Question

1. Suppose we are selling hot dogs during a basketball game. A hot dog sells for $15 each but only costs $5 each to make. If we run out of hot dogs during the game, it will be impossible to get more. On the other hand, a leftover hot dog has a value of $1 for each hot dog. Assume that we believe the fans would buy 100 hot dogs with probability 3/10, 110 hot dogs with probability 2/10, 120 hot dogs with probability 2/10, 130 hot dogs with probability 1/10, and 140 hot dogs with probability 2/10. (a) Find the expected demand. (b) If 120 hot dogs are prepared, what is the expected profit? (c) What is the corresponding expected (potential) cost?

Answer 1

the probability distribution of the demand is

Demand (x)	100	110	120	130	140
Probability $P(x_i)$	0.3	0.2	0.2	0.1	0.2

a) the expected demand is $E[X]=\sum x_iP(x_i)=100(0.3)+110(0.2)+120(0.2)+130(0.1)+140(0.2)=117$

b)A Sold hot dog will earn a profit of $15-$5=$10

An unsold hotdog will be a loss of $5-$1=$4

If the hotdogs are finished and demand is more that too will be a potential loss and it will be a loss of a potential profit of $10

and If 120 hot dogs are prepared than

the profit will also have probability distribution according to the distribution of demand

If the demand is 100, than 20 hot dogs will be left and the profit is $100*$10 -20*$4 =$920$ dollars

This occurs with probability 0.3. Similarly

If the demand is 110, than 10 hot dogs will be left and the profit is $110*10 -10*$4 =$1060$ dollars

This occurs with probability 0.2.

Thus the expected profit is $920*0.3+1060*0.2+1200*0.2+1200*0.1+1200*0.2=1088$ dollars

c) Expected potential cost = Expected demand * the cost =117*$5=$585