The operations manager for a local bus company wants to decide whether he should purchase a small, medium, or large new bus for his company. He estimates that the annual profits (in $000) will vary depending upon whether passenger demand is low, moderate, or high, as follows:

(a) If he uses the maximin criterion, which size bus will he decide to purchase?
(b) If he uses the minimax regret criterion, which size bus will he decide to purchase?
(c) If he feels the chances of low, moderate, and high demand are 30%, 30%, and 40% respectively, which size bus will he decide to purchase and what is the expected annual profit for this option?

Bus size Demand
Low Medium High
Small 50 60 70
Medium 40 80 90
Large 20 50 120

a) the small bus

b) the medium or the larger bus

c) the small bus

Explanation:

a) under the maximin criterion the manager will want to maximise the minimum possible profit (pessimistic criterion) , thus the profits will be

Bus size Demand

Low Medium High

Small 50 60 70  → min profit = 50

Medium 40 80 90  → min profit = 40

Large 20 50 120 → min profit = 20

therefore the option with maximum minimum profit is the small size

b) under the maximin criterion the manager will want to minimise the maximum possible loss , thus the losses respect with the best profit option will be

Bus size Demand

Low Medium High  

Small 0 -20 -50  → max loss= -50

Medium -10 0 -30  → max loss = -30

Large -30 -30 0 → max loss = -30

therefore the option with minimum maximum loss is the medium or the large one

c) for the expected value of each option

small = 30/100*50 +30/100*60 +40/100*70 = 61

medium = 30/100*40 +30/100*80 +40/100*90 = 72

large = 30/100*20 +30/100*50 +40/100*120 = 69

then the small bus option is the better one, since it has the maximum expected profit

idk sorry

explanation: