ITPEC FE Morning October 2022 Question 79
When the estimated probabilities for the number of units that are sold per day for a product are as listed in the table below and the profit per unit is 10 dollars, what is the number of purchased units where the expected value for profit becomes a maximum? Here, a disposal loss of 3 dollars per unit is incurred if the product remains unsold on the day it was purchased.
Explanation
6 units, with the highest expected profit of 48.3 dollars — for each purchase quantity, weight the profit of every sales outcome by its probability.
The profit of an outcome is 10 dollars per unit sold minus 3 dollars per unit left unsold. Using the probabilities in each purchase row:
- •Buy 4: 1.0 × 40 = 40.0
- •Buy 5: 0.3 × (40 − 3) + 0.7 × 50 = 11.1 + 35 = 46.1
- •Buy 6: 0.3 × (40 − 6) + 0.3 × (50 − 3) + 0.4 × 60 = 10.2 + 14.1 + 24 = 48.3
- •Buy 7: 0.3 × (40 − 9) + 0.3 × (50 − 6) + 0.3 × (60 − 3) + 0.1 × 70 = 9.3 + 13.2 + 17.1 + 7 = 46.6
Answer (c)
Why not others:
- •(a) 4 — a certain 40 dollars, the lowest expected profit; it never earns from a fifth unit, which sells 70% of the time when 5 are bought
- •(b) 5 — 46.1 dollars, below 48.3
- •(d) 7 — 46.6 dollars: the seventh unit sells only 10% of the time, while in every other outcome more units are discarded at 3 dollars each
Key rule: Compute Σ probability × payoff for each option, with the loss on unsold stock inside each payoff, and pick the largest; neither the safest nor the largest quantity is automatically best.
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