the following matrix gives the expected profit in thousands

Styles Page 9 of 10 15. The following decision evaluation matrix gives the expected savings in maintenance costs of dollars) for three policies of preventive maintenance and three levels of (in thousands operation of equipment. Given the probabilities of each level of operation, P, = 0.3, P2-0.25, and Ps = 0.45, determine the best policy based on the most probable future criterion. Policy Level of Operation Li 10 30 M2 26 40 30 Also, determine the best policy under uncertainty, using the Laplaco rule. the Maximax rule, and the Hurwicz rule with 0.2

Solution

Question 14:

Maximin Rule (Wald Rule) -
Here the decision make chooses the minimum(worst case) payoff for each alternative and then choose the alternative that provides the best (maximum) of the worst case scenario (minimum).

Here, worst case scenario of profits for each marketing strategy are,
M1: $10,000 (worst among 10, 20, 30, 40, 50)
M2: $20,000 (worst among 20, 25, 25, 30, 35)
M3: $ 5,000 (worst among 50, 40, 5, 15, 20)
M4: $25,000 (worst among 40, 35, 30, 25, 25)
M5: $10,000 (worst among 10, 20, 25, 30, 20)
The best (Max) strategy among above worst case scenario is $25,000, which is M4.
Thus the Maximin strategy among above alternatives is M4.

Maximax Rule:
Here the decision maker is extremely optimistic, who chooses the best (Max) among the best case scenario for each strategy. Here best case scenario of profits for each marketing strategy are,
M1: $50,000 (best among 10, 20, 30, 40, 50)
M2: $35,000 (best among 20, 25, 25, 30, 35)
M3: $50,000 (best among 50, 40, 5, 15, 20)
M4: $40,000 (best among 40, 35, 30, 25, 25)
M5: $30,000 (best among 10, 20, 25, 30, 20)
The best (Max) strategy among above best case scenario is $50,000, which is M1 or M3.
Since there is a tie, let us compare the second best profit scenario for each strategy. For them both, it is $40,000. Continuing with the tie-breaker, the third best scenario for M1 is $30,000 and for M3 it is $20,000. So we choose M1.
Thus the Maximax strategy among above alternatives is M1.

Hurwicz Rule:
Here we utilize the alpha index to find a weightage average of best and worst scenario for each strategy and then choose the strategy with the best weightage average.
Here = 0.4; (1- ) = 0.6.
Weighted averages for each strategy are,
M1 (best 50, worst 10): 50(0.4) + 10(0.6) = 26
M2 (best 35, worst 20): 35(0.4) + 20(0.6) = 26
M3 (best 50, worst 5): 50(0.4) + 5(0.6) = 23
M4( best 40, worst 25): 40(0.4) + 25(0.6) = 31
M5( best 30, worst 10): 30(0.4) + 10(0.6) = 18

Thus the best marketing strategy as per Hurwicz rule (with =0.4) is M4.

Question 15:

Making decision under Known Probability/ Risk (P1 = 0.3, P2 = 0.25, P3 = 0.45):
EMV (expect monetary value = expected maintenance cost savings) for each maintenance policy:
M1: 10x0.3 + 20x0.25 + 30x0.45 = 21.5
M2: 22x0.3 + 26x0.25 + 26x0.45 = 24.8
M3: 40x0.3 + 30x0.25 + 15x0.45 = 26.25

Thus the best policy is M3.

Laplace Rule:
Here they assume all scenarios to have equal probability. And compare weighted averages for each policy.
M1: (10+20+30)/3 = 20
M2: (22+26+26)/3 = 24.67
M3: (40+30+15)/3 = 28.33
Best policy is M3.

Maximax Rule:
M1: $30,000 (best of 10, 20, 30 K$)
M2: $26,000 (best of 22, 26, 26 K$)
M3: $40,000 (best of 40, 30, 15 K$)
The M3 is the best Maximax policy.

Hurwicz Rule ( = 0.2):
M1: 30(0.2) + 10(0.8) = 23
M2: 26(0.2) + 22(0.8) = 22.8
M3: 40(0.2) + 15(0.8) = 20
The best policy as per Hurwicz rule ( =0.2) is M1.