PQT UNIT-4 1

Unit-IV Queueing Models d-capacity of the system,

1. Explain about queueing theory e-queue discipline,
The theory which deals with the study of waiting lines such f-calling source
as waiting to use a telephone booth, machines to be repaired.
9. Define Transient state.
2. Define Deterministic Queueing system. A queueing system is said to be in steady state when its
Here the customers arrive at regular intervals and the operating characteristics (input, output, mean queue length)
service time for each customer is known and constant. are dependent upon time.

3. Define Probabilistic Queueing system. 10. Define Steady state.

Here the arrival rates and service rates are both variable and A queueing system is said to be in steady state when its
uncertain. operating characteristics are independent upon time.

4. Define a queue 11. Explain the characteristics of queueing model.

The flow of customers from finite or infinite population
towards the service facility is called a queue or waiting line a) Arrival pattern- It represents the pattern in which
eg- persons waiting at a ticket counter the number of customers arrives for service. It is
measured either by mean arrival rate or inter-arrival
5. Define a customer time , mostly arrival rates follows Poisson distribution
The arriving unit that requires some service to be performed and /or the inter-arrival time follows exponential
is known as a customer distribution.

6. Define a server. b) Service pattern-It represents the pattern in which

The process that performs the services to the customer is the number of customers gets serviced. Service time is
termed as server or service facility or service channel. the time interval from the commencement of service to
the completion of service. Generally service time is
7. Define a Poisson queue. measured in terms of exponential distribution.
The queues that follow the Poisson arrivals (Exponential
inter arrival times) and Poisson services(Exponential service c) Service facility-
time) are called Poisson queue. i) Single server- Customers may wait in a single
queue till the server is ready for servicing.
8. Write Kendall’s notation. Customers may wait in several queues but single
It is (a/b/c): (d/e/f) where server is available
a-the arrival pattern, ii) Multi server-Customers may wait in a single
b-the service pattern , queueuntil one of the server is ready for servicing.
c-service facility Customers may wait in Parallel channels which
 PQT UNIT-4 2

gives identical service facilities. Customer may b) Non Pre-emptive priority-Customer of high priority
wait is Series channelsand passes through all the goes ahead in the queue, but his service is started after
channels in a sequence before the service is the low priority customerwho is in emergency.
completed. Eg:-Patient who is in serious condition needs treatment.
iii) Finite number of servers-Customers may be
served according to a specific order. 12. Explain the behavior of a customer.
iv) Infinite number of servers-Customers may be a) Patient customer- A customer may decide to wait no
served instantaneously upon arrival. matter how long the queue is.
v) Bulk service system- Customers may be served b) Impatient customer-A customer may decide not to
in bulk or in batches wait or not to enter the queue of certain reasons.
i) Balking-Customer may not enter the queue if it is
d) Capacity of the system- too long.
i) Finite capacity-There is a finite limit to the ii) Reneging-Due to impatience a customer leaves the
maximum queue size. queue.
ii) Infinite capacity-There is no finite limit for iii) Jockeying-Customer who waits in one queue may
customers. join the other queue which moves faster in order to
reduce the waiting time.
e) Queue discipline-
i) FIFO-Fist In First Out-Customers served in a 13.Define a Pure Birth process.
strict order. The process in which only arrivals are counted and no
Eg:-Patients waiting in a clinic. departures takes place is called a pure birth process.
ii) LIFO-Last In First Out-The last arrival is served
first. 13. Define a Pure Death process.
Eg:-Rice bags in a godown. The process in which only departures are counted and no
iii) SIRO-Service In Random Order-The arrivals other arrivals are allowed is called a pure death process.
are serviced randomly irrespective of their arrival.
Eg:-Customers get coffee, tea, snacks at a railway 14. Define a Pure Birth-Death process.
station in which a train stops for few minutes. It arises when additional customer increases the arrival in
iv) Priority- The service is done to a customer in the system and decrease of a serviced customer (regarded as
preference over the other. death) from the system.
a) Pre-emptive priority- Customer of high
priority are given service over the low priority
customer
Eg:-VIP dharsan in a temple.
 3
Model I – (M/M/1)(∞/FIFO)Problems the sets in the order in which they arrive and expects that the time
1) In a self service store with one cashier, 9 customers arrive on an required to repair a set has an ED with mean 30mins
average of every 5 minutes. And the cashier can serve 10 in 5 minutes. 1
Service rate 𝜇 = 1/30𝑚𝑖𝑛𝑠 ⟹ 𝜇 = 30 / 𝑚𝑖𝑛
Assuming Poisson distribution for arrival rate and Exponential 𝜆 1/48 5
distribution for service rate, Find (i) Average number of customers in the 𝜌 = ⟹𝜌= ⟹𝜌=
system, (ii) Average number of customers in queue or average queue 𝜇 1/30 8
length, (iii) Average time a customer waits before being served and
5 3
𝑃0 = 1 − 𝜌 ⟹ 𝑃0 = 1 − ⟹ 𝑃0 =
(iv)Average time a customer spends in the system. 8 8
2
Solution: 5
𝜌2 ( )
8 customers arrive on an average of every 5 minutes 𝐿𝑞 = ⟹ 𝐿𝑞 = 8 ⟹ 𝐋𝐪 = 𝟏. 𝟎𝟒𝟏𝟕 𝐬𝐞𝐭𝐬
9 1−𝜌 5
Arrival rate 𝜆 = 9/5𝑚𝑖𝑛𝑠 ⟹ 𝜆 = 5 / 𝑚𝑖𝑛 1−
8
The cashier can serve 10 in 5 minutes (i) repairman’s expected idle time each day
Service rate 𝜇 = 10/5 𝑚𝑖𝑛𝑠 ⟹ 𝜇 = 2 / 𝑚𝑖𝑛 5
𝑃0 ×8ℎ𝑟/𝑑𝑎𝑦 = (1 − 𝜌)×8 = (1 − ) ×8 =3 ℎ𝑟𝑠/𝑑𝑎𝑦
𝜆 9/5 9 8
𝜌 = ⟹𝜌= ⟹𝜌= (ii) Number of jobs are ahead of the average set just brought in
𝜇 2 10
9 1 Average No. of T.V. sets in queue to be repaired 𝐋𝐪 = 𝟏. 𝟎𝟒𝟏𝟕 𝐬𝐞𝐭𝐬
𝑃0 = 1 − 𝜌 ⟹ 𝑃0 = 1 − ⟹ 𝑃0 =
10 10
9 2 3) In a railway marshalling yard, goods trains arrive at a rate of 30
𝜌2 (10) trains per day. Assuming that the inter arrival time follows an
𝐿𝑞 = ⟹ 𝐿𝑞 = ⟹ 𝐋𝐪 = 𝟖. 𝟏 𝐜𝐮𝐬𝐭𝐨𝐦𝐞𝐫𝐬
1−𝜌 1−
9 Exponential distribution and the service time distribution is also
10 exponential with an average 36 minutes .Calculate the following.
(i) Average number of customers in the system 𝐿𝑠 (i) The mean queue size,(ii)The probability the queue size exceeds 10
𝜆 9/5
𝐿𝑠 = 𝐿𝑞 + ⟹ 𝐿𝑠 = 8.1 + ⟹ 𝐿𝑠 = 8.1 + 0.9 ⟹ 𝐋𝐬 = 𝟗 𝐜𝐮𝐬𝐭𝐨𝐦𝐞𝐫𝐬 and (iii) If the input of trains increases to an average 33 per day,
𝜇 2 What will be the change in(i) and(ii).
(ii) Average number of customers in queue (average queue length) 𝐿𝑞 Solution:
𝐋𝐪 = 𝟖. 𝟏 𝐜𝐮𝐬𝐭𝐨𝐦𝐞𝐫𝐬 Goods trains arrive at a rate of 30 trains per day
(iii) Average time a customer waits before being served. 𝑊𝑞 30
Arrival rate 𝜆 = 30/𝑑𝑎𝑦 ⟹ 𝜆 = 24𝑥60 / 𝑚𝑖𝑛 ⟹ 𝜆 = 48 /𝑚𝑖𝑛
1
𝐿𝑞 8.1
𝑊𝑞 = ⟹ 𝑊𝑞 = ⟹ 𝐖𝐪 = 𝟒. 𝟓 𝐦𝐢𝐧𝐮𝐭𝐞𝐬 service time distribution is exponential with an average 36 minutes
𝜆 9/5 1
(iv)Average time a customer spends in the system 𝑊𝑠 Service rate 𝜇 = 1/36 𝑚𝑖𝑛𝑠 ⟹ 𝜇 = / 𝑚𝑖𝑛
36
𝐿𝑞 1 8.1 1 𝜆 1/48 3
𝑊𝑠 = + ⟹ 𝑊𝑠 = + ⟹ 𝑊𝑠 = 4.5 + 0.5 ⟹ 𝐖𝐬 = 𝟓 𝐦𝐢𝐧𝐮𝐭𝐞𝐬 𝜌 = ⟹𝜌= ⟹𝜌=
𝜆 𝜇 9/5 2 𝜇 1/36 4
3 1
𝑃0 = 1 − 𝜌 ⟹ 𝑃0 = 1 − ⟹ 𝑃0 =
2) A T.V. repairman finds that the time spent on his jobs has an 4
2
4
exponential distribution with mean 30 minutes. If he repairs sets in 3
𝜌2 ( )
the order in which they came in and if the arrival of sets is 𝐿𝑞 = ⟹ 𝐿𝑞 = 4 ⟹ 𝐋𝐪 = 𝟐. 𝟐𝟓 𝐭𝐫𝐚𝐢𝐧𝐬
1−𝜌 3
approximately Poisson with an average rate of 10 per 8 hour day. 1−4
(i) What is the repairman’s expected idle time each day? (i) The mean queue size 𝐿𝑞
(ii)How many jobs are ahead of the average set just brought in? 𝐋𝐪 = 𝟐. 𝟐𝟓 𝐭𝐫𝐚𝐢𝐧𝐬
Solution: (ii)The probability the queue size exceeds 10 𝑃[𝑛 ≥ 10] = 𝜌10
T.V. sets arrive in a Poisson fashion at an average rate of 10/8 hr day 3 10
10 10 1
Arrival rate 𝜆 = 8ℎ𝑟 ⟹ 𝜆 = 8𝑥60 /𝑚𝑖𝑛 ⟹ 𝜆 = 48 / 𝑚𝑖𝑛 𝜌10 = ( ) = 0.0563
4
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(iii) If the input of trains increases to an average 33 per day, What The amount to be budgeted by the clinic to decrease the average size
will be the change in(i) and(ii). 1
of the queue from 1 patients to ½ a patient.
33 1 3
Arrival rate 𝜆 = 33/𝑑𝑎𝑦 ⟹ 𝜆 = 24𝑥60 / 𝑚𝑖𝑛 ⟹ 𝜆 = 480 /𝑚𝑖𝑛 1 𝜌2 1
𝜆 11/480 33 To find the service time , 𝐿𝑞 = ⟹ =
𝜌 = ⟹𝜌= ⟹𝜌= 2 1−𝜌 2
𝜇 1/36 40 𝜆 2 1
33 7 (𝜇) 1 𝜆2 1 1
𝑃0 = 1 − 𝜌 ⟹ 𝑃0 = 1 − ⟹ 𝑃0 = ⟹ = ⟹ = ⟹ 225 =
40 40 𝜆 2 𝜇(𝜇 − 𝜆) 2 𝜇(𝜇 − 1/15) 2
33 2 1−
𝜇
𝜌2 ( ) 1 1
𝐿𝑞 = ⟹ 𝐿𝑞 = 40 ⟹ 𝐋𝐪 = 𝟑. 𝟖𝟖𝟗𝟑 𝐭𝐫𝐚𝐢𝐧𝐬 ⟹ =
1−𝜌 33 15𝜇(15𝜇 − 1) 2
1 − 40
(i) The mean queue size 𝐿𝑞 ⟹ 2 = 225𝜇2 − 15𝜇
𝐋𝐪 = 𝟑. 𝟖𝟖𝟗𝟑 𝐭𝐫𝐚𝐢𝐧𝐬 225𝜇2 − 15𝜇 − 2 = 0
(ii)The probability the queue size exceeds 10 𝑃[𝑛 ≥ 10] = 𝜌10 2
solving we get 𝜇 = 2/15 𝑖. 𝑒 𝜇 = / 𝑚𝑖𝑛
33 10 15
10
𝜌 = ( ) = 0.1461 mean service time = 7.5 mins
40 Amount budgeted= Rs 100+( decreased mean service time)x Rs10
= 100 + (10 − 7.5)×10
4) On an average 96 patients per 24 hour day require the service of an = 𝑅𝑠 125
emergency clinic. Also on an average, a patient requires 10 minutes of
active attention. Assume that the facility can handle only one 5) Customers arrive at a one-man barbershop according to a Poisson
emergency at a time. Suppose that it costs the clinic Rs.100 per process with a mean inter arrival time of 12 minutes. Customers spend
patient treated to obtain an average servicing time of 10 minutes, and an average of 10 minutes in the barber’s chair.
that each minute of decrease in this average time would cost Rs.10 (i)What is the expected number of customers in the barber shop and in
per patient treated. How much would have to be budgeted by the the queue,
1
clinic to decrease the average size of the queue from 1 patients to ½ (ii) Calculate the percentage of time an arrival can walk straight into the
3
a patient. barber’s chair without having to wait,
(iii) What is the probability that more than 3 customers are in the
Solution:
system?
On an average 96 patients per 24 hour day
96 1 (iv)What is the average time spend in the queue ,
Arrival rate λ = 96/24 hr day ⟹ 𝜆 = 24𝑥60 / 𝑚𝑖𝑛 ⟹ 𝜆 = 15 /𝑚𝑖𝑛 (v)What is the probability that the waiting time in the system is greater
patient requires 10 minutes of active attention than 30 minutes ,
1
Service rate 𝜇 = 1/10 𝑚𝑖𝑛𝑠 ⟹ 𝜇 = 10 / 𝑚𝑖𝑛 (vi)Calculate the percentage of customers who have to wait prior to
getting into the barber’s chair and
𝜆 1/15 2
𝜌 = ⟹𝜌= ⟹𝜌= (vii) How much time can a customer expect to spend in the barber’s shop?,
𝜇 1/10 3 (viii)Management will provide another chair and hire another barber,
2 1 when a customer’s waiting time in the shop exceeds 1.25 h. How much
𝑃0 = 1 − 𝜌 ⟹ 𝑃0 = 1 − ⟹ 𝑃0 =
3 3 must the average rate of arrivals increase to warrant a second barber?
2
2 Solution:
𝜌2 ( )
𝐿𝑞 = ⟹ 𝐿𝑞 = 3 ⟹ 𝐋𝐪 = 𝟏. 𝟑𝟑 𝐩𝐚𝐭𝐢𝐞𝐧𝐭𝐬 Customers arrive at a one-man barbershop according to a Poisson
1−𝜌 2
1−3 process with a mean inter arrival time of 12 minutes
1
Clinic cost Rs 100 and each minute of decrease in this average time Arrival rate 𝜆 = 1/12𝑚𝑖𝑛𝑠 ⟹ 𝜆 = 12 / 𝑚𝑖𝑛
would cost Rs.10 per patient Customers spend an average of 10 minutes in the barber’s chair.
 5
Service rate 𝜇 = 1/10𝑚𝑖𝑛𝑠 ⟹ 𝜇 = / 𝑚𝑖𝑛
1
Arrival rate is increased in order to justify a second booth.
10
𝜆 1/12 Model II - (M/M/s)(∞/FIFO)Problems
𝜌 = ⟹ρ= ⟹𝛒 == 𝟓/𝟔 1) A petrol pump station has 4 pumps. The service times follow the
𝜇 1/10
5 1 exponential distribution with a mean of 6 minutes and cars arrive for
𝑃0 = 1 − 𝜌 ⟹ 𝑃0 = 1 − ⟹ 𝑃0 = service in a Poisson process at the rate of 30 cars per hour.
6 6
5 2 (i)What is the probability that an arrival would have to wait in line?
𝜌2 ( ) (ii)Find the average waiting time, average time spent in the system
𝐿𝑞 = ⟹ 𝐿𝑞 = 6 ⟹ 𝐋𝐪 = 𝟒. 𝟏𝟔𝟔𝟕 𝐜𝐮𝐬𝐭𝐨𝐦𝐞𝐫𝐬 and the average number of cars in the system.
1−𝜌 5
1−6 (iii)For what percentage of time would a pump be idle on an average?
(i) Expected number of customers in the shop 𝐋𝐬 and in the queue𝐋𝐪 , Solution:
𝜆 1/12 Petrol pump station has 4 pumps servers 𝑠 = 4
𝐿𝑠 = 𝐿𝑞 + ⟹ 𝐿𝑠 = 4.1667 + ⟹ 𝑳𝒔 = 𝟓 𝐜𝐮𝐬𝐭𝐨𝐦𝐞𝐫𝐬
𝜇 1/10 service times follow exponential distribution with a mean of 6 minutes
1
𝐋𝐪 = 𝟒. 𝟏𝟔𝟔𝟕 𝐜𝐮𝐬𝐭𝐨𝐦𝐞𝐫𝐬 Service rate 𝜇 = 1/6 𝑚𝑖𝑛𝑠 ⟹ 𝜇 = / 𝑚𝑖𝑛
6
(ii) Percentage of time an arrival can walk straight into the barber’s Cars arrive for service in a Poisson process at the rate of 30 cars per
1
chair without having to wait = 100×𝑃0 = 100× = 16.67% hour
6
30 1
(iii) Probability that more than 3 customers are in the system Arrival rate 𝜆 = 30/hr ⟹ 𝜆 = 60 /𝑚𝑖𝑛 ⟹ 𝜆 = 2 /𝑚𝑖𝑛
3+1 4
5 4 𝜆 1/2
𝑃[𝑛 > 3] = 𝜌 = 𝜌 = ( ) = 0.4823 𝜌 = ⟹ρ= ⟹𝛒 = 𝟑/𝟒
6 𝜇𝑠 (1/6)4
(iv) Average time spend in the queue −1
𝐿𝑞 4.1667 𝑠−1 −1 4−1 3 𝑛 3 4
𝑊𝑞 = = = 50.0004 𝑚𝑖𝑛𝑠 (𝑠ρ) 𝑛
𝑠 ρ
𝑠 s (4× ) 4
4 +4 4 ] ( )
𝜆 1/12 𝑃0 = [∑ + ] ⟹ 𝑃0 = [∑
𝑛! 𝑠! 1 − ρ 𝑛! 4! 1 − 3
(v) Probability that the waiting time in the system is greater than 30 𝑛=0 𝑛=0 4
1 1 −1
−( − )30
minutes 𝑃[𝑊𝑠 > 𝑡] = 𝑒 −(𝜇−𝜆)𝑡 = 𝑃[𝑊𝑠 > 30] = 𝑒 10 12 = 0.6065 3
(3)𝑛 81 3 32 33 27
−1

(vi) Percentage of customers have to wait prior to getting into the ⟹ 𝑃0 = [∑

𝑛!
+
1
] ⟹ 𝑃0 = [1 + + + + ]
1! 2! 3! 2
5
barber’s chair = 100P[arrival will have to wait] = 100×𝜌 = 100× 6 = 83.33% 𝑛=0 24× 4
(vii) The time can a customer expect to spend in the barber’s shop 9 9 27 −1 53 −1
𝐿 1 4.1667 1 ⟹ 𝑃0 = [1 + 3 + + + ] ⟹ 𝑃0 = [ ] ⟹𝐏𝟎 = 𝟎. 𝟎𝟑𝟕𝟕
𝑊𝑠 = 𝑞 + ⟹ 𝑊𝑠 = + ⟹ 𝑊𝑠 = 60.0004 𝑚𝑖𝑛𝑠 2 2 2 2
𝜆 μ 1/12 10 4
3 3
(viii)Management will provide another chair and hire another barber, (𝑠𝜌)𝑠 𝜌 (4× 4)
4
when a customer’s waiting time in the shop exceeds 1.25 h. We have 𝐿𝑞 = 𝑃 ⟹ 𝐿𝑞 = 0.0377
𝑠! (1 − 𝜌)2 0 4! 3 2
to find the average rate of arrivals increase to warrant a second (1 − 4)
barberWs ≥ 1.25 hrs ⟹ Ws ≥ 75 mins 81 3/4
ρ2 λ2 ⟹ Lq = × ×0.0377 ⟹𝐋𝐪 = 𝟏. 𝟓𝟐𝟔𝟗
24 1/16
𝐿𝑞 1 1−ρ 1 μ(μ − λ) 1 (i)The probability that an arrival would have to wait in line
⟹ + ≥ 75 ⟹ + ≥ 75 ⟹ + ≥ 75
𝜆 μ λ μ λ μ 3 4
λ 1 λ 1 100λ (𝑠𝜌)𝑠 𝑃0 (4× 4) 0.0377
⟹ + ≥ 75 ⟹ + ≥ 75 ⟹ + 10 ≥ 75 P[arrival will have to wait] = = = 0.5090
μ(μ − λ) μ 1 1 1 1 − 10λ 𝑠! 1 − 𝜌 4! 1 − 3/4
( − λ)
10 10 10
65 13 (ii) The average waiting time 𝑊𝑞 , average time spent in the system𝑊𝑠
⟹ 100λ + 10 − 100λ ≥ 75 − 750λ ⟹ 750λ ≥ 65 ⟹ λ ≥ ⟹λ≥ and the average number of cars in the system 𝐿𝑠
750 150
1
Old arrival rate = /min = 0.0833/ min, New arrival rate =
13
/min = 𝐿𝑞 1.5269
12 150 𝑊𝑞 = = ⟹ 𝑾𝒒 = 𝟑. 𝟎𝟓𝟑𝟖 𝒎𝒊𝒏𝒔
0.0867/min 𝜆 1/2
 6
𝐿𝑞 1 1.5269 1 27/8 1/2
𝑊𝑠 = + = + ⟹ 𝑾𝒔 = 𝟗. 𝟎𝟓𝟑𝟖 𝒎𝒊𝒏𝒔 ⟹ Lq = × ×0.0377 ⟹𝐋𝐪 = 𝟎. 𝟐𝟑𝟔𝟖
𝜆 μ 1/2 1/6 6 1/4
𝜆 1/2 (i) Average number of customer in the system 𝐿𝑠
𝐿𝑠 = 𝐿𝑞 + ⟹ 𝐿𝑠 = 1.5269 + ⟹ 𝑳𝒔 = 𝟒. 𝟓𝟐𝟔𝟗 𝜆 1/10
𝜇 1/6
𝐿𝑠 = 𝐿𝑞 + ⟹ 𝐿𝑠 = 0.2368 + ⟹ 𝑳𝒔 = 𝟏. 𝟕𝟑𝟔𝟖
(iii) Percentage of time would a pump be idle on an average 𝜇 1/15
3 (ii) Average number of customers waiting to be serviced 𝐿𝑞
= 100×(1 − 𝜌) = 100× (1 − ) = 𝟐𝟓%
4 𝐋𝐪 = 𝟎. 𝟐𝟑𝟔𝟖
(iii) Average waiting time a customer spends in the system 𝑊𝑠
2. A tax consulting firm has 3 counters in its office to receive people 𝐿𝑞 1 0.2368 1
who have problems concerning their income, wealth and sales taxes. 𝑊𝑠 = + = + ⟹𝑾𝒔 = 𝟏𝟕. 𝟑𝟔𝟖 𝒎𝒊𝒏𝒔
𝜆 μ 1/10 1/15
On the averages 48 persons arrive in an 8 hour day. Each tax advisor (iv) Average waiting time of a customer 𝑊𝑞
spends 15 minutes on the average on an arrival. If the arrivals are 𝐿𝑞 0.2368
Poisson distributed and service times are according to exponential 𝑊𝑞 = ⟹ 𝑊𝑞 = ⟹𝐖𝐪 = 𝟐. 𝟑𝟔𝟖𝐦𝐢𝐧𝐮𝐭𝐞𝐬
𝜆 1/10
distributions find (i) average number of customer in the system. (ii)
(v) The number of hours in each week a tax advisor spends performing
Average number of customers waiting to be serviced. (iii) Average 1
waiting time a customer spends in the system.( iv) Average waiting his jobs = 𝜌×8ℎ𝑟𝑠×5 𝑑𝑎𝑦𝑠 = ×8×5 = 20 ℎ𝑜𝑢𝑟𝑠
2
time of a customer. (v) The number of hours in each week a tax (vi) The expected number of idle tax advisors at any specified time
𝑠−1 3−1
advisor spends performing his jobs. (vi) The expected number of idle
tax advisors at any specified time. (vii) The probability that a ∑(𝑠 − 𝑛)𝑃𝑛 = ∑(3 − 𝑛)𝑃𝑛 = 3𝑃0 +2𝑃1+𝑃2
customer has to wait before he gets service. 𝑛=0 𝑛=0
𝑠𝑛 𝑛
Solution: 𝜌 𝑃0 , 0 < 𝑛 < 𝑠
Tax consulting firm has 3 counters 𝑠 = 3 = 3𝑃0 +2(1.5)𝑃0 +(1.5)2 /2! 𝑃0 𝑤ℎ𝑒𝑟𝑒 𝑃𝑛 = { 𝑛! 𝑠
𝑠 𝑛
𝜌 𝑃0 , 𝑛 ≥ 𝑠,
On the averages 48 persons arrive in an 8 hour day 𝑠!
Arrival rate 𝜆 = 48/8hrday ⟹ 𝜆 =
48
/𝑚𝑖𝑛 ⟹ 𝜆 = /𝑚𝑖𝑛
1 = 7.1250𝑃0 = 7.1250(0.2105) = 1.4998
8×60 10 (vii) The probability that a customer has to wait before he gets service
Each tax advisor spends 15 minutes on the average
1 1 3
Service rate 𝜇 = 1/15 𝑚𝑖𝑛𝑠 ⟹ 𝜇 = / 𝑚𝑖𝑛 (𝑠𝜌)𝑠 𝑃0 (3× 2) 0.2105
15 P[arrival will have to wait] = = = 0.2368
𝜆 1/10 𝑠! 1 − 𝜌 3! 1 − 1/2
𝜌 = ⟹ρ= ⟹𝛒 = 𝟏/𝟐
𝜇𝑠 (1/15)3
𝑠−1 −1 3−1 1 𝑛 1 3
−1 3. Given an average arrival rate of 20 per hour, is it better for a
(𝑠ρ) 𝑛
𝑠 ρ
𝑠 s (3× 2) 3 (2)
3
customer to get service at a single channel with mean service rate of
𝑃0 = [∑ + ] ⟹ 𝑃0 = [∑ + ] 22 customers per hour or at one of the two channels in parallel with
𝑛! 𝑠! 1 − ρ 𝑛! 3! 1 − 1
𝑛=0 𝑛=0 2 mean service rate of 11 customers per hour for each of the two
2 1 −1
−1 channels. Assume both queues to be of Poisson type.
(3/2)𝑛 9 (8) 3/2 (3/2)2 9 Solution: To find the better channel we should find 𝑊𝑞
⟹ 𝑃0 = [∑ + ] ⟹ 𝑃0 = [1 + + + ]
𝑛! 2 1 1! 2! 8 Average arrival rate of 20 per hour
𝑛=0 2 20 1
3 9 9 −1 38 −1 Arrival rate 𝜆 = 20/hr ⟹ 𝜆 = 60 /𝑚𝑖𝑛 ⟹ 𝜆 = 3 /𝑚𝑖𝑛
⟹ 𝑃0 = [1 + + + ] ⟹ 𝑃0 = [ ] ⟹𝐏𝟎 = 𝟎. 𝟐𝟏𝟎𝟓 (i)At a single channel with mean service rate of 22 customers per
2 8 8 8
3 22 11
1 1 hour Service rate μ = 22/hr ⟹ 𝜇 = 60 / 𝑚𝑖𝑛 ⟹ 𝜇 = 30 / 𝑚𝑖𝑛
(𝑠𝜌)𝑠 𝜌 (3× 2)
𝐿𝑞 = 𝑃 ⟹ 𝐿𝑞 = 2 0.2105 one of the two channels in parallel with mean service rate of 11
𝑠! (1 − 𝜌)2 0 3! 1 2
−
(1 2) customers per hour for each of the two channels
 7
11 14
6 7
Service rate 𝜇 = 11/ℎ𝑟 ⟹ 𝜇 = / 𝑚𝑖𝑛 For S1 Arrival rate 𝜆 = 14/hr
6 ⟹ 𝜆 = /𝑚𝑖𝑛 ⟹ 𝜆 = 15 4
/𝑚𝑖𝑛
60 60 30
(i) For single server 𝜌 =
𝜆 1/3
⟹ ρ = 11/30 ⟹𝛒 = 𝟏𝟎/𝟏𝟏 Depositors are found to arrive with mean arrival rate of 164 per hour
𝜇 16
4 47
𝜆 1/3 For S2 Arrival rate 𝜆 = 16/hr
4 ⟹ 𝜆 = 60 /𝑚𝑖𝑛 ⟹ 𝜆 = 15 /𝑚𝑖𝑛
(i) For Parallel server 𝜌 = 𝜇𝑠
⟹ ρ = (11/60)2 ⟹𝛒 = 𝟏𝟎/𝟏𝟏 𝜆 4/15
30

10 For (With drawers )S1 𝜌 = ⟹ρ= ⟹𝛒 = 𝟒/𝟓

(i) For single server 𝑃0 = 1 − ρ ⟹ 𝑃0 = 1 − ⟹𝐏𝟎 = 𝟏/𝟏𝟏 𝜇 1/3
11 𝑃0 = 1 − 𝜌 ⟹ 𝑃0 = 1 − 4/5 ⟹ 𝑃0 = 1/5
(𝑠ρ)𝑛 𝑠𝑠 ρs −1
(i) For Parallel server 𝑃0 = [∑𝑠−1
𝑛=0 𝑛!
+ 𝑠! 1−ρ] 𝜌2 (4/5)2
−1 −1
𝐿𝑞 = ⟹ 𝐿𝑞 = ⟹ 𝐋𝐪 = 𝟑. 𝟐
2−1 10 𝑛 10 2 20 𝑛 1 100 1−𝜌 1 − 4/5
(2× ) 22 (11) ) ( (
) 𝜆 7/30
⟹ 𝑃0 = [∑ 11 + ] ⟹ 𝑃0 = [∑ 11 + 2 121 ] For (Depositors) S2 𝜌 = ⟹ρ= ⟹𝛒 = 𝟕/𝟏𝟎
𝑛! 2! 1 − 10 𝑛! 1 𝜇 1/3
𝑛=0 𝑛=0
11 11 𝑃0 = 1 − 𝜌 ⟹ 𝑃0 = 1 − 7/10 ⟹ 𝑃0 = 3/10
20/11 200 −1 20 200 −1 𝜌2 (7/10)2
⟹ 𝑃0 = [1 + + ] ⟹ 𝑃0 = [1 + + ] 𝐿𝑞 = ⟹ 𝐿𝑞 = ⟹ 𝐋𝐪 = 𝟏. 𝟔𝟑𝟑𝟑
1! 11 11 11 1−𝜌 1 − 7/10
−1
231
⟹ 𝑃0 = [ ] ⟹𝑷𝟎 = 𝟎. 𝟎𝟒𝟕𝟔 (i) Effect of average waiting time for the customers if each teller could
11 handle both withdrawal and deposits (i.e ) comparison of 𝑊𝑞 for S1 ,
10 2
𝜌2 ( )
i) For single server 𝐿𝑞 = 1−𝜌 ⟹ 𝐿𝑞 = 11
10 ⟹ 𝐋𝐪 = 𝟗. 𝟎𝟗𝟎𝟗 S2 with parallel channel(both jobs)
1− 𝐿 3.2
11
10 2 10 For (With drawers )S1 𝑊𝑞 = 𝑞 ⟹ 𝑊𝑞 = ⟹𝐖𝐪 = 𝟏𝟐 𝐦𝐢𝐧𝐮𝐭𝐞𝐬
𝜆 4/15
(i) For Parallel server 𝐿𝑞 = (𝑠𝜌)
𝑠 𝜌 (2× )
(1−𝜌)2
𝑃0 ⟹ 𝐿𝑞 = 11 11
10 2
0.0476 ⟹ 𝐋𝐪 = 𝟖. 𝟔𝟓𝟒𝟓 𝐿𝑞 1.6333
For (Depositors) S2 𝑊𝑞 =
𝑠! 2! (1− )
𝐿𝑞
11
𝜆
⟹ 𝑊𝑞 = 7/30
⟹𝐖𝐪 = 𝟕 𝐦𝐢𝐧𝐮𝐭𝐞𝐬
9.0909
i) For single server 𝑊𝑞 = ⟹ 𝑊𝑞 = ⟹𝐖𝐪 = 𝟐𝟕. 𝟐𝟕𝟐𝟕 𝐦𝐢𝐧𝐮𝐭𝐞𝐬 1
𝜆 1/3 Parallel channel(both jobs) Service rate 𝜇 = 1/3𝑚𝑖𝑛 ⟹ 𝜇 = 3 / 𝑚𝑖𝑛
𝐿𝑞 8.6545
(i) For Parallel server𝑊𝑞 = ⟹ 𝑊𝑞 = ⟹𝐖𝐪 = 𝟐𝟓. 𝟗𝟔𝟑𝟔 𝐦𝐢𝐧𝐮𝐭𝐞𝐬 30
Arrival rate 𝜆 = (14 + 16)/hr ⟹ 𝜆 = 60 /min ⟹ 𝜆 = 2 /min
1
𝜆 1/3
Here Waiting time of Parallel channel is less, so it is the better one 𝜆 1/2
𝜌 = ⟹ρ= ⟹𝛒 = 𝟑/𝟒
𝜇𝑠 (1/3)2
4. A bank has two tellers working on savings accounts the first teller −1 3 𝑛 3 2
−1
𝑠−1 2−1
handles withdrawals only the second teller handle deposits only. It (𝑠ρ) 𝑛
𝑠 ρ
𝑠 s (2× )
4 +2 4 ]
2 ( )
has been found that the service time distributions for both deposits 𝑃0 = [∑ + ] ⟹ 𝑃0 = [∑
𝑛! 𝑠! 1 − ρ 𝑛! 2! 1 − 3
and withdrawals are exponential with mean service time 3 minutes 𝑛=0 𝑛=0 4
per customer. Depositors are found to arrive in a Poisson fashion 1 9 −1

throughout the day with mean arrival rate of 16 4 per hour. With (3/2)𝑛 ( ) 3/2 9 −1
⟹ 𝑃0 = [∑ + 2 16 ] ⟹ 𝑃0 = [1 + + ]
drawers also arrive in Poisson fashion with arrival rate of 14
6 per 𝑛! 1 1! 2
𝑛=0 4
hour. (i) What would be the effect of average waiting time for the
customers if each teller could handle both withdrawal and deposits 3 9 −1
⟹ 𝑃0 = [1 + + ] ⟹ 𝑃0 = [7]−1 ⟹𝐏𝟎 = 𝟎. 𝟏𝟒𝟐𝟗
Solution: 2 2
first teller handles withdrawals only(single server)S1 3 2 3
(𝑠𝜌)𝑠 𝜌 (2× 4)
the second teller handle deposits only(single server)S2 𝐿𝑞 = 𝑃 ⟹ 𝐿𝑞 = 4 0.1429 ⟹ 𝐿𝑞 = 1.9292
service time for both deposits and withdrawals are exponential with 𝑠! (1 − 𝜌)2 0 2! 3 2
(1 − 4)
mean service time 3 minutes per customer 𝐿𝑞 1.9292
1
For S1 Service rate 𝜇 = 1/3𝑚𝑖𝑛 ⟹ 𝜇 = 3 / 𝑚𝑖𝑛 𝑊𝑞 = ⟹ 𝑊𝑞 = ⟹𝐖𝐪 = 𝟑. 𝟖𝟓𝟖𝟑 𝐦𝐢𝐧𝐮𝐭𝐞𝐬
1
𝜆 1/2
For S2 Service rate 𝜇 = 1/3𝑚𝑖𝑛 ⟹ 𝜇 = 3 / 𝑚𝑖𝑛 If both teller do both jobs(parallel channel) then both withdrawers
With drawers arrive with arrival rate of 146 per hour and depositors get benefited since 𝑊𝑞 is less than that of both S1 & S2
 8
5. A general insurance company has three claim adjusters in its office. mean 10 minutes per car. The facility cannot handle more than one
People with claims against the company are found to arrive in Poisson car at a time and has a total of 5 parking spaces.(i) Find the effective
fashion at an average rate of 20 per 8 hour day. The amount of time arrival rate, (ii) What is the probability that an arriving car will get
that an adjuster spends with a claimant is found to have a negative service immediately upon arrival? and (iii) Find the expected number
exponential distribution with mean service time of 40 minutes. of parking spaces occupied.
Claimants are processed in the order of their appearance. (i) How Solution:
many hours a week can an adjuster expect to spend with Claimants? Cars arrive for service with mean 5 per hour
1
and (ii)How much time., on the average, does a claimant spend in the Arrival rate 𝜆 = 5/ℎ𝑟 ⟹ 𝜆 = / 𝑚𝑖𝑛
12
branch office. The time for washing and cleaning with mean 10 minutes per car
Solution: 1
Service rate 𝜇 = 1/10𝑚𝑖𝑛𝑠 ⟹ 𝜇 = / 𝑚𝑖𝑛
insurance company has three claim adjusters 𝑠 = 3 10
The facility cannot handle more than one car at a time and has a total of
People arrive at an average rate of 20 per 8 hour day
20 1 5 parking spaces
Arrival rate 𝜆 = 20/8hrday ⟹ 𝜆 = /𝑚𝑖𝑛 ⟹ 𝜆 = /𝑚𝑖𝑛 Single server queuing system with k = W + S = 5 + 1 = 6
8×60 24
Adjuster spends with mean service time of 40 minutes 𝜌=
𝜆
,𝜆 ≠ 𝜇 ⟹ 𝜌 =
1/12
⟹𝜌 =
5
1 𝜇 1/10 6
Service rate 𝜇 = 1/40 𝑚𝑖𝑛𝑠 ⟹ 𝜇 = 40 / 𝑚𝑖𝑛
1−ρ 1 − 5/6
𝜆 1/24 𝑃0 = ⟹ 𝑃0 = ⟹𝑷𝟎 = 𝟎. 𝟐𝟑𝟏𝟐
𝜌 = ⟹ρ= ⟹𝛒 = 𝟓/𝟗 1 − ρk+1 1 − (5/6)6+1
𝜇𝑠 (1/40)3 ρ2 [1 − kρk−1 + (k − 1)ρk ]
−1
𝑠−1 −1 3−1 5 𝑛 5 3 𝐿𝑞 =
(1 − ρ)(1 − ρk+1 )
(𝑠ρ)𝑛 𝑠 𝑠 ρs (3× ) 3 3 ( )
9 + 9 ]
𝑃0 = [∑ + ] ⟹ 𝑃0 = [∑ (5/6)2 [1 − 6(5/6)6−1 + (6 − 1)(5/6)6 ]
𝑛! 𝑠! 1 − ρ 𝑛! 3! 1 − 5 ⟹ 𝐿𝑞 =
𝑛=0 𝑛=0
9 [1 − (5/6)][1 − (5/6)6+1 ]
125 −1 (25/36)[1−6(5/6)5 +5(5/6)6 ]
2
(5/3)𝑛 9 (729) 5/3 (5/3)2 125
−1 ⟹ 𝐿𝑞 = (1/6)[1−(5/6)7 ]
⟹ 𝐿𝑞 = 1.5213 cars
⟹ 𝑃0 = [∑ + ] ⟹ 𝑃0 = [1 + + + ] 1
𝑛! 2 4 1! 2! 72 λ′ = λ(1 − PK ) ⟹ λ′ = λ(1 − P0 ρk ) ⟹ λ′ = [1 − 0.2312 (5/6)6 ]
𝑛=0
9 12
5 25 125 −1 417 −1 ⟹ λ′ = 0.0769/ min 𝑜𝑟 4.6129/ℎ𝑟
⟹ 𝑃0 = [1 + + + ] ⟹ 𝑃0 = [ ] ⟹𝐏𝟎 = 𝟎. 𝟏𝟕𝟐𝟕 (i)Effective arrival rate 𝛌′ = 𝟎. 𝟎𝟕𝟔𝟗/ 𝐦𝐢𝐧 𝒐𝒓 𝟒. 𝟔𝟏𝟐𝟗/𝒉𝒓
3 18 72 72
5 3 5 (ii)Probability that an arriving car will get service immediately upon
(𝑠𝜌)𝑠 𝜌 (3× 9) arrival (system empty) 𝑷𝟎 = 𝟎. 𝟐𝟑𝟏𝟐
𝐿𝑞 = 𝑃 ⟹ 𝐿𝑞 = 9 0.1727 ⟹𝑳𝒒 = 𝟎. 𝟑𝟕𝟒𝟖
𝑠! (1 − 𝜌)2 0 3! 5 2 (iii)Expected number of parking spaces occupied (queue) 𝐿𝑞 = 1.5213 cars
(1 − 9)
(i) Number of hours a week can an adjuster expect to spend with 2. In a single server queuing system with Poisson input and
5
Claimants = 8 ℎ𝑟𝑠×5 𝑑𝑎𝑦𝑠×ρ = 8×5× 9 = 22.2222 ℎ𝑜𝑢𝑟𝑠 Exponential service times, If the mean arrival rate is 3 calling units
(ii) Average time a claimant spend in the branch office 𝑊𝑠 per hour, the expected service time is 0.25 hour and the maximum
𝐿𝑞 1 0.3748 1 possible number of calling units in the system is 2, Find Pn (n ≥ 0),
𝑊𝑠 = + = + Average number of calling units in the system and in the queue and
𝜆 μ 1 1
24 40 average waiting time in the system and in the queue.
⟹𝑾𝒔 = 𝟒𝟖. 𝟗𝟗𝟓𝟐 𝒎𝒊𝒏𝒔 Solution:
Model III - (M/M/1)(K/FIFO)Problems Mean arrival rate is 3 calling units per hour
1. In a car wash facility, cars arrive for service according to a Poisson 1
Arrival rate 𝜆 = 3/ℎ𝑟 ⟹ 𝜆 = 20 / 𝑚𝑖𝑛
distribution with mean 5 per hour. The time for washing and cleaning
Expected service time is 0.25 hour
each car varies but is found to follow an exponential distribution with
 9
1
Service rate 𝜇 = 1/15 𝑚𝑖𝑛𝑠 ⟹ 𝜇 = / 𝑚𝑖𝑛 in the system. Also find the average waiting time of a new train
15
Maximum possible no. of calling units in the system is 2 (k=2) coming into the yard. If the handling rate is doubled, how will the
Single server queuing system with k=2 above results get modified?
𝜆 1/20 3 Solution:
𝜌=
𝜇
,𝜆 ≠ 𝜇 ⟹ 𝜌 =
1/15
⟹𝜌 =
4 The railway yard is sufficient only for two trains to wait, while the other
3 is given signal to leave the station Capacity 𝑘 = 𝑠 + 𝑤 = 1 + 2 = 3
1−ρ 1− Trains arrive at the station at an average rate of 6 per hour
𝑃0 = ⟹ 𝑃0 = 4 ⟹𝑷 = 𝟎. 𝟒𝟑𝟐𝟒
𝟎 1
1−ρ k+1
3 2+1 Arrival rate 𝜆 = 6/ℎ𝑟 ⟹ 𝜆 = / 𝑚𝑖𝑛
1−( ) 10
4 railway station can handle them on an average of 6 per hour
ρ2 [1 − kρk−1 + (k − 1)ρk ] 1
𝐿𝑞 = Service rate 𝜇 = 6/ℎ𝑟 ⟹ 𝜇 = / 𝑚𝑖𝑛
(1 − ρ)(1 − ρk+1 ) 10
Here 𝜆 = 𝜇
(3/4)2 [1 − 2(3/4)2−1 + (2 − 1)(3/4)2 ] 𝜆 1/10
⟹ 𝐿𝑞 = 𝜌= ,𝜆 = 𝜇 ⟹ 𝜌 = ⟹𝜌 = 1
[1 − (3/4)][1 − (3/4)2+1 ] 𝜇 1/10
1 1
(9/16)[1−2(3/4)+(3/4)2 ] 𝑃0 = ⟹ 𝑃0 = ⟹𝑃0 = 0.25
⟹ 𝐿𝑞 = ⟹ 𝐿𝑞 = 0.2432 calling units k+1 3+1
(1/4)[1−(3/4)3 ] k(k−1) 3(3−1)
𝐿𝑞 = 2(k+1) ⟹ 𝐿𝑞 = 2(3+1) ⟹ 𝐿𝑞 = 0.75 trains
λ′ = λ(1 − PK ) ⟹ λ′ = λ(1 − P0 ρk ) k 1 3
λ′ = λ ( ) ⟹ λ′ = ( ) ⟹ 𝛌′ = 𝟎. 𝟎𝟕𝟓𝟎/ 𝐦𝐢𝐧 𝐨𝐫 𝟒. 𝟓/𝐡𝐫
k+1 10 3+1
1 1
⟹ λ′ = [1 − 0.4324 (3/4)2 ] (i)Probabilities for the numbers of trains in the system 𝑃𝑛 = = 0.25
k+1
20
ii)Average waiting time of a new train coming into the yard 𝑊𝑞
⟹ λ′ = 0.0378/ min or 2.2703/hr Lq 0.75
(i)Steady state probability distribution of the number of calling units 𝑊𝑞 = ′ ⟹ 𝑊𝑞 = ⟹ 𝑊𝑞 = 10 mins
λ 0.0750
in the system 𝑃0 𝑎𝑛𝑑 𝑃𝑛 1
iii)If the handling rate is doubled(𝜇 = 12/ℎ𝑟 = /𝑚𝑖𝑛) , find 𝑃𝑛 𝑎𝑛𝑑 𝐿𝑞
𝑃0 = 0.4324 , 𝑃𝑛 = P0 ρn = 0.4324(3/4)𝑛 5
𝜆 1/10 1
(ii) Average number of calling units in the system 𝜌= ,𝜆 ≠ 𝜇 ⟹ 𝜌 = ⟹𝜌 =
𝜇 1/5 2
λ′ 0.0378 1−ρ 1 − 1/2
𝐿𝑠 = Lq + ⟹ 𝐿𝑠 = 0.2432 + ⟹𝑳𝒔 = 𝟎. 𝟖𝟏𝟎𝟐 𝒄𝒂𝒍𝒍𝒊𝒏𝒈 𝒖𝒏𝒊𝒕𝒔 𝑃0 = ⟹ 𝑃0 = ⟹𝑃0 = 0.5333
μ 1/15 1 − ρk+1 1 − (1/2)3+1
(iii)Average number of calling units in the queue 𝐋𝐪 = 𝟎. 𝟐𝟒𝟑𝟐 1 2
1 3−1 1 3
ρ2 [1 − kρk−1 + (k − 1)ρk ] ( ) [1 − 3 ( ) + (3 − 1) ( ) ]
(iv)Average waiting time in the system 𝑊𝑠 2 2 2
𝐿𝑞 = ⟹ 𝐿𝑞 = 3+1
𝐿𝑞 1 0.2432 1 (1 − ρ)(1 − ρk+1 ) 1 1
𝑊𝑠 = ′ + = + ⟹𝑾𝒔 = 𝟐𝟏. 𝟒𝟑𝟑𝟗 𝒎𝒊𝒏𝒔 [1 − ( )] [1 − ( ) ]
2 2
𝜆 μ 0.0378 1/15 (1/4)[1 − 3(1/2)2 + 2(1/2)3 ]
(v)Average waiting time in the queue ⟹ 𝐿𝑞 = ⟹ 𝐿𝑞 = 0.2667 𝑡𝑟𝑎𝑖𝑛𝑠
𝐿𝑞 0.2432 (1/2)[1 − (1/2)4 ]
𝑊𝑞 = ′ = ⟹𝑾𝒒 = 𝟔. 𝟒𝟑𝟑𝟗 𝒎𝒊𝒏𝒔 1
𝜆 0.0378 λ′ = λ(1 − PK ) ⟹ λ′ = λ(1 − P0 ρk ) ⟹ λ′ = [1 − 0.5333 (1/2)3 ]
10
⟹ 𝛌′ = 𝟎. 𝟎𝟗𝟑𝟑/ 𝐦𝐢𝐧 𝐨𝐫 𝟓. 𝟔/𝐡𝐫
3. At a railway station, only one train is handled at a time. The 1 n
railway yard is sufficient only for two trains to wait, while the other is (i)Probabilities for the no.of trains in the system 𝑃𝑛 = 𝑃0 ρn = 0.5333 ( )
2
given signal to leave the station. Trains arrive at the station at an (ii)Average waiting time of a new train coming into the yard 𝑊𝑞
average rate of 6 per hour and the railway station can handle them on 𝐿𝑞 0.2667
an average of 6 per hour. Assuming Poisson arrivals and Exponential 𝑊𝑞 = = ⟹𝑾𝒒 = 𝟐. 𝟖𝟓𝟖𝟓 𝒎𝒊𝒏𝒔
𝜆′ 0.0933
service distribution, Find the probabilities for the numbers of trains
 10
Model IV - (M/M/S)(K/FIFO)Problems 33
⟹ λ′ = 1 [1 − 0.009 (2)7 ]
1. A car servicing station has 3 stalls where service can be offered 3!
simultaneously. The cars wait in such a way that when a stall ⟹ λ′ = 0.4816/ min or 28.8960/hr
becomes vacant, the car at the head of the line pulls up to it. The (i) The average number of cars in the service station 𝐿𝑠
station can accommodate at most four cars waiting (seven in the λ′ 0.4816
station) at one time. The arrival pattern is Poisson with a mean of one 𝐿𝑠 = Lq + ⟹ 𝐿𝑠 = 3.1752 + ⟹𝑳𝒔 = 𝟔. 𝟎𝟔𝟒𝟖
car per minute during the peak hours. The service time is exponential μ 1/6
with mean 6 minutes. Find (i) the average number of cars in the (ii) The average waiting time 𝑊𝑞
service station during peak hours, (ii) the average waiting time and 𝐿𝑞 3.1752
𝑊𝑞 = ′ = ⟹𝑾𝒒 = 𝟔. 𝟓𝟗𝟑𝟎 𝒎𝒊𝒏𝒔
(iii) the average number of cars per hour that cannot enter the station 𝜆 0.4816
because of full capacity. (iii) The average number of cars per hour that cannot enter the
Solution: station because of full capacity (overflow 𝑃7 )
Car servicing station has 3 stalls 𝑠 = 3 𝑠𝑠 33
𝑃𝑘 = P0 ρk ⟹ 𝑃7 = 0.0009(2)7 ⟹ 𝑃7 = 0.5184
The station can accommodate at most four cars waiting (seven in the 𝑠! 3!
station) 𝑘 = 𝑠 + 𝑤 ⟹ 𝑘 = 3 + 4 ⟹ 𝑘 = 7
The arrival pattern is Poisson with a mean of one car per minute 2. At a port there are 6 unloading berths and 4 unloading crews.
Arrival rate 𝜆 = 1/min When all the berths are full, arriving ships are diverted to an overflow
The service time is exponential with mean 6 minutes facility 20 kilometers down the river. Tankers arrive according to a
1 Poisson process with a mean of 1 every 2 hour. It takes an unloading
Service rate 𝜇 = 1/6 𝑚𝑖𝑛𝑠 ⟹ 𝜇 = 6 / 𝑚𝑖𝑛
crew, on the average,10 hour to unload a tanker ,the unloading time
𝜆 1/1 following an Exponential distribution .Find(i)How many are the port
𝜌 = ⟹ρ= ⟹𝛒 = 𝟐
𝜇𝑠 (1/6)3 on the average?(ii)How long does a tanker spend at the port on the
−1
𝑠−1
(𝑠ρ)𝑛 𝑠𝑠
𝑘 3−1
(3×2)𝑛 33
7 −1
average? And (iii) What is the average arrival rate at the overflow
𝑃0 = [∑ + ∑ ρn ] ⟹ 𝑃0 = [∑ + ∑ 2n ] facility?
𝑛! 𝑠! 𝑛! 3!
𝑛=0 𝑛=𝑠 𝑛=0 𝑛=3 Solution:
−1
2
6𝑛 9
7
6 62
9
−1 6 unloading berths (place)and 4 unloading crews(men)
⟹ 𝑃0 = [∑ + ∑ 2n ] ⟹ 𝑃0 = [1 + + + (23 + 24 + 25 + 26 + 27 )] 𝑘 = 6 places or berths , 𝑠 = 4 men or crews
𝑛! 2 1! 2! 2
𝑛=0 𝑛=3
−1 −1 Tankers arrive with a mean of 1 every 2 hour
9 9 1 1
⟹ 𝑃0 = [1 + 6 + 18 + (8 + 16 + 32 + 64 + 128)] ⟹ 𝑃0 = [25 + (248)] Arrival rate 𝜆 = 1/2hr ⟹ 𝜆 = /𝑚𝑖𝑛 ⟹ 𝜆 = /𝑚𝑖𝑛
2 2 2×60 120
⟹ 𝑃0 = [25 + 1116]−1 ⟹ 𝑃0 = [1141]−1 ⟹𝐏𝟎 = 𝟎. 𝟎𝟎𝟎𝟗 on the average,10 hour to unload a tanker
1 1
(𝑠𝜌)𝑠 𝜌 Service rate 𝜇 = 1/10ℎ𝑟 ⟹ 𝜇 = /𝑚𝑖𝑛 ⟹ 𝜇 = /𝑚𝑖𝑛
𝐿𝑞 = 𝑃 {1 − (k − s + 1)ρk−s + (k − s)ρk−s+1 } 10×60 600
𝑠! (1 − 𝜌)2 0 𝜆 1/120
𝜌 = ⟹ρ= ⟹𝛒 = 𝟓/𝟒
(3×2)3 2 𝜇𝑠 (1/600)4
⟹ 𝐿𝑞 = 0.0009{1 − (7 − 3 + 1)27−3 + (7 − 3)27−3+1 } −1 5 𝑛
−1
3! (1 − 2)2 𝑠−1 𝑘 4−1 6
(𝑠ρ)𝑛 𝑠
𝑠 n (4× )
4 44 5 n
⟹ 𝐿𝑞 = 36×2×0.0009{1 − (5)(16) + 4(32)} 𝑃0 = [∑ +∑ ρ ] ⟹ 𝑃0 = [∑ +∑ ( ) ]
𝑛! 𝑠! 𝑛! 4! 4
⟹ 𝐿𝑞 = 36×2×0.0009×49 𝑛=0 𝑛=𝑠 𝑛=0 𝑛=4
−1 −1
3 6
⟹𝑳𝒒 = 𝟑. 𝟏𝟕𝟓𝟐 5𝑛 32 5
n
5 52 53 32 5
4
5
5
5
6
⟹ 𝑃0 = [∑ + ∑ ( ) ] ⟹ 𝑃0 = [1 + + + + [( ) + ( ) + ( ) ]]
𝑠𝑠 𝑛! 2 4 1! 2! 3! 3 4 4 4
λ′ = λ(1 − PK ) ⟹ λ′ = λ (1 − P ρk ) 𝑛=0 𝑛=4
𝑠! 0 25 125 32 625 3125 15625 −1
⟹ 𝑃0 = [1 + 5 + + + ( + + )]
2 6 3 256 1024 4096
 11
25 125 625 3125 15625 −1 customers arrive in a Poisson fashion at a rate of 5 per hour
⟹ 𝑃0 = [1 + 5 + + + + + ] ⟹𝐏𝟎 = 𝟎. 𝟎𝟎𝟕𝟐 5 1
2 6 64 96 384 Arrival rate 𝜆 = 5/hr ⟹ 𝜆 = /𝑚𝑖𝑛 ⟹ 𝜆 = /𝑚𝑖𝑛
(𝑠𝜌)𝑠 𝜌 60 12
𝐿𝑞 = 𝑃 {1 − (k − s + 1)ρk−s + (k − s)ρk−s+1 } each barber services with mean of 15 minutes
𝑠! (1 − 𝜌)2 0
1
5 4 5 Service rate 𝜇 = 1/15𝑚𝑖𝑛 ⟹ 𝜇 = 15 /𝑚𝑖𝑛
(4× ) 5 6−4 5 6−4+1
4 4
⟹ 𝐿𝑞 = 0.0072 {1 − (6 − 4 + 1) ( ) + (6 − 4) ( ) } 𝜆 1/12
4! 5 2 4 4
(1 − ) 𝜌 = ⟹ρ= ⟹𝛒 = 𝟓/𝟖
4 𝜇𝑠 (1/15)2
625 25 125 −1
⟹ 𝐿𝑞 = ×20×0.0072 {1 − (3) ( ) + 2 ( )} 𝑠−1 𝑘 −1 5 𝑛
2−1 5 n
(𝑠ρ)𝑛 𝑠𝑠 (2× ) 2
24 16 64 𝑃0 = [∑ + ∑ ρn ] ⟹ 𝑃0 = [∑ 8 + ∑ 2 (5) ]
625 7 𝑛! 𝑠! 𝑛! 2! 8
⟹ 𝐿𝑞 = ×20×0.0072× 𝑛=0 𝑛=𝑠 𝑛=0 𝑛=2
24 32 𝑛 −1 −1
5 1 5
⟹𝑳𝒒 = 𝟎. 𝟖𝟐𝟎𝟑 ( ) 5 n 5/4 5 2 5 3 5 4 5 5
⟹ 𝑃0 = [∑ 4 + 2 ∑ ( ) ] ⟹ 𝑃0 = [1 + + 2 [( ) + ( ) + ( ) + ( ) ]]
𝑠𝑠 𝑛! 8 1! 8 8 8 8
λ′ = λ(1 − PK ) ⟹ λ′ = λ (1 − P ρk ) 𝑛=0 𝑛=2

𝑠! 0
5 25 125 625 3125 −1
′
44 5 6 ⟹ 𝑃0 = [1 + + 2 ( + + + )] ⟹𝐏𝟎 = 𝟎. 𝟐𝟒𝟗𝟎
⟹ λ = 1 [1 − 0.0072 ( ) ] 4 64 512 4096 32768
4! 4
(𝑠𝜌)𝑠 𝜌
⟹ λ′ = 0.0059/ min or 0.3540/hr 𝐿𝑞 = 𝑃 {1 − (k − s + 1)ρk−s + (k − s)ρk−s+1 }
𝑠! (1 − 𝜌)2 0
(i) Number of crews are at the port on the average𝐿𝑠 5 2 5
′ (2× ) 5 5−2 5 5−2+1
λ 0.0059 ⟹ 𝐿𝑞 = 8 8
2 0.2490 {1 − (5 − 2 + 1) ( ) + (5 − 2) ( ) }
𝐿𝑠 = Lq + ⟹ 𝐿𝑠 = 0.8203 + ⟹𝑳𝒔 = 𝟒. 𝟑𝟔𝟎𝟑 2!
(1 − )
5 8 8
μ 1/600 8
(ii) The time a tanker spend at the port on the average𝑊𝑠 25 40 125 625
⟹ 𝐿𝑞 = × ×0.2490 {1 − (4) ( ) + 3( )}
𝐿𝑞 1 0.8203 1 32 9 512 4096
𝑊𝑠 = ′ + = + ⟹𝑾𝒔 = 𝟕𝟑𝟗. 𝟎𝟑𝟑𝟗 𝒎𝒊𝒏𝒔 25 40 1971
𝜆 μ 0.0059 1/600 ⟹ 𝐿𝑞 = × ×0.2490×
32 9 4096
(iii) The average arrival rate at the overflow facility 𝜆𝑃𝑘 = ⟹𝑳𝒒 = 𝟎. 𝟒𝟏𝟔𝟎
𝑠𝑠 1 44 5 6 𝑠𝑠
𝜆 ( 𝑠! P0 ρk ) ⟹ 𝜆𝑃𝑘 = 120 4! 0.0072 (4) ⟹ 𝜆𝑃6 = 0.0024/𝑚𝑖𝑛 λ′ = λ(1 − PK ) ⟹ λ′ = λ (1 − P ρk )
𝑠! 0
22 5 5
⟹ λ′ = 1 [1 − 0.2490 ( ) ]
3. A barber shop has two barbers and three chairs for waiting 2! 8
customers. Assume that customers arrive in a Poisson fashion at a ′
⟹ λ = 0.0794/ min or 4.7625/hr
rate of 5 per hour and that each barber services customers according (i) steady state probabilities 𝑃0 , 𝑃𝑛 𝐏𝟎 = 𝟎. 𝟐𝟒𝟗𝟎
to an exponential distribution with mean of 15 minutes. Further if a n
s n 2n 5 n
ρ P0 , 0 ≤ n < s ( ) (0.2490), 0 ≤ n < 2
customer arrives and there are no empty chairs in the shop, he will Pn = { n! , P = n! 8
ss n n
22 5 n
leave. (i)Find the steady state probabilities, (ii) what is the probability ρ P0 , s ≤ n ≤ k
s! { 2! (8) (0.2490) ,2 ≤ n ≤ 5
that the shop is empty? and (iii) what is the expected number of
(ii) the probability that the shop is empty 𝐏𝟎 = 𝟎. 𝟐𝟒𝟗𝟎
customers in the shop?
(iii) expected number of customers in the shop 𝐿𝑠
Solution:
A barber shop has two barbers and three chairs for waiting customers
λ′ 0.0794
𝐿𝑠 = Lq + ⟹ 𝐿𝑠 = 0.4160 + ⟹𝑳𝒔 = 𝟏. 𝟔𝟎𝟕𝟎
𝑘 = 𝑠 + 𝑤 ⟹ 𝑘 = 3 + 2 ⟹ 𝑘 = 5 , 𝑠 = 2 barbers μ 1/15
 12
(𝐌/𝐌/𝟏): (∞/𝐅𝐈𝐅𝐎) (𝐌/𝐌/𝐒): (∞/𝐅𝐈𝐅𝐎) (𝐌/𝐌/𝟏): (𝐊/𝐅𝐈𝐅𝐎) (𝐌/𝐌/𝐒): (𝐊/𝐅𝐈𝐅𝐎)
λ λ λ λ
𝛒 ρ= ρ= ρ= ρ=
μ μs μ μs
s−1 −1 1−ρ −1
(sρ)n s s ρs ,λ ≠ μ s−1 k
1 − ρk+1 (sρ)n ss
𝐏𝟎 P0 = 1 − ρ P0 = [∑ +( )] P0 = P0 = [∑ + ∑ ρn ]
n! s! 1 − ρ 1 n! s!
n=0
{ k + 1 ,λ = μ n=0 n=s
n
s n sn n
ρ P0 , 0 ≤ n < s ρn P0 , λ ≠ μ ρ P0 , 0 ≤ n < s
𝐏𝐧 Pn = P0 ρn Pn = { n! s Pn = { 1 Pn = { n!
s n ,λ = μ ss n
ρ P0 , n ≥ s k+1 ρ P0 , s ≤ n ≤ k
s! s!
ρ2 [1 − kρk−1 + (k − 1)ρk ]
2 (sρ)s ρ ,λ ≠ μ (sρ)s ρ
ρ (1 − ρ)(1 − ρk+1 )
𝐋𝐪 Lq = Lq = P Lq = Lq = P [1 − (k − s + 1)ρk−s + (k − s)ρk−s+1 ]
1−ρ s! (1 − ρ)2 0 K(K − 1) s! (1 − ρ)2 0
,λ = μ
{ 2(K − 1)
λ λ λ′ λ′
𝐋𝐬 Ls = Lq + Ls = Lq + Ls = Lq + Ls = Lq +
μ μ μ μ
Lq 1 Lq 1 Lq 1 Lq 1
𝐖𝐬 Ws = + Ws = + Ws = + Ws = +
λ μ λ μ λ′ μ λ′ μ
Lq Lq Lq Lq
𝐖𝐪 Wq = Wq = Wq = Wq =
λ λ λ′ λ′
(sρ)s P0 (sρ)s P0
Busy

P(n − 1 > 0) = ρ P(n − 1 > 0) = P(n − 1 > 0) = 1 − P0 , λ ≠ μ or λ = μ P(n − 1 > 0) = [1 − ρk−s+1 ]

s! 1 − ρ s! 1 − ρ

Non empty Queue Lw Non empty Queue Lw 𝐄𝐟𝐟𝐞𝐜𝐭𝐢𝐯𝐞 λ(1 − PK ) ,λ ≠ μ

1 ρ 𝐚𝐫𝐫𝐢𝐯𝐚𝐥 𝜆′ 𝑜𝑟 𝜆𝑒𝑓𝑓 = { λk Effective arrival rate 𝜆′ 𝑜𝑟 𝜆𝑒𝑓𝑓 = λ(1 − PK )
E(Lq /q > 0) = E(Lq /q > 0) = ,λ = μ
1−ρ 1−ρ 𝐫𝐚𝐭𝐞 k+1
P[waiting time exceeds Expected number of idle servers Overflow facility PK = ρk P0 ss k
′𝐭′ mins in the system] Overflow facility PK = ρ P0
s!
𝐏(𝐖𝐬 > 𝑡) = 𝐞(𝛌−𝛍)𝐭 ∑s−1
n=0 (s − n) Pn
′
P[waiting time exceeds Average arrival rate at overflow facility Average arrival rate at overflow facility λ PK or λeff PK
′𝐭′ mins in the queue]
𝛌 λ′ PK or λeff PK
𝐏(𝐖𝐪 > 𝑡) = 𝐞(𝛌−𝛍)𝐭
𝛍