## Stochastic Models and Simulation

##### Queuing

A **single server** follows the first-come, first-served rule and serves clients one at a time from the head of the line. When the service is finished, the customer moves down the line and the system has one less customer.

When adopting this type of queuing model, following assumptions are made:

- Customers are limitless and patient,
- customer arrivals and service rates also follow exponential distributions,
- The waiting line is managed on a first-come, first-served basis.

The likelihood that the waiting area may in fact be constrained is an evident constraint. Another scenario is that the arrival rate varies by state. That example, if they see a big line, potential customers will be deterred from joining it.

**Multiple Server model**: Multiple service facilities are operating concurrently and provide the same service in a multi-server queue. More than one station can provide service to every customer in the queue. Following poison and exponential distribution are the arrival and service times.

**Solved Example: 9134-01**

Customers arrive at a reception counter at an average interval rate of 10 minutes and the receptionist takes an average of 6 minutes for one customer. Determine the average queue length.

A. 9/10

B. 7/10

C. 11/10

D. 3/10

Correct Answer: **A**

**Solved Example: 9134-02**

The term 'Jockeying' in queuing theory refers to:

A. Not entering the long queue

B. Leaving the queue

C. Shifting from one queue to another parallel queue

D. None of the above

Correct Answer: **C**

**Solved Example: 9134-03**

Cars arrive at a service station according to Poisson's distribution with a mean rate of 5 per hour. The service time per car is exponential with a mean of 10 minutes. At steady state, the average waiting time in the queue is:

A. 10 minutes

B. 20 minutes

C. 25 minutes

D. 50 minutes

Correct Answer: **D**

##### Markov Processes

**Solved Example: 9135-01**

The probability density function of a Markov process is:

A. p(x1,x2,x3.......xn) = p(x1)p(x2/x1)p(x3/x2).......p(xn/xn-1)

B. p(x1,x2,x3.......xn) = p(x1)p(x1/x2)p(x2/x3).......p(xn-1/xn)

C. p(x1,x2,x3......xn) = p(x1)p(x2)p(x3).......p(xn)

D. p(x1,x2,x3......xn) = p(x1)p(x2 *x1)p(x3*x2)........p(xn*xn-1)

Correct Answer: **A**