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e-Notes by P.V.Raveendra, MSRIT, Bangalore

O5 MBA 21 QUANTITATIVE TECHNIQUES FOR MANAGEMENT

MODULE 1 INTRODUCTION TO OPERATIONS REASERACH Origin of Operations Research was during the Second world war. The

military management in England called upon a team ( inter disciplinary ) of

scientists to study the strategic and tactical problems related to air and land

defense of the country. Since they were having very limited military resources, it

was necessary to decide upon the most effective utilization of them e.g., the

efficient ocean transport, effective bombing. The OR teams were not actually

engaged in military operation and in fighting the war. But, they were only advisors

and significantly instrumental in winning the war to the extent that the scientific

and systematic approaches involved in OR provided a good intellectual

support to the strategic initiates of the military commands. Hence OR can

be associated “ an art of winning the war without actually fighting it”. The OR

teams in U.S. are known as Operational analysis, operations evaluation, system

analysis operations research and management science.

Following the end of war, the success of military teams attracted the attention of

industrial manager who were seeking solution to their complex executive –type

problems. The most common problem was: what methods should be adopted so

that the total cost is minimum or total profits maximum? Today, the impact of OR

can be felt in man areas apart from military and business application. Which

include transportation system, libraries, hospitals, city planning, financial

institutions etc.

OR provides a quantitative technique or a scientific approach to the

executives for making better decisions for operation under their control. In

other words the OR provides a scientific approach to problem solving for

executive management.

While making use of the techniques of OR, a mathematical model of the

problem is formulated. This model is actually a simplified representation of the

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problems in which only the most important features are considered for reasons of

simplicity. Then an optional or most favorable solution is found. Since the model

is an idealized instead, the solutions requires further improvement. The apparent

weaknesses in the initial solutions are used to suggest improvement in the

model, its input data and the solution procedure. A new solution is thus obtained

and the process is repeated until the further improvement in the succeeding

solutions becomes negligible.

If the model is carefully formulated and tested, the resulting solution

should reach to be good approximation to the ideal course of action for the real

problems. Although, we may not get the best answers, but definitely we are able

to find the bad answers where worse exist. The OR techniques are always able

to save us from worse situations of practical life.

Definitions:

“OR is the systematic method oriented study of the basic structure,

characteristics, functions and relationships of an organization to provide

the executive with a sound, scientific and quantita tive basis for decision

making”

“OR is the application of scientific methods, techn iques and tools to

problems involving the operations of system so as t o provide these in

control of the operations with optimum solutions to the problems”

“OR is concerned with scientifically deciding how t o best design and

operate man-machine systems usually requiring the a llocation of scarce

resources.

Characteristics of OR: OR approaches problem solving and decision making

from a total system’s perspective. It is interdisciplinary. Model building and

mathematical manipulation provide the methodology which has been the key

contribution of OR.

OR is for operations economy. The primary focus is on decision making and

computers are used

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PHASES OF OPERATIONS RESEARCH STUDY

PHASE 1; Formulation The Problem:

Formulate the problem in the form of appropriate model by using the following

information

1. Who has to take the decision?

2. What are the objectives?

3. What are the ranges of controlled variables?

4. What are the uncontrolled variable that may affect the possible solutions?

5. What are the restrictions or constraints on the variable?

PHASE II Constructing a Mathematical method:

Reformulate the problem in an appropriate form with suitable identification of

both static and dynamic structural elements. A mathematical model should

include the following three important basic factors.

i) Decision variable and parameters

ii) Constraints or restrictions

iii) Objective function.

PHASE III : Deriving the solutions from the model:

Compute those values of decision variable that maximize /minimize the

objective function. Such solution is called optimal solution which is always in the

best interest of the problem under consideration. The general techniques for

deriving the solution of OR model includes Distribution models ( LPP,

transportation and assignment), Queuing models ,Network analysis, Job

sequencing , Replacement models, Simulation models and etc.

PHASEIV: Testing the model and solution ( updating the model)

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After completing the model, it is once again tested as a whole for the

errors if any. A valid and good model can provide a reliable prediction of the

system’s performance and applicable for present and future specifications of the

problem.

PHASEV: Controlling the solution:

This phase establishes controls over the solution with any degree of

satisfaction. The model requires immediate modification as the controlled

variables change significantly, otherwise the model goes out of control. As the

conditions are constantly changing in the world, the model and the solution may

not remain valid for a long time.

PHASE VI: Implementing the solution:

Finally, the tested results of the model are implemented to work. This

phase is primarily executed with cooperation of OR experts and those who are

responsible for managing and operating the systems.

PRINCIPLE OF MODELLING

1. Do not build up a complicated model when simple one will suffice

2. Beware of molding the problem to fit the techniques

3. The deduction phase of modeling must be conducted rigorously

4. models should be validated prior to implementation

5. A model should never be taken too literally

6. A model should neither be pressed to do, nor criticized for failing to

do that for which it was never intended

7. Beware of over selling a model

8. Some of the primary benefits of modeling are associated with

process of developing the model

9. A model can not be any better than the information that goes into it

10. Models cannot replace decision makers.

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Techniques of OR

1. Distribution ( allocation) models: Distribution models are concerned

with the allotment of available resources so as to minimize cost or

maximize profit subject to prescribed restrictions. Methods for solving

such type of problems are known as mathematical programming

techniques. Which can be distinguished as linear a(LPP) and non-liner

programming models(transportation and Assignment models)

2. Waiting Line( queuing) models: In using models an attempt is made to

predict

i) How much average time will be spent by the customer in a queue

ii) What will be an average length of waiting line or queue?

iii) What will be the traffic intensity of a queuing system? Etc

This models provides us methods to minimize the some of the cost of

providing service and cost of obtaining service which are primarily associated

with value of time spent by the customer in a queue.

3. Production/Inventory models. These are concerned with determination

of optimal( economic order quantity and ordering interval considering the

factors such a s demand per annum, cost of placing orders, inventory

carrying cost. Such models also helpful in dealing with quantity discounts

and multiple products.

4. Competitive strategy models( Games theory) these models are used to

determine the behavior of decision making under competition or conflict.

Methods for solving such models have not been found suitable for

industrial application, mainly because they are referred to idealistic world

neglecting any essential features of reality.

5. Network analysis. These models are applicable in large projects

involving complexities and interdependencies of activates. PERT and

CPM are used for planning scheduling and controlling complex project

which can be characterized by networks.

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6. Job Sequencing model: These models involve the selection is such a

sequence of performing a series of jobs to be done on service

facilities(machines) that optimize the efficiency measure of performance of

the system.

7. Replacement models: these deal with determinations of optimum

replacement policy in situation that arise when some items of machinery

need replacement by a new one. Individual and group replacement

policies can be used in the case of such equipment that fail completely

and instantaneously.

8. Markovian models: These models are applicable in such situating where

the state of the system can defined by some description measure of

numerical value and where the system moves from one state to another

on a probability basis. Brand switching problems considered in marketing

studies ins an example of this model.

9. Simulation models: these models are more useful when following

problem arises. 1) the number of variables and constraints relationship

may be so large that it is not computationally feasible to pursue such

analysis. 2. this model may be much away from the reality that no

confidence can be placed on the computational results. In fact such

models are solved by simulation techniques where no other methods are

available for its solution.

Scope of Operations Research:

OR is useful in the following various important fields

In Agriculture : With explosion of population and consequent shortage of food

every country is facing the problem of

a) Optimum allocation of land to various corps in accordance with e climatic

conditions

b) Optimum distribution of water from various resources like can for

irrigation purposes.

In Finance : OR techniques can be applied

i) to maximize the per capita income with minimum resources

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ii) to find out the profit plan for the company

iii) to determine the best replacement policies

In Industry : OR is applicable in deciding optimum allocation of various limited

resources such as men, machine , material ,money, time to arrive at the optimal

decision.

In Marketing : OR can be applied

i) Where to distribute the products for sale so that the total cost of

transportation is minimum

ii) The minimum per unit sale rice

iii) The size of the stock to met the future demand

iv) How to select the best advertising media with respect to time, cost etc.

v) How, when and what to purchase at the minimum possible cost.

5. In Personnel management : OR can be applied

i) To appoint the most suitable persons on minimum salary

ii) to determine the best age of retirement for the employees

iii)To find out the number of persons to be appointed on full time basis when the

workload is seasonal.

In Production Management : A production manager can use OR techniques

i) To find out number and size of the items to produced

ii) In scheduling and sequencing the production run by proper allocation of

machines

iii) In calculating the optimum product mix

iv) To select, locate and design the sites for production plants.

Finally we can say: wherever there is a problem there is OR. The application of

OR covers the whole extern of any thing.

LIMITATIONS:

• Magnitude of computation :

• Absence of Quantification:

• Distance between managers and OR experts.

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REPLACEMENT THEORY PROBEMS

Problem 1: The cost of a machine is Rs 6100 & its scrap value is only Rs

100. Maintenance costs are followed from the experience.

Year 1 2 3 4 5 6 7 8

Maintenance

cost in Rs 100 250 400 600 900 1250 1600 2000

When should machine be replaced?

Solution:

year Maintenance

cost

Total

Maintenance

cost

Difference

B/N cost-

scarp value

Total cost Average

cost

1 100 100 6000 6100 6100

2 250 350 6000 6350 3175

3 400 750 6000 6750 2250

4 600 1350 6000 7350 1837.5

5 900 2250 6000 8250 1650

6 1250 3500 6000 9500 1583.33

7 1600 5100 6000 11100 1585.7

8 2000 7100 6000 12100 1637.5

As the average cost is lowest in the 6th year machine should replaced at the end

of 6th year or at the beginning of the 7th year.

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Problem 2: A machine owner finds from his past records that cost per year

of maintaining a machine whose purchase price is Rs 6000 are given below,

Year 1 2 3 4 5 6 7 8

Maintenance

cost in Rs 1000 1200 1400 1800 2300 2800 3400 4000

Resale price 3000 1500 750 375 200 200 200 200

Determine at what each is a replacement due.

Solution:

year Maintenance

cost

Total

Maintenance

cost

Difference

B/N cost-

scarp value

Total cost Average

cost

1 1000 1000 3000 4000 4000

2 1200 2200 4500 6700 3350

3 1400 3600 5250 8850 2950

4 1800 5400 5625 11025 2756

5 2300 7700 5800 13500 2700

6 2800 10500 5800 16300 2717

7 3400 13900 5800 19700 2814

8 4000 17900 5800 23700 2962.5

As the average cost is less in the 5th year machine should replaced at the end of

the 5th year or at the starting of the 6th year.

Problem 3. The cost of a new machine is Rs.5000. The maintenance cost

during the nth year is given by Rn = 500 X (n-1), n=1,2,…Suppose that the

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discount rate per year is 0.05 . After how many years it will be economical to

replace the machine by a new one?

Solution:

Since the discount rate of money per year is 0.05, the present worth of the

money to be spent over in a period of one year is

v = 1/(1 + o.o5) = 0.9523

For the best replacement age, we compute the following table for the given

machine by using present worth or one rupee to be spent in n year from now

onwards

Computation of weighted average cost for machine

Year Maintenan

ce

Cost Rn

(2)

Discount

Factor

V=(1/1+r)n

(3)

Discounted

Maitenanc

e

Cost

4 =

(2) X (3)

Discounted

cumulative

Cost

C+∑RnV n-

1

(5)

Cumulati

ve

Discount

ed

Cost

∑ V n-1

(6)

Weighted

average

annual cost

(6)/(5)

1 0 1.0000 0 5000 1.000 5000

2 500 0.9523 476 5476 1.9523 2805

3 1000 0.9070 907 6383 2.8593 2232

4 1500 0.8638 1296 7679 3.7231 2063

5 2000 0.8227 1645 9324 4.5458 2051

6 2500 0.7835 1959 11283 5.3293 2117

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From the above table, we find tht the minimum value of column ( 7) occurs after 5

years, hence it will be economical to replace the machne at the end of the 5 th

year.

Problem 4: The management of a large hotel is considering the periodic

replacement of light bulbs fitted in ins rooms. There are 500 rooms in the hotel

and each room has 6 bulbs. The management is now following the policy of

replacing the bulbs as they fail at a total cost of Rs.3 per bulb. The management

feels that this cost can be reduced to Rs.1 by adopting the periodic replacement

method. On the basis of the information give below, evaluate the alternative and

make a recommendation to the management.

Months of use 1 2 3 4 5

% of bulbs failing by that

month

10 25 50 80 100

Solution :

Let Pi be the probability that a light bulb, which was new when installed fails

during the ith month of its life. This value is the difference between the

proportion alive at the beginning of ith month and proportion alive at the end of ith

month. Thus we have

P1 = 10/100= 0.10

P2 = 25 – 10 /100 =0.15

P3 = (50 –25)/100 =0.25

P4 = (80 –50)/100 = 0.30

P5 =(100 –80)/100 = 0.20 ( since the sum of al probabilities can never be greater

than unity, all the probabilities beyond P5 should be taken as zero)

Let “No” denotes the number of original bulbs and “Fi” denotes the number of

bulbs replacement made at the end of ith month. Then we have

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No = 500 X 6 = 3000

F1 = No X P1 = 3000 X 0.1 = 300 i.e. the number of bulbs being replaced by the

end of the first month.

F2 = No. of bulbs replaced by the end of second month

= ( number of bulbs in the beginning )( probability that these bulbs in the

second month) + ( number of bulbs replaced in the first month) ( probability that

these fails during second month)

= No X P2 + F1 X P1 = ( 3000 X 0.15) + ( 300 X 0.1) = 480

F3 =No X P3 + F1 X P2 + F2 XP1

= ( 3000 X 0.25) + ( 300 X 0.15) + (480 X 0.10) = 843

F4 = ( 3000 X 0.3) + ( 300 X 0.25) + (480 X 0.15) + (843 X 0.10)

= 1131

similarly F5 = 1049

Calculation of Average cost per month in group replacement policy

End of

the

month

Total

number

Bulbs

failed

Cumulativ

e

Number

of

failures

Cost of

Replacem

ent

(

Rs.3/bulb)

Cost of

Group

Replaceme

nt

(

Re.1?bulb)

Total

Cost

(in Rs.)

Averag

e cost

per

month

1 300 300 900 3000 3900 3900

2 480 780 2340 3000 5340 2670

3 843 1623 4869 3000 7869 2623

4 1131 2754 8262 3000 11262 2815.5

5 1049 3803 11409 3000 14409 2881.8

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The average cost of is minimum in third month.

Individual replacement policy

Expected life of each bulb = ∑ i X Pi

= 1 X 0.1 + 2 X 0.15 + 3 X 0.25 + 4 X 0.3 + 5 X 0.2

= 3.35 MONTHS.

The average number of replacement will be = N / mean life

= 3000 /3.35 = 896bulbs

individual replacement = 896 X 3 = 2688.

Decision:

Thus it is better to have a individual replacement policy for first two months and

group replacement after end of third month.

Problem 5: Following mortality rates have been observed for a certain type of

fuses

Week 1 2 3 4 5

%of failure at the end of

week

5 15 35 75 100

There are 1,000 fuses in use and it costs Rs.5 to replace an individual fuse. If all

fuses were replaced simultaneously if would cost Rs.1.25 per fuse. It is

proposed to replace all fuses at fixed intervals of time, whether or not they have

burnt out, and to continue replacing burnt-out fuses as they fail. At what

intervals, the group replacement should be made? Also prove that this optimal

policy is superior to the straight forward policy of replacing each fuse only when it

fails.

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Solution :Let Pi be the probability that a fuse, which was new when

installed fails during the ith week of its life. This value is the difference between

the proportion alive at the beginning of ith week and proportion alive at the end of

ith week. Thus we have

P1 = 5/100= 0.05

P2 = 15 –5 /100 =0.10

P3 = (35 –15)/100 =0.20

P4 = (75 –35)/100 = 0.40

P5 =(100 –75)/100 = 0.25 ( since the sum of al probabilities can never be greater

than unity, all the probabilities beyond P5 should be taken as zero)

Let “No” denotes the number of original fuses and “Fi” denotes the number of

fuses replacement made at the end of ith week. Then we have

No = 1000

F1 = No X P1 = 1000 X 0.05 = 50 i.e. the number of fuses being replaced by the

end of the first week.

F2 = No. of fuses replaced by the end of second week

= ( number of fuses in the beginning )( probability that these fuses in the

second week) + ( number of fuses replaced in the first week) ( probability that

these fails during second week)

= No X P2 + F1 X P1 = ( 1000 X 0.1) + ( 50 X 0.1) = 102

F3 =No X P3 + F1 X P2 + F2 XP1

= ( 1000 X 0.20) + ( 50 X 0.10) + (102 X 0.05) = 210

F4 = 210

similarly F5 = 430

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Calculation of Average cost per week in group replacement policy

End of the week

Total number fuses failed

Cumulative Number of failures

Cost of Replacement ( Rs.5/fuse)

Cost of Group Replacement ( re.1/fuse)

Total Cost (in Rs.)

Average cost per week

1 50 50 250 1250 1500 1500

2 102 152 760 1250 2010 1005

3 210 362 1810 1250 3060 1020

4 430 792 3960 1250 5210 1302.5

5 333 1125 5625 1250 6875 1375

The average cost of is minimum in second week.

Individual replacement policy

Expected life of each bulb = ∑ i X Pi

= 1 X 0.05 + 2 X 0.10 + 3 X 0.20 + 4 X 0.40 + 5 X 0.25

= 3.7weeks.

The average number of replacement will be = N / mean life

= 1000 /3.7 = 270 fuses

Individual replacement = 270 X 5 = 1350.

Decision:

Thus it is better to have an individual replacement policy for first week and group

replacement after end of second week.

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Problem 6: A computer contains 10,000 resistors. When any of the register

fails it is replaced. The cost of replacing a single resistor is Rs.10 only. If all the

resistors are replaced at the same time, the cost per resistor would be reduced to

Rs.3.50. The per cent surviving by the end of the month t is as follows. What is

the optimum plan.

month 0 1 2 3 4 5 6

%surviving

at the end of

month

100 97 90 70 30 15 0

Here % of survival rate is given with the help of it we can find the failure rate as

follows.

Failure in first month 100-97/100 =3

Failure in second month 100 – 90 /100 = 10

Similarly in the third month =30

Failure in the fourth month =70

Failure in the 5th month =85

Failure in the six month = 100

With the help of above failure rate one can find probabilities and continue in the

similar manner( as above two problems).

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Notes for Assignment models

Dr. P.V.Raveendra

Professor, M.S.R.I.T.

SOLUTION TO THE PROBLEM 1

Programmers A B C D

1 120 100 80 90

2 80 90 110 70

3 110 140 120 100

4 90 90 80 90

1. CHECK THE OBJECTIVE FUNCTION ( GIVEN PROBLEMS IS

MINIMISATION)

2. CHECK FOR THE BALANCE ( GIVEN PROBLEM IS BALANCED ONE)

3. NOW APPLY HUNGARIAN METHOD

Row conversion matrix

A B C D

1 120-80=40 20 0 10

2 80-70=10 20 40 0

3 110-100=10 40 20 0

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90-80=10 10 0 10

COLUMN CONVERSION MATRIX

A B C D

1 40-10=30 20-10=10 0 10

2 0 10 40 0

3 0 30 20 0

4 0 0 0 10

ALLOCATION ACCORDING TO THE METHOD

A B C D

1 40-10=30 20-10=10 0 10

2 0 10 40 0 x

3 0 x 30 20 0 4 0 x 0 0 x 10

Allocation 1- C 2-A 3-D 4 –B 80 + 80 + 100 + 90 = 350

SOLUTION TO THE PROBLEM 2

1. CHECK THE OBJECTIVE FUNCTION ( GIVEN PROBLEMS IS

MINIMISATION)

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2. CHECK FOR THE BALANCE ( GIVEN PROBLEM IS BALANCED ONE)

3. NOW APPLY HUNGARIAN METHOD

I II III IV

A 44 80 52 60

B 60 56 40 72

C 36 60 48 48

D 52 76 36 40

ROW REDUCTION

I II III IV

A 0 36 8 16

B 20 16 0 32

C 0 24 12 12

D 16 40 0 4

COLUMN REDUCTION

I II III IV

A 0 20 8 12

B 20 0 0 28

20

C 0 8 12 8

D 16 24 0 0

ALLOCATION ACCORDING TO HUNGARIAN METHOD

I II III IV

A 0 20 8 12

B 20 0 0 28

C 0 X 8 12 8

D 16 24 0 0 X

Mark the unallocated rows ( i.e. C) and mark the columns where there are zeros

in the marked rows. (i.e.). Mark the rows where allocation is there in the marked

columns (i.e. A). Now draw the lines through unmarked rows i.e.B and D and

marked columns (i.e.)

Selected the minimum element(8) among uncovered elements ( where lines are

not passing) and deduct it from the rest of the uncovered elements .Add the

same where ever there is intersection of the lines i.e. cell (2,1) and cell (4,1)

The revised table is as follows with allocations

I II III IV

A 0 12 0 X 4

B 20+8=28 0 0 X 28

C 0 X 0 X 4 0

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D 16+8=24 24 0 0 X

THUS A –I , B – II, C—IV, D – III WITH COST OF ( 44+56+48+36)

SOLUTION TO THE PROBLEM 3

1. CHECK THE OBJECTIVE FUNCTION ( GIVEN PROBLEMS IS

MINIMISATION)

2. CHECK FOR THE BALANCE ( GIVEN PROBLEM IS BALANCED ONE)

3. NOW APPLY HUNGARIAN METHOD

M1 M2 M3 M4 M5

301 ∞ 4 2 ∞ 1

302 1 1 5 1 2

303 2 ∞ 1 4 ∞

304 3 2 3 3 3

305 ∞ 3 4 2 ∞

ROW REDUCTION

M1 M2 M3 M4 M5

301 ∞ 3 1 ∞ 0

302 0 0 4 0 1

303 1 ∞ 0 3 ∞

304 1 0 1 1 1

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305 ∞ 1 2 0 ∞

COLUMN REDUCTION

(IT WILL BE THE SAME ABOVE AS THERE IS A ZERO IS EACH COLUMN)

M1 M2 M3 M4 M5

301 ∞ 3 1 ∞ 0

302 0 0 4 0 1

303 1 ∞ 0 3 ∞

304 1 0 1 1 1

305 ∞ 1 2 0 ∞

NOW ALLOCATING ACCORDING TO HUNGARIAN METHOD

M1 M2 M3 M4 M5

301 ∞ 3 1 ∞ 0 302 0 0 X 4 0 X 1

303 1 ∞ 0 3 ∞

304 1 0 1 1 1

305 ∞ 1 2 0 ∞

ALLOCATION 301 - M5 =1

23

302 – M1 = 1

303- M3 =1

304 - M2 = 2

305- M4 =2

SOLUTION TO PROBLEM 4

THIS IS PROBLEM IS MAXIMASATION CASE

Salesmen D1 D2 D3 D4 D5

S1 40 46 48 36 48

S2 48 32 36 29 44

S3 49 35 41 38 45

S4 30 46 49 44 44

S5 37 41 48 43 47

CONCVERTING INTO MINIMIZATION CASE BY SELECTION THE MAXIMUM

NUMBER(49) FROM THE MATRIX AND DEDUCTING IT FROM THE CELL

ELEMENTS ( LOSS MATRIX) WILL BE AS FOLLOWING ONE

Salesmen D1 D2 D3 D4 D5

S1 49-40=9 3 1 13 1

S2 1 17 13 20 5

24

S3 0 14 8 11 4

S4 19 3 0 5 5

S5 12 8 1 6 2

2. CHECK FOR THE BALANCE ( THE ABOVE ONE IS BALANCED ONE AS IT

IS SQUARE MATRIX)

AS THIS PROBLEM IS MINIMISATION CASE WE CAN APPLY HUNGARIAN

METHOD

ROW CONVERSION

Salesmen D1 D2 D3 D4 D5

S1 8 2 0 12 0

S2 0 16 12 19 4

S3 0 14 8 11 4

S4 19 3 0 5 5

S5 11 7 0 5 2

COLUMN REDUCTION

Salesmen D1 D2 D3 D4 D5

S1 8 0 0 7 0

S2 0 14 12 14 4

25

S3 0 12 8 6 4

S4 19 1 0 0 5

S5 11 5 0 0 2

ALLOCATION

Salesmen D1 D2 D3 D4 D5

S1 8 0 0 X 7 0 X

S2 0 14 12 14 4

S3

0 X

12 8

6

4

S4 19 1 0 0 X 5

S5 11 5 0 X 0 2

Put a tick mark for unallocated rows( i.e. S4 ) AND in the marked row mark the

column where there is zero ( i.e. D1) . In the marked column put a tick mark to

the row where there is an allocation ( i.e. S2).

Then draw the line through unmarked rows (i.e.S!,S4 and S5) and marked

columns ( i.e.D1)

Select the smallest element from the uncovered elements( where lines are not

passing ) i.e 4 and deduct it from the rest of uncovered elements only. Add 4

where ever there is intersection of the lines ( i.e. cell (1,1) and (1,4) and (1,5))

Rewrite the table as follows and allocate according to the method.

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Salesmen D1 D2 D3 D4 D5

S1 8+4=12 0 0 X 7 0 X

S2 0 14-4=10 12-4=8 14-4=10 0 X

S3

0 X

12-4=8 8-4=4

6-4=2

0

S4 19+3=23 1 0 0 X 5

S5 11+4=15 5 0 X 0 2

NOW WE HAVE MADE ALLOCATION IN THE LOSS MATRIX. TAKE THESE

ALLOCATION TO MAXIMISATION MATRIX AS SHOWN BELOW

Salesmen D1 D2 D3 D4 D5

S1 40 46 48 36 48

S2 48 32 36 29 44

S3 49 35 41 38 45 S4 30 46 49 44 44

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S5 37 41 48 43 47

THUS S1 TO D2 ,S2-D1,S3-D5,S4-D3 AND S5-D4 ( 46 +48+45+49+43) SALES.

SOLUTION TO PROBLEM 5

ASSUMING A TO A DISTANCE IS ∞ and so on.

From/to A B C D E

A ∞ 4 7 3 4

B 4 ∞ 6 3 4

C 7 6 ∞ 7 5

D 3 3 7 ∞ 7

E 4 4 5 7 ∞

Given problem is minimization case and balanced one. We can apply Hungarian

method

Row conversion matrix

From/to A B C D E

A ∞ 1 4 0 1

B 1 ∞ 3 0 1

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C 2 1 ∞ 2 0

D 0 0 4 ∞ 4

E 0 0 1 3 ∞

Column conversion matrix

From/to A B C D E

A ∞ 1 3 0 1

B 1 ∞ 2 0 1

C 2 1 ∞ 2 0

D 0 0 3 ∞ 4

E 0 0 0 3 ∞

Allocation matrix

From/to A B C D E

A ∞ 1 3 0 1

B 1 ∞ 2 0 X 1

C 2 1 ∞ 2 0

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D 0 0 X 3 ∞ 4

E 0 X 0 X 0 3 ∞

As explained in the above cases the improved matrix will be

From/to A B C D E

A ∞ 0 2 0 X 0 X

B 0 X ∞ 1 0 0 X

C 2 1 ∞ 3 0 D 0 0 X 3 ∞ 4

E 0 X 0 X 0 4 ∞

IN THE ABOVE SOLUTION D IS ASSIGNED TO A WHICH MEANS SALESMAN

IS COMING BACK TO A WITHOUT VISITING ALL THE CITIES.. But the

condition is that salesman should not come to home with out visiting to all the

cities ( this condition is given priority compared to the cost , so reallocating as

below the cost may increase by one rupee but the first condition is satisfied)

From/to A B C D E

A ∞ 0 2 0 0 X

B 0 X ∞ 1 0 X 0 X

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C 2 1 ∞ 3 0 D 0 0 3 ∞ 4

E 0 0 X 0 X 4 ∞

HERE ALLOCATION IS MADE TO CELL (2,3) WHERE COST IS 1 BECASUSE

SALEMAN SHOULD NOT RETURN HOME WITHOUT VISITOING ALL THE

CITIES.

TOTAL COST IS A –D-B—C—E—A Rs. (3+3+6+5+4)

SOLUTION TO PROBLEM 6

We have to calculate the time required to complete the job P by A = 199/12 =17

hours. For each hour he will charge 5 rupees. So the cost involved in completion

of the job P by A is given by 17 x 5 = 85.

typist P Q R S T

A 199/12=17

X5=85

75 65 85 75

B 90 78 66 90 78

C 75 66 57 75 69

D 80 72 60 80 72

E 70 64 56 72 68

The above problem is minimization case and balanced one. We can apply

Hungarian method and find the solution as usual

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SOLUTION TO PROBLEM 7

Let us assume that all the crew is based at Hyderbad. Then lay-over times for

various combinations of flights as shown below. For example the flight A1 which

start from Hyderabad at 6 a m , reached Delhi at 8 a m. If it is to return as B1,

schedule time for which is 8 a m, they it can do so after 24 hours, since a

minimum layover time of 5 hours is required. Similarly, lay over times for other

flight combinations can be obtained.

Layover time -- crew based at Hyderabad

Flight B1 B2 B3 B4

A1 24 25 6 11

A2 22 23 28 9

A3 16 17 22 27

A4 10 11 16 21

The lay-over times for various flight combinations, when crew is assumed to be

based at Delhi, are similarly calculated as shown below

Lay-over time –crew based at Delhi

Flight B1 B2 B3 B4

A1 20 19 14 9

A2 22 21 16 11

A3 28 27 22 17

A4 10 9 28 23

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Now, since the crew can based at either of the stations, minimum lay-over times

can be obtained for different flight combinations by selecting the corresponding

lower value out of the above two tables. For instance in combining A1-B1, we

select 20, which is lower of the two values 24 and 20.thus rewriting the table.

Layover times crew based at Hyderabad/Delhi

Flight B1 B2 B3 B4

A1 20* 19* 6 9*

A2 22 21* 16* 9

A3 16 17 22 17*

A4 10 9* 16 21

• when crew is based at Delhi.

Now we can apply the Hungarian method to minimize the layover time and the

allocation will be as follows

Flight FLIGHT NO. LAYOVER

TIME

CREW IS

BASED AT

A1 B3 6 HYDERABAD

A2 B4 9 HYDERABAD

A3 B1 16 HYDERABAD

A4 B2 9 DELHI.

** *************************************************************

Note: IF THE GIVEN PROBLEM IS UNBALANCED i.e. the number of row is not

equal to the number of columns then we can add the column or rows depending

on the requirement using a dummy row/column using zero as the cost. Then

apply the Hungarian method.

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Queuing Models Queuing System: General Structure

• Arrival Process

• According to source • According to numbers • According to time

• Service System • Single server facility • Multiple, parallel facilities with single queue • Multiple, parallel facilities with multiple

queues • Service facilities in a parallel

• Queue Structure • First come first served • Last come first served • Service in random order • Priority service

Model 1: Poisson-exponential single server model – infinite population Assumptions:

� Arrivals are Poisson with a mean arrival rate of, say λ

� Service time is exponential, rate being µ � Source population is infinite � Customer service on first come first served basis � Single service station

For the system to be workable, λ ≤ µ Model 2: Poisson-exponential single server model – finite population Has same assumptions as model 1, except that population is finite

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• Model 3: Poisson-exponential multiple server model – infinite population

Assumptions � Arrival of customers follows Poisson law, mean

rate λ � Service time has exponential distribution, mean

service rate µ � There are K service stations � A single waiting line is formed � Source population is infinite � Service on a first-come-first-served basis � Arrival rate is smaller than combined service rate

of all service facilities Model: 1Operating Characteristics

a) Queue length � average number of customers in queue waiting to

get service b) System length

� average number of customers in the system c) Waiting time in queue

� average waiting time of a customer to get service d) Total time in system

� average time a customer spends in the system e) Server idle time

� relative frequency with which system is idle

• Measurement parameters • λ= mean number of arrivals per time period (eg. Per

hour) • µ = mean number of customers served per time period • Probability of system being busy/traffic intensity

ρ = λ / µ • Average waiting time system Ws = 1/(µ- λ) • Average waiting time in queue

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Wq= λ/ µ(µ- λ) • Average number of customers in the system

Ls = λ/ (µ- λ) • Average number of customers in the queue

Lq = λ2/ µ(µ- λ) • Probability of an empty facility/system being idle

P(0) = 1– P(w) • Probability of being in the system longer than time (t)

P(T>t)= e –(µ- λ)t Probability of customers not exceeding k in the system P (n.≥k) = ρk P( n>k) = ρ(k+1) Probability of exactly N customers in the system P(N) = ρN (1-ρ) Problem 1. Customers arrive at a booking office window, being manned by a single individual rate of 25per hour. Time required to serve a customer has exponential distribution with a mean of 120 seconds. Find the mean waiting time of a customer in the queue Solution: Given λ =25 customer/hr µ = 30 customer/hr Mean waiting time in Queue = λ/ µ (µ- λ) = 25/30 (30-25) = 0.167 hr Mean waiting time in system =1/ µ- λ =1/30-25 =0.2 hr Problem 2: A repairman is to be hired to repair machines which breakdown at a n average rate of 6 per hour. The breakdowns follow Poisson distribution. The non-production

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time of a machine is considered to cost Rs. 20 per hour. Two repairmen Mr. X and Mr.Y have been interviewed for this purpose. Mr. X charges Rs.10 per hour and he service breakdown machines at the rate of 8 per hour. Mr. Y demands Rs.14 per hour and he services at an average of 12 per hour. Which repairman should be hired? ( Assume 8 hours shift per day) solution: Given λ =6 /hr µx=8/hr µy=12/hr Given no of machine cost at idle = 20Rs/hr No of machine in X Lsx = λ/ µx - λ =6/8-6 =3 machines Total no of machines =3*8=24 machines Total cost=hiring charges of x + cost of idle machine =10*8 + 24*20 =Rs.560 No of machine in y Lsy = λ/ µy – λ =6/12-6 =1 Total no of machine = 1*8=8 machines Total cost= hiring charges of y + cost of idle machine = 14*8 + 20*8 = Rs.272 We chose Mr.Y since cost is lower than Mr. X Problem 3: A warehouse has only one loading dock manned by a three person crew. Trucks arrive at the loading dock at an average rate of 4 trucks per hour and the arrival rate is Poisson distributed. The loading of a truck takes 10 minutes on an average and can be assumed to be exponentially

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distributed . The operating cost of a truck is Rs.20 per hour and the members of the crew are paid @ Rs.6 each per hour. Would you advise the truck owner to add another crew of three persons? Solutions: 1) λ=4 truck/ hr

µ=10min=60/10=6truck/hr No of trucks in system, Ls= λ/ (µ- λ) =4/6-4 =2 trucks Total cost = cost of maintaining trucks + cost of crew =20*20 + 3*6 =40 + 18 = Rs.58

2) λ =4trucks /hr If we double the crew, µ= 12 trucks/ hr

No of trucks in system, Ls= λ/ (µ- λ) =4/12-4 =0.5 trucks Total cost = cost of maintaining trucks + cost of crew =0.5*20 + 6*6 =10 + 36 =Rs.46 We will advice the owner to add 3 persons for loading cost the total cost for 3 extra persons is less than previous. Problem 4 At a service counter of fast-food joint, the customers arrive at the average interval of six minutes whereas the counter clerk takes on an average 5 minutes for preparation of bill and delivery of the item. Calculate the following

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a. counter utilization level b. average waiting time of th4e customers at the fast food joint c. Expected average waiting time in the line d. Average number of customers in the service counter area e. average number of customer in the line f. probability that the counter clerk is idle g. Probability of finding the clerk busy h. chances that customer is required to wait more than 30 minutes in the system i. probability of having four customer in the system J) probability of finding more than 3 customer in the system Solutions: Given λ = 60/10 = 10 customer/hr µ = 12 customer/hr

a) ρ = λ / µ = 10/12 = 0.833

b) Ws = 1/ µ- λ = 1/12-10 = 0.5 hr

c) Wq = λ/ µ (µ- λ) = 10/12(12-10) = 0.416 hr

d) Ls= λ/ (µ- λ) = 10/12-10 = 5 customers

e) Lq = λ2 / µ (µ- λ) = 102 /12(12-10) = 4.167 customers

f) 1- ρ = 1- λ / µ = 1- 10/12 = 0.167

g) ρ = λ / µ = 10/12 = 0.833

h) chances of probability that customer wait more than 30min = 30/60 = 0.5 hrs

P (T>t) = e – (µ- λ) t

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P (T>0.5) = e – (12- 10) 0.5 = 0.368 i) P (N) = ρN (1-ρ) P (4) = ρ4 (1-ρ) = (0.833)4(1-0.833) = 0.0806 j) P (n>k) = ρ (k+1) P (n>3) = ρ (3+1) = (λ / µ) 4= (10/12) 4 = 0.474 Problem 5: Customers arrive at a one-window drive-in bank according to a Poisson distribution with mean 10 per hour. Service time per customer is exponential with mean 5 minutes. The space in front of the window including that for the serviced car accommodate a maximum of 3 cars. Other cars can wait outside the space. Calculate

• A) what is the probability that an arriving customer can drive directly to the space in front of the window.

• B) what is the probability that an arriving customer will have to wait outside the indicated space

• C) How long is arriving customer expected to wait before stating the service.

• D) How many spaces should be provided in front of the window so that all the arriving customers can wait in front of the window at least 20% of the time.

solution: Given λ = 10 /hr µ = 5min = 60/5 = 12customer/hr

ρ = λ / µ = 10/12 = 0.83 We know that P (N) = ρN (1-ρ)

a) P = Po + P1 + P2 = ρ0 (1-ρ) + ρ1 (1-ρ) + ρ2 (1-ρ) = (1- ρ) (1+ ρ + ρ2) = (1- 0.83) (1+ 0.83 + 0.832)

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= 0.428 b) P = 1 – (Po + P1 + P2 + P3 )

= 1 – [0.42 + ρ3 (1-ρ)] = 1 – [0.42 + 0.833 (1-0.83)]

= 0.482 c) Wq = λ/ µ (µ- λ) = 10/12(12-10) = 0.416 hr d) Po = ρ0 (1-ρ) 1-10/12 = 0.16

e) P1 = ρ1 (1-ρ) = 10/12(1-10/12) = 0.14

Probability that there will be no or one car in the space which cover = Po + P1 = 0.30 = 30%, hence there should at least 1 car space for waiting at the window space at least 20% of the time. Problem 6 Customers arrive at the first class ticket counter of a theatre at a rate of 12 per hours. There is one clerk serving the customers at a rate of 30 per hour. Assuming the conditions for use of the single channel queuing model, evaluate

a) The probability that there is no customer at the counter (i.e. that the system is idle)

b) The probability that there are more than 20 customers at the counter

c) The probability that there is no customer waiting to be served

d) The probability that a customer is being served and no body is waiting.

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Solution: Given λ = 12 customers /hr µ = 30 customers/hr 1) Ideal =1- ρ = 1- λ / µ = 1- 12/30 = 0.6 2) At least 3 customers at counter P (n>k) = ρ (k+1) = ρ (3+1) = (12/30)4 = 0.025 3) Probability that no customers waiting to be served P (at least 1 customer at the counter)

= Po + P1 = ρ0 (1-ρ) + ρ1 (1-ρ) = (1- ρ) (1 + ρ) = (1- ρ2 ) = (1- (12/30)2 = 0.84

4) Probability that a customers being served and nobody waiting = P1 = ρ1 (1-ρ) = 12/30(1-12/30) = 0.24

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