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MB363 Management Decision Tools Assignment 1
Prepared By
: Kor Xian Thong Ronnie _(081723A15)
Sub-Group
:3
Supervisor
: A/P Li Zhi-Feng, Michael
Date
: 8 October 2009
th
Table of Contents 1)
2)
3)
Chapter 2: Linear Programming I (a)
Problem 2.16
2
(b)
Problem 2.21
3
Chapter 3: Linear Programming Programming II (a)
Problem 3.5
4
(b)
Problem 3.16
7
(c)
Problem 3.24
9
Chapter 5: Sensitivity Analysis for LP (a)
4)
5)
Page
Problem 5.2
11
Chapter 8: Nonlinear Programming (a)
Problem 8.4
17
(b)
Problem 8.10
19
Chapter 9: Decision Analysis (a)
Problem 9.7
22
(b)
Problem 9.18
26
1|Page
Chapter 2: Linear Programming I Q1(a) Problem 2.16 Nutri-Jenny Weight Management Centre (a) Linear Programming Model on Spreadsheet
(b) Notation: B: Beef Tips G: Gravy P: Peas C: Carrots D: Dinner Roll Choose the values of B, P, J, A, M and C so as to minimize Total Cost, C=$(0.40B+0.35G+0.15P+0.18C+0 C=$(0.40B+0.35G+0.15P+0.18C+0.10D) .10D) Subject to satisfying all the following constraints: co nstraints: 280 ≤ 54B + 20G + 15P + 8C + 40D ≤ 320 19B + 15G + 10D ≤ 96 15P + 350C ≤ 600 G + 3P + C ≤ 10 8B + P+ C + D ≤ 30 B ≥ 2 G ≥ 0.5B
2|Page
Q1(b) Problem 2.21 Learning Centre (a) Linear Programming Model on Spreadsheet
(b) Notation: B: Bread P: Peanut Butter J: Jelly A: Apple M: Milk C: Cranberry Juice Choose the values of B, P, J, A, M and C so as to minimize Total Cost, C=$(0.06B+0.05P+0.08J+0.35A+0.20M + 0.40C) Subject to satisfying all the following constraints: 300 ≤ 80B + 100P + 70J + 90A + 120M +110C ≤ 500 15B + 80P + 60M ≤ 0.30(80B + 100P + 70J + 90A + 120M +110C) 4J + 6A + 2M + 80C ≥ 60 4B + 3J + 10A + C ≥ 10 B ≥ 2 P ≥ 1 J ≥ 1 M + C ≥ 1
3|Page
Chapter 3: Linear Programming II Q2(a) Problem 3.5 Omega Manufacturing Company (a) Resource Allocation Problem The activities under consideration are: Activity 1: Produce Product 1 Activity 2: Produce Product 2 Activity 3: Produce Product 3 The resources to be allocated to these activities are: Resource 1: Available time (machine-hours per week) of Milling Machine Resource 2: Available time (machine-hours per week) of Lathe Resource 3: Available time (machine-hours per week) of Grinder This problem has 3 resource constraints: Constraint 1: Total Available time of Milling Machine = 500 Machine-Hours per Week Constraint 2: Total Available time of Lathe = 350 Machine-Hours per Week Constraint 3: Total Available time of Grinder = 150 Machine-Hours per Week Side Constraint: Constraint 4: Sales Potential for Product 3 = 20 units per week
(b) With the 3 products under consideration, these are the decisions to be made: Decision 1: P1 = Number of units of Product 1 to produce Decision 2: P2 = Number of units of Product 2 to produce Decision 3: P3 = Number of units of Product 3 to produce Overall Measure of Performance: The objective is to maximize total profits earned from producing and selling certain product-mix of Product 1, 2 and 3 per week.
4|Page
(c) Decisions: P1: Number of Product 1 produced and sold P2: Number of Product 2 produced and sold P3: Number of Product 3 produced and sold Resource Constraints: Total number of machine-hours per week used for Milling machine ≤ 500 Total number of machine-hours per week used for Lathe ≤ 350 Total number of machine-hours per week used for Grinder ≤ 500 Side Constraint: Total units of Product 3 produced per week ≤ Sales potential for Product 3 (20 units/week) Overall Measure of Performance: Maximize Total Profit = Sum of Profits earned from selling certain product mix of Product 1,2 and 3. Total Profit = 50P 1 + 20P2 + 25P3 (d)
Excel Equation for Output Cell (Milling Machine): SUMPRODUCT (Machine Hours for milling machine used by each product, Units Produced) Excel Equation for Output Cell (Lathe): SUMPRODUCT (Machine Hours for L athe used by each product, Units Produced) Excel Equation for Output Cell (Grinder): SUMPRODUCT (Machine Hours for Grinder used by each product, Units Produced)
5|Page
(e) Maximize Profit = 50P1 + 20P2 + 25P3 Subject to Milling Machine: 9P1 + 3P2 + 5P3 ≤ 500 Lathe: 5P1 + 4P2 ≤ 350 Grinder: 3P1 + 2P3 ≤ 150 And P3 ≤ 20
6|Page
Q2(b) Problem 3.16 Fagersta Steelworks (a) Notation: M1S1: Amount of iron ore shipped per month from Mine M1 to Storage Facility S1 M1S2: Amount of iron ore shipped per month from Mine M1 to Storage Facility S2 M2S1: Amount of iron ore shipped per month from Mine M2 to Storage Facility S1 M2S2: Amount of iron ore shipped per month from Mine M2 to Storage Facility S2 S1P: Amount of iron ore shipped per month from Storage Facility S1 to Steel Plant P S2P: Amount of iron ore shipped per month from Storage Facility S2 to Steel Plant P Resources: Resource 1: Amount of iron ore produced by Mine M1 = 40 tons Resource 2: Amount of iron ore produced by Mine M2 = 60 tons Fixed Requirement Constraints: Requirement 1: Total amount of iron ore shipped out of Mine M1 = Total Produced by Mine M1 Requirement 2: Total amount of iron ore shipped out of Mine M2 = Total Produced by Mine M2 Requirement 3: Steel Plant P must receive 100 tons of iron ore. Resource Constraints: Constraint 1: Amount of iron ore shipped from Mine M1 to Storage Facility S1 ≤ 30 tonnes Constraint 2: Amount of iron ore shipped from Mine M1 to Storage Facility S2 ≤ 30 tonnes Constraint 3: Amount of iron ore shipped from Mine M2 to Storage Facility S1 ≤ 50 tonnes Constraint 4: Amount of iron ore shipped from Mine M2 to Storage Facility S2 ≤ 50 tonnes Constraint 5: Amount of iron ore shipped from Storage Facility S1 to Steel Plant P ≤ 70 tonnes Constraint 6: Amount of iron ore shipped from Storage Facility S2 to Steel Plant P ≤ 70 tonnes
7|Page
(b)
(c) Minimize Total Shipping Cost = 2000M1S1 + 1700M1S2 + 1600M2S1 + 1100M2S2 + 400S1P + 800S2P Subject to the following constraints: 1. Fixed Requirement Constraints: M1S1 + M1S2 = 40 M2S1 + M2S2 = 60 S1P + S2P = 100 2. Resource Constraints: M1S1 ≤ 30, M1S2 ≤ 30 (Mine M1) M2S1 ≤ 50, M2S2 ≤ 50 (Mine M2) S1P ≤ 70 (Storage Facility S1) S2P ≤ 70 (Storage Facility S2) 3. Nonnegativity Constraints: M1S1 ≥ 0, M1S2 ≥ 0, M2S1 ≥ 0, M2S2 ≥ 0, S1P ≥ 0, S2P ≥ 0 8|Page
Q2(c) Problem 3.24 Day Care for Preschoolers (a) Resource Constraints: Requirement 1: Each food choice needs to provide no more than 600 calories. Requirement 2: Each food choice should not have more than 30% of the calories that comes from fats. Benefits Constraints: Requirement 3: Each foold choice needs to provide at least 400 calories. Requirement 4: Each food choice needs to have at least 60mg of Vitamin C. Requirement 5: Each food choice needs to provide at least 12g of protein. Requirement 6: Each food choice needs to have at least twice as much peanut butter as jelly. Requirement 7: Each food choice should have at least 1 cup of liquid (milk and/or juice). Fixed Requirement Constraints: Requirement 7: Each food choice should have exactly 2 slices of bread. (b)
9|Page
(c) Notation: B: Bread P: Peanut Butter S: Strawberry Jelly G: Graham Cracker M: Milk J: Juice Choose B, P, S, G, M, J to minimize Total Cost C = 0.05B + 0.04P + 0.07S + 0.08G + 0.15M + 0.35J Subject to the following constraints: 1. Resource Constraints: 70B + 100P + 50S + 60G + 150M + 100J ≤ 600 10B + 75P + 20G + 70M ≤ 0.30(70B + 100P + 50S + 60G + 150M + 100J) 2. Benefits Constraints: 70B + 100P + 50S + 60G + 150M + 100J ≥ 400 3S + 2M + 120J ≤ 60 3B + 4P + G + 8M + J ≤ 12 P ≥ 2S M + J ≥ 1 3. Fixed Requirement Constraints: B=2
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Chapter 5: Sensitivity Analysis for LP Q3 Problem 5.2 (a)
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(b) With the unit profit of activity 2 at a constant $5 per unit,
When the unit profit changes for activity 1 changes from $2 to $1, the optimal solution changes from (6,2) to (0,4).
When the unit profit changes for activity 1 changes from $2 to $3, the optimal solution changes from (6,2) to (10,0).
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(c) With the unit profit of activity 1 at a constant $2 per unit,
When the unit profit changes for activity 2 changes from $5 to $2.50, the optimal solution changes from (6,2) to (10,0).
When the unit profit changes for activity 2 changes from $5 to $7.50, the optimal solution changes from (6,2) to (0,4).
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(d) Unit Profit for Activity 1 $1.00 $1.20 $1.40 $1.60 $1.80 $2.00 $2.20 $2.40 $2.60 $2.80 $3.00
Optimal Units Used Activity 1 Activity 2 6 2 0 4 0 4 0 4 0 4 6 2 6 2 6 2 6 2 10 0 10 0 10 0
Total Profit $22.00 $20.00 $20.00 $20.00 $20.00 $20.80 $22.00 $23.20 $24.40 $26.00 $28.00 $30.00
Unit Profit for Activity 2
Optimal Units Used Activity 1 Activity 2
Total Profit
$2.50 $3.00 $3.50 $4.00 $4.50 $5.00 $5.50 $6.00 $6.50 $7.00 $7.50
6 10 10 10 6 6 6 6 0 0 0 0
2 0 0 0 2 2 2 2 4 4 4 4
$22.00 $20.00 $20.00 $20.00 $20.00 $21.00 $22.00 $23.00 $24.00 $26.00 $28.00 $30.00
For Activity 1, the allowable range for the unit profit of Activity 1 is from $1.80 to $2.40. For Activity 2, the allowable range for the unit profit of Activity 2 is from $4.00 to $5.50.
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(e)
(f)
From the Graphical Linear Programming and Sensitivity Analysis module, when the unit profit of Activity 2 is kept constant at $5.00, the allowable range for unit profit of Activity 1 is from $1.67 to $2.50. When the unit profit of Activity 1 is kept constant at $2.00, the allowable range for unit profit of Activity 1 is from $4.00 to $6.00. 15 | P a g e
(g) Units Used
Unit Profit for Activity 2 (6,2)
$2.50
$3.00
$3.50
$4.00
$4.50
$5.00
$5.50
$6.00
$6.50
$7.00
$7.50
$1.00
(6,2)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
$1.20
(6,2)
(6,2)
(6,2)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
$1.40
(10,0)
(6,2)
(6,2)
(6,2)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
Unit Profit
$1.60
(10,0)
(10,0)
(6,2)
(6,2)
(6,2)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
for Activity 1
$1.80
(10,0)
(10,0)
(10,0)
(6,2)
(6,2)
(6,2)
(0,4)
(0,4)
(0,4)
(0,4)
(0,4)
$2.00
(10,0)
(10,0)
(10,0)
(6,2)
(6,2)
(6,2)
(6,2)
(0,4)
(0,4)
(0,4)
(0,4)
$2.20
(10,0)
(10,0)
(10,0)
(10,0)
(6,2)
(6,2)
(6,2)
(6,2)
(6,2)
(0,4)
(0,4)
$2.40
(10,0)
(10,0)
(10,0)
(10,0)
(10,0)
(6,2)
(6,2)
(6,2)
(6,2)
(6,2)
(0,4)
$2.60
(10,0)
(10,0)
(10,0)
(10,0)
(10,0)
(10,0)
(6,2)
(6,2)
(6,2)
(6,2)
(6,2)
$2.80
(10,0)
(10,0)
(10,0)
(10,0)
(10,0)
(10,0)
(10,0)
(6,2)
(6,2)
(6,2)
(6,2)
$3.00
(10,0)
(10,0)
(10,0)
(10,0)
(10,0)
(10,0)
(10,0)
(6,2)
(6,2)
(6,2)
(6,2)
(h) From the plot formed in the Graphical Linear Programming and Sensitivity Analysis module, it can be seen that for any objective function line to cross the optimum point, the gradient of the function has to lie between and not inclusive of 0.333, to and not inclusive of 0.50. From the data obtained from using two-dimensional Solver Table, the highlighted data are the unit profit of activity 1 and activity 2 that corresponds to the optimal solution of (6,2), and this values of respective unit profit satisfies the condition of 0.333≤Gradient≤0.50. Outside of this range, the optimal solution changes. Hence the data obtained from the two-dimensional Solver Table is proved to be correct.
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Chapter 8: Nonlinear Programming Q4(a) Problem 8.4 (a) Production Rate (R)
Profit per Day (P)
P=$100R-$5R^2
% Error
0 1 2 3 4
$0.00 $95.00 $184.00 $255.00 $320.00
$0.00 $95.00 $180.00 $255.00 $320.00
0 0 2.1739 0 0
(b) Production Rate (R) 0 1 2 3 4
Profit per Day (P) $0.00 $95.00 $184.00 $255.00 $320.00
P=$104R-$6R^2 $0.00 $98.00 $184.00 $258.00 $320.00
% Error 0 3.15789 0 1.17647 0
(c) From the calculations, it can be seen that the approximation P=$100R - $5R2 fits closely to the actual data (Profit per Day), except at R=2 with a % difference of 2.1739%. Hence it fits the graph for 80% of the data and thus provides a close approximation of P. As for the approximation P=$104R - $6R2, it fits the data at 3 close points and does not fit at 2 points, namely at R=1 with a % difference of 3.158% and R=3 with a % difference of 1.176%. Overall % difference is slightly larger at 4.334%. Also, It only fits the graph for 60% of the data. Thus, it does not provide a close approximation of P. Hence it can be concluded that the approximation P=$100R - $5R2 provides better fit to all the data.
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(d) y = -5.5714x2 + 102.29x – 0.3429
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Q4(b) Problem 8.10 Dorwyn Company (a)
(b) Production Rate
Door 0 1 2 3
Gross Profit ($’000) 0 4 8 12
Marketing Costs ($’000) 0 1 8 27
Profit ($’000) 0 3 0 -15
Incremental Profit ($’000) 3 -3 -15
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(c) Production Rate Window 0 1 2 3
Gross Profit 0 6 12 18
Marketing Costs 0 2 8 18
Profit 0 4 4 0
Incremental Profit 4 0 -4
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(d)
Dorwyn management should produce the product mix of producing 1 door and 1 window. (e) The solution based on separable programming approximation from part d is the same as the solution obtained in part a for exact nonlinear programming.
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Chapter 9: Decision Analysis Q5(a) Problem 9.7 (a) ������� ��������� ����� �� ������ ���������
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From Maximum Likelihood criterion, Alternative A1 should be chosen because State S1 has the highest prior probabilities, and A1 has the maximum payoff for that state.
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(d) ������ �������� ���� ������ �����
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From Bayes’ Decision Rules, Alternative A1 should be chosen as it has larger expected payoff than A2. (e)
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(f)
From TreePlan, Alternative A1 should be chosen. (g) ����������� �� �� ������� ������
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From Sensitivity Analysis using Solver Table, it can be seen that the action changes from Alternative A2 to Alternative A1 at values of probabilities of S1 above 0.4. Hence the best alternative changes to A1 as probabilities of S1 increases above 0.40. By redoing the analysis with an increment of 0.01, it is found that the best alternative changes when the probability of S1 increases above 0.43. 24 | P a g e
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From Sensitivity Analysis using Solver Table, it can be seen that the action changes from Alternative A2 to Alternative A1 at values of probabilities of S1 above 0.5. Hence the best alternative changes to A1 as probabilities of S1 increases above 0.50. By redoing the analysis with an increment of 0.01, it is found that the best alternative changes when the probability of S1 increases above 0.51. (i) ����������� �� �� ������� ������
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From Sensitivity Analysis using Solver Table, it can be seen that the action changes from Alternative A2 to Alternative A1 at values of probabilities of S2 above 0.1. Hence the best alternative changes to A1 as probabilities of S2 increases above 0.10. By redoing the analysis with an increment of 0.01, it is found that the best alternative changes when the probability of S1 increases above 0.13. (j) Based on the 3 sensitivity analysis done above, the desired alternative changes from Alternative 1 to Alternative 2 when the prior probabilities of the given state of nature deviates more than 0.10 from the original value. Since the true probabilities of the states of nature should within 10% of given prior probabilities, I would choose Alternative 1.
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Q5(b) Problem 9.18 Telemore Company (a) Decision Alternatives: •
Develop and market the new product
•
Not to develop and market the new product
States of Nature: •
Successful launch of new product
•
Unsuccessful launch of new product
(b)
From Bayes’ Decision Rule, it is found that launching the new product has a higher expected payoff of $400 000, hence the decision chosen is to launch the new product.
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(c)
EP (without marketing survey) = 0.667 x $400,000 + 0.333 x $0 = $266,667 EP (with marketing survey) = $1,000,000 EVPI = EP (with marketing survey) - EP (without marketing survey = $1,000,000 - $266,667 = $7,333,333 Since Cost ($100,000) < EVPI ($7,333,333), it is worthwhile to conduct marketing survey. (d)
P(S|FSS) = P(S)•P(FSS|S) / [P(S)•P(FSS|S) + P(US)•P(FSS|US)] = 0.8(2/3) / [0.8(2/3) + 0.2(1/3)] = 16/19 (0.889) P(US|FSS) = P(US)•P(FSS|US) / [P(S)•P(FSS|S) + P(US)•P(FSS|US)] = 0.2(1/3) / [0.8(2/3) + 0.2(1/3)] = 3/19 (0.111) P(S|USS) = P(S)•P(USS|S) / [P(S)•P(USS|S) + P(US)•P(USS|US)] = 0.3(2/3) / [0.3(2/3) + 0.7(1/3)] = 4/11 (0.462) P(US|USS) = P(US)•P(USS|US) / [P(S)•P(USS|S) + P(US)•P(USS|US)] = 0.7(1/3) / [0.3(2/3) + 0.7(1/3)] = 7/11 (0.538)
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(e) Optimal Policy: Conduct the marketing survey. If the result is favourable, launch the new product. If the result is unfavourable, do not launch the product. The expected payoff (which includes the cost of marketing survey) is $520 000.
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