This document discusses queuing theory and provides responses to questions about key concepts. It explains that 'in the line' refers to those waiting in the queue, while 'in the system' includes those being served. It also discusses transient vs steady-state solutions, finite vs infinite sources, different queue disciplines like shortest operation time, and how simulation can be used to analyze queuing systems through random number tables. The optimal number of spare parts or service channels is determined to be three based on analyzing costs for different cases.

1. CHAPTER 7: Queuing Theory
Responses to Questions:
1. ‘In the line’: those waiting in the queue.
‘In the system’: those ‘in the line’ plus those that are being worked upon i.e.
being served.
2. ‘Transient’: time-dependent. Solutions given in the chapter are the steady-
state solutions.
3. When the source (i.e. population) from which the arrivals originate is finite,
the characteristics of the arrival process get affected by the number in the
queueing system – the greater the number waiting, the lower the arrival rate
as there are fewer remaining ‘customers’. Generally, with a number in the
source above 30, the assumption of infinite source holds quite well.
However, with a finite source, the formulae are different and the
mathematics of it can get complex. Finite queuing tables will help in such
cases. These tables, for instance, can give ‘average number of arrivals
waiting for service’, ‘percent waiting in line’, etc for different population sizes
and different number of service channels.
4. Instead of First Come First Served (FCFS) queue discipline used in the
formulae given in the Chapter, if one had ‘Shortest Operation Time’ (SOT)
queue discipline then all arrivals with long operation times will get pushed to
the back of the line and keep waiting. Some may keep waiting almost
indefinitely. However, the customers (i.e. arrivals) with short operations
times will be served faster and get pushed out of the system quickly.
Therefore, overall efficiency may be high but some customers (long
operation times) would suffer.
To manage the above situation, we may ‘truncate’ the SOT queue discipline
and stipulate that after every so many applications of SOT discipline, for a
few (stipulated number) customers we shall use the ‘longest operation time’
(LOT) rule. This may help in clearing some of the backlog of long operation
time customers. Imagine, otherwise, in a barber shop only the moustache
trimming and hair trimming jobs will get done and customers who need full
works (hair-cut, hair-streaking) will keep waiting for a very long time.
5. As the server utilization increases, the average waiting time for customers
decreases. However, this can happen upto the point that all the idle capacity
is utilized. Beyond that, if the arrivals come at any faster rate, the average
waiting time would only increase. Practically, this means the server
utilization or rate of service cannot be pushed beyond a point without
affecting the service provided to the customers.
6. We assume that the breakdown and service rates follow a Poison distribution
and that the cost of capital is 25 percent.

2. 2
Mean of arrival rates = λ = 2 arrivals/week, and Mean of servicing rates = µ
= 3 per week.
Since there are 30 aircraft each with four engines, i.e. 120 engines, which is
a large number, we shall use the infinite source assumption. The problem
concerns the determination of the number of spares in the spares-bank
which can be regarded as the servers or channels in the queue system.
Let us analyze the different values of s, the number of channels, and
calculate the relevant characteristics of the queuing system for each case.
Case I: s = 2
Using formulae from Table 7.2, we have the following:
Utilization factor ρ = __λ__ = 2__ = 0.333
s . µ (2) (3)
P0 = 1 ÷ 1 + (0.333 x 2) + (0.333 x 2)2
2 x 0.667
= 1 ÷ (1 + 0.666 + 0.333)
= 0.500
Ns = _ (0.666)3
___ x 0.500 + 0.666
1 x (2 – 0.666)2
= 0.295 x 0.500 + 0.666
1.780
= 0.083 + 0.666
= 0.749
Nw = Ns - λ = 0.083
µ
Ts = Nw + 1 = 0.083 + 1 = 0.375 week
λ µ 2 3
Tw = Nw = 0.042 week
λ

3. 3
The cost of downtime per year = (avg. time an arrival waits) x
(No. of avg. arrivals per week) x (No. of weeks in a year) x
(Cost of downtime per week per arrival)
= 0.042 x 2 x 52 x Rs 25 lakh
= Rs 1.092 crore per year
The cost of carrying the two spares, at 25 per cent cost of capital =
Rs 2 crore x 0.25 = Rs 0.50 crore.
Therefore, Total cost = 1.092 + 0.50 = Rs 1.592 crore per year
Case II: s = 3
We refer to a table for Multiple Channel Single Service Queuing; the table gives
values of Nw for various (λ/µ) and ‘s’ values.
For r = λ = 0.666 and s = 3, we have Nw = 0.0093
µ
Tw = Nw = 0.0093 = 0.0047 week∗
λ 2
The cost of downtime per year = 0.0047 x 2 x 52 x 0.25 crore
= Rs 0.1209 crore per year
The cost of carrying the three spares, at 25 per cent cost of capital =
Rs 3 crore x 0.25 = 0.75 crore per year
Therefore, Total cost = 0.1209 + 0.75 = Rs 0.8709 crore
∗ We could have used Table 7.2 formulae, instead. Then ρ = 2/(3x3) = 2/9
P0 = 1 1+ (2/9x3)1
+ (2/9x3)2
+ (2/3)3
__ = 0.5124
1 2 6x (1- 2/9)
Nw = (2/3)4
__ x 0.5124 = 0.00927
2 x (3 - 2/3)2

4. 4
Case III: s = 1
Again from above mentioned table, for r = 0.666 and s = 1,
we have: Nw = 1.343
Tw = 1.343 = 0.6715 week
2
Cost of downtime per year in Rs crore = (0.6715) (2) (52) (0.25)
= Rs. 17.459 crore
Cost of carrying the spare = Rs 0.25 crore
Therefore, total cost = Rs 17.459 + 0.250 crore
= Rs 17.709 crore
Case IV: s = 4
If the number of service channels, i.e. the number of spares in a spares
bank is four, the value of waiting time becomes almost negligible.
The total cost for this policy = Cost of carrying four spares
= Rs 1.00 crore per year
Thus, carrying three (3) spares is an optimal policy as the total costs are
the least of all the options considered.
7&8. The mean number in the waiting line for different number of servers
(service channels) and corresponding to different values of λ/µ are
available in readymade tables.
These are not available in the present chapter. Otherwise, there is trial-
and-error involved, using the formulae as suggested in response to
Question # 6. This is cumbersome and hence not done here. Interested
readers are requested to refer to the book: Chary S.N., “Theory &
Problems in Production and Operations Management”, Tata McGraw-Hill,
New Delhi, 1995.

5. 5
9. Simulation exercise requires our picking numbers from the random
number table. For instance, we can arrange the interviewing period as
follows:
Interview period
(minutes)
Frequency Cum. Probability Associated
Random Numbers
15 40 0.40 00-40
20 20 0.60 41-60
25 10 0.70 61-70
10 20 0.90 71-90
05 10 1.00 91-99
Also, the arrivals for the interview can be arranged as follows:
Arrival Time Frequency Cum-Probability Associated
Random Numbers
10 minutes before 20 0.20 00-20
05 minutes before 20 0.40 21-40
On time 30 0.30 41-70
05 minutes late 20 0.20 71-90
absent 10 0.10 91-99
We start the procedure at 9:00 a.m. We pick the 1st
random number for the
arrival of the candidate. Say the number is 67, which means the candidate
arrives ‘on time’. The second random number is say 84, which means the
interview lasts for only 10 minutes i.e. from 9:00 a.m. to 9:10 a.m.
Let us pick the next random number for arrival. It is say 01. This means, the
candidate has arrived 10 minutes before his scheduled time of 9:15 a.m.
Hence, the 2nd
candidate is available for interview at 9:10 (when the 1st
candidate finishes). How long will the interview last would again be decided
from a random number, say 77 (refer to Random Number Table). This
means, interview lasts for 10 minutes and gets over by 9:20 a.m. We are
now ready for the 3rd
candidate.
Next random number is 90. So, the arrival of the 3rd
candidate (scheduled at
9:30) will be 5 minutes late i.e. he will arrive at 9:35. The other random,
number is 14, which means the interview will last for 15 minutes, from 9:35
am to 9:50 am.
The next draw of the number is 15, i.e. 4th
candidate arrives 10 minutes
early (i.e. arrives at 9:50 am). The interview for him can start at 9:50 am.
The next draw of number gives 74 i.e. an interview period is of 10 minutes.
So, the interview will be over at 10:00 am (i.e. from 9:50 to 10:00).
We can give the rest of the draws and results.

6. 6
Candidate Schedule Random
Numbers
Arrives
actually at
Interview
starts
Interview
ends
1 9:00 67 & 84 9:00 9:00 9:10
2 9:15 01 & 77 9:05 9:10 9:20
3 9:30 90 & 14 9:35 9:35 9:50
4 9:45 15 & 74 9:35 9:50 10:00
5 10:00 44 & 77 10:00 10:00 10:10
6 10:15 25& 38 10:10 10:10 10:25
7 10:30 66& 43 10:30 10:30 10:50
8 10:45 24 & 33 10:40 10:50 11:05
COFFEE TIME : 11:05 to 11:20
9 11:15 08 & 22 11:05 11:20 11:35
10 11:30 70 & 15 11:30 11:35 11:50
11 11:45 17 & 92 11:35 11:50 11:55
12 12:00 25 & 35 11:55 11:55 12:10
13 12:15 32 & 96 12:10 12:10 12:15
14 12:30 80 & 73 12:35 12:35 12:45
15 12:45 60 & 21 12:45 12:45 13:00
LUNCH TIME : 13:00 to 14:00
16 14:00 75 & 04 14:05 14:05 14:20
17 14:15 91 & 05 Absent
18 14:30 11 & 68 14:20 14:20 14:45
19 14:45 48 & 02 14:45 14:45 15:00
20 15:00 80 & 67 15:05 15:05 15:30
21 15:15 27 & 26 15:10 15:30 15:45
22 15:30 29 &87 15:25 15:45 15:55
23 15:45 72 & 79 15:50 15:55 16:05
Thus, in Round 1 of simulation, Prof. Chary will not be able to leave by
16:00 hrs.
Nine more rounds of simulations will be performed in a similar manner. The
number of times Prof. Chary is delayed beyond 16:00 hrs will be found.
However, the simulation has proved that there is always a chance that the
professor will be delayed.
The simulation presented here does not suggest any change in the queue
discipline, which would avoid the delay beyond 16:00 hours. The only time
two candidates are present at the same time is for candidates 3 & 4. But,
choosing either of them leads to the same result. Thus, any modification in
the queue discipline does not seem to help.
Perhaps, the professor should allow no candidate to arrive later than his
appointed/scheduled time. If a candidate arrives late, he should not be
interviewed. However, such a rule may not strictly fall under ‘queue
discipline’.

7. 7
10. Identify variables, which have an impact on the project’s NPV. Portray the
risk in terms of a probability distribution for each of the variables. Choose, at
random, a value for each of these variables. Based on these values,
compute the NPV. Repeat this procedure for a large number of times.
Obtain the mean and the standard deviation of the NPVs.
The NPVs under different simulations can be plotted to give a frequency
distribution. One may check as to with what probability the NPV may exceed
a pre-fixed value (for capital budgeting purposes).

8. 8
CHAPTER 7: Queuing Theory
Objective Questions:
1. Even when the service rate is faster than the rate of arrival of customers, a
queue of the customers can develop because :
√a. the service and arrival rates are not constant, but subject to
statistical distributions.
b. the source from which the arrivals come is infinite.
c. a & b
d. none of the above.
2. One of the practical problems in applying theoretical models to waiting line
situations in operations is :
a. arrival and service rates are not constant; they are probalistic in
nature.
√ b. arrivals may depend upon the state of the queue system; this is a
behavioral problem.
c. a & b
d. none of the above.
3. In a railway counter two clerks are issuing tickets each at the same average
rate of 40 passengers per hour. The mean arrival rate is 64 passengers per
hour. The utilization factor for this queueing system is :
a. 0.625
b. 1.00
c. 1.60
√ d. none of the above
4. The assumption of infinite source can generally hold good in a queueing
system with a number in the source :
√a. above 30
b. above 50
c. above 100
d. above 250
5. In a Job-shop system of manufacturing the jobs arrive at machines that can
be seen as servers. The objective of this queueing system could be :
a. maximize machinery/labour utilization
b. minimize total flow time
c. minimize percentage of orders completed late
√ d. all of the above.

9. 9
6. The names Peck and Hazelwood are associated with :
a. Developing Erlangian arrival/service rates
b. Behavioral aspects of the Waiting Line problem
√ c. Finite queuing tables
d. Pioneering the ‘simulation’ approach
7. Following is important for a queue system :
a. arrival and service rates
b. queue discipline
√ c. a & b
d. none of the above
8. Applicants arrive at a rate of 10 per hour and are seen by the officer at an
average time of 5 minutes per applicant. If the arrival rate and service rate
are both Poisson, the average number of applicants waiting in line would be
approximately :
a. 1
b. 2
c. 3
√ d. 4
9. For the above problem, the averaging waiting time would be:
a. 15 minutes
√ b. 25 minutes
c. 40 minutes
d. 50 minutes
10. ‘Monte Carlo’ is associated with :
a. Game theory
b. Time-series
c. Moving averages
√ d. Simulation
11. Dispatch rule in a Job-shop production is :
a. the loading of jobs to machines.
b. the route scheduling rule for vehicles.
√ c. the queue discipline.
d. the pattern of arrival of jobs.

10. 10
12. If ‘Queueing theory’ is used to analyze the ‘Rotable spares’ then:
√ a. the defective equipment are the arrivals and the rotable spares’
bank are the servers
b. the defective equipment are the arrivals and the gang of repair-
men are the servers.
c. The breakdowns are the servers and the spares’ bank are the
arrivals.
d. The breakdowns are the servers and the number of cycles of
repair are the arrivals.
13. Simulation can be used for:
a. incorporating risk into the capital budgeting decisions under
uncertainty.
b. complex queuing situations with complex distribution of inter-
arrival times.
c. complex queuing situations with complex distribution of service
times.
√ d. all of the above
14. ‘Arrivals’ in a queueing system could depend upon:
a. the length of the queue.
b. the characteristics of preceding arrivals.
c. the service characteristics.
√ d. all of the above.