Question

4. Suppose we are selling lemonade during a football game. The lemonade sells for $18 per gallon butonly costs $3 per gallon to make. If we run out of lemonade during the game, it will be impossibleto get more. On the other hand, leftover lemonade has a value of $1. Assume that we believe thefans would buy 10 gallons with probability 1/10, 11 gallons with probability 2/10, 12 gallons withprobability 4/10, 13 gallons with probability 2/10, and 14 gallons with probability 1/10.(a) What is the mean demand?(b) If 11 gallons are prepared, what is the expected profit?(c) What is the best amount of lemonade to order before the game?(d) In what sense is that amount optimal? In other words, what was your objective function?(e) Instead, suppose that the demand was normally distributed with mean 100 gallons and variance100 gallons2. How much lemonade should be ordered? (Despite the demand follows a Normaldistribution can take negative values, the probability of it is extremely small with our assumptionsof the parameters and can be ignored.)

Answer 1

The given information and expected gallons are given below (Expected gallons=probability *gallons of lemonda)

Gallons of lemonade	Probability	Expected Gallons
10	0.1	1
11	0.2	2.2
12	0.4	4.8
13	0.2	2.6
14	0.1	1.4
Total	1	12

A. Mean demand= 10*.1+....+14*.1=12

B. If 11 gallons is prepared

Profit=11*18-11*3

Profit=198-33

Profit=165.

C. The optimal quantity is 12 gallons as that would maximize our profit as there will be no leftovers.

Profit= 12*18-12*3=216-36=180.

D. It is optimal cause it maximizes profit. In other words, it optimizes the following function

Maximize (18Q_d+(Q_s-Q_d)*1-3Q_s)

Where Q_s=Quantity supplied

Q_d=Quantity demanded