Question

Indiana Bell customer service representatives receive an average of 1700 calls per hour. The time between calls follows an exponential distribution. A customer service representative can now handle an average of 35 calls per hour. The time required to handle a call is also exponentially distributed. Indiana Bell can put up to 25 people on hold. If 25 are on hold, a call is lost to the system. Indiana Bell hired 5 extra service representatives and now has 80 service representatives.

1)What fraction of the time are all operators busy?

2)   What fraction of all calls is lost to the system?

Answer 1

Firstly we need to compute the compute the steady-state probabilities for this birth–death process.

Let state i at any time equal the number of callers whose calls are being processed or are on hold.

Now i = 0, 1, 2, . . . ,104,105  

Also =1,700. The fact that any calls received when 80 +25 = 105 calls are in the system are lost to the system implies that =0. Then no state i > 105 can occur.

Also U₀ = 0

We have :

= 35i for i = 1,2,3,.......,80    

= 35 * (80) = 2800 for i > 80

Now to calculate the steady state probabilities = Fraction of time the state is i

Here C_j is calculated as

C₁ =

C₂ = C_1*

and so on

1)What fraction of the time are all operators busy?

We have to find

= 0.00002329

2)   What fraction of all calls is lost to the system?

An arriving call is turned away if the state equals 105. A fraction

=.0000000000349712

of all arrivals will be turned away. Thus, the phone company is good at its service.