The Truck Company needs some help in buying supplies for its truck which is park

Question

The Truck Company needs some help in buying supplies for its truck which is parked outside of MAC D every day - especially bread. From its experience, it needs 500 loaves of fresh bread on an average day to satisfy demand. The standard deviation of the number of loaves it needs is 200 loaves. Truck Company must pay $.50 to the bakery for each loaf it buys. They estimate that when loaves are turned into sandwiches and sold, the revenue attributed to each loaf is $1.50. Any bread unsold at the end of a day is worthless. Assume that all probability distributions are normal.

A.) How many loaves should they buy each day to maximize expected profit?

B.) Instead of throwing away the loaves at the end of the day, Truck Company realized they can be sold at a discounted price of $0.25 per loaf at store closing. What quantity should they now order to maximize profit? (Round to the nearest integer)

C) Continuing from Part B, if now the demand distribution is a Poisson distribution with ?=400, what is the optimal order quantity to maximize profit? (Round to the nearest integer)

Explanation / Answer

Here the mean demand = 500

standard deviation of demand = 200

Here,

Buying price = $ 0.50

Here revenue or selling price of each loaf = $ 1.50

so here to calculate the amount of loaves per day that will maximize expected profit is Q0

F(Q0) = (1.50 - 0.50)/ (1.50 - 0) = 1/1.50 = 2/3

so here the cumulative distribution at Q0 would be 2/3

so, the z value for the given cumulative p - value is

Z = 0.43073 = (Q0 - 500)/200

Q0 = 500 + 200 * 0.43073 = 586.15 loaves

(B) Here the discounted price = $ 0.25 per loaves

so here to calculate the amount of loaves per day that will maximize expected profit is Q0

F(Q0) = (1.50 - 0.50)/ (1.50 - 0.25) = 0.8

so here the cumulative distribution at Q0 would be 0.8

so, the z value for the given cumulative p - value is

Z = 0.842 = (Q0 - 500)/200

Q0 = 500 + 200 * 0.8416 =668.32 loaves

(C) Here if the distribution is poisson with paameter = 400

so here also F(Q0) = 0.8

so here, as per poisson distribution to be approximated

so,

(Q0 -400)/20 = 0.842

Q0 = 400 + 20 * 0.842 = 416.84

Here Q0 = 417

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