MB363 Assignment 1 Report

MB363 Management Decision Tools

Assignment 1

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NN NNNN NN NN NNYY YYYY YY YY YY

AA AAAA AA AA AANN NNNN NN NN NN

GG GGGG GG GG GG TT TTTT TT TT TT

EE EEEE EE EE EECC CCCC CC CC CC

HH HHHH HH HH HHNN NNNN NN NN NN

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GG GGGG GG GG GGII IIII II II II CC CCCC CC CC CC

AA AAAA AA AA AALL LLLL LL LL LL UU UUUU UU UU UU

NN NNNN NN NN NNII IIII II II II VV VVVV VV VV VV

EE EEEE EE EE EERR RRRR RR RR RR

SS SSSS SS SS SSII IIII II II II TT TTTT TT TT TT

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Prepared By : Kor Xian Thong Ronnie

_(081723A15)

Sub-Group : 3

Supervisor : A/P Li Zhi-Feng, Michael

Date : 8th

October 2009

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Table of Contents Page

1)

Chapter 2: Linear Programming I

(a) Problem 2.16

2

(b) Problem 2.21

3

2) Chapter 3: Linear Programming II

(a) Problem 3.5

4

(b) Problem 3.16

7

(c) Problem 3.24

9

3) Chapter 5: Sensitivity Analysis for LP

(a) Problem 5.2

11

4) Chapter 8: Nonlinear Programming

(a) Problem 8.4

17

(b) Problem 8.10

19

5) Chapter 9: Decision Analysis

(a) Problem 9.7

22

(b) Problem 9.18

26

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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+0.10D)

Subject to satisfying all the following constraints:

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

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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

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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



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



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.

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(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 = 50P1 + 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 Lathe used by each product, Units Produced) Excel Equation for Output Cell (Grinder): SUMPRODUCT (Machine Hours for Grinder used by each product, Units Produced)

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(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

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Q2(b) Problem 3.16 Fagersta Steelworks

(a) Notation:

M1S1: Amount of iron ore shipped per month from Mine M1 to Storage Facility S1




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


tonnes


tonnes


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

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(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

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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)

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(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


When the unit profit changes for activity 2 changes from $5 to $7.50, the optimal solution


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(d)

Unit Profit Optimal Units Used Total

for Activity 1 Activity 1 Activity 2 Profit

6 2 $22.00

$1.00 0 4 $20.00

$1.20 0 4 $20.00

$1.40 0 4 $20.00

$1.60 0 4 $20.00

$1.80 6 2 $20.80

$2.00 6 2 $22.00

$2.20 6 2 $23.20

$2.40 6 2 $24.40

$2.60 10 0 $26.00

$2.80 10 0 $28.00

$3.00 10 0 $30.00

Unit Profit Optimal Units Used Total

for Activity 2 Activity 1 Activity 2 Profit

6 2 $22.00

$2.50 10 0 $20.00

$3.00 10 0 $20.00

$3.50 10 0 $20.00

$4.00 6 2 $20.00

$4.50 6 2 $21.00

$5.00 6 2 $22.00

$5.50 6 2 $23.00

$6.00 0 4 $24.00

$6.50 0 4 $26.00

$7.00 0 4 $28.00

$7.50 0 4 $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.

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(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 $0.00 $0.00 0

1 $95.00 $95.00 0

2 $184.00 $180.00 2.1739

3 $255.00 $255.00 0

4 $320.00 $320.00 0

(b)

Production Rate (R) Profit per Day (P) P=$104R-$6R^2 % Error

0 $0.00 $0.00 0

1 $95.00 $98.00 3.15789

2 $184.00 $184.00 0

3 $255.00 $258.00 1.17647

4 $320.00 $320.00 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 Gross Profit

($’000) Marketing Costs

($’000) Profit ($’000)

Incremental Profit ($’000)

0 0 0 0 -

1 4 1 3 3

2 8 8 0 -3

3 12 27 -15 -15

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(c)

Production Rate

Window Gross Profit Marketing Costs Profit Incremental Profit

0 0 0 0

1 6 2 4 4

2 12 8 4 0

3 18 18 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)

Maximax Criterion

State of Nature

Alternate S1 S2 S3

Maximum in

Row

A1 220 170 110

220 ← Maximax

A2 200 180 150

200

From Maximax criterion, Alternative A1 should be chosen because because it has the maximum of the maximum payoffs from each alternative.

(b)

Maximin Criterion

State of Nature

Alternate S1 S2 S3

Minimum in

Row

A1 220 170 110

110

A2 200 180 150

150 ←Maximin

From Maximin criterion, Alternative A2 should be chosen because it has the maximum of the minimum payoffs from each alternative.

(c)

Maximum Likelihood Criterion

State of Nature

Alternate S1 S2 S3

A1 220 170 110 ←2. Maximum

A2 200 180 150

Prior Probabilities 0.6 0.3 0.1

↑

1.

Maximum

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)

Bayes' Decision Rule

Payoff Table State of Nature

Expected

Alternate S1 S2 S3

Payoff

A1 220 170 110

194

A2 200 180 150

189

PriorProbabilities 0.6 0.3 0.1

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)

Probability of S3 remains fixed:

Probability of S1 Action Expected Payoff

A1 194

0.3 A2 183

0.35 A2 184

0.4 A2 185

0.45 A1 186.5

0.5 A1 189

0.55 A1 191.5

0.6 A1 194

0.65 A1 196.5

0.7 A1 199

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.

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(h)



A1 194

0.3 A2 174

0.35 A2 176.5

0.4 A2 179

0.45 A2 181.5

0.5 A2 184

0.55 A1 188.5

0.6 A1 194

0.65 A1 199.5

0.7 A1 205






(i)



A1 194

0 A2 180

0.05 A2 181.5

0.1 A2 183

0.15 A1 185

0.2 A1 188

0.25 A1 191

0.3 A1 194

0.35 A1 197

0.4 A1 200






(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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(f) Spider Graph

Tornado Diagram

MB363 Assignment 1 Report

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