CS6704 RESOURCES MANAGEMENT TECHNIQUES

VELTECH HIGHTECH Dr.RANGARAJAN Dr.SAKUNTHALA ENGINEERING COLLEGE DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING

VELTECH Dr.RANGARAJAN Dr.SAKUNTHALA ENGINEERING COLLEGE Page 1

CS6704_RESOURCES MANAGEMENT TECHNIQUES

UNIT I

Resources management techniques or Operations Research deals with different

techniques for optimal use of resources and to ultimately optimize the Objective

function.

The natural resources available are limited except the air. The question how

judiciously they have to be utilized to maximize the Objective function. Since

RMT is mostly applied to Business the Objectives are maximization of Profit,

Income and minimization of Loss, expenditure. Etc.

RMT is applied in department as:

1. Allocation and Distribution

2. Production and Facility Planning

3. Procurement

4. Marketing

5. Finance

6. Personnel and

7. Research and Development.

One of the definitions of OR or RMT RUNS AS FOLLOWS

It is aid for the Executive in making his decisions by providing him with the

needed quantitative information based on the scientific method of analysis.

This was given by C.Kittel.

RMT is the science of efficient way of managing the resources for optimization

of objectives against certain constraints. By resource me mean some of them

like man, machine, money, land, time, raw materials and other resources.

Therefore Decision making is very important skill a person must possess to be

successful. Thus it is a decision science that helps a management to make

better decisions.



The essential characteristics of decision making are

1. Objectives

2. Alternatives

3. Influencing factors( constraints)

Phases of RMT

1 Formulating the problem

2 Constructing the model to represent the system under study

3 Deriving solution from the model

4 Testing the model with obtained solution.

5 Establishing control over the solution

6 Implementation.

CONSTRUCTION OF MODEL is an important phase of the RMT to optimize the

objective function.

Definition of a model:

A model is a reasonably simplified representation of a real world situation. It is an

abstraction of reality.

Models can be classified as

(i) Iconic models

(ii) Analogue models

(iii) Deterministic models

(iv) Stochastic models.

1. Iconic models: This is a physical or pictorial representation of various aspects

of a system.



e.g. Toy, miniature model of a building, the bar graph of a company, etc.

2. Analogue models: This uses one set of properties to represent another set of

properties which a system under study has.

e.g. A network of water pipes to represent the flow of current in an electrical net

work or graphs , organizational charts ,etc.

3. Mathematical symbolic model: This uses a set of mathematical symbols ( letter,

numbers, etc) to represent the decision variables of a system under

consideration. These variables are related by mathematical equations or in

equations which describe the property of the system.

e.g A linear Programming Problem (LPP), Non linear Programming problem, etc.

4. Static model: This is a model which does not take account time as a

parameter. It assumes that the value of the variables do not change with time

during a period of time horizon.

e.g. LPP. Assignment problem, Transportation problem. Etc.

5. Dynamic model: Here time is considered as one of the important variables.

e.g. Dynamic Programming Problem , Replacement Problem, etc.

6 Deterministic model: Here uncertainty is not taken into account,

e.g. LPP. AP. TP. Etc.

7. Stochastic models: This considers uncertainty as important aspect of the

problem

e.g. Queuing models, inventory of stochastic nature, etc.

CHARACTERISTIC of a good model:

1 Must be simple

2. Must be designed to accept any changes.



3. Assumptions must be as small as possible.

4. Number of variables must be small.

5. Must be open to parametric treatment.

Principles pf modeling:

1. Do not build complicated models.

2. Beware of moulding the problems to fit a technique.

3. Careful deductions are necessary.

4. Validated prior to implementation

5. Solution can’t be accurate unless variables are properly chosen.

6. Must act as Aid for decision making.

7. Must be as accurate as possible.

The various methods of solving a model are :

1. Analytic method

2. Iterative method

3. Monte-Carlo technique.

The main Phases of RMT are:

1. Formulation of the problem

2. Construction of mathematical model

3. Solving the constructed model

4. Controlling and updating

5. Testing the model and its solution.

6. Implementation.

LIMITATION of RMT:

RMT does not take into account quantitative or emotional or some human

factors which are essential for decision making.

FORMULATION OF LPP:



Given a problem, identify the objective function, the decision variables and the

constraints, the mathematical formulation is :

Optimize Z = f( x1 , x2, x3, ……….. xn)

Subject to constraints:

gj(x1 , x2, x3, ……….. xn ) ≤ = ≥ bj j= 1,2, ………………m.

xi ≥ 0 for all i = 1,2,……………n

Z is the objective function, it depends on the decision variables xi ≥ 0 for all I =

1,2,……………n for its optimization

gj(x1 , x2, x3, ……….. xn ) ≤ = ≥ bj are the m constraints of the LPP.

Note that , here f( x1 , x2, x3, ……….. xn) = c1x1 + c2x2 +…………+ cnxn

c1, c2,…………,cn are assumed to be known in advance.

b1, b2,…………,bn the constants on the rhs of the constraints are assumed to be

known ( availability)

The coefficient matrix A = (ai j ), which are the coefficients of the decision

variables in the constraints are assumed to be known.

[In ordinary course the above need not be true ]

The mathematical formulation can also be given as :

Optimize z = c1x1 + c2x2 +…………+ cnxn


a11x1 + a12x2 +…………+ a1nxn ≤ = ≥ b1

a21x1 + a22x2 +…………+ a2nxn ≤ = ≥ b2

a31x1 + a32x2 +…………+ a3nxn ≤ = ≥ b3 (I)

………………………………………………………..



…………………………………………………………

Am1x1 + am2x2 +…………+ amnxn ≤ = ≥ bm

xi ≥ 0 for all i = 1,2,……………n which cal also be written as

Optimise Z = CX subject

AX = b and X ≥ 0 where C , b are as explained and the A is the matrix of the

coefficients of the decision variables in the constraints.

1. Solution of an LPP: By solution of an LPP we mean finding the values of the

decision variables which satisfy all the constraints given in the LPP.

2. Feasible solution: A solution of an LPP is said to be feasible if they satisfy

the non negative constraints i.e all the decision variables must greater than

or zero.

3. Optimal solution: By optimal solution of an LPP we mean the feasible

solution of the LPP optimizing the objective function.

An LPP of the form (I) is said to be in CANNONICAL form if optimization is

MAXIMIZATION and all the constraints are LESS THAN OR EQUAL TO .

An LPP of the form (I) is said to be in STANDARD form if optimization is

MAXIMIZATION and all the constraints are EQUAL TO. The decision variables

are non negative.

The solution to an LPP with two decision variables can be solved by graphical

method.

1. We draw the equations of the constraints and they are straight lines.

2. Keeping these as reference lines identify the regions the in equations

define, whether above the line or below the line and mark it with a shade.

3. Every constraint will define a region that they may all bound a common

region , which contains all those points that satisfy all the constraints given

in the problem. We call this region as FEASIBLE REGION.



4. Therefore a feasible region is one , bounded by all the lines of constraints

and contains all those points that satisfy all the constraints given in the

problem.

5. Optimal value of objective function occurs at the vertices of the

polyhydron formed by the lines.

6. For maximisation problem the optimal vertex ( vertices) will be away from

the origin whereas for a minimisation problem the vertex(vertices) will be

closer to the origin.

Solve the following LPP using graphical method:

Maximize Z = 3x1 + 4x2

Subject to constraints :

x1 + x2 ≤ 450

2x1 + x2 ≤ 600 , x1 , x2 ≥ 0

In the above graph OABC is the region of feasible solution and the solutions lie

at the vertices of the polyhydron namely at O or A or B or C where B is the



point of intersection of the two lines. The following table gives the value of Z

at all vertices and the one which gives the maximum for Z is the optimal of Z

VERTEX VALUE OF Z

O(0 , 0 ) 0

A ( 300 , 0 ) 900 B ( 150, 300 ) 1650

C ( 0, 450 ) 1800

The maximum is obtained for Z at x1 = 0 and x2 = 450 and max Z = 1800

This problem can be viewed as problem of product mix . The function Z the

overall profit on two products say A and B . Let the profit one unit of A be Rs.3

and that of B be Rs 4.Let us suppose that one unit of of a particular raw

material is required to produce one unit of A and for B also one unit. Let us

assume that 2 unit of time is required to produce one unit of A and one unit of

time is required to produce one unit of B . The total availability of raw

material is 450 units and total time available is 600 units per week. The

question is how many units of A and B can be produced ( product mix) so that

the overall profit is maximum.

There are LPPs which may have

1. Unique solutions.

2. Infinite number of optimal solutions.

3. Unbounded solution.

4. No solution.

Advantages of LPP:.

1. Provides insightness into the problem

2. Gives scientific and mathematical analysis of the problem.



3. Deals with changing situations.

4. Gives the best solution possible

Limitation of LPP:

1. Assumption of linear relationship only.

2. All parameters assumed to be known.

3. More number of variables and constraints make more complicated and

difficult to solve.

4. Always deals with one objective only.

Note that a feasible region is a convex region. A convex region is one in which

the join any two random points lie in the same region.

The basic assumptions made in Lapp are :

1. Proportionality: When the proportion of input increases the out put will

also increase in the corresponding proportion.

2. Divisibility: The values of decision variables can be non integers.

3. Coefficients of decision variables in objective function and in constraints

are assumed to be known.

4. The availability of resources are assumed to be known.

5. Variables and constraints are finite in number.

6. Optimality. Objective function is optimised.

SLACK Variable.

∑ 𝑎𝑖𝑗𝑛𝑗=1 𝑥𝑖𝑗 ≤ bi i = 1, 2, ……….k

∑ 𝑎𝑖𝑗𝑛𝑗=1 𝑥𝑖𝑗 + si = bi i = 1, 2, ……….k

si ≥ 0 called slack variable and it signifies unused resources.

SURPLUS Variable:

∑ 𝑎𝑖𝑗𝑛𝑗=1 𝑥𝑖𝑗 ≥ bi i = k, k+1,………..

The above can be written as ∑ 𝑎𝑖𝑗𝑛𝑗=1 𝑥𝑖𝑗 - si = bi , i = k, k+1,………..



Here si ≥ 0 is called surplus variable and signifies excess used resource.

Unrestricted variable:

Avariable x is said to be unrestricted variable if it can be expressed as

x = x ’ + x” and x ’, x” ≥ 0

Simple method:

This is also called iteration method, this method is applied when the number

of constraints are large or the number of decision variables are more than two.

In this method we first find the initial solution called basic solution and using

iterative method improve the solution under certain optimality conditions.

Certain Definitions:

1. Basic solution: If there are n variables and m equations ( n > m ), n – m

variables are made as zeros and they are called as non basic variables and

the m variables are called basic variables and is called the basis of the

problem.

2. Non degenerate basic solution: It is a basic solution if none of the basic

variables are zero.

3. Degenerate basic solution: It is a basic solution in which one or more basic

variable are zero.

4. Basic feasible solution: A feasible solution which is also basic is called basic

feasible solution.

Note

1. we use simplex method to solve Maximization of LPP, Minimization is

solved by converting the minimization to maximization problem as:

This method is also called iteration method. A basic solution is found and Is

improved by the process of iteration under the condition of optimality. A

solution is said to be obtained for the LPP if optimality condition is satisfied

or that it becomes certain that no solution exists for the problem.



Simplex method is employed when the number of constraints or there are

more than two variables in the LPP.

Min Z = - Max( -Z) = - Max ( Z*) where Z* = -Z. Now we solve for Max Z* and

after solving and finding the optimal solution , we get the solution to the

original problem by “ – MaxZ* = - Max ( -Z ) for the same values of decision

variables.[ Z* = -f(x) ]

2. When constraints are ≥ or = , then the initial solution that is the basic

solution will be negative. Normally the basic solution should never be negative

( if so solved by dual simplex method ) , to overcome situations of that type we

introduce Artificial variables and use them to solve the problem but they form

a part of the solution.

Procedure to solve using simplex method.:

Given an LPP, in cannonical form convert into standard form , if minisation

convert to maximization.

Make n – m variables as zeros and with m basic variables for the iteration to

improve the basic solutions. Solution is optimized with an optimality condition.

The iterations are carried out till the optimality condition is satisfied. By when

we would have arrived at the solution or confirmed that solution does not

exist.

In the case of Big M method or Big M technique or Charnes method to get rid

of artificial variables we assign a very big penalty ( –M for maximization and

+M for minimization problems.)as the coefficients of artificial variables in the

objective function.

While solving note the following:

1. If no artificial variables remain in the basis ( under XB ) and optimality

condition is satisfied , then we have got optimal basic feasible solution.



2. If atleast one artificial variable remains in the basis in zero level and

optimality condition is satisfied , then current solution is an optimal basic

feasible solution though it is degenerated.

3. If atleast one artificial variable remains in the basis in non zero level and

optimality condition is satisfied , then the original problem does not have

feasible solution. The solution satisfies the constraints but does not

optimise the objective function as it contains a very large M and is called

Pseudo- optimal solutiois an optimal basic feasible solution though it is

degenerated.

Example: Solve the following Lpp using simplex method:

Max z = 5X1 + 12X2 + 8X3

subject to:

3X1 + 4X2 + 5X3 ≤ 60

5X1 + 4X2 + 4X3 ≤ 72

2X1 + 4X2 + 5X3 ≤ 100, X1 ,X2 ,X3 ≥ 0

SOLUTION:

The given problem is in cannonical form. First we convert to standard form

using slack variables as:



There are 6 variables and 3 equation therefore 6 – 3 = 3, 3 variables have to be

made as zeros and the are x1 , x2, x3 = 0 which when substituted in (ii) we get

the basic solution as s1 = 60. s2 = 72, s3 = 100 called the basic solution and

is the basis of the problem. We us iterative method to solve now:

Initial iteration:

Using the method as explained before we can get the first oteration as :



We find that optimality condition is not satisfied. That is c – z ≤ 0 in all columns

are positive. The column with most positive value is chosen as key column and

the coresponding variable is entering variable. Here X2 is the entering variable

and the leaving variable is one with minimum θ value. Here it is s1. We find pivot

equation and other new equations using the usual method. We get the iteration 2

as follows:



We observe that C – Z ≤ 0 in all columns, therefore optimal solution has been

obtained. Solution is Max Z = 5 x 0 + 12 x 15 + 8 x 0 = 180.

Sensitivity Analysis: This is the study of changes in the optimal solution

when

1. Ther is deletion and adition of new constraints.

2. Change in the values of coefficients in the objective function

3. Coefficients of variables in the constraints change.

4. The right handside constants of the constraints change..

Variation affecting the feasibility :

Two types of changes that could affect the current feasibility solution are:

1. Change in resources availability

2. Addition of constraints.

3. Any change in b will not affect the ZJ - cJ as it is optimal condition not

involving b’s. XB = B- (b) depends upon b therefore change in b will afect

feasibility of current solution

4. Addition of new constraints cannot improve the optimal value of any LPP

5. If new constraint is satisfied by current optimal solution , then the new

constraint is called REDUNTANT constraint as its addition will not change

the solution i.e current solution is feasiblblw and optimal.

UNIT II

If the given LPP is called as PRIMAL ( dual ) problem then this changed to

another problem with optimisation opposite to the given problem and

constraints changed is called the Dual problem( primal )

If the given problem is maximisation ( minimisation) then its dual will be

minimisation ( maximisation )



If there are n variables and m equations in primal then in its dualthere will be

m variables and n equations.

In the primal maximisation ( minimisation ) all the constraints have to be ≤ ( all

the constraints have to be ≥ ) . Its dual minisation ( maximisation ) will have all

its constraints ≥ (all the constraints have to be ≤ )

If in the primal x3 is unrestricted then in the dual third constraint is a

equality.If in the primal 4th constraint is equality then in dual fourth variable is

unrestricted.

Differentiate between Simplex method and dual simplex method.

1. In Simplex method feasible solution may be ther but optimality condition

wll not be satisfied.

In Dual simplex method the solution is be infeasible but optimality

condition is not satisfied.

2. W e find the entering variable first in the iteration table in the case of

simplex method by considering the most negetive of Zj - Cj , called the key

column and then the leaving variable by considering the minimum of θ ,

where θ = Min [𝑋𝐵𝑖

𝑎𝑖𝑟 ] where XBi refers to ith row of XB column and air refers

to element in XBi th row in the key column. The variable with respect to

this element is the entering variable and the row is called the key row.

3. W e find the leaving variable first in the iteration table in the case of dual

simplex method by considering the most negetive of XBi ‘s , called the key

row and then the entering variable by considering the maximum of θ ,

where θ = Max [𝑍𝑗− 𝐶𝑗

𝑎𝑘𝑗 ] where XBi refers to ith row of XB column and akj

refers to element in XBi th row of Zj - Cj th column . The variable with most

negetive in XB column is leaving variable and entering variable is with max

of θ.

Transportation Problem:



A transportation is a minimisation problem and is an LPP. There are m

sources and n destination and commodities are transported from every source

to every other destination. The assumptions are :

1. The availability of commodities in each source and the requirement in

every destination

2. The unit cost of transportation from each source to every other destination.

If Cij refers to unit cost of transportation from ith source to jth destination

and xij refers to alloction ith source to jth destination , the total cost of

transportation from ith source to jth destination is Cij xij . The RIM condition is

that sum of supplies in any source (row) is equal to the availability in that

source and the sum of demands in any destinnation (column ) is equal to the

requirement at that destination. With this as constraint the LPP of TP is given

by:

Minimisation Z = ∑ ∑ 𝑐𝑖𝑗𝑥𝑖𝑗𝑛1

𝑚1

Subject to conditions ∑ 𝑥𝑖𝑗𝑛1 = ai I = 1,2,…………..m

∑ 𝑥𝑖𝑗𝑚11 = bj j = 1,2,…………..n

Xij ≥ 0

Points to be remembered while working out a problem on TP:

1. Check whether the TP is balanced in the sense that Sum of the supplies is

equal to sum of the demands i.e ∑ 𝑎𝑖𝑚1 = ∑ 𝑏𝑗

𝑛1

2. Find the solution to the TP which is nothing but finding the allocations from

sources to destinations. The solutions so obtained should satisfy the RIM

conditions namely:

∑ 𝑥𝑖𝑗𝑛1 = ai I = 1,2,…………..m

∑ 𝑥𝑖𝑗𝑚11 = bj j = 1,2,…………..n. The number of allocations should not be

more m+ n+ 1 where m refers to number of rows and n refers to number of

columns.



3. By feasible solution of a TP we mean the non negetive allocations that

satisfy rim conditions. If the number of allocations do not exceed m+ n+ 1

where m refers to number of rows and n refers to number of columns. i.e

Xij ≥ 0 for all i and j then it is called as basic feasible solution

4. Optimal solution of a TP: It is the feasible solution of the TP which minises

the obejective function viz the cost of transportation.

5. An initial basic fesible solution(ibfs) is said to be non degenerative or the

given TP is said to be a problem of non-degeneration if the number of

allocations is exactly equal to m+ n+ 1.

6. The ibfs of a TP is said to be degenerative or it is a problem of

degeneration if the number of allocation is less than m+ n+ 1.

7. The method to overcome problem of degeneracy: This degeneracy problem

is overcome by assigning the very small quantity ( allocation ) ∈ ≥ 0 to any

one of the independent cells. After finding the ibfs , allow ∈ to tend to zero

and obtain the cost of TP.

8. We say the allocations are in independent cells if it is impossible to increase

or decrease any allocation without either changing the position of

allocation orviolating the rim conditions.

9. The three methods employed to solve a TP are:

(i). North West Corner Method ( nwcm)

(ii) Least Cost Method (lcm)

(iii) Vogel’s Approximation Method (VAM)

North West Corner Method:

1. Choose Cell (1 1) , the nwc cell . Allocate to this cell Min( supply demand ).

Close that row or column if full availability or full demand is met. Round off

the full allocate figure, simultanesoult reduce the higher figure to that



exten. ( in case both are same both the rows and columns, which will result

in degeneracy )

2. Rewrite the remaining matrix and repeat the process as in 1.

3. Continue this till full allocations are made . These allocations form the ibfs.

The initial cost is obtained by adding the allocations with their respective

unit cost of transportation

Least Cost method:

The cell with the least transportation cost is chosen. As in NWCM the

allocation is made. After closing the row or column ( or both – degeneracy ) the

remaining matrixis formed and the process is repeated till all the alocations are

made.

Vogel’s Approximation Method (VAM): This is called as penalty method. The

penalties are nothing but the difference between the least cost in a row(or

column ) and the next least cost. Penalty is found for every row and coulmn in the

cost matrix. The row ( or coumn ) with maximum penalty is chosen. As in

previous two methods , the least cost cell is chosen in that row( or column ) and

allocated to this cell the minimum of supply and demand and that row or column

is closed . The full allocation is rounded off and the higher figure is reduced to

that extent. The process of finding penaltyand allocation accordingly are carried

out till all allocations are done.

We may notice that the initial basic transportation using VAM < the initial basic

transportation using LCM < the initial basic transportation using NWCM.

MODIFIED DISTRIBUTION METHOD or called as Modi method is u – v method:

Steps involved are:

1. This method is applied in order to find the optimal solution to the given TP.

After finding the ibfs, we assign variables u1, u2, ……………. Un.to rows and

v1, v2, ……………. Vm. to columns. The ui or vj is assigned the value zero

whose row or column has the maximum number of allocations.



2. Now keeping allocated cells as reference , the values for the remaining u’s

and v’s are found using the formula uij + vij = Cij for all i and j.

3. For the unallocated cells two quantities are determined. One is uij + vij =

Cij and the other is dij = Cij - ( uij + vij ) for all I and j

OBSERVATIONS on dij :

(i) If dij ≥ 0 then the solution obtained is optimal.

(ii) If dij = 0 for some i and j the solution is optimal but alternative

solution exists.

(iii) If dij < 0 atleast for one i and j, then the solution is not optimal and

this cell will be allocated the maximum by a looping method which is

explained as below:

Assign the maximum +θ to this cell. Join this cell with other allocated

cells using horizontal and vertical lines. Assign -θ, +θ, -θ, +θ

alternately to cells which form the corners. Choose the minimum of

all allocations in –θ cells . Add this with the allocation in the +θ cell

allocation and subtract this from allocation in –θ cell. Note that the

reallocation is made without disturbing the rim conditions.

Perfom now the MODI test If all dij ≥ 0 then the solution obtained is

optimal if not repeat the process to get optimal solution

Example: Solve the following TP:



ASSIGNMENT PROBLEM:



This is similar to TP. Here there are n jobs and n persons . All the n persons can

perform all the n jobs . This is one job one person optimisation problem . Here it

is assignment of jobs to persons in such a way that the overall time consumed for

completion of jobs orthe overall cost for performing the job is kept

minimum.Unlike the TP assignment problem is always a problem of degeneration

as it is n xn matrix . in TP the number of allocations are 2n – 1 for degneration but

the number of solutions for AP is n only .

The method employed is Hungarian method:

1. Check whether the problem is balanced , if not balance i.e the number of

jobs must be equal to number of persons.

2. Choose the minimum of every row , subtract it from all other elements in

that row, thereby there will be atleast one zero in every row.

3. Check whether every column has atleast a zero, if not repeat the process

as explained in step2.

4. Now bracket a zero for assignment provided it is the only zero in that row.

If more than one zero are there , defer, go to the next and assign. While

making assignment , the other zeros lying up and down , left and right are

marked. This process is carried out for rows and then the same procedure

is followed for columns. The we would notice all the zeros are either

marked or assigned. If all the jobs are all persons the problem is over and

we get optimal solution.If not the following steps are followed:

(i) Tick the unassigned row( rows)

(ii) Move along that row (rows) and tick that column ( columns ) which

has marked zero.

(iii) Move along the ticked column ( columns ) and tick that row ( rows )

which has assignment.

(iv) This process is repeated till all assigned rows and marked columns

are ticked.

(v) Lines are drawn over unticked rows and ticked columns thereby we

get minimum number of lines drawn over all the zeros whose

number is equal to the present assignment.



(vi) The minimum of open numbers is chosen. This is subtracted from all

other open numbers and added with those numbers at the

intersection of lines. The other numbers are left undisturbed.

(vii) The assignments are made in the same way as explained above.If all

the jobs are assigned to all the persons the problem is over or we

pass onto next iteration starting from step (v) till we get all the jobs

assigned to persons .

The masthematical formulation of the Assignment Problem is

Minimisation Z = ∑ ∑ 𝑐𝑖𝑗𝑥𝑖𝑗𝑛1

𝑚1

Subject to conditions ∑ 𝑥𝑖𝑗𝑛1 = 1 I = 1,2,…………..n

∑ 𝑥𝑖𝑗𝑚11 = 1 . j = 1,2,…………..n and Xij ≥ 0 for all I and j.

Here Xij refers to assignment of ith job to jth person and Cij refers to cost of

performing ith job by jth person or time required for doing ith job by the jth

person. Assignment matrix is called the effective matrix as it can be cost or

time matrix. Note that Xij = 0 if ith job cannot be performed by jth person and

Xij = 1 if ith job can be performed by jth person therefore Xij = 0 or Xij = 1.

Therefore AP is an Integer Programming problem, likewise TP also.

Comparison between Transportation Problem and assignment problem:

Note that both TP and AP minimisation problems. In TP it is allocation of

commodities from sources to destinations. In AP it is assignment of jobs to

different persons.

Commodities can be supplied from a source to any number of destination but

sum of allocation must be equal to its availability.Similarly a destination can

get allocation from any number of sources but the sum of demands received

must be equal to its requirement. This is RIM condition. In AP it is one job one

person that sum of assignmenta in any row or column is equal to one.



The TP should not be a problem of degeneration whereas AP is always a

problem of degeneration because the number of assignments is always leaa

than m+ n+ 1.

Travelling Salesman Problem (TSP):

This is also assignment problem. A salesman is expected to visit n number of

places for doing business. While going from place to place he is not permitted

to visit a place more than once except the place of start or the Head

Quarters.He is supposed to take the shortes route possible. This can be

achieved by assignment only. As and when a person visits a palce , the

question is which is his next place of visit and which is nearest. This will take

care of distance and time.

(a)The salesman has to go through every city exactly once except starting city.

(b)The salesman starts from one city ( head quarters ) and comes back to

same city ( head quarters )

Hence going from one city to same city is not allowed. ( no assignmnts along

the diagonal of the matrix.)

( a) and (b) are usually called the route conditions.

Example:

Solve the following TSP:

Example:

Solve the following AP:



Solution : The given ptoblem is unbalanced. It has to be balanced first and then

only can be solved.. We create fictitious job 6 with zero time for each worker

to finish.

Using Hungarian method weget first assignment as follows:



Solution is not optimal as all workers are not assigned jobs .We then procedd

as follows:

Using the Hungarain method as explained above, we get



UNIT III

An Interger programming problem can be Pure Inter Programming Problem and

Mixed Inter Programming Problem.

Inter programming problem is also a LPP . The decision variables are non

negetive intergers and is given by:



Maximize Z = f( x1 , x2, x3, ……….. xn)


gj(x1 , x2, x3, ……….. xn ) ≤ = ≥ bj j= 1,2, ………………m.

xi ≥ 0 for all i = 1,2,……………n and are integers.

A mixed ipp is :

Maximize Z = f( x1 , x2, x3, ……….. xn)


gj(x1 , x2, x3, ……….. xn ) ≤ = ≥ bj j= 1,2, ………………m.

xi ≥ 0 for all i = 1,2,……………n and are integers.

Some are variables are restricted to integers and rest can be fractions or

integers.

The pitfalls of rounding decision variable to integer if obtained as decimals.

If the optimal solution obtained as non integer and corrected to nearest in

teger to the lower or higher side then it is likely that it may affect the optimal

solution drastically or the optimal solution will fail to satisfy some of the

constraints.

Two methods are employed to solve ipps in order to overcome such pitfalls.

The methods are:

1. Cutting method

2. Search method

In cutting method or cutting plane method additional constraints are constructed

and introduced with the help of a source row to eliminate the region that does



not contain integer solution. As many constraints as required are introduced to

eliminate regions which do not give integer solution and solved using dual simplex

method to get integer solution.

Gomorian constraints are introduced systematically using the source row whose

XB value has the maximum fractional part.

The procedure:

1. Convert minimisation LPP to maximisation problem.

2. Solve the LPP using simplex method for optimal solution.

3. If all vaues of decision variables are non negetive integers ,the problem is

over and we have the required solution.

4. If 𝑋𝐵1, 𝑋𝐵2, 𝑋𝐵3, ….. 𝑋𝐵𝑛, are all non integers, the one with maximum

faractional part is chosen as source row say 𝑋𝐵𝑘 = [𝑋𝐵𝑘 ] + fk where [𝑋𝐵𝑘 ]

Is the integral part of 𝑋𝐵𝑘 and fk is the fractional part and 0 < fk < 1. If

the problem is pure IPP, then the GOMORIAN constraint is constructed as

fllows:

Express each of the negetive fractions if any in source row as a negetive

integer plus a non negetive fraction.

∑ 𝑎𝑘𝑗𝑥𝑗 = 𝑋𝐵𝑘

∑{[𝑎𝑘𝑗 ] + 𝑓𝑘𝑗}𝑥𝑗 = [𝑋𝐵𝑘 ] + fk

In the form

∑ 𝑓𝑘𝑗𝑥𝑗 ≥ 𝑓𝑘

Or - ∑ 𝑓𝑘𝑗𝑥𝑗 + g1 = - 𝑓𝑘 , g1 ≥ o caled the Gomorian slack and the

constraint so formed is called the gomorian constraint.

This Constraint is inserted as additiona row in the final iteration where we

found the fractional optimal solution. Now the problem has become dual

simplex problem with infeasible solution and optimality condition is

reached. Now we solve the problem using dual simplex method . As many

Gomorian constraints as required are introduced and the iterations are

solved using dual simplex method till we get integer values for all variables.



The method employed to solve a MIXED INTER PROGRAMMING PROBLEM

is the same except for construction of Gomorian constraint.

5. Convert minimisation LPP to maximisation problem.

6. Solve the LPP using simplex method for optimal solution.

7. If all vaues of decision variables are non negetive integers ,the problem is

over and we have the required solution.

8. If 𝑋𝐵1, 𝑋𝐵2, 𝑋𝐵3, ….. 𝑋𝐵𝑛, are all non integers, the one with maximum

faractional part is chosen as source row say 𝑋𝐵𝑘 = [𝑋𝐵𝑘 ] + fk where [𝑋𝐵𝑘 ]

Is the integral part of 𝑋𝐵𝑘 and fk is the fractional part and 0 < fk < 1. If

the problem is Mixed IPP, then the GOMORIAN constraint is constructed as

fllows:

Express each of the negetive fractions if any in source row as a negetive

integer plus a non negetive fraction.

∑ 𝑎𝑘𝑗𝑥𝑗 = 𝑋𝐵𝑘

∑{[𝑎𝑘𝑗 ] + 𝑓𝑘𝑗}𝑥𝑗 = [𝑋𝐵𝑘 ] + fk

In the form ∑ 𝑓𝑘𝑗𝑥𝑗 ≥ 𝑓𝑘

∑ 𝑓𝑘𝑗𝑥𝑗𝑗∈𝑗+ + [𝑓𝑘

𝑓𝑘−1 ] ∑ 𝑓𝑘𝑗𝑥𝑗𝑗∈𝑗− ≥ 𝑓𝑘

-∑ 𝑓𝑘𝑗𝑥𝑗𝑗∈𝑗+ - [𝑓𝑘

𝑓𝑘−1 ] ∑ 𝑓𝑘𝑗𝑥𝑗𝑗∈𝑗− ≤ - 𝑓𝑘

-∑ 𝑓𝑘𝑗𝑥𝑗𝑗∈𝑗+ - [𝑓𝑘

𝑓𝑘−1 ] ∑ 𝑓𝑘𝑗𝑥𝑗𝑗∈𝑗− + g1 = - 𝑓𝑘

Where g1 ≥ 0 and the constraint is Gomorian constraint

Where j+ refers to terms which are positive and j- are terms which are

negetive.

Search method: In this the branch and bound method is applied to get the

integer solutions either by graphical method ( if two variables ) and simplex

method otherwise.

In branch and bound method for e.g x1 = 2.75 , it means 2 ≤ x1 ≤ 3.



It is clear that there is no integer solution in 2< x1 < 3. Therefore we search

for integer solution in regions x1 ≤ 2 and x1 ≥ 3 and these two regions are

mutually disjoint.

Now we have two sub problems with newly formed constraints and we

have to solve the two sub problems . The braching process is continued

until each problems terminates with either interger optimal solution or

there is evidence that it can not yield a better solution. The integer valued

solution among al the subproblems which gives the most optimum value

of sthe objective function is then selected as the optimum solution.

Main disadvantage of this method is that it requires the optimum solution

of each sub –problem. In large problems this could be very tedious job. But

the computational efficiency osf this method is increased by using the

concept of BOUNDING. By this concept whenever the continuous optimum

solution of a sub-problem yields a value of the objective function lower

than that of the best available solution ( for maximisation problem ) it is

useless to explore the problem any further. This problem is said to be

fathomed and is dropped from further consideration. Thus once a feasible

inter solution is obtained , its associated objective function can be taken

as the LOWER BOUND ( for maximisation problem ) to give inferior sub

problems. Hence efficiency of a branch and bound method depends upon

the successive sub problems are found.

For minimisation problem instead of lower bound upp bound is

considered.

Example: Use Branch and Bound technique to solve the following LPP for

integer values:

2.) Solve the following IPP:

Min Z = -2X1 - 3X2 , subject to:

2X1 + 2X2 ≤ 57

X1 ≤ 2, X2 ≤ 2, X1 , X2 ≥ 0 and are integers.

Solution: Convert to Maximisation problem as :



Max Z* = 2X1 +3X2 subject to :

2X1 + 2X2 ≤ 57

X1 ≤ 2, X2 ≤ 2, X1 , X2 ≥ 0 and are integers.

First we solve the above using simplex method and the optimal is obtained in the second

iteration :

Solution:

The standard for is :

Max Z* = 2X1 +3X2 + 0 s1 + 0s2 + 0s3 subject to :

2X1 + 2X2 + s1 + 0s2 + 0s3 = 5 7

X1 + 0X2 + 0s1 + s2 + 0s3 = 2

0X1 + X2 + 0s1 + 0s2 + s3 = 2

X1 , X2 , s1 ,s2, ,s3 ≥ 0 and X1 ,X2 are integers.

First we solve LPP using simplex method and the optimal is obtained in the second iteration :

Cj 2 3 0 0 0

CB YB XB X1 X2 s1 s2 S3

2 X1 3/2 1 0 1/2 0 -1

0 s2 1/2 0 0 -1/2 1 1

3 X1 2 0 1 0 0 1

Z*j - Cj 9 0 0 1 0 1

We find that X1 = 3/2 , a non integer so with this as source row Gomorian constraint is given as

follows:

0X1 + 0X2 -1/2 s1 +0 s2 +0 s3 + g1 = -1/2, where g1 ≥ 0 is called Gomorialn slack and the

constraint is called Gomorian constraint, the iteration becomes :



Cj 2 3 0 0 0 0

CB YB XB X1 X2 s1 s2 s3 g1

2 X1 3/2 1 0 1/2 0 -1 0

0 s2 1/2 0 0 -1/2 1 1 0

3 X1 2 0 1 0 0 1 0

0 g1 -1/2 0 0 -1/2 0 0 1

Z*j - Cj 9 0 0 1 0 1 0

The leaving variable is g1 and entering variable is s1. Using the usual simplex method we get the

next iteration as :

3rd iteration

Cj 2 3 0 0 0 0 0

CB YB XB X1 X2 s1 s2 s3 g1

2 X1 1 1 0 0 0 -1 1

0 s2 1 0 0 0 1 1 -1

3 X2 2 0 1 0 0 1 0

0 s1 1 0 0 1 0 0 -2

Z*j - Cj 8 0 0 0 0 1 2

Since Z*j - Cj ≥ 0 for all I and j we have obtained the optimal solution namely Z*B = 8 with X1 =1

and X2 =2 and both are non negetive integers.

The solution for the original problem is MinZ = - { Max Z* } = -(8) = -8, therefore

MinZ](1, 2 ) = -2 x 1 – 3x2 = -8

DYNAMIC PROGRAMMING PROBLEMS (DPP)

This is a m mathematical technique of optimisation using multistage decision

process.

The dynamic programming technique decomposes the original problem into n

subproblems ( stages ) each in one variable. The solution is obtained in an



ordinary manner by starting from one stage to next and is completed after the

final is reached.

Bellman’s principle of optimality:

An optimal policy ( set of decisions ) has the property that whatever be the

initial state and initial decisions, the remaining decisions must constitute an

optimal policy for the state resulting from the first decision.

It states that given the initial state of the system an optimal policy for the

subsquent stages does not depend upon the policy adopted at the preceeding

stages.

Characteristics of DPP:

1. Problem can be divided into stages , with policy decision required at each

stage.

2. Sttes are associated with each stage may be finite or infinite.

3. Of policy decision at a stage is change the state to next state in next

stage.

4. Current situation is described by a set of secision variables.

5. Given the current state or optimal policy for remaining stages is

independent of the policy adopted in previous stages.

6. The solution procedure begins by finding the optimal policy for each state

of the last stage.

7. A recursive relation ( recurrsive relation or functional equation ) which

identifies the optimal policy for each state with n stages remaining given

the optimal policy for each state with ( n – 1 ) stages left.

8. Using this recursive relationship the solution procedure moves backward

stage by stage, each time finding the policy when starting at the initial

stage.

DPP algorithm:

1. Identify decision variables and the objective function

2. Number of sub problems and identify that state variables at each state.



3. Write the general recursive relationship for computing optimal policy.

4. Write the reltion giving the optimal decision function for one stage sub

problem and solve it.

5. Solve the optimal decision function for 2-stage, 3-stage….. ( n-1 ) stage and

n- stage problem.

If the DPP is solved using the recursive equation . starting from the first

through the last stage i.e obtaining a sequence

f1 → f2 → f3 ………………………………………….. → fn of the optimal solution . This

equation is called the forward computational procedure.

If the recursive equation are formulated to obtain a sequence as gibrn

here fn → fn-1 → fn-2 ………………………………………….. → f1 then the computation is

called backward computational procedure.

Applications of DPP:

1. It is applied in production area- production sheduling and employment

smoothening problems .

2. In deterministic inventory problems.

3. Allocation of different salesman to different places for optimisation of

sales.

4. To determine the optimal combination of advertising media (TV, Radio.

Paper)and the frequency of advertising.

Example:

Optimal sub division problem:

Divide the quantity c into n parts such a way that their product is maximum. That divide c into

y1, y2, ………………….., yn parts such that

Maximum Z = y1 X y2 X …………………..X yn subject to constraints:

y1 + y2 + …………………..+ yn = c

Solution:

We develop a recursive equation to solve the problem.



The problem is split into n stages with optimal decision function for the ( n – 1 ) stage sub

problem

Let yi ith part of c then it is ith stage , since it assumes non negetive value y1 + y2 + …………………..+

yn = c

This means yi is continuous. Therefore we can use idea of differentiation for finding the

optimal value.

Let fn ( c ) denote the Max of y1 X y2 X …………………..X yn when c is divided into n parts.. Hence it is a

function of n now.

Let n =1 , then

f1( c ) = c ( trivial ) …………………………………(1)

Let n = 2 ( 2nd stage ) . Let y1 = x, then y2 = (c – x) and y1 + y2 = c, then

f2( c ) = 𝑀𝑎𝑥0 ≤𝑥 ≤𝑐 { y1 , y2 } = 𝑀𝑎𝑥0 ≤𝑥 ≤𝑐 { x ( c – x ) } = 𝑀𝑎𝑥0 ≤𝑥 ≤𝑐 { x f1( c – x ) } ….(2)

when n = 3 ,stage 3, let y1 = x, then y2 + y3 = (c – x) and y1 + y2 + y3 = c, Here c is divided into

3 parts as y1 ,y2 and y3

Now we have

f3( c ) = 𝑀𝑎𝑥0 ≤𝑥 ≤𝑐 { y1 , y2 , y3 } = 𝑀𝑎𝑥0 ≤𝑥 ≤𝑐 { x f1( c – x ) }

In general we have for n stage problem:

fn( c ) = 𝑀𝑎𝑥0 ≤𝑥 ≤𝑐 { y1 , y2 , y3 ….. yn} = 𝑀𝑎𝑥0 ≤𝑥 ≤𝑐 { x fn-1 ( c – x } ……………….(3)

Let us solve the recursive equation (2)

f2( c ) = 𝑀𝑎𝑥0 ≤𝑥 ≤𝑐 { x f1( c – x ) }

= 𝑀𝑎𝑥0 ≤𝑥 ≤𝑐 { x ( c – x ) }

Applying the method of finding maxima and minima for 𝑀𝑎𝑥0 ≤𝑥 ≤𝑐 { x ( c – x ) } , we get

𝑑[ 𝑥 (𝑐−𝑥)]

𝑑𝑥 = 0 , gives x =0 or x= c/2, therefore f2( c ) = c/2 xc/2 = [

𝑐

2 ]2

Similarly we can get f3( c ) = c/2 xc/2xc/2 = [ 𝑐

3 ]3

Proceeding in this way we can get for n =m



fm( c ) = c/2 xc/2xc/2xc/2x……..mtimes = [ 𝑐

𝑚 ]m

now for n = m + 1

fm+1( c ) = 𝑀𝑎𝑥0 ≤𝑥 ≤𝑐 { x fm ( c – x) } = 𝑀𝑎𝑥0 ≤𝑥 ≤𝑐 { x [(𝑐−𝑥)

𝑚] m } (4)

the function x [(𝑐−𝑥)

𝑚] m attains its its maximum when x =

𝑐

𝑚+1 , using this in (4) , we get

fm+1( c ) = { 𝑐

𝑚+1 }m + 1

hence by mathematical induction, we get

optimal policy as ( c/n, c/n, …. c/n ) and fn( c ) = { 𝑐

𝑛 }n

UNIT IV

An unconstrained optimisation problem is one where you only have to be

concerned with the objective function that you try to optimise. An objective

function is one that you want to optimise and none of the variables in the

objective function are constrained.

Let F(x) is the given function. Le t F’(x) = 0 and at x = x* F” (x) > 0 . then

1. F(x*) = 𝑀𝑖𝑛⏟𝑥

F(x) x* is local Min of F(x)

2. F” (x) < 0 , F(x*) = 𝑀𝑎𝑥⏟𝑥

F(x) x* is local Max of F(x)

3. F” (x) = 0, x* is critical point F(x)

Example for local minimum:

Let F(x) = x2 +2. Now F’(x) = 2x , F’(x) = 0 ⟹ x = 0 ⟹ x = 0 is local

minimum for F(x) = x2 +2.and F("x) =2 > 0

Global maximum and global minimum:

A global maximum isa point that f(x*) ≥ f(x) for all x in that domain and

is global minimum if f(x*) ≤ f(x) for all x in that domain of f

Definition of Concave and Convex functions:



When f”(x) ≥ 0 for all x then f(x) is a convex function . Here local

minimum of x* is the global minimum of f(x) .

When f”(x) ≤ 0 for all x then f(x) is a concave function . Here local

maximum f x* is the global maximum of f(x)

The conditions for a minimum or a maximum value of a function of

several variables :

Let f(x) be a function of several independent variables of x .

1. Find ∇ f(x) =0

2. Solve to get the solution vector x*

3. Calculate ∇2 f(x)

4. Evauate at x*

5. Inspect Hessian matrix at x*

H(x) = ∇2 f(x)

The above is symetric matrix .

OBSERVATION and Inferences:

(i) If Hessian matrix H(x*) is positive definite then at x* f(x) has local

minimum at x*.

(ii) If Hessian matrix H(x*) is negetive definite then at x* f(x) has local

maximum at x*.

(iii) If Hessian matrix H(x*) is indefinite matrix then at x* f(x) has

neither local minimum nor local maximum..

If a maximization problem is given as

Max Z = f( x1 , x2, x3, ……….. xn)

Subject to

g( x1 , x2, x3, ……….. xn) ≤ b

xi ≥ 0 for all i = 1,2,……………n the necessary condition for maximisation

is



Example:

Find the extrmum points of the function





UNIT V

A project is defined as a combination of interrelated activities all of which

must be executed in a certain order to achieve goal.

A small project involving very few activities can be efficiently done with

help of experience gained in that field over a period of time. But for a big

project such thumb cannot be used , experience cannot help. A scientific

and mathematical approach is required.

Projects are of two types . They are deterministic projects and the other is

probabilistic projects.

A number of methods applying network scheduling techniques has been

developed.

Programme Evaluation Technique (PERT) and Critical Path Method (CPM)

are two of maany network techniques which commonly used for Plannin,

Scheduling and controlling large complex projects.

The three main manaferial functions for any project are:

1. Planning

2. Scheduling

3. Control.

Planning: This phase involves listing of tasks for the project. We determine

money , men and other resources required for the project. The cost of

project and duration of project is also determined.

Scheduling : We determine in this phase the activities involved in the

project with thwie sequencial order of exection and also the duration of the

project.

Control: This phase consists of reviewing the progress of the project and if

deviations crop they are rectified immediately.Remedial measures are

taken wherever possible without loss of time.



Activity: An activity consumes men, money, time and other resources and is

shown by an arrow line connecting two nodes in the network diagram.

Capital letters are used to denote activities as A,B, P,Q….

Event: Events are shown by nodes or points which form the end points of

an activity. They DO NOT CONSUME ANY RESOURECE.

A, B and C are called the activities and 1, 2, 3 and 4 are the nodes. Activities

are join of nodes.. A is called the preceeding activity of B and is called the

predecessor. B is called called the successor of A and is called the ssucceeding

activity.

Procedure for drawing net work diagram:

1. A net work moves from left to right. It starts with initial node, marked as 1.

2. Subsquent activities are drawn one after another as per the scheduling in

the problem.

3. Dummy activities are drawn wherever necessary.

4. There should not be any dangling activity.

5. There should not be any loop formation in the net work drawing.

6. No two activities must cross each other.

DUMMY ACTIVITY:

This is an imaginary activity which not consume any resource and which

serves the purpose of indicating the predecessor or successor relationship

1 2

A

C

B



clearly in any activity on arrow diagram. It is merely drawn for techical

completition of the project.

Project Network or Arrow Diagram: Tge diagram denoting all the activities of a

project by arrows taking into account the technological sequence of the

activities is called the project network or arrow diagram.

Fulkerson’s rule for numbering the nodes.

1. Number the start node which has no predecessor activity as 1.

2. Delete all the activities emanating from this node 1.

3. Number all the resulting start nodes without any predecessor as 2,3,..

4. Delete all the activities originating from the start nodes 2,3…in step 3.

5. Number all the resulting new start nodes without any predecessor next to

the last number used in step3.

6. Repeat the process until the terminal node without any successor activity is

reached and number this terminal node suitably.

NETWORK COMPUTATIONS.

We use forward pass calculation to determine Earliest Start Time (EST) . The

EST of an activity i → j in a project network is given by :

ESJ = Max [ESi + tij ] where ESi denotes the earliest start time of all activities

emanating from the node i and tij is the estimated duration of the activity

i → j

We use backward pass calculation to determine Lates Finish Time (LFT) and

Latest Start Time ( LST ) . The LST of an activity i → j in a project network is

given by :

LSi = Min [[LSj - tij ] where tij is the estimated duration activity i → j

Definition of Critical Path: This is a path connecting the initial node to the very

last terminal node , of longest duration in any project net work . ( It is also

defined as path of zero float.)



Various floats for activities of a project are determined using forward and

backward pass calcularions namely EST AND EFT and LFT and LST.

Total Float of activity i → j is given by:

{ TF ] i→ j = [ LF - EF ] i→ j = [ LS - ES ] i→ j

The Total Float is the amount of time by which the particular activity may be

delayed without affecting the duration of the project.

If TF is greater than zero ( positive quantity )it may mean that the resources

doe that activity is more than enough .

If Total Float of an activity is zero it may mean that the resources available for

the activity is just enough.

If Total Float of an activity is less than zero ( negetive ) it may mean that the

resources available for the activity is just insufficient.

Free Float ( FF ) of activity i → j is given by:

{ FF ] i→ j = [ TF ] i→ j - ( L – E ) of the event j

= [ TF ] i→ j - slack at head j

The significance of Free Float of an activity is that . that particular activity can

be delayed by that much of float without affecting the succeeding activity.

Independent Float ( IF ) of activity i → j is given by:

{ IF ] i→ j = [ FF ] i→ j - ( L – E ) of the event i

= [ FF ] i→ j - slack at tail i

The significance of Independent Float of an activity is that , that particular

activity can be delayed by that much of float without affecting the preceeding

activity.

It is be noted that



IF ≤ FF ≤ TF

Uses of floats: Floats ar useful in resource levelling and resource allocation

problems. Floats give some flexibility in resheduling

Programme Evaluation Review Technique:

This technique is aplplied to probabilistic projects to determine the critical

path and the different floats.

There are three time estimates, using these we can find one expected time

estimate for every activity . The expected duration of the project is the critical

path. The expected standard deviation of every activity is determined . Using

the standard deviation of critical path we will be determine the probability of

finishing the project before or after the expected duation time.

PERT calculations depend on

1. Optimistic (least) time estimate: ( to or a ) is the duration of any activity

when everything goes on well during the project. That is there wont be any

delay or any difficulty in getting any resource required for the project.

Scarcity of raw materials will not there.

2. Pessimistic ( greatest ) time estimate: ( tp or b ) This is the duration of any

activity when every thing goes against the will of the person doing the

project. He faces all types of difficulties during project execution.

3. Most Likely time estimate: (tm or m ) This is time duration of a project

when things go sometimes well and sometimes bad during the project

execution.

Two main assumptions made in PERT calculation are:

1. The duration of activities are independent.i.e the duration of one

activity does not have a bearing on any other activity.

2. The activity duration follows β – distribution.

The procedure for solving a problem in PERT.

1. Draw the project net work.



2. Compute the duration of every activity using te = (𝑎 + 𝑏 + 4𝑚 )

6

3. Compute EST, EFR; LST and LFT.

4. Determine the Critical path.

5. Compute the expected variance of activity using σ2 = ( 𝑎−𝑏 )2

36

6. Compute the expected standard deviation of pfoject length σ2c and

calculate the standard normal variate 𝑇𝑆 − 𝑇𝐸

𝜎𝐶

WHERE TS IS sprcified time to

complete the priject, TE expected duration of the project for completion

and σc expected S.D of the project length.

7. We can use to calculate completion of project before or after time using

area under normal curve.

Differences between PERT and CPM:

There are three time estimates in PERT whereas in CPM there is only

one time estimate.

PERT is event oriented whereas CPM is activity oriented.

PERT is appliied to R & D projects whereas CPM is applied to projects

involving well known activities of repetitive n nature. ( applied to

construction buikdins, bridges whose time of completion well

determined. )

Example:

The three time estimates ( in months ) of all actvities of a project are

given below:

For the following network shedule problem



The longest path is the CPM , it is :











1. Find the expected duration and S.S of each activity.

2. Construct project net work.

3. Determine Critical path.

4. What is the probability that the project will be completed two

months later than expected .

We have already seen that in project shedule, we determine different

floats like free float, independent float and total float and use these

floats for resiources levelling and resource s allocation.

The procedure for Resource levelling is :

1. Draw the early start schedule graph.

2. Draw the corresponding manpower chart,

3. Identify the activities with slack.

4. Adjust the activities identified in 3 and adjust them to level the peak

resource requirement.

The procedure for Resource allocation program is :

1. Allocate resources serially in time .

2. If several jobs compete for the same resources give preference to the

jobs with the smallest slack.

3. Reshedule non critical jobs , if possible so as to make available tne

needed resources for resheduling critical jobs.

Reference: 1. Operations Research by Premkumar Gupta and D.S.HIRA, 2.

Resourcesmanagement techniques by Prof. Ssundaresan, K.S. Ganapathy

Subramaniam and K.Ganesan.



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CS6704 RESOURCES MANAGEMENT TECHNIQUES