Travel-Cheap agency is planning to open a ticket desk in a new shopping plaza, staffed by one ticket agent. It is estimated that requests for tickets and information will average 15 per hour, and requests will have a Poisson distribution. Service time is assumed to be exponentially distributed. Previous experience with similar satellite operations suggests that mean service time should average about three minutes per request. Determine each of the following:

a. System utilization
b. Percent of time the server (agent) will be idle
c. The expected number of customers waiting to be served
d. The average time customers will spend in the system
e. The probability of zero customers in the system
f. The probability of four customers in the system

(a) 75% (b) 25% (c) 2.25 customers (d) 12 minutes (e) 0.25 (f)0.237

Explanation:

Solution

Given that:

The Arrival rate at Poisson distribution = 15 per hour = λ

The Service rate at exponential distribution = 20 per hour = μ

(a) System utilization = λ/μ = 15 / 20 = 0.75 = 75%

(b) The Probability of zero requests in server = 1 - λ/μ = 1 - 0.75 = 0.25

or

The percentage of time server will be idle = 25%

(c)The expected number of customers waiting to be served = Average number of customers in line = λ^2/μ (μ-λ ) = 225 /20(20-15) = 45 /20 = 2.25

Therefore, it is expected that on an average, 2.25 customers are waiting in line to get served.

(d)The average time customers will spend in system = 1/(μ-λ )

=1/(20 - 15) = 1/5 hours = 12 minutes

(e)The probability of zero customer in system = 1 -λ/μ = 1 - 0.75 = 0.25

(f) The probability of more than 4 customers in the system = (λ/μ)^ 4+1

=  (15/20)5 = 0.237

statement of owner"s equity;

d, rjskdndjsndbdnsjdbxbc