Lecturenotes Krs OR7

8/13/2019 Lecturenotes Krs OR7

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Operation Research /Operational Research (OR)

Chapter 6- Queuing theory

Definition:- Queuing theory is the mathematical study of waiting lines

or queues. It is branch of applied probability theory.

Queuing Theory can be used to describe these real world queues, ormore abstract queues, such as are often found in many branches ofcomputer science, for example in operating system design.

It relies heavily on mathematics, especially statistics.

Examples

Banks/supermarkets - waiting for service

Computers - waiting for a response

Failure situations - waiting for a failure to occur e.g. in a piece ofmachinery

public transport - waiting for a train or a bus

Need for Queuing theory : - Optimum utilization of resource consideringthe better customer service.

Costs:- Co = Cost of operating the service facility

Cw = Cost of waiting customers per unit line

Total cost = Co +Cw. When Co is less, Cw increases and vice versa. This isshown in the below figure.


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Terminologies Queuing theory :- Customer, Server are principal player

Source:- from where customers are generated. Finite & Infinite

Facility:- Where the service is provided to the customer

Queue:- Customer in the line waiting for the service at facility

Interarrival time: - time between the arrival of two successive customers. Helpanalyze the arrival process.

Service time :- The time required to complete the service at facility. Helps toanalyze the service.

Queue size: - Number of customer s in the queue. Can be infinite as in mail ofacility.

Queue discipline (FCFS, LCFS, SIRO) :- Order in which customer are selectefrom the queue. Imp factor in analysis of models. Main are

FCFS: - First come, First Service


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LCFS: - Last come, First Service

SIRO: - Service in Random order

Priority, Round Robin

Jockey:- Behavior of human from changing his/hers queue

Balk: - Reject from joining a queue altogether

Renege: - Break up from a queue

Pure Birth & Death Models

Pure Birth Model Pure Death Model

Only arrival is allowed Only departure is allowed

Arrival rate in exponential distribution

At = inter arrival time

- Service rate in exponential distribution

Dt = inter departure time

Customer at t =0 is 0 Customer at t=0 is N

p0 (t) Probability of no arrivals during tipN(t) Probability of no departure during t


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p0 (t) = p(At>t) = 1- p(At


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Limh0 [pn(t+h)- pn(t)]/h

= - pn(t) + pn-1(t) = pn(t)

i.e. 1st derivative of pn(t)

Limh0 [pn(t+h)- pn(t)]/h

= - pn(t) pn(t) + pn+1(t) = pn(t)

derivative of pn(t)

Limh0 [p0(t+h)- p0(t)]/h

= - p0(t) = p0(t)

i.e. 1st derivative of p0(t)

Limh0 [pn(t+h)- pN(t)]/h

= - pN(t) = pN(t)

i.e. 1st derivative of pN(t)

p0(t)= p1(t)

pn(t) =[( t)n e- t ]]/ n! is a Poisson distribution

p0(t)= e- t

pn(t) =[( t)N-n e- t ]]/ (N-n)! is a Pois

pN(t)= e- t

p0(t)= 1- pn(t) (from n=1 to n=N)

Exponential Poisson

Random variable Time between successivarrivals, t

Number of arrivals n, during a


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Range t> 0 n=0.1.2.3..

Density function f(t)= e- t pn(T) =[( T)n e- T ]]/ n!

Mean Value 1/ time units T arrivals during T

Cumulative probability p(t< A)=1- e- A pnA) = e- A p0(A)=e- A

Problem :-

In a state, rate of birth of babies is one in every 12 minutes. Time between birth is inexponential in nature. Find the following

a) The average number of births per year.

b) The probability that no births in any one day

c) The probability of issuing 50 Birth certificates in 3 hrs given that 40 certificat

were issued during the first 2 hrs of the 3 hr period.

Solution:-


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= 24*60/12 = 120 births/day

a)Total number of births/year = t = 120* 365 = 43,800 birth/yr

b) Prob of no births in any one day = p0(1)= e- t = e- 120*1 = 0

c) Remaining certificates =50- 40 =10

Remaining time = 3-2 = 1hr

So reqd to find prob issuing 10 certificates in one hour

Convert in /hour unit , = 120/24 = 5 birth /hour.

So p10(1)= (5*1)10 e- 5 / 10! = 0.0183

Note :-pn(T) =[( T)n e- T ] ]/ n!

Problem :-

The florist section stocks 18 dozen roses at the beginning of each week. Per day sale

average 3 dozen. (One dozen at a time). Actual demand follows Poisson distributionWhen stock level reaches 5 dozens, new stock of 18 dozens will be placed for delivthe beginning of the following week.. All roses left at t he end of the week are dispooff. Determine

a) The prob of placing order in any one day of the week.

b) The average no of dozen roses that will be discarded at the end of the week.

Solution:-

= 3 dozens/day


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Prob of placing an order by the end of the day (t)

=pn


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Generalized Poisson Queuing Model:-

Includes both arrivals and departures based on the Poisson distribution with tin exponential distribution

Based on the steady state behavior of the queue (long run) and not considerinthe transit ( warm up behavior)

Assumes both arrival rate and departure rate are state dependent ( and arecontinuously changing and depends on the number of customers in the facilit

This model derives as a function ofn and n. And using pn , determine the othemeasures like

Average Queue length


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Average waiting time

Average utilization of the facility

Transition rate diagram

Steady state number in the Queue is n

For any small interval h only one event can occur


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If at time t, number =n, for t+h, n can change to n+1 with one arrival at the raand n-1 with one departure n

State 0 can be changed to 1 when = 0

is undefined as system is empty and no departure.

Under steady state conditions, for n >0 ,

Expected rate of arrival =Expected rate of of departures

Expected rate of flow into system with state n

= n-1 pn-1 + n+1 pn+1 Expected rate of flow out of system with state n

=( n+ n) pn

Equating the above two equations, we get balance equation

n-1 pn-1 + n+1 pn+1 =( n+ n) pn where n=1,2,3

When n=0, we get 0p0 =1p1

Or p1 = 0p0 /1 or ( 0 / 1)*p0

When n =1, we get 0p0 +2p2 =( 1+ 1) p1

So p2 ={( 1+ 1) p1 - 0p0} /2

Sub substituting for p1

we get p2 ={( 1+ 1) ( 0 / 1)*p0 - 0p0} /2

= p0{( 1 0 / 1) + 0 - 0)}/2 = ( 1 0 / 2 1 ) p0

In general,


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pn = {( n-1 n-2 n-3.. 0) /(n+1 n n-1.. 1 )} p0

Where P0 = 1- Pn where n= 1 to infinite

Problem :- A grocery shop has 3 counters and condition for operation of each counbased on the number of customers and given as below

Customers 1 to 3 . 1 counter in operation.

Customers 4 to 61 & 2 are in operation

Customers > 6 All 3 in operation

Arrival of customer to the shop in Poisson distribution with mean rate as 10/hr. andaverage departure time is in exponential and average or mean is 12 minutes. Determthe steady state probability pn of n customers in the queue

Solution :-

n = = 10 customer/hour.

For n =1,2,3 only 1 counter operates Son = 60/12 = 5 customers /hour

For n= 4,5,6 2 counter operates and Son = 5*2= 10 customers /hour

For n = 7,8 3 counters operates . So n = 5*3 = 15 customers /hour

N Pn Equation

1 P1 ( 0/1)p0 = 10/5p0 =2p0


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2 P2 ( 1 0/ 21)p0 =10*10/(5*5)p0 =4p0

3 p3 ( 2 1 0/ 3 21)p0 =10*10*10/(5*5*5)p0=8p0

N Pn Equation

4 P4 ( 3 2 1 0/ 4 3 21)p0 =104/( 10*53)p0 =8p0

5 p5 ( 4 3 2 1 0/5 4 3 21)p0 =105/( 102*53)p0 =8p0

6 P6 ( 5 4 3 2 1 0/65 4 3 21)p0 =106/( 103*53) p0 = (10/5)3* (1=8p0

n pn ( n-1 n-2.. 5 4 3 2 1 0/n n-1.. 6 5 4 3 21)p0 = (10/5)3*(10/15)n-6p0

=8(2/3)n-6p0

To find p0 :-


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p0+p0(2+4+4*8+ 8*(2/3) + 8*(2/3)2 + 8*(2/3)3 . = 1

p0(1+2+4+4*8+ 8*(2/3) + 8*(2/3)2 + 8*(2/3)3 . = 1

p0{31+8[(2/3) + (2/3)2 +(2/3)3..]} =1

p0 {31+8[ 8/(1-2/3)]} = 1

p0 (55)=1 or p0 =1/55

Note Geometric sum seriesxi =1 /(1-x), wheremod x


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Characteristics of the Queuing models :-

a = Input arrival distribution

b = Departure or service time distribution

c = Number of parallel server

d = Queue or service discipline

e = maximum length of queue allowed (in queue + service)

f = Size of the calling source (population)

Representation of the Queuing models :-

Devised by D.G Kendall in 1953 Devised by A.M Lee d&e in 1953, f in

Standard Notations for arrival & departure distribution i.e a & b are

M=Monrovian or Poisson arrivals or departure distribution ( exponential inter arrivservice time distribution)

D = Constant time (Deterministic)

Ek = Erlang or Gamma distribution (or sum of independent exponential distribution

GI = General distribution of Interarrival time


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G = General Distribution of Service time

Standard notation for queue discipline i.e d are FCFS, LCFS, SIRO, GD (general

distribution)c = any number

e & f = finite number or infinite

Steady state measure of performance

Ls= Expected number of customers in system

Lq = Expected number of customers in queueWs = Expected waiting time in system

Wq = Expected waiting time in queue

C bar = Expected number of busy server

Ls = npn for n=1 to infinite

Lq = (n-c)pn for n=c+1 to infinite

Ls= eff Ws

Lq = eff Wq

Ws = Wq +1/

By multiplying the above eqn by eff on both sides we get Ls = Lq+ eff /

Average number of busy server C bar = Ls Lq = eff /

Single Server Models:- c=1

M/M/1: GD/infinite/infinite


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Per generalized solution, n= andn = for a for n =0,1,2

effective = and lost = 0

= / =Arrival rate/Service rate = Traffic intensity

pn = np0 where n= 0,1,2

p0 (1+ + 2+ 3 .)=1

Assume


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Let = t1 + t2 +tn+1

where

t1 is the time required for the customer in service to complete it

t2+ t3+..tn are service times for n-1 customers in the queue.

tn+1 represents the service time for arriving customer.

Then w( / n+1) is the conditional density function of given n customers in the systeahead of the arriving customer

w() = ! (n+1)pn for n =0 to infinity= ( ) e ( ) where >0

And Ws = 1/ ( )

M/M/I: GD/N/infinite:-

Here Q length is lm\limited to N-1 as one is in the service.

n = when n =0,1,2,N-1

=0 when n= N, N+1

So = / , then pn(t) =np0 for n< N

=0 for n>N

p0 = (1-)/ (1-)N+1 for not equal to 1

= 1/(N+1) for =1

lost = pN

eff = 1- lost = (1-pN)


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and eff


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Number of hour for which the repairman remains busy in an 8- hour day isgiven by

8(/) =8*(5/8)=5hoursHence, the idle time for repairman in an 8-hour day will

Be 8-5=3hours.

b) expected number of TV sets in systemLs= (/- ) = (5/4)/2-(5/4) =5/3=2 TV sets.

2) At a public telephone booth in an post-office arrivals are consideredto be Poisson with an average inter-arrival time of 12mins.the lengthof phone call may be assumed to be distributed exponentially withan average of 4mins calculate the following

a) What is the probability that a fresh arrival will not have to wait for the phone?

b) What is the probability that arrival will not to wait more than 10minsbefore phone is free

c) What is the average length of queue that form from time to time

Solution: From the data of problem


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Weve =1/12=0.085permin and =1/4=0.025permin; /=4/12=0.33

a) Probability that the arrival will not have wait is given by 1-p(w>0)=1-=1-0.33=0.67

b) Probability that the arrival will have wait for at least is given by

p(w10)= - )e -(- )tdt= 0.165)e -

(0.165)tdt=0.063

This shows that 6.3% of arrivals on an average will have to wait for10mins or more before can use the phone

c) The average of length of the queue form from time is

Ls= ( /- ) = (0.25/0.25-0.085)=0.25/0.165=1.5=2customers

3) In the production shop of a company, the break down of themachine is found to be Poisson distributed with average rate of3machines per hour. The break down at one machine cost Rs 40per


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hour to the company. There are 2 choices before the company forhiring the repairman. One of the repairman is slow but cheap; theother is fast but expensive. The slow-cheap repairman demands Rs20 per hour and will repair broken down machines exponentially atthe rate of 4per hour. The fast expensive repairman demands Rs30per hour and will repairs machines exponentially at rate of 6perhour.

Which repairman should be hired?

Solution: the total expected hourly cost for both the repairmen is equal tothe total wages paid plus cost due to machine break down (non-productivemachine hours)

Cost of non-productive time=average number of machine in the systeminto cost of idle machine

= Ls*(Rs 40per hour) = (/- )*40

For slow cheap repairman: weve =3 machines per hours: =4machines per hours

Ls=(/- )=3/4-3=3machine

Cost of non-productive machine time=40*3=Rs 120

Total cost of cheap repairman= (40*3) +20=140


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For fast expensive repairman: weve =3 machines per hours: =6machines per hours

Ls=(/- )=3/6-3=1machinecost of non-productive machine time=40*1=Rs 40

Total cost of expensive repairman= (40*1)+30=Rs 70

The company should hire fast but expensive repairman.

4) The shipping company has a single unloading berth with shipsarriving in a Poisson fashion at an average rate of 3per day theunloading time distribution for an ship with n unloading crews isfound to be exponential with average unloading time 1/2n days. Thecompany has a labor with out regular working hour and to avoid longwaiting lines the company has a policy of using as many unloadingcrews an ship as there are ships waiting in line or being unloaded.


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a) Under this condition what will be the average number of unloadingcrews working at any time?

b) What is the probability that more 4crews will be needed?

Solution: given that =3ships per day.

Let us assume that at any time, there are n-ships in the system (waiting inqueue & being unloaded).since the service rate depended on waiting line,

Therefore, n=2n ships per day or 2 ships per day (mean service rate withone unloading crews)

Since is constant & service rate increases with increases in queuelength

n= n when there n-ships in the queue & pn= n e (/)

a) Ls= pn= n e (/) = =3/2=1.5 crews

b) The probability that more than 4crews will be needed is same the

probability that there at least 5ships in system at any given time & is givenby


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n= n- n=1-(p0+ p1 +p2+ p3 +p4)

=1-e-3/2(563/128)=0.019

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