Quantitative Analysis BA 452 Homework 1 Questions
Quantitative Analysis BA 452 Homework 1 Questions
Quantitative Analysis BA 452 Homework 1 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 maythere just in case you find them helpful. are help you understand fine points. Some Some supplemental questions refer to a section of a chapter textbook (for supplemental questions refer to a section of a chapter in the in the textbook (for example, Section 2.1 means Chapter 2your textbook). example, Section 2.1 means Chapter 2 Section 1 of Section 1 of your textbook). Homework 1 Supplemented Answers listing answers to all 58 Homework 1 Supplemented Answersquestions. questions excluding your 4 homework listing answers to all 58 questions including your 4 homework questions.
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. Learn how to develop linear programming models for simple problems. Be able to identify the special features of a model that make it a linear programming model. Learn how to solve two variable linear programming models by the graphical solution procedure. Understand the importance of extreme points in obtaining the optimal solution.
2. 3.
4.
5.
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. Understand the following terms: problem formulation constraint function objective function solution optimal solution nonnegativity constraints mathematical model linear program linear functions feasible solution feasible region slack variable standard form redundant constraint extreme point surplus variable alternative optimal solutions infeasibility unbounded
9.
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.
3. Show a separate graph of the constraint lines and the solutions that satisfy each of the following constraints: a. 3 + 2 8 b. 12 + 8 480 c. 5 + 10 = 200
4. Show a separate graph of the constraint lines and the solutions that satisfy each of the following constraints: a. 3 4 60 b. 6 + 5 60 c. 5 2 0
5. Show a separate graph of the constraint lines and the solutions that satisfy each of the following constraints: a. 0.25( + ) b. 0.10( + ) c. 0.50( + )
6. Three objective functions for linear programming problems are 7 + 10, 6 + 4, and 4 + 7. Show the graph of each for objective function values equal to 420. 7. Identify the feasible region for the following set of constraints: 0.5 + 0.25 30 1 + 250 0.25 + 0.5 50 , 0 4
1 + 2 6 5 + 3 15 , 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 5 + 5 s.t. 1 100 1 80 2 + 4 400 , 0
12. Consider the following linear programming problem: Max 3 + 3 s.t. 2 + 4 12 6 + 4 24 , 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?
a. Show the feasible region. b. What are the extreme points of the feasible region? c. Find the optimal solution using the graphical procedure.
5 1 4 2 + 2 = 12 , 0
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 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. RMCs 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: What is the linear programming model for the problem? 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? Is there any unused material? If so, how much? Are any of the constraints redundant? Is do, which ones?
a. b. c. d.
15. Refer to the Par, Inc., problem described in Section 2.1 (Chapter 2 Section 1 of your text). Suppose that Pars 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?
17. Write the following linear program in standard form: Max 5 + 2 s.t. 1 2 420 2 + 1 610 6 1 125 , 0 18. For the linear program Max 4 + 1 s.t. 10 + 2 30 3 + 2 12 2 + 2 10 , 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 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?
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? 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 , 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?
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.
10
11
12
13
14
a. b. c. d. e.
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. 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. Use the graphical solution procedure to determine the feasible region. What are the coordinates of the extreme points? Compute the total cost at each extreme point. What is the optimal solution? 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? 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 clients 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 Inniss 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, Inniss 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?
15
43. Does the following linear program involve infeasibility, unbounded, and/or alternative optimal solutions? Explain. . . 8 + 6 24 2 4 , 0 16
a. What is the optimal solution for this problem? b. Suppose that the objective function is changed to 1 + 1. Find the new optimal solution. 45. Consider the following linear program: . . 4 + 3 3 1 1 3 , 0 1 2
5 + 3 15 3 + 5 15 , 0
a. b. c. d.
Graph the feasible region for the problem. Is the feasible region unbounded? Explain. Find the optimal solution. 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 managements 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.
17
19
A linear program for this situation is as follows: Let x = the amount of product 1 to produce x = the amount of product 2 to produce Maximize 33 + 24 . . 1.0 + .30 100 . 30 + .12 36 . 15 + .56 50 , 0
Mr. Kartick (the owner) used trial and error with a spreadsheet model to arrive at a solution. His proposed solution is x=75 and x=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 consultants 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. 20
b.
B 8
4
A
c.
B 8 Points on line are only feasible points
21
0
b.
B (0,60)
(6,0)
A 0 (40,0)
c.
B Points on line are only feasible solutions (0,20) A (40,0)
22
(20,0)
(0,-15)
b.
B
(0,12)
(-10,0)
c.
B (10,25)
Note: Point shown was used to locate position of the constraint line
23
c
200
6.
7A + 10B = 420 is labeled (a) 6A + 4B = 420 is labeled (b) -4A + 7B = 420 is labeled (c)
B 100 80 60 (b)
(c)
40 20
(a)
A
-100
-80
-60
-40
-20
20
40
60
80
100
24
50
250
8.
B 200
133 1/3
(100,200)
25
B (150,225) 200
100
(150,100)
-100
-200
26
Optimal Solution A = 12/7, B = 15/7 Value of Objective Function = 2(12/7) + 3(15/7) = 69/7
0 1 2 3 4 5 6
2B = 3B =
6 15
From (1),
27
0 100 200
28
(0,0) 1 2 3 4 (4,0) 5 6
b.
B 3 Optimal Solution A = 0, B = 3 Value of Objective Function = 18
(0,0) 1 2 3 4 5 6 7 8 9 10
c.
There are four extreme points: (0,0), (4,0), (3,1,5), and (0,3).
29
My answer to this homework question will B 8 go here. To find your answer, you may want 6 to study the answers to some of the similar Feasible Region questions. 4 consists of this line
segment only 2 0 2 4 6 8 A
b. c.
B 8
Optimal Solution A = 2, B = 4
0 2 4 6 8
30
c. d.
31
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. c. Department Cutting and Dyeing Sewing Finishing Inspection and Packaging 17. Max s.t. 1A 2B + 1S1 + 1S2 + A, B, S1, S2, S3 0 = = 1S3 = 420 610 125 5A + 2B + 0S1 + 0S2 + 0S3 Hours Used 1(540) = 540 5 /6(540) = 450 2 /3(540) = 360 1 /4(540) = 135 Max. Available 630 600 708 135 Slack 90 150 348 0 Optimal Solution is S = 0 and D = 540.
2A + 3B 6A 1B
32
4A +
1B + 0S1
+ 0S2
+ 0S3
= 30 + 1S2 + 1S3 = 12 = 10
12
10
0 2 4 6 8 10
c.
S1 = 0, S2 = 0, S3 = 4/7
33
b.
B 14 (3) 12
10
(1)
2 (2) 0 2 4 6 8 10 12 A
c.
34
c.
80
70 Constraint 2
60
50
Optimal Solution
40
Constraint 3
Constraint 1
30
20
Feasible Region
10
2A + 3B = 60 A 10 20 30 40 50 60 70 80 90 100
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
36
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.
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3000
2500
Inspection and Packaging Cutting and Dyeing 4 Feasible Region Sewing 3 5A + 4C = 4000 2 1 500 1000 1500 2000 2500 Number of All-Pro Footballs 3000
2000 5 1500
1000
500
b. Extreme Point 1 2 3 4 5 Coordinates (0, 0) (1700, 0) (1400, 600) (800, 1200) (0, 1680) Profit 5(0) + 4(0) = 0 5(1700) + 4(0) = 8500 5(1400) + 4(600) = 9400 5(800) + 4(1200) = 8800 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
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Max s.t.
2E +
E, L 0
b. c. 24. a. The binding constraints are the manufacturing time and the assembly and testing time. Let R = number of units of regular model. C = number of units of catchers model. Max s.t. 5R 1R 1/ R 2 + + + 8C 3/ C 2 1/ C 3 39 900 300 Cutting and sewing Finishing
b.
C 1000
600
C&
400
P&
c. d.
e. Department C&S F P&S Capacity 900 300 100 Usage 725 300 100 Slack 175 hours 0 hours 0 hours
40
26.
a.
a. Let N = amount spent on newspaper advertising R = amount spent on radio advertising Max s.t. 50N + 80R N + N N N, R 0 R = 1000 Budget R -2R 250 Newspaper min. 250 Radio min. 0 News 2 Radio
b.
R
1000
Radio Min
Budget
500
41
My answer to this homework question will go here. 6ITo find your answer,moderate investor want you may + 4B 240 Maximum risk for a I, B to study the0 answers to some of the similar questions.
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.
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W, M 0 Note: units for constraints are ounces b. Optimal solution: W = 560, M = 240 Value of optimal solution is 860
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Buffalo Dayton
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 s.t. 2000B 3000B B + 600D + 2000D = 2400 1 D 1 B, D 0
44
Optimal Solution: B = .8, D = 0 Optimal Production Plan Buffalo - Component 1 Buffalo - Component 2 Dayton - Component 1 Dayton - Component 2 .8(2000) = 1600 .2(1000) = 200 0(600) = 0 1(1400) = 1400
45
b.
c. d.
There are four extreme points: (375,400); (1000,400);(625,1000); (375,1000) Optimal solution is E = 625, C = 1000 Total return = $27,375
46
Feasible Region
A 0 2 Optimal Solution A = 3, B = 1 4 6 3A + 4B = 13
47
A B
48
x1 A 0 2 4 6
Optimal Solution: A = 3, B = 1, value = 5 b. (1) (2) (3) (4) c. B 3 + 4(1) = 7 2(3) + 1 = 7 3(3) + 1.5 = 10.5 -2(3) +6(1) = 0 Slack = 21 - 7 = 14 Surplus = 7 - 7 = 0 Slack = 21 - 10.5 = 10.5 Surplus = 0 - 0 = 0
49
(4,1) x1 A 0 1 2 3 4 5 6
b. c. 35. a.
There are two extreme points: (A = 4, B = 1) and (A = 21/4, B = 9/4) The optimal solution is A = 4, B = 1
Min s.t.
6A
4B
0S 1 S1
0S2
0S 3 = 12 10 4
2A 1A
+ +
1B 1B 1B
S2 + S3
= =
50
T + 3T + T, P 0 b.
c.
51
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 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 = 0.20 R + 0.20 Z (0.93 R + 0.96 Z) + (0.20 R + 0.20 Z) 1.13 R + 1.16 Z 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 s.t. 0.82 R + 0.60 R + 0.15 R + R, Z 0 Optimal Solution: R = 9600, Z = 5200, profit = 0.82(9600) + 1.04(5200) = $13,280 1.04 Z 0.45 Z 0.30 Z 8100 3000 Mild Extra Sharp
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c. Extreme Point (15, 15) (10, 20) Cost 7.50(15) + 9.00(15) = 247.50 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%). 53
1,200,000 Funds available 60,000 Annual income 3,000 Minimum units in money market
20000
10000
5000
Optimal Solution: S = 4000, M = 10000, value = 62000 b. c. Annual income = 5(4000) + 4(10000) = 60,000 Invest everything in the stock fund.
54
55
Optimal Solution: 40,000 gallons of regular gasoline, and 10,000 gallons of premium gasoline Total profit contribution = $17,000 c. Constraint 1 2 3 Value of Slack Variable 0 0 10,000 Interpretation All available grade A crude oil is used Total production capacity is used Premium gasoline production is 10,000 gallons less than the maximum demand
56
10
12
43.
x2 B 4 3 2 1 x1 Unbounded
44.
a.
x2 B Objective Function 4 Optimal Solution (30/16, 30/16) Value = 60/16
2 0 2 4 x1 A
b.
b. c.
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.
58
Extreme Point 1 2 3
G 20 20 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.
59
s ce s P ro e Tim ing
Alternate optima (125,225) (250,100) x1 0 100 200 300 400
500
400
300
200
100
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.
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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.
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Optimal Objective Value 5200.00000 Variable P T Value 90.00000 160.00000 Reduced Cost 0.00000 0.00000
Constraint 1 2 3
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 85 175 Attrition 10 30 Optimal Values 90 160 New Hires Required 15 15
Pharmacist s Technician s
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.
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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.
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1,200,00 Budget 0
64
4800 80
65
My answer to this homework question will Max 1.2R + N go here. Tos.t. find your answer, you may want + R N 80 25R + 8N to study the answers to some of the similar 800 -0.6R + 0 N questions. R, N, 0
Optimal Objective Value 90.00000 Variable R N Value 50.00000 30.00000 Reduced Cost 0.00000 0.00000
Let R = time allocated to regular customer service N = time allocated to new customer service
b.
Constraint 1 2 3
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.
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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 s.t. 160 M1 + M1 M1 40 M1 + M1, M2 0 b. Optimal Objective Value 5450.00000 Variable M1 M2 Value 12.50000 10.00000 Reduced Cost 0.00000 145.00000 345M2 M2 M2 50 M2 15 10 5 5 1000 M-100 maximum M-200 maximum M-100 minimum M-200 minimum Raw material available
Constraint 1 2 3 4 5
The optimal decision is to schedule 12.5 hours on the M-100 and 10 hours on the M-200.
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Original Problem:
Modified Problem:
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