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**5-1 – Queueing Theory
- 1 Introduction:** A flow of customers from finite or infinite population towards the service facility forms a queue (waiting line) an account of lack of capability to serve them all at a time. In the absence of a perfect balance between the service facilities and the customers, waiting time is required either for the service facilities or for the customers arrival. In general, the queueing system consists of one or more queues and one or more servers and operates under a set of procedures. Depending upon the server status, the incoming customer either waits at the queue or gets the turn to be served. If the server is free at the time of arrival of a customer, the customer can directly enter into the counter for getting service and then leave the system. In this process, over a period of time, the system may experience “ Customer waiting” and /or “Server idle time” 5. 1. 2 Queueing System: A queueing system can be completely described by (1) the input (arrival pattern) (2) the service mechanism (service pattern) (3) The queue discipline and (4) Customer’s behaviour 5. 1 .3. The input (arrival pattern) The input described the way in which the customers arrive and join the system. Generally, customers arrive in a more or less random manner which is not possible for prediction. Thus the arrival pattern can be described in terms of probabilities and consequently the probability distribution for inter-arrival times (the time between two successive arrivals) must be defined. We deal with those Queueing system in which the customers arrive in poisson process. The mean arrival rate is denoted by .
5. 1. 4 The Service Mechanism:- This means the arrangement of service facility to serve customers. If there is infinite number of servers, then all the customers are served instantaneously or arrival and there will be no queue. If the number of servers is finite then the customers are served according to a specific order with service time a constant or a random variable. Distribution of service time follows ‘ Exponential distribution’ defined by f(t) = e - t , t > 0 The mean Service rate is E(t) = 1/ 5. 1. 5 Queueing Discipline:- It is a rule according to which the customers are selected for service when a queue has been formed. The most common disciplines are
- First come first served – (FCFS)
- First in first out – (FIFO)
- Last in first out – (LIFO)
- Selection for service in random order (SIRO) 5. 1. 6 Customer’s behaviour
- Generally, it is assumed that the customers arrive into the system one by one. But in some cases, customers may arrive in groups. Such arrival is called Bulk arrival.
- If there is more than one queue, the customers from one queue may be tempted to join another queue because of its smaller size. This behaviour of customers is known as jockeying.
- If the queue length appears very large to a customer, he/she may not join the queue. This property is known as Balking of customers.
- Sometimes, a customer who is already in a queue will leave the queue in anticipation of longer waiting line. This kind of departare is known as **reneging.
- 7 List of Variables** The list of variable used in queueing models is give below: n - No of customers in the system C - No of servers in the system Pn (t) – Probability of having n customers in the system at time t. Pn - Steady state probability of having customers in the system P 0 - Probability of having zero customer in the system Lq - Average number of customers waiting in the queue. Ls - Average number of customers waiting in the system (in the queue and in the service counters) Wq - Average waiting time of customers in the queue. Ws - Average waiting time of customers in the system (in the queue and in the service counters) - Arrival rate of customers - Service rate of server - Utilization factor of the server eff - Effective rate of arrival of customers M - Poisson distribution N - Maximum numbers of customers permitted in the system. Also, it denotes the size of the calling source of the customers. GD - General discipline for service. This may be first in first – serve (FIFS), last-in-first serve (LIFS) random order (Ro) etc.
Lq =
2
Aver 1 -^ ^
s in the system
(in the queue and in the service station) = Ws
= Ws = Ls = (1 - ) = x 1 = 1 1 - (1 - ) = 1 -^ (Since = ) = 1 -
Ws = 1
Wq = Average waiting time of customers in the queue.
= Lq / = [ 1 / ] [
2
/ [1- ]]
= 1 / [
2
/ [1- ]]
= Since = -^ Example 1:
Wq =
The arrival rate of customers at a banking counter follows a poisson distibution with a mean of 30 per hours. The service rate of the counter clerk also follows poisson distribution with mean of 45 per hour. a) What is the probability of having zero customer in the system? b) What is the probability of having 8 customer in the system? c) What is the probability of having 12 customer in the system ? d) Find Ls, Lq, Ws and Wq Solution Given arrival rate follows poisson distribution with mean = 30 = 30 per hour Given service rate follows poisson distribution with mean = 45 = 45 Per hour Utilization factor = / = 30 / 45 = 2 / 3 = 0. 67 a) The probability of having zero customer in the system P 0 = ^0 (1- ) = 1- = 1 - 0. (^67 )
age waiting time of customer
b) The probability of having 8 customers in the system P 8 = 8 (1- ) = ( 0. 67 ) 8 (1- 0 .67) = 0. 0406 x 0. 33 = 0. 0134 Probability of having 12 customers in the system is P 12 = ^12 (1- ) = ( 0. 67 ) 12 (1- 0. 67 ) = 0. 0082 x 0. 33 = 0. 002706 Ls = = 0. 67 1 - 1-0. 67 = 0. 67 = 2. 03
- 33 = 2 customers Lq = ^2 = (0. 67 )^2 = 0. 4489 1- 1-0. 67 0. 33 = 1. 36 = 1 Customer Ws = 1 = 1 = 1 - 45-30 15 = 0. 0666 hour Wq = = 0. 67 = 0. 67 - 45-30 15 Example 2 : = 0. 4467 hour
At one-man barbar shop, customers arrive according to poisson dist with mean arrival rate of 5 per hour
and the hair cutting time was exponentially distributed with an average hair cut taking 10 minutes. It is
assumed that because of his excellent reputation, customers were always willing to wait. Calculate the
following:
(i) Average number of customers in the shop and the average numbers waiting for a haircut. (ii) The percentage of time arrival can walk in straight without having to wait. (iii) The percentage of customers who have to wait before getting into the barber’s chair. Solution:- (i) Average number of customers in the system (numbers in the queue and in the service station) Ls = / 1- = 0. 83 / 1 - 0. 83 = 0. 83 / 0. 17 = 4. 88 = 5 Customers (ii) The percentage of time arrival can walk straight into barber’s chair without waiting is Service utilization = % = / % = 0. 833 x 100 = 83. 3 Given mean arrival of customer = 5 / 60 = 1 / 12 and mean time for server = 1 / 10 = / = [ 1 /12] x 10 = 10 / 12 = 0. 833
(v) Maximum number of customers permitted in the system is infinite Then the steady state equation to obtain the probability of having n customers in the system is Pn = ^ n Po , o ^ n ^ C n! = n Po for n > c Where / c < 1 C n-c C! C- Where [ / c] < 1 as = / P 0 ={[ n /n!] + c / (c! [ 1 - /c])}
n = 0 where c! = 1 x 2 x 3 x …………….. upto C Lq = [ c+^1 / [c-1! (c - )] ] x P 0 = (c Pc) / (c - )^2 Ls = Lq + and Ws = Wq + 1 / Wq = Lq / Under special conditions Po = 1 - and Lq = C+!^ / c 2 Where < 1 and Po = (C-) (c – 1)! / c c and Lq = / (c- ), where / c < 1 Example 1: At a central warehouse, vehicles are at the rate of 24 per hour and the arrival rate follows poisson distribution. The unloading time of the vehicles follows exponential distribution and the unloading rate is 18 vehicles per hour. There are 4 unloading crews. Find (i) Po and P 3 (ii) Lq, Ls, Wq and Ws
Solution:
Arrival rate = 24 per hour Unloading rate = 18 Per hour No. of unloading crews C= 4 = / = 24 / 1 8 = 1. 33 C- 1 (i) P 0 ={[ n/n!] + c^ / (c! [ 1 - /c])}- 1 n = 0 3 ={[ (1. 33 )n/n!]+ ( 1. 33 )^4 /(4! [1 - ( 1. 3 3)/ 4])}-^1 n = 0 ={ ( 1. 33 )^0 / = 0! + ( 1. 33 )^1 / 1! + ( 1. 33 )^2 / 2! + (1. 33 )^3 / 3!
- ( 1. 3 3)^4 / 24! [1 - (1. 3 3)/ 4] }- =[1 + 1. 33 + 0. 88 + 0. 39 + 3. 129 / 16 .62] -^1 =[3. 60 + 0. 19 ]
- = [3. 79 ] - 1 = 0. 264 We know Pn = ( n^ / n!) Po for 0 n c P 3 = ( ^3 / 3!) Po Since 0 3 4 = [(1. 33 )^3 / 6 ] x 0. 264 = 2. 353 x 0. 044 = 0. 1035 (ii) Lq = =
C+ 1 X P 0 (C – 1)! (C-)^2 (1. 33 )^5 X^0.^264 3! X (4 – 1. 33 )^2 = (4. 161 6) X 0. 264 6 X ( 2. 77 )^2
Percentage of time the service remains idle = 67 % approximately c) The expected length of waiting time (w/w>0) = 1 / (c - ) = 1 / [(1 / 2) – (1 / 6) ] = 3 minutes Examples 3 : A petrol station has two pumps. The service time follows the exponential distribution with mean 4 minutes and cars arrive for service in a poisson process at the rate of 10 cars per hour. Find the probability that a customer has to wait for service. What proportion of time the pump remains idle?
Solution: Given C= 2 The arrival rate = 10 cars per hour.
= 10 / 60 = 1 / 6 car per minute Service rate = 4 minute per cars. Ie = ¼ car per minute. = / = ( 1 /6) / ( 1 /4) = 2 / 3 = 0. 67 Proportion of time the pumps remain busy = / c = 0. 67 / 2 = 0. 33 = 1 / 3 The proportion of time, the pumps remain idle = 1 – proportion of the pumps remain busy = 1-1 / 3 = 2 / 3 C- P 0 ={[ n/n!] + c^ / (c! [ 1 - /c])}- n = 0 =[ ( 0. 67 )^0 / 0!) + ( 0. 67 )^1 / 1!) + ( 0. 67 )^2 / 2!)[ 1 - ( 0. 67 / 2)^1 ]- =[1 + 0. 67 + 0. 4489 / (1. 3 3)]- =[1 + 0. 67 + 0. 33 ]- =[ 2 ]- = 1 / 2 Probability that a customer has to wait for service = p [w>0] = ^ c x P 0 = (0. 67 ) 2 x 1 / 2 [c [1 - / c] [2![1 – 0. 67 /2] = 0. 4489 = 0. 4489 1 .33x2 2. 66 **= 0. 1688
- 2 Simulation :** Simulation is an experiment conducted on a model of some system to collect necessary information on the behaviour of that system. 5. 2. 1 Introduction :
The representation of reality in some physical form or in some form of Mathematical equations
are called Simulations.
Simulations are imitation of reality. For example :
- Children cycling park with various signals and crossing is a simulation of a read model traffic system
- Planetarium
- Testing an air craft model in a wind tunnel. 5. 2 .2. Need for simulation :
Consider an example of the queueing system, namely the reservation system of a transport corporation.
The elements of the system are booking counters (servers) and waiting customers (queue). Generally
the arrival rate of customers follow a Poisson distribution and the service time follows exponential
distribution. Then the queueing model (M/M/1) : (GD/ / ) can be used to find the standard results.
But in reality, the following combinations of distributions my exist.
- Arrived rate does not follow Poisson distribution, but the service rate follows an exponential distribution.
- Arrival rate follows a Poisson distribution and the service rate does not follow exponential distribution.
- Arrival rate does not follows poisson distribution and the service time also does not follow exponential distribution. In each of the above cases, the standard model (M/M/1) : (G/D/ / ) cannot be used. The last resort to find the solution for such a queueing problem is to use simulation. 5. 2 .3. Some advantage of simulation :
- Simulation is Mathematically less complicated
- Simulation is flexible
- It can be modified to suit the changing environments.
- It can be used for training purpose
- It may be less expensive and less time consuming in a quite a few real world situations. 5. 2 .4. Some Limitations of Simulation :
- Quantification or Enlarging of the variables maybe difficult.
- Large number of variables make simulations unwieldy and more difficult.
- Simulation may not. Yield optimum or accurate results.
- Simulation are most expensive and time consuming model.
- We cannot relay too much on the results obtained from simulation models. 5. 2 .5. Steps in simulation :
- Identify the measure of effectiveness.
- Decide the variables which influence the measure of effectiveness and choose those variables, which affects the measure of effectiveness significantly.
- Determine the probability distribution for each variable in step 2 and construct the cumulative probability distribution.
- Choose an appropriate set of random numbers.
- Consider each random number as decimal value of the cumulative probability distribution.
- Use the simulated values so generated into the formula derived from the measure of effectiveness.
- Repeat steps 5 and 6 until the sample is large enough to arrive at a satisfactory and reliable decision. 5. 2 .6. Uses of Simulation Simulation is used for solving 1 .Inventory Problem
- Queueing Problem
- Training Programmes etc. Example :
2. Vehicles pass through a toll gate at a rate of 90 per hour. The average time to pass through the gate is
36 seconds. The arrival rate and service rate follow poisson distribution. There is a complaint the
vehicles wait for long duration. The authorities are willing to install one more gate to reduce the
average time to pass through the toll gate to 30 seconds if the idle time of the toll gate is less than 1 0%
and the average queue length at the gate is more than 5 vehicles. Vehicle whether the installation of
second gate is justified?
3. At a central ware house, vehicles arrive at the rate of 24 per hours and the arrival rate follows poisson
distribution. The unloading time of the vehicles follows exponentional distribution and the
unloading rate is 18 vehicles per hour. There are 4 unloading crews. Find the following.
a) P 0 and P 3
b) Lq, Ls, Wq and Ws
4. Explain Queneing
Discipline
5. Describe the Queueing models (M/M/1) : (GD/
and (M/M/C) : (GD/ / )
6. Cars arrive at a drive-in restaurant with mean arrival rate of 30 cars per hors and the service
rate of the cars is 22 per hors. The arrival rate and the service rate follow poisson distribution. The
number parking space for cars is only 5. Find the standard results.
( Ans Lq = 2. 38 cars, Ls = 3. 3133 Cars, Wq = 0. 116 hors and
Ws = 0. 1615 hors)
7. In a harbour, ship arrive with a mean rate of 24 per week. The harbour has 3 docks to
handle unloading and loading of ships. The service rate of individual dock is 12 per week. The arrival
rate and the service rate follow poisson distribution. At any point of time, the maximum No. of ships
permitted in the harbour is 8. Find P 0 , Lq, Ls, Wq, Ws
(Ans P 0 = 0. 1998 , Lq = 1. 0371 ships, Ls = 2. 9671 ships,
Wq = 0. 04478 week and Ws = 0. 1281 week)
8. Define simulation and its
advantages.
9. Discuss the steps of
simulation.
5 .3. Replacement models
5. 3. 1. Introduction:
The replacement problems are concerned with the situations that arise when some items such as men,
machines and usable things etc need replacement due to their decreased efficiency, failure or
breakdown. Such decreased efficiency or complete breakdown may either be gradual or all of a sudden.
If a firm wants to survive the competition it has to decide on whether to replace the out dated
equipment or to retain it, by taking the cost of maintenance and operation into account. There are two
basic reasons for considering the replacement of an equipment.
They are
N
M
The points M and N denote optimal level of maintenance and optimal cost respectively
5. 3. 3 Types of replacement problem
The replacement problem can be classified into two categories.
i) Replacement of assets that deteriorate with time (replacement due to gradual failure, due to
wear and tear of the components of the machines) This can be further classified into the
following types.
a) Determination of economic type of an asset.
b) Replacement of an existing asset with a new asset.
ii) simple probabilistic model for assets which will fail completely (replacement due to sudden
failure).
5. 3. 4. Determination of Economic Life of an asset
Any asset will have the following cost components
i) Capital recovery cost (average first cost), Computed form the first cost (Purchase price) of the
asset.
ii) Average operating and maintenance cost.
iii) Total cost which is the sum of capital recovery cost (average first cost) and average
operating and maintenance cost.
A typical shape of each of the above cost with respect to
life of the asset is shown below
From figure, when the life of the machine increases, it is clear that the capital recovery cost (average
first cost) goes on decreasing and the average operating and maintenance cost goes on
increasing. From the beginning the total cost goes on decreasing upto a particular life of the asset and
then it starts increasing. The point P were the total cost in the minimum is called the Economic life of
the asset. To solve problems under replacement, we consider the basics of interest formula.
Present worth factor denoted by (P/F, i,n). If an amount P is invested now with amount earning interest
at the rate i per year, then the future sum (F) accumulated after n years can be obtained.
P - Principal sum at year Zero
F - Future sum of P at the end of the n
th
year i - Annual interest rate
n - Number of interest periods.
Then the formula for future sum F = P ( 1 + i )
n
P = F/(1 +i)
n
= Fx (present worth factor)
If A is the annual equivalent amount which occurs at the end of every year from year one through n
years is given by
A = P x i ( 1 +i)
n
( 1 +i)
n
= P ( A / P, i, n )
= P x equal payment series capital recovery factor
Example:
A firm is considering replacement of an equipment whose first cost is Rs. 1750 and the scrap value is
negligible at any year. Based on experience, it is found that maintenance cost is zero during the first
year and it increases by Rs. 100 every year thereafter.
(i) When should be the equipment replaced if a)
i = 0%
b) i = 12%
Solution :
Given the first cost = Rs 1750 and the maintenance cost is Rs. Zero during the first years and then
increases by Rs. 100 every year thereafter. Then the following table shows the calculation.
Calculations to determine Economic life
(a) First cost Rs. 1750 Interest rate = 0 %
End of
year (n)
Maintenan
ce cost at
end of year
Summation
of
maintenanc
e
Average cost
of
maintenance
through the
Average first
cost if
replaced at the
given
Average
total cost
through the
given year
Identify the end of year for which the annual equivalent total cost is minimum in column. In this
problem the annual equivalent total cost is minimum at the end of year hence the economics life of the
equipment is 7 years.
5. 3 .5. Simple probabilistic model for items which completely fail
Electronic items like bulbs, resistors, tube lights etc. generally fail all of a sudden, instead of gradual
failure. The sudden failure of the item results in complete breakdown of the system. The system ma y
contain a collection of such items or just an item like a single tube-light. Hence we use some
replacement policy for such items which would minimize the possibility of complete breakdown. The
following are the replacement policies which are applicable in these cases.
i) Individual replacement policy :
Under this policy, each item is replaced immediately after failure.
ii) Group replacement policy :
Under group replacement policy, a decision is made with regard the replacement at what equal internals,
all the item are to be replaced simultaneously with a provision to replace the items individually which
fail during the fixed group replacement period.
Among the two types of replacement polices, we have to decide which replacement policy we have to
follow. Whether individual replacement policy is better than group replacement policy. With regard
to economic point of view. To decide this, each of the replacement policy is calculated and the most
economic one is selected for implementation.
Exercise :
1. List and explain different types of maintenance
2. Discuss the reasons for maintenance.
3. Distinguish between breakdown maintenance and preventive maintenance.
4. Distinguish between individual and group replacement polices.
5. A firm is considering replacement of an equipment whose first cost is Rs. 4000 and the scrap value
is negligible at the end of any year. Based on experience, it has been found that the
maintenance cost is zero during the first year and it is Rs. 1000 for the second year. It increase by Rs. 300
every years thereafter.
a) When should the equipment be replace if i = 0%
b) When should the equipment be replace if i = 12%
Ans. a) 5 years b) 5 years
6. A company is planning to replace an equipment whose first cost is Rs. 1 , 00 ,000. The operating and
maintenance cost of the equipment during its first year of operation is Rs. 10 , 000 and it increases by
Rs. 2 , 000 every year thereafter. The release value of the equipment at the end of the
MBA-H 2040 Quantitative
Techniques for Managers first year of its operation is Rs. 65 , 000 and it decreases by Rs. 10 , 000
every year thereafter. Find the economic life of the equipment by assuming the interest
rate as 12 %.
[Ans : Economic life = 13 years and the corresponding annual equivalent cost = Rs.
34 ,510]
7. The following table gives the operation cost, maintenance cost and salvage value at
the end of every year of machine whose purchase value is Rs. 12 ,000. Find the
economic life of the machine assuming.
a) The interest rate as 0%
b) The interest rate as 15%
End of
year
Operation cost at
the end of year
(Rs)
Maintenance cost
at the end of year
(Rs)
Salvage value at the end
of year (Rs)
Ans :
a) Economic life of the machine
= 2 years b) Economic life of the
machine = 2 years
