SOLVING THE PROBLEM USING COMPUTER WITH TORA - Quantitative Techniques for management

Example above is solved using computer with TORA.
Enter the values λ = 12
μ = 18
No. of server = 1
The input screen is shown in Figure.

Queuing Analysis Using TORA (Input Screen)

Queuing Analysis Using TORA (Input Screen)

Press Solve to get the output screen and select scenario 1 option in the select output option menu. The output screen for the problem is displayed as shown in Figure.

Queuing Analysis Using TORA (Output Screen)

Queuing Analysis Using TORA (Output Screen)

The values are
(a) P0 = 0.3333 (for n = 0)
(b) Lq = 1.33
(c) Wq = 0.1111
(d) Pb (or) r/C= 0.66667

In the same problem, to determine the probability that there are 2 trucks in the system, we use the formula,

probability

This can also be read in the output screen for n=2 the probability Pn = 0.14815, similarly, the probabilities for different values of n can be directly read.

Example: A TV repairman finds that the time spent on his jobs has a exponential distribution with mean 30 minutes. If he repairs TV sets in the order in which they come in, and if the arrivals follow approximately Poisson distribution with an average rate of 10 per 8 hour day, what is the repairman's expected idle time each day? How many jobs are ahead of the average with the set just brought in?

Solution: Given λ = 10 TV sets per day.

μ = 16 TV sets per day.

(i) The Probability for the repairman to be idle is,

P0 = 1 – ρ

We know, ρ = λ / 30 = 10 / 16 =0.625

P0 = 1 – ρ

= 1 – 0.625 = 0.375

Expected idle time per day = 8 × 0.375= 3 hours.

(ii) How many jobs are ahead of the average set just brought in

average set

Example : Auto car service provides a single channel water wash service. The incoming arrivals occur at the rate of 4 cars per hour and the mean service rate is 8 cars per hour. Assume that arrivals follow a Poisson distribution and the service rate follows an exponential probability distribution. Determine the following measures of performance:

(a) What is the average time that a car waits for water – wash to begin?
(b) What is the average time a car spends in the system?
(c) What is the average number of cars in the system?

Solution: Given λ = 4 cars per hour, μ = 8 cars per day.

(a) Average time that a car waits for water - wash to begin,

Wq= l/ l (m-l)
= 4/ 8(8-4)
= 0.125 hours or 7.5 mins

(b) Average time a car spends in the system,

Ws= 1/ m-l
= 1/ 8-4
= 1/4 = 0.25 hours or 15 mins.

(c) Average number of cars in the system,

Ls= l/ m-l
= 4/8-4
= 4/4 = 1 car

Arrivals at a telephone booth are considered to be Poisson

Example : Arrivals at a telephone booth are considered to be Poisson distributed with an average time of 10 minutes between one arrival and the next. The length of phone call is assumed to be distributed exponentially, with mean 3 minutes.

(i)What is the probability that a person arriving at the booth will have to wait?
(ii)The telephone department will install a second booth when convinced that an arrival would expect waiting for at least 3 minutes for phone call. By how much should the flow of arrivals increase in order to justify a second booth?
(iii)What is the average length of the queue that forms from time to time?
(iv)What is the probability that it will take him more than 10 minutes altogether to wait for the phone and complete his call?
(v)What is the probability that it will take him more than 10 minutes altogether to wait for the phone and complete his call?

Solution: Given λ= 1/10 = 0.10 person per minute.
μ = 1/3 = 0.33 person per minute.

(i) Probability that a person arriving at the booth will have to wait,
P (w > 0) = 1 – P0
= 1 – (1 - λ / μ) = λ / μ
= 0.10/0.33 = 0.3

(ii) The installation of second booth will be justified if the arrival rate is more than the waiting time.
Expected waiting time in the queue will be,

Wq = l/ m (m-l)

Where, E(w) = 3 and λ = λ (say ) for second booth. Simplifying we get

λ = 0.16

Hence the increase in arrival rate is, 0.16-0.10 = 0.06 arrivals per minute.

(iii) Average number of units in the system is given by,

Ls= ρ/1- ρ
= 0.3/1-0.3
= 0.43 customers

(iv) Probability of waiting for 10 minutes or more is given by

telephone booth are considered to be Poisson

This shows that 3 percent of the arrivals on an average will have to wait for 10 minutes or more before they can use the phone.

Example : A bank has decided to open a single server drive-in banking facility at its main branch office. It is estimated that 20 customers arrive each hour on an average. The time required to serve a customer is 3 minutes on an average. Assume that arrivals follow a Poisson distribution and the service rate follows an exponential probability distribution.

The bank manager is interested in knowing the following:

(a) What will be the average waiting time of a customer to get the service?
(b) The proportion of time that the system will be idle.
(c) The space required to accommodate all the arrivals, on an average, the space taken by each car is 10 feet that is waiting for service.

Solution: λ = 20 Customers per hour, μ = 60/25 = 2.4 customers per hour.

(a) Average waiting time of a customer to get the service,
Wq = l/ m (m-l)
= 20/ 24 (24-20) = 20/96
= 0.208 hour or 12.5 mins.

(b) The proportion of time that the system will be idle,

P0 = 1- l/ m
= 1- 20/24
= 0.166 hours or 10 mins.

(c ) Average number of customers waiting in the queue,

Lq = l2/ (m-l)
= 20 2/ 24 (24-20)
= 400/96
= 4.66 customers

10 feet is required for 1 customer. Hence, for 4.66 customers, the space required is
10 × 4.66 = 46.6 feet.

Example: In a Bank, customers arrive to deposit cash to a single counter server every 15 minutes. The bank staff on an average takes 10 minutes to serve a customer. The manager of the bank noticed that on an average at least one customer was waiting at the counter.

To eliminate the customer waiting time, the manager provided an automatic currency counting machine to the staff. This decreased the service time to 5 minutes on an average to every customer. Determine whether this rate of service will satisfy the manager's interest. Also use computer with TORA for solving the problem.

Solution:

Case 1: l = 60/15 = 4 customers per hour, m = 60/10 = 6 customers per hour.
Average number of customers in the system,
Ls = l/ (m-l)
= 4/6-4 = 4/2 = 2 customers.

Case 2: l = 4, m = 60/15 = 12 customers per hour.
Average number of customers in the system,
Ls= 4/12-4
=4/8 =1/2 = 0.5, say, 1 customer.
Average number of customers in the queue

Lq= l2/ m (m-l)
= 42/12 (12-4)
= 16/96 = 0.01 customers.

Since no customers are standing in the queue the manager's interest is satisfied. The problem is worked out using TORA. Enter the values as shown in the input screen below in Figure.

Queuing Analysis Using TORA (Input Screen)

Queuing Analysis Using TORA (Input Screen)

Press Solve and go to output screen. Select comparative analysis option in the queuing output analysis menu. The following output screen is displayed.

Comparative Analysis of Queuing Output Analysis Using TORA (Output Screen)

Comparative Analysis of Queuing Output Analysis Using TORA (Output Screen)

Now, on comparing scenario 1 and scenario 2, under Ls i.e., the average number of customers in the system is 2 and 0.5 respectively. In the first scenario, it means that in the entire system, one customer will be waiting in the queue while others are being served. In scenario 2, only one customer is in the system and being served, where on an average no customer will be waiting.

Example: 12 counters are available in a computerized railway reservation system. The arrival rate during peak hours is 90 customers per hour. It takes 5 minutes to serve a customer on an average. Assume that the arrivals joining in a queue will not be jockeying (i.e., move to another queue). How many counters have to be opened if the customers need not to wait for more than 15 minutes?

Solution: The problem is to be solved as one system comprising of 'n' number of single server queuing model.

Arrival rate, l =90 customers per hour
Service rate, μ = 60/5 =12 per hour

Average waiting time, Wq = 15/60 = 0.25 hours

Average waiting time, Wq = l/ m (m-l)

i.e., 0.25 = l/ m(m l) ................................(i)

Let, number of counters be x,
Considering the single server queuing system, the number of counters required to serve 90 arrivals per hour, l = 90/x substituting l = 90/x in equation (i),

single server queuing system

Hence, 10 counters are required so that an average arrival will wait less than 15 minutes.

Example: In a single pump petrol station, vehicles arrive at the rate of 20 customers per hour and petrol filling takes 2 minutes on an average. Assume the arrival rate is Poisson probability distribution and service rate is exponentially distributed, determine

(a) What is the probability that no vehicles are in the petrol station?
(b) What is the probability that 1 customer is filling and no one is waiting in the queue?
(c) What is the probability that 1 customer is filling and 2 customers are waiting in the queue?
(d) What is the probability that more than 2 customers are waiting?

Solution : l = 20 vehicles per hour, μ = 60/2 = 30 vehicles per hour.

Poisson probability distribution and service rate is exponentially distributed

Poisson probability distribution and service rate is exponentially distributed

Tora output screen


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Quantitative Techniques for management Topics