Business Decision Making: Multiple Objective Analysis
VerifiedAdded on  2020/10/05
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Homework Assignment
AI Summary
This assignment solution delves into multiple objective decision-making, covering various problems and analytical techniques. It includes the application of minimax regret, Expected Value of Perfect Information (EVPI), and Expected Payoff Cost (EPC) analysis to determine optimal choices under uncertainty. The solution incorporates pairwise comparisons, priority vectors, and the Simplex Tableau method for linear programming. Graphical presentations are used to visualize constraints and solutions, and the assignment provides detailed calculations for each problem, offering a comprehensive approach to business decision-making, including cost analysis, reliability, and delivery considerations, and the final recommendation for Gill to buy from Picobuy based on cost and reliability scores.

Multiple Objective Decision
Making
Making
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TABLE OF CONTENTS
Problem 1.........................................................................................................................................1
Problem 2.........................................................................................................................................2
Problem 4.........................................................................................................................................4
A) Linear Programming.........................................................................................................4
B) Graphical presentation.......................................................................................................5
C) Total Hours........................................................................................................................5
Problem 5.........................................................................................................................................6
Problem 1.........................................................................................................................................1
Problem 2.........................................................................................................................................2
Problem 4.........................................................................................................................................4
A) Linear Programming.........................................................................................................4
B) Graphical presentation.......................................................................................................5
C) Total Hours........................................................................................................................5
Problem 5.........................................................................................................................................6

Problem 1
Demand Prob. Manufacture Profit Buy Profit Decision
Low 0.35 -20 -7 10 3.5 Buy
Medium 0.35 40 14 45 15.75 Buy
High 0.3 100 30 70 21 Make
(a): From the above calculation, it has been found that Kitchen Aid’s toaster can buy because
they are getting reliable profit of 3.5 in low or medium demand. While, they can make products
in case of high demand.
(b):
Minimax regret table
Alternative Low Medium High Maximum
Manufacturing -7 14 30 30
Buy 3.5 15.75 21 15.75
According to the above table, the one which is showing minimum output is taken into
consideration. Thus, 15.75 is the low as per the minimax regret table so, Kitchen aids can have
the option to buy the component.
(c):
Demand Prob. Manufacture EVPI
Low 0.35 -20 -7
Medium 0.35 40 14
High 0.3 100 30
Estimated cost of uncertainty (EMV) 37
Demand Prob. Manufacture
Low 0.65 -20 -13
Medium 0.65 40 26
High 0.7 100 70
Expected payoff cost(EPC) 83
EPC-EMV 46
Demand Prob. Buy EVPI
Low 0.35 10 3.5
Medium 0.35 45 15.75
High 0.3 70 21
40.25
Demand Prob. Buy
Low 0.65 10 6.5
1
Demand Prob. Manufacture Profit Buy Profit Decision
Low 0.35 -20 -7 10 3.5 Buy
Medium 0.35 40 14 45 15.75 Buy
High 0.3 100 30 70 21 Make
(a): From the above calculation, it has been found that Kitchen Aid’s toaster can buy because
they are getting reliable profit of 3.5 in low or medium demand. While, they can make products
in case of high demand.
(b):
Minimax regret table
Alternative Low Medium High Maximum
Manufacturing -7 14 30 30
Buy 3.5 15.75 21 15.75
According to the above table, the one which is showing minimum output is taken into
consideration. Thus, 15.75 is the low as per the minimax regret table so, Kitchen aids can have
the option to buy the component.
(c):
Demand Prob. Manufacture EVPI
Low 0.35 -20 -7
Medium 0.35 40 14
High 0.3 100 30
Estimated cost of uncertainty (EMV) 37
Demand Prob. Manufacture
Low 0.65 -20 -13
Medium 0.65 40 26
High 0.7 100 70
Expected payoff cost(EPC) 83
EPC-EMV 46
Demand Prob. Buy EVPI
Low 0.35 10 3.5
Medium 0.35 45 15.75
High 0.3 70 21
40.25
Demand Prob. Buy
Low 0.65 10 6.5
1
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Medium 0.65 45 29.25
High 0.7 70 49
84.75
EPC-
EMV 44.5
From the above calculation of EVPI, the difference between the EPC and EMV of
optimal expected profit is taken into account. this will be foregone due to the uncertainty and
equal to the EVPI. It is interesting to be taken into account that EVPI is equal to the EOL of the
optimal expected profit. Manufacturing of element is must not be so effective because it is
providing EVPI of 46. Henceforth, EVPI of market is 44.5 which is more appropriate for the
company.
(d):
Demand Prob. Manufacture Profit Buy Profit Decision
Low 0.35 -20 -7 10 3.5 Buy
Medium 0.35 40 14 45 15.75 Buy
According to the table, it seen that in case of low demand with the probability of 0.35
Kitchen aids can easily for buying the components from the market rather than manufacture it. In
case the probability goes ups then in those situation, expected profit would also increase but
there is not any effect on the decision as company can still have the option to buy the component.
Problem 2
(a): Priority vector for Cost
Cost
C B p
C 1 1/3 6
B 3 1 7
p 1/6 1/7 1
Cost
C B p
C 1 1/3 6
B 3 1 7
p 1/6 1/7 1
SUM 25/6 31/21 14
Cost
C B p
2
High 0.7 70 49
84.75
EPC-
EMV 44.5
From the above calculation of EVPI, the difference between the EPC and EMV of
optimal expected profit is taken into account. this will be foregone due to the uncertainty and
equal to the EVPI. It is interesting to be taken into account that EVPI is equal to the EOL of the
optimal expected profit. Manufacturing of element is must not be so effective because it is
providing EVPI of 46. Henceforth, EVPI of market is 44.5 which is more appropriate for the
company.
(d):
Demand Prob. Manufacture Profit Buy Profit Decision
Low 0.35 -20 -7 10 3.5 Buy
Medium 0.35 40 14 45 15.75 Buy
According to the table, it seen that in case of low demand with the probability of 0.35
Kitchen aids can easily for buying the components from the market rather than manufacture it. In
case the probability goes ups then in those situation, expected profit would also increase but
there is not any effect on the decision as company can still have the option to buy the component.
Problem 2
(a): Priority vector for Cost
Cost
C B p
C 1 1/3 6
B 3 1 7
p 1/6 1/7 1
Cost
C B p
C 1 1/3 6
B 3 1 7
p 1/6 1/7 1
SUM 25/6 31/21 14
Cost
C B p
2
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C 6/25 7/31 6/14
B 18/25 21/31 7/14
p 1/25 3/31 1/14
SUM 1 1 1
C 0.897
B 1.898
P 0.2074
(b) Priority vector for delivery
Delivery
C B p
C 1 8 1
B 1/8 1 1
p 1 8 1
Delivery
C B p
C 1 8 1
B 1/8 1 1
p 1 8 1
Sum 10/8 17 3
Delivery
C B p
C 8/10 17/8 3/1
B 1/10 17/1 3/1
p 8/10 17/8 3/1
Sum 1 1 1
C 1.7
B 21.25
P 9
(c): Priority vector for 3 case
Reliability
C B p
C 1 7 2
B 1/7 1 5
p 1/6 1/5 1
3
B 18/25 21/31 7/14
p 1/25 3/31 1/14
SUM 1 1 1
C 0.897
B 1.898
P 0.2074
(b) Priority vector for delivery
Delivery
C B p
C 1 8 1
B 1/8 1 1
p 1 8 1
Delivery
C B p
C 1 8 1
B 1/8 1 1
p 1 8 1
Sum 10/8 17 3
Delivery
C B p
C 8/10 17/8 3/1
B 1/10 17/1 3/1
p 8/10 17/8 3/1
Sum 1 1 1
C 1.7
B 21.25
P 9
(c): Priority vector for 3 case
Reliability
C B p
C 1 7 2
B 1/7 1 5
p 1/6 1/5 1
3

Cost
C B p
C 1 7 2
B 1/7 1 5
p 1/6 1/5 1
Sum 55/42 41/5 8
Cost
C B p
C 42/55 35/41 2/8
B 6/55 5/41 5/8
p 7/55 1/41 1/8
Sum 1 1 1
C 0.99
B 0.99
P 0
(d):
Criteria Weight C B P
Cost 261.8 89 144 29
Reliability 0.92 0.6 0.2 0.2
Delivery 227.05 15 149 63
Score 489.77 105 293 92
From the above table, it is suggested that gill would need to buy from Picobuy (P)
because reliability is high and cost is low.
Problem 4
A) Linear Programming
ST Total Limit
Molding 1 1.5 15 <= 900
Painting 0.5 0.33 4.14 <= 300
Packaging 0.125 0.25 2.375 >= 100
Cell Name
Original
Value Final Value
$B$4 Max Z= Profit 0 79
4
C B p
C 1 7 2
B 1/7 1 5
p 1/6 1/5 1
Sum 55/42 41/5 8
Cost
C B p
C 42/55 35/41 2/8
B 6/55 5/41 5/8
p 7/55 1/41 1/8
Sum 1 1 1
C 0.99
B 0.99
P 0
(d):
Criteria Weight C B P
Cost 261.8 89 144 29
Reliability 0.92 0.6 0.2 0.2
Delivery 227.05 15 149 63
Score 489.77 105 293 92
From the above table, it is suggested that gill would need to buy from Picobuy (P)
because reliability is high and cost is low.
Problem 4
A) Linear Programming
ST Total Limit
Molding 1 1.5 15 <= 900
Painting 0.5 0.33 4.14 <= 300
Packaging 0.125 0.25 2.375 >= 100
Cell Name
Original
Value Final Value
$B$4 Max Z= Profit 0 79
4
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Cell Name
Original
Value Final Value Integer
$B$8 Values X1 0 3 Contin
$C$8 Values X2 0 8 Contin
Cell Name Cell Value Formula Status Slack
$D$11
Constraint 1
Total 38 $D$11<=$F$11
Not
Binding 2
$D$12
Constraint 2
Total 42 $D$12<=$F$12 Binding 0
$D$13
Constraint 3
Total 3 $D$13>=$F$13 Binding 0
B) Graphical presentation
C) Total Hours
ST Hours
Moulding 2250
Painting 249
Packaging 37.5
Problem 5
Simplex Tableau Method
5
Original
Value Final Value Integer
$B$8 Values X1 0 3 Contin
$C$8 Values X2 0 8 Contin
Cell Name Cell Value Formula Status Slack
$D$11
Constraint 1
Total 38 $D$11<=$F$11
Not
Binding 2
$D$12
Constraint 2
Total 42 $D$12<=$F$12 Binding 0
$D$13
Constraint 3
Total 3 $D$13>=$F$13 Binding 0
B) Graphical presentation
C) Total Hours
ST Hours
Moulding 2250
Painting 249
Packaging 37.5
Problem 5
Simplex Tableau Method
5
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x1 x2 x3 s1 s2 z b
1 3 12 1 0 0 2800
1 2 1 0 1 0 6000
-5 -5 -24 0 0 1 1200
6
1 3 12 1 0 0 2800
1 2 1 0 1 0 6000
-5 -5 -24 0 0 1 1200
6
1 out of 8
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