Question

Six months before its annual convention, the American Medical Association must determine how many rooms to reserve. At this time, the AMA can reserve rooms at a cost of $50 per room. All rooms that are reserved must be paid for even if they are not used. The AMA believes the number of doctors attending the convention will be normally distributed with a mean of 5000 and a standard deviation of 1000. If the number of people attending the convention exceeds the number of rooms reserved, extra rooms must be acquired at a cost of $80 per room. Use a simulation model to determine the expected cost of the rooms for the convention for the following reservation levels: 2000, 3000, 4000, 5000, 6000, 7000, and 8000. From the above list of reservation levels, select the optimal number of rooms to reserve. Use a simulation with 2000 replicates and answer the questions below:

a) What is the optimal number of rooms to reserve?  (Click to select)  2000  3000  4000  5000  6000  7000  8000

(Round the following answers to two decimals)

b) What is the expected minimum cost? $

b) Construct a 95% confidence interval for the expected cost for the optimal number of rooms to reserve:

Answer 1

Let X be the number of doctors attending the convention. X has a normal distribution with mean =5000 and standard deviation =1000

If Q is the number of rooms reserved,

The cost of the rooms is

50Q+(X-min(X,Q))*80

We will use the excel function =NORM.INV(RAND(),5000,1000) to simulate from normal distribution

We will conduct the simulation in excel as below

Set the sheet as below

columns A:D

The rest

replicate the rows to make 2000 trials

Paste the random numbers as values to avoid the changes

get the following

We get the expected cost and standard deviations

Get this

We can see that the lowest cost is when the number of rooms reserved is 5000

a) ans: the optimal number of rooms to reserve is 5000

b) The expected minimum cost is $282,509.46

c) The expected cost is and the sample standard deviation of the cost is s=47625.5646

We will estimate the population standard deviation of cost using the sample.

The standard error of the expected cost is

where n=2000 is the number of replicates.

Finally 95% confidence interval gives the level of significance as . The critical value of z is

this is same as

Using excel function =NORM.INV(0.975,0,1) or from standard normal table we get the critical value of z as 1.96

Finally the 95% confidence interval is

ans: 95% confidence interval for the expected cost for the optimal number of rooms to reserve: is [$280,422.18,$284,596.74]