Question

Students arrive at the Administrative Services Office at an average of one every 24 minutes, and their requests take on average 20 minutes to be processed. The service counter is staffed by only one clerk, Judy Gumshoes, who works eight hours per day. Assume Poisson arrivals and exponential service times.

a. What percentage of time is Judy idle? (Round your answer to 1 decimal place.)

Percentage of time --------------%

b. How much time, on average, does a student spend waiting in line? (Do not round intermediate calculations. Round your answer to 1 decimal place.)

Average time ------------minutes

c. How long is the (waiting) line on average? (Round your answer to 2 decimal places.)

Average length of the waiting line ----------------customers

d. What is the probability that an arriving student (just before entering the Administrative Services Office) will find at least one other student waiting in line? (Do not round intermediate calculations. Round your answer to 2 decimal places.)

Probability ------------%

Answer 1

a)

Percentage of time is Judy idle= ( 1 - T / A )

T - Time taken to process a request

A - Time after which next student arrive

T = 20

A = 24

Percentage of time is Judy idle= ( 1 - 20/24 ) = 16.667%

b)

Lets say T/A = P

Time a student spend waiting in line = P / ( 1 - P ) * (Ci^2 + Cp^2) / 2 * T

Ci = 1

Cp = 1

Time a student spend waiting in line = 0.8333 / ( 1 - 0.8333 ) * (1^2 + 1^2) / 2 * 20

Time a student spend waiting in line = 5 * 20 = 100 minutes

c)

Length of waiting line on average = Time a student spend waiting in line / A

Length of waiting line on average = 100 / 24 = 4.1666 students

d)

At least one other student waiting in line means that there will be atleast 2 students are there in the system. So

Probability = ( T / A )^2 = (20/24)^2 = 0.69444