Question

A customer service center has customers call with questions about the use of a product. The mean arrival rate of customer calls is 20 per hour and follows a Poisson distribution. The mean service rate (how long it takes the call center employee to answer the customer’s question) is 30 customers per hour (most customer questions can be answered rather quickly) and follows an Exponential distribution.

What is the mean (average) time in hours that a customer spends in the system (waiting on hold, then having their question answered)?

What is the mean utilization percentage of the call center employee answering the phones (the server in this system)?

What is the mean number of customers that are on hold, waiting for their call to be answered?

d.   If the company decides to hire a second person to answer calls (there are now two “servers” in this system), the waiting time should be faster. On average, how long will a customer now wait on hold for their question to be answered?

Answer 1

We can model this situation as an M/M/1 queuing system where,
L = average arrival rate = 20 per hour
M = average service rate = 30 per hour

(a)

Ws = average time a customer spends in the system = 1 / (M - L) = 1/(30-20) = 0.10 hours

(b)

Utilization (U) = L/M = 20/30 = 0.67

(c)

Lq = average number of customers on hold = L² / (M*(M - L)) = 400/(30*(30-20)) = 1.33 callers

(d)

This will now be an M/M/2 system
S = no. of servers = 2 which is > U=0.67
L=20; M=30
P₀ = Probability of empty system is calculated by the following formula -

and Lq = average number of customers on hold is given by -

Implementing this formulation, we will get L_q = 0.0833 callers