget N ≥ 250 Newspaper min. R ≥ 250 Radio min. N -2R ≥ 0 News ≥ 2 Radio

N, R ≥ 0 b.

500

500

R

N0

1000

1000

Optimal SolutionN = 666.67, R = 333.33

Value = 60,000

Newspaper Min

Budget

Radio Min

Feasible regionis this line segment

N = 2 R


42

27. a. Let I = Internet fund investment in thousands, and B = Blue Chip fund investment in thousands Max 0.12I + 0.09B

s.t. 1I + 1B ≤ 50 Available investment funds 1I ≤ 35 Maximum investment in the internet fund 6I + 4B ≤ 240 Maximum risk for a moderate investor

I, B ≥ 0

Internet fund $20,000 Blue Chip fund $30,000 Annual return $5,100 b. The third constraint for the aggressive investor becomes 6I + 4B ≤ 320

This constraint is redundant; the available funds and the maximum Internet fund investment constraints define the feasible region. The optimal solution is:

Internet fund $35,000 Blue Chip fund $15,000 Annual return $5,550

The aggressive investor places as much funds as possible in the high return but high risk Internet fund. c. The third constraint for the conservative investor becomes 6I + 4B ≤ 160 This constraint becomes a binding constraint. The optimal solution is

Internet fund $0 Blue Chip fund $40,000 Annual return $ 3,600

The slack for constraint 1 is $10,000. This indicates that investing all $50,000 in the Blue Chip fund is still too risky for the conservative investor. $40,000 can be invested in the Blue Chip fund. The remaining $10,000 could be invested in low-risk bonds or certificates of deposit.



43

28. a. Let W = number of jars of Western Foods Salsa produced M = number of jars of Mexico City Salsa produced

Max 1W + 1.25M s.t. 5W 7M ≤ 4480 Whole tomatoes 3W + 1M ≤ 2080 Tomato sauce 2W + 2M ≤ 1600 Tomato paste

W, M ≥ 0 Note: units for constraints are ounces b. Optimal solution: W = 560, M = 240 Value of optimal solution is 860


44

29. a. Let B = proportion of Buffalo's time used to produce component 1 D = proportion of Dayton's time used to produce component 1

Maximum Daily Production Component 1 Component 2 Buffalo 2000 1000 Dayton 600 1400

Number of units of component 1 produced: 2000B + 600D Number of units of component 2 produced: 1000(1 - B) + 600(1 - D)

For assembly of the ignition systems, the number of units of component 1 produced must equal the number of units of component 2 produced. Therefore, 2000B + 600D = 1000(1 - B) + 1400(1 - D) 2000B + 600D = 1000 - 1000B + 1400 - 1400D 3000B + 2000D = 2400

Note: Because every ignition system uses 1 unit of component 1 and 1 unit of component 2, we can maximize the number of electronic ignition systems produced by maximizing the number of units of subassembly 1 produced. Max 2000B + 600D In addition, B ≤ 1 and D ≤ 1. The linear programming model is:

Max 2000B + 600D s.t. 3000B + 2000D = 2400

B ≤ 1 D ≤ 1 B, D ≥ 0

The graphical solution is shown below.


45

Optimal Solution: B = .8, D = 0 Optimal Production Plan Buffalo - Component 1 .8(2000) = 1600 Buffalo - Component 2 .2(1000) = 200 Dayton - Component 1 0(600) = 0 Dayton - Component 2 1(1400) = 1400 Total units of electronic ignition system = 1600 per day.


46

30. a. Let E = number of shares of Eastern Cable C = number of shares of ComSwitch

Max 15E + 18C s.t. 40E + 25C ≤ 50,000 Maximum Investment 40E ≥ 15,000 Eastern Cable Minimum 25C ≥ 10,000 ComSwitch Minimum 25C ≤ 25,000 ComSwitch Maximum

E, C ≥ 0

b. c. There are four extreme points: (375,400); (1000,400);(625,1000); (375,1000) d. Optimal solution is E = 625, C = 1000 Total return = $27,375


47

31.

Objective Function Value = 13

0

2

4

B

2 4

Optimal Solution

A

A = 3, B = 1

6

63A + 4B = 13

FeasibleRegion


48

32.

Extreme Points

Objective Function Value

Surplus Demand

Surplus Total

Production

Slack Processing Time

(A = 250, B = 100) 800 125 — — (A = 125, B = 225) 925 — — 125 (A = 125, B = 350) 1300 — 125 —

A

A

B

B

A


49

33. a.

Optimal Solution: A = 3, B = 1, value = 5 b.

(1) 3 + 4(1) = 7 Slack = 21 - 7 = 14 (2) 2(3) + 1 = 7 Surplus = 7 - 7 = 0 (3) 3(3) + 1.5 = 10.5 Slack = 21 - 10.5 = 10.5 (4) -2(3) +6(1) = 0 Surplus = 0 - 0 = 0

c.

Optimal Solution: A = 6, B = 2, value = 34

x2

x10 2 4 6

2

4

6

B

A

B

A


50

34. a.

b. There are two extreme points: (A = 4, B = 1) and (A = 21/4, B = 9/4) c. The optimal solution is A = 4, B = 1 35. a.

Min 6A + 4B + 0S1

+ 0S2 + 0S3

s.t. 2A + 1B - S1 = 12 1A + 1B - S2 = 10 1B + S3 = 4

A, B, S1, S2, S3 ≥ 0

b. The optimal solution is A = 6, B = 4. c. S1 = 4, S2 = 0, S3 = 0.

0 1 2 3 4 5

1

2

3

4

6

x2

x1

(21/4, 9/4)

(4,1)

Feasible Region

B

A


51

36. a. Let T = number of training programs on teaming P = number of training programs on problem solving

Max 10,000T + 8,000P

s.t. T ≥ 8 Minimum Teaming P ≥ 10 Minimum Problem Solving

T + P ≥ 25 Minimum Total 3 T + 2 P ≤ 84 Days Available

T, P ≥ 0 b.

c. There are four extreme points: (15,10); (21.33,10); (8,30); (8,17)

d. The minimum cost solution is T = 8, P = 17 Total cost = $216,000


52

37. Regular Zesty

Mild 80% 60% 8100 Extra Sharp 20% 40% 3000

Let R = number of containers of Regular Z = number of containers of Zesty Each container holds 12/16 or 0.75 pounds of cheese Pounds of mild cheese used = 0.80 (0.75) R + 0.60 (0.75) Z = 0.60 R + 0.45 Z Pounds of extra sharp cheese used = 0.20 (0.75) R + 0.40 (0.75) Z = 0.15 R + 0.30 Z Cost of Cheese = Cost of mild + Cost of extra sharp = 1.20 (0.60 R + 0.45 Z) + 1.40 (0.15 R + 0.30 Z) = 0.72 R + 0.54 Z + 0.21 R + 0.42 Z = 0.93 R + 0.96 Z Packaging Cost = 0.20 R + 0.20 Z Total Cost = (0.93 R + 0.96 Z) + (0.20 R + 0.20 Z) = 1.13 R + 1.16 Z Revenue = 1.95 R + 2.20 Z Profit Contribution = Revenue - Total Cost = (1.95 R + 2.20 Z) - (1.13 R + 1.16 Z) = 0.82 R + 1.04 Z

Max 0.82 R + 1.04 Z s.t. 0.60 R + 0.45 Z ≤ 8100 Mild 0.15 R + 0.30 Z ≤ 3000 Extra Sharp

R, Z ≥ 0 Optimal Solution: R = 9600, Z = 5200, profit = 0.82(9600) + 1.04(5200) = $13,280


53

38. a. Let S = yards of the standard grade material per frame P = yards of the professional grade material per frame

Min 7.50S + 9.00P

s.t. 0.10S + 0.30P ≥ 6 carbon fiber (at least 20% of 30 yards)

0.06S + 0.12P ≤ 3 kevlar (no more than 10% of 30 yards)

S + P = 30 total (30 yards) S, P ≥ 0 b.

c.

Extreme Point

Cost

(15, 15) 7.50(15) + 9.00(15) = 247.50

(10, 20) 7.50(10) + 9.00(20) = 255.00

The optimal solution is S = 15, P = 15

d. Optimal solution does not change: S = 15 and P = 15. However, the value of the optimal solution is reduced to 7.50(15) + 8(15) = $232.50.

e. At $7.40 per yard, the optimal solution is S = 10, P = 20. The value of the optimal solution is

reduced to 7.50(10) + 7.40(20) = $223.00. A lower price for the professional grade will not change the S = 10, P = 20 solution because of the requirement for the maximum percentage of kevlar (10%).


54

39. a. Let S = number of units purchased in the stock fund M = number of units purchased in the money market fund

Min 8S + 3M

s.t. 50S + 100M ≤ 1,200,000 Funds available 5S + 4M ≥ 60,000 Annual income M ≥ 3,000 Minimum units in money market

S, M, ≥ 0

Optimal Solution: S = 4000, M = 10000, value = 62000 b. Annual income = 5(4000) + 4(10000) = 60,000 c. Invest everything in the stock fund.

.

x2

x1

8x1 + 3x2 = 62,000

0 5000 10000 15000 20000

20000

15000

10000

5000

Units of Stock Fund

Uni

ts o

f Mon

ey M

arke

t Fun

d

Optimal Solution

M

S

8S + 3M = 62,000


55

40. Let P1 = gallons of product 1 P2 = gallons of product 2

Min 1P1 + 1P2

s.t. 1P1 + ≥ 30 Product 1 minimum 1P2 ≥ 20 Product 2 minimum 1P1 + 2P2 ≥ 80 Raw material

P1, P2 ≥ 0

Optimal Solution: P1 = 30, P2 = 25 Cost = $55


56

41. a. Let R = number of gallons of regular gasoline produced P = number of gallons of premium gasoline produced

Max 0.30R + 0.50P s.t. 0.30R + 0.60P ≤ 18,000 Grade A crude oil available

1R + 1P ≤ 50,000 Production capacity

1P ≤ 20,000 Demand for premium

R, P ≥ 0 b.

Optimal Solution: 40,000 gallons of regular gasoline, and 10,000 gallons of premium gasoline Total profit contribution = $17,000 c.

Constraint

Value of Slack

Variable

Interpretation

1 0 All available grade A crude oil is used 2 0 Total production capacity is used 3 10,000 Premium gasoline production is 10,000 gallons less

than the maximum demand


57

d. Grade A crude oil and production capacity are the binding constraints. 42.

43.

0 1 2 3

1

2

3

4x2

x1

Unbounded

44. a.

b. New optimal solution is A = 0, B = 3, value = 6.

x2

x10 2 4 6 8 10

2

4

6

8

10

12

14

12

Satisfies Constraint #2

Satisfies Constraint #1

Infeasibility

2

4

2 40

Optimal Solution(30/16, 30/16)Value = 60/16

Objective Functionx2

x1

R

B

A

B

A

B

A


58

45. a.

b. Feasible region is unbounded. c. Optimal Solution: A = 3, B = 0, z = 3.

d. An unbounded feasible region does not imply the problem is unbounded. This will only be the case when it is unbounded in the direction of improvement for the objective function.

B

A

A B


59

46. Let N = number of sq. ft. for national brands G = number of sq. ft. for generic brands Problem Constraints:

N + G ≤ 200 Space available N ≥ 120 National brands G ≥ 20 Generic

Extreme Point N G 1 120 20 2 180 20 3 120 80

a. Optimal solution is extreme point 2; 180 sq. ft. for the national brand and 20 sq. ft. for the

generic brand. b. Alternative optimal solutions. Any point on the line segment joining extreme point 2 and

extreme point 3 is optimal.

c. Optimal solution is extreme point 3; 120 sq. ft. for the national brand and 80 sq. ft. for the generic brand.


60

47.

Alternative optimal solutions exist at extreme points (A = 125, B = 225) and (A = 250, B = 100). Cost = 3(125) + 3(225) = 1050 or Cost = 3(250) + 3(100) = 1050 The solution (A = 250, B = 100) uses all available processing time. However, the solution (A = 125, B = 225) uses only 2(125) + 1(225) = 475 hours.

Thus, (A = 125, B = 225) provides 600 - 475 = 125 hours of slack processing time which may be used for other products.

600

400

200

200 400

x2

x1

100

300

500

100 3000

(125,225)

(250,100)

Processing Time

Alternate optima

B

A


61

48.

Possible Actions: i. Reduce total production to A = 125, B = 350 on 475 gallons.

ii. Make solution A = 125, B = 375 which would require 2(125) + 1(375) = 625 hours of processing time. This would involve 25 hours of overtime or extra processing time. iii. Reduce minimum A production to 100, making A = 100, B = 400 the desired solution.


62

49. a. Let P = number of full-time equivalent pharmacists T = number of full-time equivalent physicians The model and the optimal solution are shown below: MIN 40P+10T S.T. 1) P+T >=250 2) 2P-T>=0 3) P>=90

Optimal Objective Value 5200.00000

Variable Value Reduced Cost

P 90.00000 0.00000 T 160.00000 0.00000

Constraint Slack/Surplus Dual Value

1 0.00000 10.00000 2 20.00000 0.00000 3 0.00000 30.00000

The optimal solution requires 90 full-time equivalent pharmacists and 160 full-time equivalent technicians. The total cost is $5200 per hour. b.

Current Levels

Attrition Optimal Values New Hires Required

Pharmacists

85 10 90 15

Technicians

175 30 160 15

The payroll cost using the current levels of 85 pharmacists and 175 technicians is 40(85) + 10(175) =

$5150 per hour. The payroll cost using the optimal solution in part (a) is $5200 per hour.

Thus, the payroll cost will go up by $50


63

50. Let M = number of Mount Everest Parkas R = number of Rocky Mountain Parkas

Max 100M + 150R s.t.

30M + 20R ≤ 7200 Cutting time

45M + 15R ≤ 7200 Sewing time

0.8M - 0.2R ≥ 0 % requirement

Tip: Here are steps to formulating the % requirement constraint: M must be at least 20% of total production M ≥ 0.2 (total production) M ≥ 0.2 (M + R) M ≥ 0.2M + 0.2R 0.8M - 0.2R ≥ 0

The optimal solution is M = 65.45 and R = 261.82; the value of this solution is z = 100(65.45) + 150(261.82) = $45,818. If we think of this situation as an on-going continuous production process, the fractional values simply represent partially completed products. If this is not the case, we can approximate the optimal solution by rounding down; this yields the solution M = 65 and R = 261 with a corresponding profit of $45,650.


64

51. Let C = number sent to current customers N = number sent to new customers Note: Number of current customers that test drive = .25 C Number of new customers that test drive = .20 N Number sold = .12 ( .25 C ) + .20 (.20 N ) = .03 C + .04 N

Max .03C + .04N s.t. .25 C ≥ 30,000 Current Min .20 N ≥ 10,000 New Min .25 C - .40 N ≥ 0 Current vs. New 4 C + 6 N ≤ 1,200,00

0 Budget

C, N, ≥ 0


65

52. Let S = number of standard size rackets O = number of oversize size rackets

Max 10S + 15O

s.t. 0.8S - 0.2O ≥ 0 % standard

10S + 12O ≤ 4800 Time 0.125S + 0.4O ≤ 80 Alloy

S, O, ≥ 0


66

53. a. Let R = time allocated to regular customer service N = time allocated to new customer service

Max 1.2R + N s.t. R + N ≤ 80 25R + 8N ≥ 800 -0.6R + N ≥ 0

R, N, ≥ 0

b.



R 50.00000 0.00000 N 30.00000 0.00000

Constraint Slack/Surplus Dual Value

1 0.00000 1.12500 2 690.00000 0.00000 3 0.00000 -0.12500

Optimal solution: R = 50, N = 30, value = 90

HTS should allocate 50 hours to service for regular customers and 30 hours to calling on new customers.



67

54. a. Let M1 = number of hours spent on the M-100 machine M2 = number of hours spent on the M-200 machine Total Cost 6(40)M1 + 6(50)M2 + 50M1 + 75M2 = 290M1 + 375M2 Total Revenue 25(18)M1 + 40(18)M2 = 450M1 + 720M2 Profit Contribution (450 - 290)M1 + (720 - 375)M2 = 160M1 + 345M2

Max 160 M1 + 345M2 s.t. M1 ≤ 15 M-100 maximum M2 ≤ 10 M-200 maximum M1 ≥ 5 M-100 minimum M2 ≥ 5 M-200 minimum 40 M1 + 50 M2 ≤ 1000 Raw material available

M1, M2 ≥ 0 b.



M1 12.50000 0.00000 M2 10.00000 145.00000

Constraint Slack/Surplus Dual Value 1 2.50000 0.00000 2 0.00000 145.00000 3 7.50000 0.00000 4 5.00000 0.00000 5 0.00000 4.00000

The optimal decision is to schedule 12.5 hours on the M-100 and 10 hours on the M-200.


68

55. Mr. Krtick’s solution cannot be optimal. Every department has unused hours, so there are no binding constraints. With unused hours in every department, clearly some more product can be made.

56. No, it is not possible that the problem is now infeasible. Note that the original problem was

feasible (it had an optimal solution). Every solution that was feasible is still feasible when we change the constraint to less-than-or-equal-to, since the new constraint is satisfied at equality (as well as inequality). In summary, we have relaxed the constraint so that the previous solutions are feasible (and possibly more satisfying the constraint as strict inequality).


69

57. Yes, it is possible that the modified problem is infeasible. To see this, consider a redundant greater-than-or-equal to constraint as shown below. Constraints 2, 3, and 4 form the feasible region and constraint 1 is redundant. Change constraint 1 to less-than-or-equal-to and the modified problem is infeasible. Original Problem:

Modified Problem:


70

58. It makes no sense to add this constraint. The objective of the problem is to minimize the number of products needed so that everyone’s top three choices are included. There are only two possible outcomes relative to the boss’ new constraint. First, suppose the minimum number of products is <= 15, then there was no need for the new constraint. Second, suppose the minimum number is > 15. Then the new constraint makes the problem infeasible.

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