mb0048 set2

8/3/2019 MB0048 SET2

http://slidepdf.com/reader/full/mb0048-set2 1/13

8/3/2019 MB0048 SET2


MB0048 –Operations Research set-2

Q.1.Explain how to transform an unbalanced transportation

problem into a balanced transportation problem where the

demand of warehouses is satisfied by the supply of factoriesAns:-

Unbalanced T.P

is shown that a basic solution to an m-origin, n destination; transportation problem

can have at the most m+n-1 positive basic variables (non-zero), otherwise the basic

solution degenerates. It follows that whenever the number of basic cells is less than m

+ n – 1, the transportation problem is a degenerate one. The degeneracy can develop

in two ways:

Case 1: The degeneracy develops while determining an initial assignment via any one

of the initial assignment methods discussed earlier.

To resolve degeneracy, you must augment the positive variables by as many zero-

valued variables as is necessary to complete the required m + n – 1 basic variable.

These zero-valued variables are selected in such a manner that the resulting m + n – 1

variable constitutes a basic solution. The selected zero valued variables are

designated by allocating an extremely small positive value ε to each one of them. The

cells containing these extremely small allocations are then treated like any other basic

cells. The ε‟s are kept in the transportation table until temporary degeneracy is

removed or until the optimum solution is attained, whichever occurs first. At that

point, we set each ε = 0.

Case 2: The degeneracy develops at the iteration stage. This happens when the

selection of the entering variable results in the simultaneous drive to zero of two or

more current (pre-iteration) basic variables.

To resolve degeneracy, the positive variables are augmented by as many zero-valued

variables as it is necessary to complete m+n-1 basic variables. These zero-valued

variables are selected from among those current basic variables, which are

simultaneously driven to zero. The rest of the procedure is exactly the same as

discussed above in case 1.

Note: The extremely small value ε is infinitely small and it never affects the value it is

‟ in unallocated minimum cost cell to avoid

forming a loop.

8/3/2019 MB0048 SET2



The transportation problem is a special type of linear programming problem in which

the objective is to transport a homogeneous product manufactured at several plants

(origins) to a number of different destinations at a minimum total cost. In this unit,

you have learnt several different techniques for computing an initial basic feasible

solution to a transportation problem, such as north-west corner rule, matrix minimum

method and Vogel‟s approximation method. Further, you studied the degeneracy in

transportation problem with examples on obtaining an optimum basic feasible solution

Transportation model is an important class of linear programs. For a given supply at

each source and a given demand at each destination, the model studies the

minimisation of the cost of transporting a commodity from a number of sources to

several destinations.

The transportation problem involves m sources, each of which has available ai (i = 1,

2… m) units of homogeneous product and n destinations, each of which requires bj (j =

1, 2…., n) units of products. Here ai and bj are positive integers. The cost cij of

transporting one unit of the product from the ith source to the jth destination is given

for each i and j. The objective is to develop an integral transportation schedule that

meets all demands from the inventory at a minimum total transportation cost.

The condition (1) is guaranteed by creating either a fictitious destination with a

demand equal to the surplus if total demand is less than the total supply or a (dummy)

source with a supply equal to the shortage if total demand exceeds total supply. The

cost of transportation from the fictitious destination to all sources and from all

destinations to the fictitious sources are assumed to be zero so that total cost of

transportation will remain the same. The standard mathematical model for the

transportation problem is as follows.

Let xij be number of units of the homogenous product to be transported from source i

to the destination j Then objective is toThe first approximation to (2) is integral.

Therefore, you always need to find a feasible solution. Rather than determining a first

approximation by a direct application of the simplex method, it is more efficient to

work with the transportation table given below. The transportation algorithm is the

simplex method specialised to the format of table involving the following steps:

i) Finding an integral basic feasible solution

8/3/2019 MB0048 SET2



ii) Testing the solution for optimality

iii) Improving the solution, when it is not optimal

iv) Repeating steps

(ii) and (iii) until the optimal solution is obtained

The solution to TP is obtained in two stages.

In the first stage, you find the basic feasible solution using any of the following

methods

a) North-west corner rule

b) Matrix Minima Method or least cost method

c) Vogel‟s approximation method. In the second stage, you test the basic feasible

solution for its optimality either by MODI method or by stepping stone method.

8/3/2019 MB0048 SET2



Q.2. Explain how the profit maximization transportation problem

into a balanced transportation problem where the demand of

warehouses is satisfied by the supply of factoriesAns:-

Some assignment problems are phrased in terms of maximising the profit or

effectiveness or payoff of an assignment of people to tasks or of jobs to machines. You

cannot apply the Hungarian method to such maximisation problems. Therefore, you

need to reduce it to a minimisation problem.

It is easy to obtain an equivalent minimisation problem by converting every number in

the table to an opportunity loss. To do so, you need to subtract every value from the

highest value of the matrix and then proceed as usual

You will notice that minimising the opportunity loss produces the same assignment

solution as the original maximisation problem.

A tailoring unit has four sewing machines of different makes. Each machine is capable

of stitching all the required designs and patterns. However, the profit factor differs

for each assignment. The unit is looking at maximising profit.

To do so, the unit needs to carry out an optimal assignment exercise of assigning the

right jobs to the right machines.

This unit on assignment problems focuses on a special type of transportation problem,

where the objective is to allocate „n‟ number of different facilities to „n‟ number of

different tasks. Although an assignment problem can be formulated as a linear

programming problem, it is solved by a special method know as Hungarian method. If

the number of persons is the same as the number of jobs, the assignment problem is

said to be balanced. The unit also explains the travelling salesman problem in brief.

8/3/2019 MB0048 SET2



Q.3 Illustrate graphically the following special cases of Linear

programming problems

Q.3.a. Multiple optimal solutions

Ans:-

8/3/2019 MB0048 SET2



Q.3 (B):- No Feasible solution

Ans:-

8/3/2019 MB0048 SET2



Q.3.C. Unbounded problem

Ans:-

8/3/2019 MB0048 SET2



Q.4.A. the objective function is to be maximized? Ans:-

Maximisation in AP

Some assignment problems are phrased in terms of maximising the profit or

effectiveness or payoff of an assignment of people to tasks or of jobs to machines. You

cannot apply the Hungarian method to such maximisation problems. Therefore, you

need to reduce it to a minimisation problem.

It is easy to obtain an equivalent minimisation problem by converting every number in

the table to an opportunity loss. To do so, you need to subtract every value from the

highest value of the matrix and then proceed as usual

You will notice that minimising the opportunity loss produces the same assignment

solution as the original maximisation problem.where the objective is to allocate „n‟

number of dif ferent facilities to „n‟ number of different tasks. Although an assignment

problem can be formulated as a linear programming problem, it is solved by a special

method know as Hungarian method. If the number of persons is the same as the

number of jobs, the assignment problem is said to be balanced. The unit also explains

the travelling salesman problem in brief.

Q4.B. Some Assignments are prohibited? Ans:-

Infeasible assignments

It is sometimes possible that a particular person is incapable of doing certain work or a

specific a specific job cannot be performed on a particular machine. The solution of

the assignment problem should take into account these restrictions so that the

infeasible assignment can be avoided. This can be achieved by assigning a very high

cost (say ∞ or M) to the cells where assignments are prohibited, thereby, restricting

the entry of this pair of job – machine or resource – activity into the final solution.

the problem.

8/3/2019 MB0048 SET2


8/3/2019 MB0048 SET2



In any simulation problem, the variables to be studied will be given with associated

probabilities. The initial conditions will also be specified. You can choose random

numbers from table. However, to get uniform results, the random numbers will be

specified. The first step involves coding the data that is, you assign random numbers

to the variable. Then you identify the relationship between the variables and run the

simulation to get the results

The range of application of simulation in business is extremely wide. Unlike other

mathematical models, simulation can be easily understood by the users and thereby

facilitates their active involvement. This makes the results more reliable and also

ensures easy acceptance for implementation. The degree to which a simulation model

can be made close to reality is dependent upon the ingenuity of the OR team who

identifies the relevant variables as well as their behaviour.

You have already seen by means of an example how simulation could be used in a

queuing system. It can also be employed for a wide variety of problems encountered in

production systems – the policy for optimal maintenance in terms of frequency of

replacement of spares or preventive maintenance, number of maintenance crews,

number of equipment for handling materials, job shop scheduling, routing problems,

stock control and so forth. The other areas of application include dock facilities,

facilities at airports to minimise congestion, hospital appointment systems and even

management games.

In case of other OR models, simulation helps the manager to strike a balance between

opposing costs of providing facilities (usually meaning long term commitment of funds)

and the opportunity and costs of not providing them.

Limitations

The simulation approach is recognised as a powerful tool for management decision-

making. One should not ignore the cost associated with a simulation study for data

collection, formation of the model and the computer time as it is fairly significant

A simulation application is based on the premise that the behaviour pattern of

relevant variables is known, and this very premise sometimes becomes questionable.

Not always can the probabilities be estimated with ease or desired reliability. The

results of simulation should always be compared with solutions obtained by other

methods wherever possible, and “tempered” with managerial judgment.

8/3/2019 MB0048 SET2



Q.6. Describe Gomory’s method of solving an all -integer

programming problem.Ans:-

Gomory’s All – IPP Method

An optimum solution to an IPP is obtained by using the simplex method, ignoring the

restriction of integral values. In the optimum solution, if all the variables have integer

values, the current solution will be the required optimum integer solution.

Otherwise, the given IPP is modified by inserting a new constraint called Gomory’s

constraint or secondary constraint. This constraint represents necessary conditions for

integrability and eliminates some non-integer solution without losing any integral

solution. On addition of the secondary constraint, the problem is solved using dual

simplex method to obtain an optimum integral solution.

If all the values of the variables in the solution are integers, then an optimum inter-

solution is obtained, or else a new constraint is added to the modified LPP and the

procedure is repeated till the optimum solution is derived. An optimum integer

solution will be reached eventually after introducing enough new constraints to

eliminate all the superior non-integer solutions. The construction of additional

constraints, called secondary or Gomory’s constraints is important and needs special

attention.

Construction of Gomory’s constraints

Consider a LPP or an optimum non–integer basic feasible solution. With the usual

notations, The optimum basic feasible solution is given by

xB = [x2, x3 ] = [y10, y20]; max z = y00

Since xB is a non-integer solution, we can assume that y10 is fractional. The constraint

equation is

It reduces to

y10 = y11 x1 + x2 + y14 x4 _____ (1)

Because x2 and x3 are basic variables (which implies that y12 = 1 and y13 = 0). The

above equation can be rewritten as

x2 = y10 - y11 x1 - y14 x4

8/3/2019 MB0048 SET2



This is a linear combination of non-basic variables.

Now, since y10 0 the fractional part of y10 must also be non-negative. You can split

each of yij in (1) into an integral part Iij , and a non-negative fractional part, f1j for j

= 0,1,2,3,4.

I10 + f10 = (I11 + f11) x2 + (I14 + f14) x4

Or

f10 - f11 x2 - f14 x4 = x2 + I11 x1 + I14x4 - I10 _____ (2)

If you compare (1) and (2), you will see that if you add an additional constraint in such

a way that the left-hand side of (2) is an integer; then you will be forcing the non-

integer y10 towards an integer.

The desired Gomory’s constraint is

f10 – f11 x1 – f11 x4 ≤ 0

It is possible to have f10 – f11 x1 – f11 x4 = h where h > 0 is an integer. Then f10 = h +

f11 x1 + f14 x4 is greater than one. This contradicts that 0 < fij < 1 for j = 0, 1, 2, 3, 4.

Where Gsla (1) is a slack variable in the above first Gomory’s constraint

The additional constraint to be included in the given LPP is towards obtaining an

optimum all integer solution.

Since – f10 is negative, the optimal solution is unfeasible. Thus the dual simplex

method is to be applied for obtaining an optimum feasible solution. After obtaining

this solution, the above referred procedure is applied for constructing second

Gomory’s constraint. The process is to be continued till all the integer solution has

been obtained.

One is the cutting plane algorithm devised by Gomory and the other is the branch and

bound algorithm developed by Land Doig.

mb0048 set2

Documents