13259677 assignment problem

Prof Vinay PanditASSIGNMENT AND TRANSPORTATION THEORY

1) What is an Assignment Problem?

• The assignment problem can be stated as a problem where different jobs are to be

assigned to different machines on the basis of the cost of doing these jobs. The

objective is to minimize the total cost of doing all the jobs on different machines

• The peculiarity of the assignment problem is only one job can be assigned to one

machine i.e., it should be a one-to-one assignment

• The cost data is given as a matrix where rows correspond to jobs and columns to

machines and there are as many rows as the number of columns i.e. the number of

jobs and number of Machines should be equal

• This can be compared to demand equals supply condition in a balanced

transportation problem. In the optimal solution there should be only one

assignment in each row and columns of the given assignment table. one can

observe various situations where assignment problem can exist e.g., assignment

of workers to jobs like assigning clerks to different counters in a bank or salesman

to different areas for sales, different contracts to bidders.

• Assignment becomes a problem because each job requires different skills and the

capacity or efficiency of each person with respect to these jobs can be different.

This gives rise to cost differences. If each person is able to do all jobs equally

efficiently then all costs will be the same and each job can be assigned to any

person.

• When assignment is a problem it becomes a typical optimization problem it can

therefore be compared to a transportation problem. The cost elements are given

and is a square matrix and requirement at each destination is one and availability

at each origin is also one.

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• In addition we have number of origins which equals the number of destinations

hence the total demand equals total supply . There is only one assignment in each

row and each column .However If we compare this to a transportation problem we

find that a general transportation problem does not have the above mentioned

limitations. These limitations are peculiar to assignment problem only.

2) What is a Balanced and Unbalanced Assignment Problem?

A balanced assignment problem is one where the number of rows = the number of

columns (comparable to a balanced transportation problem where total demand =total

supply)

Balanced assignment problem: no of rows = no of columns

Unbalanced assignment is one when the number of rows not equal to the number of

columns and vice versa. e.g. The number of machines may be more than the number of

jobs or the number of jobs may be more than the number of machines.

In such a situation we introduce dummy row/column(s) in the matrix. These rows or

columns have a zero cost element. Thus we can balance the problem and then use

Hungarian method to find optimal assignment.

Unbalanced assignment problem: no of rows not equal to no of columns

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3) What is a Prohibited Assignment Problem?

A usual assignment problem presumes that all jobs can be performed by

all individuals there can be a free or unrestricted assignment of jobs and individuals. A

prohibited assignment problem occurs when a machine may not be in, a position to

perform a particular job as there be some technical difficulties in using a certain

machine for a certain job. In such cases the assignment is constrained by given facts.

To solve this type problem of restriction on job assignment we will have

to assign a very high cost M This ensures that restricted or impractical combination does

not enter the optimal assignment plan which aims at minimization of total cost.

4) What are the methods to solve an Assignment Problem (Hungarian Method)?

There are different methods of solving an assignment problem:

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DIFFERENT METHODS OF ASSIGNMENT PROBLEM

Hungarian methodSimplex Method

TransportationProblem

Complete Enumeration


1)Complete Enumeration Method: This method can be used in case of assignment.

problems of small size. In such cases a complete enumeration and evaluation of all

combinations of persons and jobs is possible.

One can select the optimal combination. We may also come across more than one optimal

combination The number of combinations increases manifold as the size of the problem

increases as the total number of possible combinations depends on the number of say,

jobs and machines. Hence the use of enumeration method is not feasible in real world

cases.

2) Simplex Method: The assignment problem can be formulated as a linear programming

problem and hence can be solved by using simplex method.However solving the problem

using simplex method can be a tedious job.

3)Transportation Method: The assignment problem is comparable to a transportation

problem hence transportation method of solution can be used to find optimum allocation.

Howver the major problem is that allocation degenerate as the allocation is on basis one

to one per person per person per job Hence we need a method specially designed to solve

assignment problems.

4 )Hungarian Assignment Method (HAM):

This method is based on the concept of opportunity cost and is more efficient in solving

assignment problems.

Method in case of a minimization problem.

As we are using the concept “opportunity this means that the cost of any opportunity that

is lost while taking a particular decision or action is taken into account while making

assignment. Given below are the steps involved to solve an assignment problem by using

Hungarian method.

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Step 1:

Determine the opportunity cost tableI

• Locate the smallest cost in each row and subtract it from each cost figure in that

row. This would result in at least one zero in each row. The new table is called

reduced cost table.

• Locate the lowest cost in each column of the reduced cost table subtract this

figure from each cost figure in that column. This would result in at least one zero

in each row and each column, in the second reduced cost table.

Step 2:

Determine the possibility of an optimal assignment:

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Step 1:

Determine the opportunity cost table

Step 2:

Determine the possibility of an optimal assignment

Step 3

Modify the second reduced cost table

Step 4:

Make the optimum assignment


• To make an optimal assignment in a say 3 x 3 table. We should be in a position to

locate 3 zero’s in the table. Such that 3 jobs are assigned to 3 persons and the total

opportunity cost is zero .A very convenient way to determine such an optimal

assignment is as follows:

• Draw minimum number of straight lines vertical and horizontal, to cover all the

zero elements in the second reduced cost table. One cannot draw a diagonal

straight line. The aim is that the number of lines (N) to cover all the zero

elements should be minimum. If the number of lines is equal to the number of

rows (or columns) (n) i.e N=n it is possible to find optimal assignment .

• Example :for a 3 x 3 assignment table we need 3 straight lines which cover all the

zero elements in the second reduced cost table. If the number of lines is less than

the number of rows (columns) N < n optimum assignment cannot be made. we

then move to the next step.

Step 3:

Modify the second reduced cost table:

• Select the smallest number in the table which is not covered by the lines. Subtract

this number from all uncovered numbers aswell as from itself.

• Add this number to the element which is at the intersection of any vertical and

horizontal lines.

• Draw minimum number of lines to cover all the zeros in the revised opportunity

cost table.

• If the number of straight lines at least equals number of rows (columns) an

optimum assignment is possible.

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Step 4:

Make the optimum assignment:

If the assignment table is small in size it is easy to make assignment after step 3.

However, in case of large tables it is necessary to make the assignments systematically.

So that the total cost is minimum. To decide optimum allocation.

• Select a row or column in which there is only one zero element and encircle it

Assign the job corresponding to the zero element i.e. assign the job to the circle

with zero element. Mark a X in the cells of all other zeros lying in the column

(row) of the encircled zero. So that these zeros cannot be considered for next

assignment.

• Again select a row with one zero element from the remaining rows or columns.

Make the next assignment continue in this manner for all the rows.

• Repeat the process till all the assignments are made i.e. no unmarked zero is left.

• now we will have one encircled zero in each row and’ each column of the cost

matrix. The assignment made in this manner is Optimal.

• Calculate the total cost of assignment from the original given cost table.

Maximization method

In order to solve a maximization type problem we find the regret values instead of

opportunity cost. the problem can be solved in two ways

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• The first method is by putting a negative sign before the values in the assignment

matrix and then solves the sum as a minimization case using Hungarian methods

as shown above.

• Second method is to locate the largest value in the given matrix and subtract each

element in the matrix from this value. Then one can solve this problem as a

minimization case using the new modified matrix.

Hence there are mainly four methods to solve assignment problem but the most efficient

and most widely used method is the Hungarian method

Q5 ) Note on Traveling Salesmen problem.

Traveling salesman problem is a routine problem. It can be considered as a typical

assignment problem with certain restrictions. Consider a salesman who is assigned the

job of visiting n different cities. He knows (is given) the distances between all pairs of

cities. He is asked to visit each of the cities only once. The trip should be continuous and

he should come back to the city from where he started using the shortest route. It does not

matter, from which city he starts. These restrictions imply.

(1) No assignment should be made along the diagonal.

(2) No city should be included on the route more than once

This type of problem is quite simple but there is no general algorithm available for its

solution. The problem is usually solved by enumeration method, where the number of

enumerations is very large.

For example for a salesman who is instructed to visit five cities we shall have to consider

more than 100 possible routes. The method is therefore impractical for large size

problems and it also implies approximations in finding route with minimum distance.

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The peculiar nature of the problem and the various restrictions imposed on resulting

solution indicate that the method of solution to a traveling salesman problem should

include:

(1) Assigning an infinitely large element M in the diagonal of the distance matrix.

(2) Solve the problem using Hungarian Method as it gives shortcut route but,

(3) Test the solution for feasibility whether it satisfies the condition of a continues route

without visiting a city more than once.

If the route is not feasible, make adjustments with minimum increase in the total distance

traveled by the salesman. This is how one can solve traveling sales man problem

6) What is a Transportation Problem?

• A transportation problem is concerned with transportation methods or selecting

routes in a product distribution network among the manufacturing plants and

distribution warehouses situated in different regions or local outlets.

• In applying the transportation method, management is searching for a distribution

route, which can lead to minimization of transportation cost or maximization of

profit.

• The problem involved belongs to a family of specially structured LPP called

network flow problems.

7) What is a Balanced and Unbalanced Transportation Problem?

Balanced transportation problem

Transportation problems that have the supply and demand equal is a balanced

transportation problem. In other words requirements for the rows must equal the

requirements for the columns.

Unbalanced transportation problem

An Unbalanced transportation problem is that in which the supply and demand are

unequal. There are 2 possibilities that make the problem unbalanced which are

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(i) Aggregate supply exceeds the aggregate demand or

(ii) Aggregate demand exceeds the aggregate supply.

Such problems are called unbalanced problems. It is necessary to balance them before

they are solved.

Balancing the transportation problem

• Where total Supply exceeds total demand .

In such a case the excess supply is, assumed to go to inventory and costs nothing for

shipping(transporting). This type of problem is balanced by creating a fictitious

destination. This serves the same purpose as the slack variable in the simplex/method A

column of slack variables is added to the transportation tableau which represents a

dummy destination with a requirement equal to the amount of excess supply and the

transportation cost equal to zero. This problem can now be solved using the usual

transportation methods.

• When aggregate demand exceeds aggregate supply in a transportation

problem

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Aggregate supply exceeds the

aggregate demand

Aggregate demand exceeds the

aggregate supply

2 Possibilities That Make The Problem Unbalanced


When aggregate demand exceeds aggregate supply in a transportation problem a dummy

row is added to restore the balance. This row has an availability equal to the excess

demand and each cell of this row has a zero transportation cost per unit. Once the

problem is balanced it can be solved by the procedures normally used to solve a

transportation problem.

8) What is a Prohibited Transportation Problem?

Sometimes in a given transportation problem some route(s) may not be available.

This could be due to a variety of reasons like-

(i) Strikes in certain region

(ii) Unfavorable weather conditions on a particular route

(iii) Entry restriction.

In such situations there is a restriction on the routes available for transportation. To

overcome this difficulty we assign a very large cost M or infinity to such routes. When a

large cost is added to these routes they are automatically eliminated form the solution.

The problem then can be solved using usual methods.

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REASONS FOR UNAVAILABILITY OF ROUTES

Strikes in certain

regionUnfavorable weather

conditions on a

particular route

Entry restriction.


9) What is Degeneracy in a Transportation Problem?

The initial basic feasible solution to a transportation problem should have a total number

of occupied cell (stone squares) which is equal to the total number of rim requirements

minus one i.e. m + n — 1. When this rule is not met the solution is degenerate.

Degeneracy may occur-

If the number of occupied cells is more than m + n — 1.

This type of degeneracy arises only in developing the initial solution. It is caused by

an improper assignment of frequencies or an error in formulating the problem. In such

cases one must modify the initial solution in order to get a solution which satisfies the

rule m + n—i.

The problem becomes degenerate at the-

(i) Initial stage

When in the initial solution the number of occupied cells is less than m + n — 1

(rim requirements minus 1) i.e. the number of stone squares in insufficient

(ii) When two or more cells are vacated simultaneously

Degeneracy may appear subsequently when two or more cells are vacated

simultaneously in the process of transferring the units, along the closed loop to

obtain an optimal solution.

When transportation problem becomes degenerate

When transportation problem becomes degenerate it cannot be tested for

optimality because it is impossible to compute u and, v values with MODI method. To

overcome the problem of insufficient number of occupied cell we proceed by assigning

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an infinitesimally small amount (close to zero) to one or more (if needed) empty cell and

treat that cell as occupied cell. This amount is represented by the Greek letter E (epsilon).

It is an insignificant value and does not affect the total cost. But it is appreciable enough

to be considered a basic variable. When the initial basic solution is degenerate, we assign

c to an independent empty cell. An independent cell is one from which a closed loop

cannot be traced. It is preferably assigned to a cell which has minimum per unit cost.

After introducing e we solve the problem using usual methods of solution.

10) Steps to solve a Transportation Problem.

A transportation problem can be solved in 2 phases

PHASE I

Step 1:

Check whether the given T.P. is balanced or not. If it is unbalanced then balance it by

adding a row or a column.

Step 2:

Develop initial feasible solution by any of the five methods:

a) North West Corner Rule (NWCR) or South West Corner Rule (SWCR)

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Step 1:

Check whether the given T.P. is balanced or not

Step 2:

Develop initial feasible solution by any of the five methods


b) Row Minima Method (RMM)

c) Column Minima Method (CMM)

d) Matrix Minima Method (MMM)

e) Vogel’s Approximation Method (VAM)

We discuss here the two commonly used methods to make initial assignments

(1) Northwest corner rule (2) Vogel’s Approximation Method (VAM)

(1) Northwest Corner Rule:

Start with the northwest corner of the transportation tableau and consider the cell in the

first column and first row. We have values a1 and b1 at the end on the first row and

column i.e. the availability at row one is a1 and requirement of column 1 is b1.

(i) If al > b1 assign quantity b1 in the cell, i.e. x1 b1. Then proceed horizontally to the

next column in the first row until a1 is exhausted i.e. assign the remaining number a1 - b1

in the next column.

(ii) If al < b1 then put Xl al and then proceed vertically down to the next row until b1 is

satisfied. i.e. assign b1 – a1 in the next row.

(iii) If a1 = b1 then put XII = a1 and proceed diagonally to the next cell or square

determined by next row and next column.

In this way move horizontally until a supply source is exhausted, and vertically down

until destination demand is completed and diagonally when a1 = b1, until the south-east

corner of the table is reached.

(2) Vogel’s Approximation Method (VAM):

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The north-west corner rule for initial allocation considers only the requirements and

availability of the goods. It does not take into account shipping costs given in the tableau.

It is therefore, not a very sound method as it ignores the important factor, namely cost

whiçh we seek to minimize. The VAM, on the other hand considers the cost in each cell

while making the allocations we explain below this method.

(i) Consider each row of the cost matrix individually and find the difference between two

least cost cells in it. Then repeat this exercise for each column. Identify the row or

column with the Largest difference (select any one in case of a tie).

(ii) Now consider the cell with minimum cost in that column (or row) and assign the

maximum units possible to that cell.

(iii) Delete the row/column that is satisfied.

(iv) Again find out the differences and proceed in the same manner as stated in earlier

paragraph and continue until all units have been assigned.

PHASE II – TEST FOR OPTIMALITY

Before we enter phase II, the following two conditions should be fulfilled in that order.

(i) Obtaining a basic feasible solution implies finding a minimum number of ij values.

This minimum number is m + n - 1. Where m is the number of origins n is the number of

destinations. Thus initial assignment should occupy m + n - I cells i.e. requirements of

demand and supply cells minus — 1.

(ii) These ij should be at independent positions

These requirements are called RIM requirements.

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Test for optimality (or improvement) :

After obtaining the initial feasible solution, the next step is to test whether it is optimal or

not.

We explain here the Modified Distribution (MODI) Method for testing the optimality.

If the solution is non-optimal as found from MODI method then we improve the solution

by exchanging non-basic variable for a basic variable. In other words we rearrange the

allocation by transferring units from an occupied cell to an empty cell that has the largest

net cost change or improvement index, and then shift the units from other related cells so

that all the rim (supply, demand) requirements are satisfied. This is done by tracing a

closed path or closed loop.

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Step 1:

Add a column u to the RHS of the transportation tableau and a row v at the bottom of the tableau.

Step 2:

Assign, arbitrarily, any value to u or v generally u = 0.

Step 3

Having determined u1 and v calculate ij = (u1 + v1) — for every

unoccupied cell.

Step 4:

If the solution is not optimal select the cell with largest positive

improvement index.

Step 5:

Test the solution again for optimality and improve fit if necessary


Step I:

Add a column u to the RHS of the transportation tableau and a row v at the bottom of the

tableau.

Step 2:

Assign, arbitrarily, any value to u or v generally u = 0. This method of assigning values

to u1 and v1 is workable only if the initial solution is non-degenerate i.e., for a table there

are exactly m + n -1 occupied cells.

Step 3:

Having determined u1 and v calculate ij = (u1 + v1) — for every unoccupied cell. This

represents the net cost change or improvement index of these cells

(1) If all the empty cells have negative net cost change ij, the solution is optimal and

unique

(2) If an empty cell has a zero Xij and all other empty cells have negative Xij the solution

is optimal but not unique.

(3) If the solution has positive Ai for one or more empty cells the solution is not optimal.

Step 4:

If the solution is not optimal select the cell with largest positive improvement index.

Then trace a closed loop and transfer the units along the route.

Tracing loop (closed path):

1) Choose the unused square to be evaluated.

(2) Beginning with the selected unused square trace a loop via used squares back to the

original unused squares. Only one loop exists for any unused square in a given solution.

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(3) Assign (+) and (—) signs alternately at each square of the loop beginning with a plus

sign at the unused square. Assign these sign in clockwise or anticlockwise direction.

These signs indicate addition or subtraction of units to a square.

(4) Determine the per unit net change in cost as a result of the changes made in tracing

the loop. Compare the addition to the decrease in cost. It will give the improvement

index. (It is equivalent to j in a LPP).

(5) Determine the improvement index for each unused square.

(6) In a minimization case. If all the indices are greater than or equal to zero, the solution

is optimal. If not optimal, we should find a better solution.

We may also note the following points:

(i) An even number of at least four cells participate in a closed loop. An occupied cell can

be considered only once.

(ii) If there exists a basic feasible solution with m + n — 1 positive variables, then there

would be one and only one closed loop for each cell.

(iii) All cells that receive a plus or minus sign except the starting empty cell, must be the

occupied cells.

(iv) Closed loops may or may not be square or rectangular in shape. They may have

peculiar configurations and a loop may cross over itself.

Step 5:

Test the solution again for optimality and improve fit if necessary. Repeat the process

until an optimum solution is obtained.

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11) SWhat are the differences between assignment problem and transportation

problem?

The differences between AP and TP are the following:

1. TP has supply and demand constraints while AP does not have the same.

2. The optimal test for TP is when all cell evaluation \s are greater than or equal to

zero whereas in AP the number of lines must be equal to the size of matrix.

3. A TP sum is balanced when demand is equal to supply and an AP sum is balanced

when number of rows are equal to the number of columns.

4. for AP. We use Hungarian method and for transportation we use MODI method

5. In AP. We have to assign different jobs to different entities while in transportation

we have to find optimum transportation cost.

Q12) What are the advantages and disadvantages of LPP?

A LPP is concerned with the use of allocation of resources such as time, capital,

materials, etc.

THE ADVANTAGES LPP ARE:

1. It helps the sale manager to negotiate prices with customers. He can price on the basis

of customer demand and price on the basis of supply and demand of them market.

2. Production manager can formulate optimal maximum product mix.

3. it helps manager improve his decision making abilities.

4. it helps make the best decision for cost minimization and profit maximization.

THE DISADVANTAGES OF LPP ARE:

1. A primary requirement for LPP is the objective function and every consistent must be

linear. In practical situation it is not possible to state all coefficients in the objective

function and constraints with certainty.

2. There is no guarantee that LPP will give an integer value solution.

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eg. a solution may call for 9.3 trucks or 8.7 units of product.

3. It does not take into consideration the effect of time and uncertainty.

4. There may be cases of infeasibility and unbounded ness.

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13259677 assignment problem

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