Respuesta :
Answer:
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
If he uses the maximin criterion, the size of the bus which he will decide to purchase is the small bus.
If he uses the minimax regret criterion, the size of the bus which he will decide to purchase would be the medium or the larger bus
The size of the bus which he will decide to purchase is the small bus if he feels the chances of low, moderate, and high demand are 30%, 30%, and 40% respectively,
Calculations and Parameters:
Using the maximin criterion, the profits would 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
Hence, the option with maximum-minimum profit is the small size bus.
Using the maximin criterion, to minimize the maximum possible loss, thus the losses respect with to the best profit option will be the medium or the larger bus.
Given that
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
Hence, the option with minimum-maximum loss is the medium or the large one
To find 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
Therefore, the small bus option is the best one, since it has the maximum expected profit.
What is Profit?
This refers to the gain made on the sale of a product in relation to its buying and selling price.
Read more about profit here:
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