Operational Management: Decision Making Function

Question:
Describe about the Report for Operational Management of Decision Making Function.

 
Answer:

Given, d1, d2, d3 and d4 are the four decision alternatives and s1, s2, s3 and s4 are the four state of nature. The profit payoff table for it is given below:

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Decision Alternative

S1

S2

S3

S4

D1

14

9

10

5

D2

11

10

8

7

D3

9

10

10

11

D4

8

10

11

13

If the decision maker has no idea about the probabilities of the four states of nature, then using optimistic approach, the decision maker will choose the alternative which will provide maximum payoff for each decision alternative. Thus in this case, for state S1 the decision maker will choose decision alternative D1. For state S2 the decision maker will choose any of the decision alternative D2, D3 or D4. For state S3 the decision maker will choose the decision alternative D4 and For state S4 the decision maker will choose the decision alternative D4.

For conservative approach, the decision maker will choose the alternative which will provide minimum payoff for each decision alternative. Thus in this case, for state S1 the decision maker will choose decision alternative D4. For state S2 the decision maker will choose the decision alternative D1. For state S3 the decision maker will choose the decision alternative D2 and for state S4 the decision maker will choose the decision alternative D1. (Kaplan Financial, 2012)

For minimax regret approach, the decision maker will try to minimize the maximum regret. The regret is defined as the opportunity loss due to the decision taken. It is calculated by |Vi –Vij|, Vij is the value for decision i and state j and Vi is the best alternative for state i 

Thus the regret table is calculated below

Regret

S1

S2

S3

S4

D1

0

1

1

8

D2

3

0

3

6

D3

5

0

1

2

D4

6

0

0

0

The decision maker will choose the alternative which will provide minimum regret for each decision alternative.

Thus in this case, for state S1 the maximum regret is 6, for state S2 the maximum regret is 1, for state S3 the maximum regret is 3, for state S4 the maximum regret is 8.

Thus the decision maker will chose decision alternative D2 to minimize the maximum regret. (Kaplan Financial, 2012)

b) I think conservative approach is better as it helps us to analyse the projects by providing the minimum return that the project will garner and help us in estimating if we should go ahead with the project. In other approaches, there is a possibility of expecting a higher return which might not be the case and result in loss.

Establishing the most appropriate approach helps the decision maker in choosing the projects and it helps the decision maker to understand whether the return from the projects are over estimated or underestimated and based on the approach he can plan ahead the future decisions regarding the project. Thus it is important to establish the most appropriate approach.

c) If the table provides the cost instead of profits then using optimistic approach, the decision maker will choose the alternative which will provide minimum cost for each decision alternative. Thus in this case, for state S1 the decision maker will choose decision alternative D4. For state S2 the decision maker will choose the decision alternative D1. For state S3 the decision maker will choose the decision alternative D2 and For state S4 the decision maker will choose the decision alternative D1.

For conservative approach, the decision maker will choose the alternative which will provide maximum cost for each decision alternative. Thus in this case, for state S1 the decision maker will choose decision alternative D1. For state S2 the decision maker will choose any of the the decision alternative D2, D3 or D4. For state S3 the decision maker will choose the decision alternative D4 and for state S4 the decision maker will choose the decision alternative D4. (Kaplan Financial, 2012)

For minimax regret approach, the regret table is given below

Regret

S1

S2

S3

S4

D1

6

0

2

0

D2

3

1

0

2

D3

1

1

2

6

D4

0

1

3

8

The decision maker will choose the alternative which will provide minimum regret for each decision alternative.

Thus in this case, for state S1 the maximum regret is 6, for state S2 the maximum regret is 1, for state S3 the maximum regret is 3, for state S4 the maximum regret is 8.

Thus the decision maker will chose decision alternative D2 to minimize the maximum regret.

The profit pay off table for decision alternatives d1 and d2 is given. The state of nature are s1 and s2.

Using graphical approach, the payoff for the two states are plotted. In x axis, the probability is given. For probability = 0, the pay off of state s1 is used and for probability = 1, the the pay off of state s2 is used. (Scgroeder, 2007)

The two decision alternatives intersect at a point P. The P can we found by solving the equations y = 10 – 9* x and y = 4 – 1* x.

Thus solving we get x = 0.75 and y = 3.25 thus for the range of probability 0 to 0.75 for the state s1, the decision alternatives will have highest expected value.

Given P(s1) = 0.2 and P(s2) = 0.8. Thus the expected value of the project is given by

Expected value = Probability * Value

Thus for decision alternative d1, the Expected value = 0.2* 10 + 0.8* 1 = 2.8

Thus for decision alternative d2, the Expected value = 0.2* 4 + 0.8* 3 = 2

Thus using the expected value approach, the decision alternative d1 is better.

The solution found in part b will be optimal if the expected value of decision 1 is greater than expected value of decision 2. Let x1 be the value of the payoff of the state 1 and x2 be the value of the payoff of the state 2. Thus for the probability of the state s1 and s2 as 0.2 and 0.8 respectively, the expected value is given by

0.2* x1 +0.8* x2 which must be greater than expected value of decision 2 .i.e. = 2

For x2 = 1, 0.2* x1 + 0.8 * 1 > 2

X1 > 6

For X1 = 10, 0.2* 10 + 0.8* x2 > 2

 X2 > 0

Thus the range of payoff for s1 state is more than 6 and range of payoff for s2 state is more than 0

The probability associated to the payoff of s2 is 0.8 while the probability associated to payoff of s1 is 0.2. Thus the solution is more sensitive to the payoff of s2.

The company Myrtle Air express is offering service between Cleveland and Myrtle Beach. The demand of the service and the price of the service are given with options of full price and discount price.

The management has to decide what pricing policy they should follow as the demand can be strong and weak. If the management chooses full price then the profits earned during the strong demand will be high but in case of weak demand the company will have loss. While if the management chooses discount price then the profits earned during the weak demand will be lower but in case of weak demand the company will have not have any loss. There are two decision alternatives full prices and discount and two possible outcomes strong demand or weak demand.

Using the optimistic approach, the decision maker will choose the alternative which will provide maximum payoff for each decision alternative. The maximum pay off is obtained when the demand is strong and management uses full price. Thus the company will choose full price

Using the conservative approach, the decision maker will choose the alternative which will provide minimum payoff for each decision alternative. The minimum pay off is obtained when the demand is weak and management uses discount price. Thus the company will choose discount price.

 
The regret table is given below

Regret

Strong

Weak

Full price

0

1180

Discount

0

350

Thus the maximum regret for full price is 1180 and the maximum regret for discount is 350. Thus using Minimax regret approach the company will minimize the maximum regret. Thus company will choose discount price.

Given P (strong) = 0.7 and P(weak) = 0.3. Thus the expected value of the project is given by Expected value = Probability * Value

Thus for decision alternative full price, the Expected value = 0.7* 960 + 0.3* (- 490) = 336

Thus for decision alternative full price, the Expected value = 0.7* 960 + 0.3* (- 490) = 565

Thus the company should use discount price as the expected value is higher in that case.

Given P (strong) = 0.8 and P(weak) = 0.2. Thus the expected value of the project is given by Expected value = Probability * Value

Thus for decision alternative full price, the Expected value = 0.8* 960 + 0.2* (- 490) = 454

Thus for decision alternative full price, the Expected value = 0.8* 960 + 0.2* (- 490) = 600

Thus the company should use discount price as the expected value is higher in that case.

Using the graphical approach, the payoff for the two decisions variables are plotted. In x axis, the probability is given. For probability = 0, the pay off of strong demand is used and for probability = 1, the the pay off of weak demand is used. (Scgroeder, 2007)

The point till which the line full price is above discount is range where Full price has higher expected value and the range where the line full price is below discount is range where Full price has lower expected value.

Solving the equation of line y = 960 -1450*x and y = 670 -350*x we get,

X = 0.26  

Thus in this case for probability between 0 and 0.26, the full price has higher expected value and for probability 0.26 to 1, discount has higher expected value.

Given, d1 and d2 are the two decision alternatives and s1, s2, s3 are the three states of nature. The probability of the three states are given as P(s1) = 0.65, P(s2) = 0.15, P(s3) = 0.20

  1. If perfect information is available then the company will choose alternative d1 for state s1, alternative d1 or d2 for state s2 and alternative d2 for state s3 as the return is higher for these alternatives.
  2. Thus the expected value is given by Expected value EVwPI = Probability * Value

EVwPI = 0.65* 250 + 0.15* 100 + 0.2* 75 = 192.50

If perfect information is not available, then expected value approach, the expected value is given by Expected value = Probability * Value

Expected value for d1 = 0.65* 250 + 0.15* 100 + 0.2* 25 = 182.50

Expected value for d1 = 0.65* 100 + 0.15* 100 + 0.2* 75 = 85

Thus Expected value without perfect information is given by EVwoPI = 182.50

The Expected value for perfect information is the additional price the company is willing to pay to achieve the information. It is the difference between the expected value with perfect information and expected value without perfect information. It is given by EVPI = EVwPI – EvwoPI = 192.5 – 182.5 = 10 (Boston University Metropolitan College., (2013))

The expected value of the project is given by Expected value = Probability * Value

Thus for decision alternative small, the Expected value = 0.1* 400 + 0.6* (500) + 0.3* 660 = 538

Thus for decision alternative medium, the Expected value = 0.1* (- 250) + 0.6* (650) + 0.3* 800 = 605

Thus for decision alternative large, the Expected value = 0.1* (- 400) + 0.6* (580) + 0.3* 990 = 605 thousand dollars

Thus using the expected value approach Lake Placid should choose medium or large size.

The risk is given by [∑ Probability* ( Payoff value – Expected value)] 5

For medium

Probability

Value

(value – Expected)^2

P*(value – Expected)^2

0.1

-250

731025

73102.5

0.6

650

2025

1215

0.3

800

38025

11407.5

Expected

605

Total

85725

Thus risk = 85725^0.5 = 292.78

For large

Probability

Value

(value – Expected)^2

P*(value – Expected)^2

0.1

-400

1010025

101002.5

0.6

580

625

375

0.3

990

148225

44467.5

Expected

605

Total

145845

Thus risk = 145845^ 0.5 = 381.89

Thus the risk in the large centre is higher. Thus the Lake Placid should choose medium size.

Expected value with perfect information = 0.1* 400 + 0.6* 650 + 0.3* 990 = 727

Expected value without perfect information = 605

Thus Expected value of perfect information = Expected value with perfect information – Expected value without perfect information = 727 – 605 = 122 thousand dollars

If the probability of worst case increase to 0.2 and the probability of base case decrease to 0.5 then, using the expected value approach

Thus for decision alternative small, the Expected value = 0.2* 400 + 0.5* (500) + 0.3* 660 = 528 thousand dollars

Thus for decision alternative medium, the Expected value = 0.2* (- 250) + 0.5* (650) + 0.3* 800 = 515

Thus for decision alternative large, the Expected value = 0.2* (- 400) + 0.5* (580) + 0.3* 990 = 507

Thus using the expected value approach Lake Placid should choose small size.

If the probability of worst case decreases to 0 and the probability of best case increase to 0.4 then, using the expected value approach

Thus for decision alternative small, the Expected value = 0* 400 + 0.5* (500) + 0.4* 660 = 514 thousand dollars

Thus for decision alternative medium, the Expected value = 0* (- 250) + 0.5* (650) + 0.4* 800 = 645

Thus for decision alternative large, the Expected value = 0* (- 400) + 0.5* (580) + 0.4* 990 = 686

Thus the expected value of the project increases from 605 thousand dollars to 686 thousand dollars. Thus profit = 686 -605 = 81 thousand dollars whereas investment made is 150 thousand dollars. Thus this will not be an good investment. (Anderson, 2012)

 
References

Scgroeder, R. (2007). Operations Management: Decision Making in the Operations Function. Mcgraw Hill.

Boston University Metropolitan College. (2013). Expected Value of Perfect Information. Retrieved on August 16, 2016 from https://onlinecampus.bu.edu/bbcswebdav/pid-843933-dt-content-rid-2221759_1/courses/13sprgmetad715_ol/module_03a/metad715_m03l01t05_expected_value_perfect_information.html

Kaplan Financial. (2012). Maximax, maximin and minimax regret. Retrieved on August 16, 2016 from https://kfknowledgebank.kaplan.co.uk/KFKB/Wiki%20Pages/Maximax,%20maximin%20and%20minimax%20regret.aspx

Anderson, D. (2012). Quantitative Methods for Business. Cengage Learning.

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