Question

A queueing system serves two types of customers. Type 1 customers arrive according to a Poisson process with a mean rate of 5 per hour. Type 2 customers arrive according to a Poisson process at a mean rate of 3 per hour. The system has two servers, both of which serve both types of customer. All service times have exponential distribution with a mean of 10 minutes. Service is provided on a first-come-first-served basis.

a. What is the probability distribution of the time between consecutive arrivals of customers of any type, what is its mean?

b. Assume that when a Type 2 customer arrives, he finds two Type 1 customers being served and no other customers in the system. What is the probability distribution of this Type 2 customer’s waiting time in the queue and its mean?

Answer 1

Answer:-

Given That:-

A queueing system serves two types of customers. Type 1 customers arrive according to a Poisson process with a mean rate of 5 per hour. Type 2 customers arrive according to a Poisson process at a mean rate of 3 per hour. The system has two servers, both of which serve both types of customer. All service times have exponential distribution with a mean of 10 minutes. Service is provided on a first-come-first-served basis.

Given data is,

A queuing system has two types of customers Type 1 customers have arrival poisson process with a mean rate of 5 per hour.

Type 2 customers have arrival poisson process with a mean rate of 3 per hour.

The system ahs two servers.

Both servers have service of exponential distribution with a mean of 10 minutes.

Therefore the service rate of system is as follows:

and then service is provided on first come first serve basis.

a. What is the probability distribution of the time between consecutive arrivals of customers of any type, what is its mean?

We have to find the probability distribution of the time between consecutive arrivals of customers of any type.

Consider the property: "uneffected by aggregation ot disaggregation"

The combined arrival process for two types of customers will be same poission process with a mean rate of

5 + 3 = 8 per hour

As there are customers are arriving per hour, the time between consecutive no. of arrivals of customers of type is as follows:

Time between consecutive arrivals = 1/8 hours

= 1/8 * 60 min

=7.5 min

Therefore, the probability distribution of the times between consecutive arrivals of customers of any type is exponential with a mean of 7.5 min.

b. Assume that when a Type 2 customer arrives, he finds two Type 1 customers being served and no other customers in the system. What is the probability distribution of this Type 2 customer’s waiting time in the queue and its mean?

We have to find the probability distribution of such type 2 customers waiting time in the queue when a type 2 customer arrives, there are two customers of type 1 being served.

Therefore the waiting time of such type 2 customer depends on the service rate of customers being served.

The probability distribution will be same with mean rate equal to minimum of two service rates.

For n 2, the service rate is exponential with a mean of 5 minutes.

Therefore The probability distribution of such type 2 customer is exponential with a mean of 5 minutes.

Plz like it.....,