unit 4 operations research

7/29/2019 UNIT 4 OPERATIONS RESEARCH

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UNIT IV

INTRODUCTION TO LINEAR PROGRAMMING

1. What is OR? Give some applications.Operations Research has been variously described as the Scienceof Use, Quantitative Common

sense, Scientific approach to decision making problem etc. But only a few are commonly used and

accepted namely

(i) Operations Research is the application of scientific methods techniques and tools toproblems involving operations of a system so as to provide those in control of the system

with optimum solutions to the problem.

(ii) Operations Research is the art of giving bad answers to problems which otherwise haveworse answers.

(iii)O.R. is a scientific method of providing executive department with a quantitative basis fordecisions regarding the operations under this control.

(iv)O.R. is applied decision theory. It use any scientific mathematical or logical means toattempt to cope with the problems that confront the executive when he tries to achieve a

thorough going rationality in dealing with his decision problems.

(v) O.R. is a scientific approach to problems solving for executive management.(vi)O.R. is a scientific knowledge through interdisciplinary team effort for the purpose of

determining the best utilization of limited resources.

2. What do you mean by general LPP?Linear Programming is a mathematical technique for choosing the best alternative from a set of

feasible alternatives, in situations where the objective function as well as the restrictions or constraints can

be expressed as linear mathematical function.

3. Give the standard form and canonical form of a LPP.Canonical Form: The general linear programming problem can always be expressed in the following

form.

Max. Z = C1X1 + C2X2 + C3X3 + . + CnXn

Subject to Constraints

a11x1 +a12x2 + a13x3+ + a1nxn b1

a21x1 +a22x2 + a23x3+ + a2nxn b2

a31x1 +a32x2 + a33x3+ + a3nxn b3

..

..

..

am1x1 + am2x2 + am3x3+ . + amn xn bm


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and non-negativity restrictions x1 , x2, x3xn 0

This form of LPP is called Canonical form of LPP. In matrix notation the Canonical form of LPP can be

expressed as

Max. Z = CX (Objective Fn.)

Subject to AX b (Constraints)

And X 0 (Non negativity restrictions)

Where C = (C1, C2, C3, C4. Cn)

a11 a12 a13 . a1n

a21 a22 a23 . a2n

A = a31 a32 a33 . a3n

. .

. .

am1 am2 am3 . amn

Standard form: The general linear programming problem in the form

Max. Z = C1x1 + C2x2 + C3x3 + .. + Cnxn

Subject to the constraints cAnd x1 , x2, x3 , .. >= 0 is known as standard form.

4. Specify the components of a LPP (OR) Specify the basic assumptions of LPP(i) Proportionality (ii) Additively (iii) Divisibility (iv) Certainty (or) Deterministic(iv) Finiteness (vi) Optimality.

5. Write any two situations where LPP is applied.Linear Programming technique is used in many industrial and economic problems. They are applied

in product mix, blending, diet, transportation and assignment problems. Oil refineries, airlines,

railways, textiles, industries, Chemical industries, steel industries, food processing industries and

defense establishments.

6. Graphical solution is not possible for LPP problem with more than two constraints True or False?Justify your answer.

Which is false. We can solve the LPP by graphical solution with more than two constraints.

7. What is the use of artificial variable in LPP?The purpose of introducing artificial variables is just to obtain an initial basic feasible solution.


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8. Define Slack, Surplus variables.Slack Variable: If the constraints of a given LPP be aij xj bi then the non - negative variable Si which

are introduced to convert the inequalities to equalities aij xj + Si = bi are called slack variables.

Surplus variable: If the constraints of a given LPP be aij xj bi then the nonnegative variable Si which

are introduced to convert the inequality constraints to the equations aij xj - Si = bi are called surplus

variables.

9. What do you mean by degenerate solution in LPP?A basic solution is said to be a degenerate basic solution if one or more of the basic variables are zero.

10.Define a feasible region in graphical method.A region or a set of points is said to be convex (or) feasible region if the line joining any two of its

points lies completely within the region.

11.What is meant by an optimal solution?Any feasible solution, which optimizes (maximizes or minimizes) the objective function of the LPP

is called its optimum solution or optimal solution.

12.What is the difference between feasible solution and basic feasible solution?Feasible solution: Any solution to a LPP, which satisfies the non-negativity restrictions of the LPP is

called its feasible solution.

Basic solution: A basic solution is said to be a degenerate basic solution if one or more of the basic

variables are zero.Basic feasible solution: A feasible solution, which is also basic, is called a basic Feasible solutions.

13.Define non-degenerate solution.A basic solution is said to be a non-degenerate basic solution if none of the basic variables is zero.

14.Define unbalanced solution and infeasible solution.Let there exists a basic feasible solution to a given LPP if for at least one j, for which aij 0 Zj Cj is

negative, and then there does not exists any optimum solution to this LPP Infeasible solution: If some

values of the set of values x1,x2,x3,x4..xn are negative which satisfies the constraints of the LPP is

called its infeasible solution.

15.Which are decision variables in the construction of OR problems?Linear programming problem deals with the optimization ie maximization or minimization of a

function of decision variable. The variables whose values determine the solution of a problem are

called decision variables of the problem.

16.How many basic feasible solutions are there to a given system of m equations in n unknowns?Here n > m. There are nCm basic solutions are possible.


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17.What is key column and how is it selected?By performing the optimality test we can find whether the current feasible solution can be

improved or not. This can be done by performing Cj Ej. If Cj Ej is positive under any column, at least

one better solution is possible. To find the incoming variable take the highest positive integer in the

row Cj Ej, the variables belongs to highest positive integer column is the incoming variable and that

column is a key column K.

18.What is key row and how is it selected?To find the outgoing variable divide the elements of the column b by the element of the key

column. The row containing the minimum positive ratio is marked. The variable belongs to that row is

the outgoing variable. That row is called key row.

19.How will you find whether a LPP has got an alternative optimal solution or not, from the optimalsimplex table?

If Cj Ej is positive under any column, the profit can be increased ie the current basic feasible

solution is not optimal and a better solution exists. When no more positive values remain in the Cj Ej

row, the solution becomes Optimal.

20.What are the advantages of duality?(i) If the primal problem contains a large number of rows (Constraints) and a smaller number of

columns (Variables), the computational procedure can be considerably reduced by converting it

into dual and then solving it. Hence it offers advantages in many applications.

(ii) It gives additional information as to how the optimal solution changes as a result of the changesin the coefficients and the formulation of the problem.

(iii)Duality in LP has certain far-reaching consequences of economic nature.(iv)Calculations of the dual checks the accuracy of the primal solution.

21.State the existence theorem of duality.If either problem has an unbounded solution then the other problem has no feasible solution.

22.State fundamental theorem of duality.If either the primal or dual problem has a finite optimal solution, then the other problem also has a

finite optimal solution and also the optimal values of the objective function in both the problem are the

same. I.e., Max Z = Min Z.

23.What are the advantages of dual simplex method?The dual simplex method is used to solve the problems, which start dual feasible i.e., the primal is

optimal but infeasible. The advantages of this Method is avoiding the artificial variables introduced in

the constraints along with the surplus variables as all >= constraints are converted into


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24.What do you meant by shadow prices?The values of the decision variable of dual of a LPP represents shadow price of a resource.

25.What is the advantage of dual simplex method?The advantage of dual simplex method is to avoid introducing the artificial variables along with the

surplus variable as the type constraint is converted into type.

PART B

1. A firm produces 3 products. These products are processed on 3 different machines. The time requiredto manufacture one unit of each of the 3 products and the daily capacity of the 3 machines are given

below:

It is required to determine the number of units to be manufactured for each product daily. The

profit per unit for product 1,2 and 3 is Rs4, Rs3, Rs6 respectively. It is assumed that all the amounts

produced are consumed in the market. Formulate the mathematical model for the problem.

Sol: Maximize Z = 4x1+3x2+6x3

Sub to the constraints

2x1+3x2+2x3 440

4x1+3x2 470

2x1+5x2 430 & x1,x2,x3 0

2. A firm produces an alloy having the following specifications:

(i) Specific gravity 0.98(ii) Chromium 8%(iii)Melting point 450C

Raw materials A, B, C having the properties shown in the table can be used to make the alloy.

MachineTime per unit (minutes) Machine

Capacity

(Minutes/day)Product1 Product2 Product3

M1 2 3 2 440

M2 4 - 3 470

M3 2 5 - 430

PropertyRaw material

A B C

Specific gravity 0.92 0.97 1.04

Chromium 7% 13% 16%

Melting point 440C 490C 480C


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Cost of the various raw materials per unit tons are: Rs.90 for A, Rs.280 for B and Rs.40 for C. Find

the proportions in which A, B and C be used to obtain an alloy of desired properties while the cost of raw

materials is minimum.

Sol: Minimize Z = 90x1 + 280x2 + 40x3

Sub to

0.92x1 + 0.97x2 +1.04x3 0.98

7x1 + 13x2 +16x3 8

440x1 +490x2 +480x3 450 & x1, x2, x3 0

3. ABC manufacturing company can make 2 products P1 and P2. Each of the product require time on acutting machine and a finishing machine relevant data are

The number of cutting hours available per week is 390 and the number of finishing hours available

per week is 810.How much of each product should be produce in order to maximize the profit ?

Sol: Max Z = 6x1 +4x2

Sub to

2x1 +x2 390

3x1 + 3x2 810 & x1, x2 0

4. Old hens can be bought at Rs.2 each and young ones at Rs.5 each. The old hens lay 3 eggs per week andthe young ones lay 5eggs per week, each egg being worth 30 paise. A hen costs Rs.1 per week to feed.

A person has only Rs.80 to spend for hens. How many of each kind should he buy to give a profit ofmore than Rs.6 per week, assuming that he cannot house more than 20 hens. Formulate this as a L.P.P.

Sol: Max Z = 0.5 x2 0.1x1

Subject to the constraints 2x1 + 5x2 80 ; x1 + x2 20 ; 0.5x2 0.1x1 6 & x1, x2 0.

5. A television company operates 2 assembly sections, section A and section B. Each section is used toassemble the components of 3 types of televisions: Colour, standard and Economy. The expected daily

production on each section is as follows :

T.V Model Section A Section BColour 3 1

Standard 1 1

Economy 2 6

Product

P1 P2Cutting hrs (per unit) 2 1

Finishing hrs (per unit) 3 3

Profit 9per unit) Rs.6 Rs.4

Max. sales (per week) - 210


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The daily running costs for 2 sections average Rs.6000 for section A and Rs.4000 for section B .It is

given that the company must produce at least 24 colours, 16 standard and 40 Economy TV sets for which

an order is pending. Formulate this as a L.P.P so as to minimize the total cost.

Sol: Minimize Z = 6000x1 + 4000x2

Subject to the constraints 3x1 + x2 24 ; x1 + x2 16 ; 2x1 + 6x2 40 & x1 , x2 0.

6. An electronics company produces three types of parts for automatic washing machine. It purchasescosting of the parts from a local foundry and then finishes the part of drilling, shaping and polishing

machines. The selling prices of parts A, B and C respectively are Rs. 8, Rs.10 and Rs.14. All parts made

can be sold. Casing for parts A, B and C respectively cost Rs.5, Rs.6 and Rs.10.The shop possesses only

one of each type of machine. Costs per hour to run each type of three machines are Rs.20 for drilling,

Rs.30 for shaping and for polishing. The capacities for each part on each machine are shown in the

following table.

Machine/Capacity per hour Part A Part B Part C

Drilling 25 40 25

Shaping 25 20 20

Polishing 40 30 40

7. Solve Graphically:(i) Maximize Z = 3x1 + 9x2

Subject to the constraints x1 + 4x2 8 ; x1 + 2x2 4 & x1, x2 0.

Sol: Z = 18, x1 = 0, x2 = 2, x3 = 0, x4 = 0.

(ii) Maximize Z = 2x1 + 4x2Subject to the constraints x1 + 2x2 5 ; x1 + x2 4 & x1, x2 0.

Sol: Alternate solution

Max Z = 10 and x1 = 0 , x2 = 5/2 & Max Z = 10 and x1 = 3, x2 = 1.

(iii)Maximize Z = 2x1

+ x2

Subject to the constraints x1 x2 10 ; 2x1 40 & x1,x2 0.

Sol: Unbounded solution.

(iv)Maximize Z = 3x1 + 2x2Subject to the constraints 2x1 + x2 2 ; 3x1 + 4x2 12 & x1,x2 0.

Sol: Infeasible solution.

(v) Solve graphically : Maximize Z = 100x1 + 40x2Subject to the constraints 5x1 +2 x2 1000 ; 3 x1 +2 x2 900 ; x1 + 2x2 500 & x1 , x2 0.


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8. Solve by Simplex Method:(i) Maximize Z = 3x1 + 9x2

Subject to the constraints x1 + 4x2 8 ; x1 + 2x2 4 & x1,x2 0.

Sol: Max Z = 18, x1 = 0, x2 = 2 , x3 = 0, x4 = 0.

(ii) Maximize Z = 2x1 + 4x2Subject to the constraints x1 + 2x2 5 ; x1 + x2 4 & x1,x2 0.

Sol: Alternate solution

Max Z = 10 and x1 = 0, x2 = 5/2 & Max Z = 10 and x1 = 3, x2 = 1.

(iii)Maximize Z = 2x1+ x2Subject to the constraints x1 x2 10 ; 2x1 40 & x1,x2 0.

Sol: Unbounded solution



(v) Maximize Z = 20 x1 + 30x2Subject to the constraints 2x1 + 3x2 120 ; x1 + x2 35 ; 2x1 + 1.5x2 90 & x1, x2 0.

Sol: Z = 1050, x1 = 0, x2 = 35.

(vi)Maximize Z = 15x1 +6 x2 +9 x3+2x4Subject to the constraints 2x1 +x2 + 5x3 +6x4 20

3x1 +x2 + 3x3 +25x4 24 ; 7x1 + x4 70 & x1, x2,x3, x4 0.

(vii)Minimize Z = 8x1 2x2Subject to the constraints -4x1 + 2x2 1 ; 5x1 4x2 3 & x1, x2 0.

Sol: Min Z = -1, x1 = 0, x2 = .

9. Use Big-M (OR) Penalty Method to solve:(i) Maximize Z = 2x1 + x2 + x3

Subject to the constraints 4x1 + 6x2 + 3x3 8 ; 3x1 6x2 4x3 1;

2x1 + 3x2 5x3 4 & x1, x2,x3 0.

Sol: Max Z = 64/21, x1 = 9/7, x2 = 10/21, x3 = 0.

(ii) Minimize Z = 4x1 + 3x2Subject to the constraints 2x1 + x2 10 ; -3x1 + 2x2 6 ; x1 + x2 6 & x1,x2 0.

Sol: Min Z = 22, x1 = 4, x2 = 2.


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10.Use Two-phase method to solve(i) Maximize Z = 5x1 + 8x2

Subject to the constraints 3x1 + 2x2 3 ; x1 + 4x2 4 ; x1 + x2 5 & x1,x2 0.

Sol: Max Z = 40, x1 = 0, x2 = 5.

(ii) Minimize Z = -2x1 x2Subject to the constraints X1 + x2 2 ; X1 + x2 4 & x1,x2 0.

Sol: Min Z = -8, x1 = 4, x2 = 0.

Variants of the Simplex Method

11.Solve the L.P.P by Simplex Method :(i) Maximize Z = x1 + 2x2 + x3

Subject to the constraints 2x1 + x2 x3 2 ; -2x1 + x2 5x3 -6 ; 4x1 + x2+ x3 6 & x1,x2,x3 0.

Sol: Max Z = 10, x1 = 0, x2 = 4, x3 = 2.

(ii) Maximize Z = 3x1 x2Subject to the constraints X1 x2 10 ; X1 20 & x1,x2 0.

Sol: Unbounded solution.

(iii)Maximize Z = 6x1 + 4x2Subject to the constraints 2x1 + 3x2 30 ; 3x1 + 2x2 24 ; x1 + x2 3 & x1,x2 0.

Sol: Max Z = 48, x1 = 8, x2 = 0 & Max Z = 48, x1 = 12/5, x2 = 42/5



(v) Maximize Z = 2x1 + 3x2Subject to the constraints -x1 + 2x2 4 ; x1 + x2 6 ; x1 + 3x2 9 & x1, x2 are unrestricted.

Sol: Max Z = 27/2, x1 = 9/2, x2 = 3/2.

12.Solve by Dual Simplex Method:(i) Minimize Z = 3x1 + x2

Subject to the constraints 3x1 + x2 3 ; 4x1 + 3x2 6 ; x1 + x2 3 & x1,x2 0.

Sol: Min Z = 21/5, x1 = 3/5, x2 = 6/5.

(ii) Minimize Z = 2x1 + 3x2Subject to the constraints 2x1 + 2x2 30 ; x1 + 2x2 10 & x1, x2 0.

13.Write down the dual of the following LPP and solve it.Max Z = 4x1 + 2x2

Subject to the constraints -x1 x2 -3 ; -x1 + x2 -2 & x1, x2 0.



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14.Use duality to solve the following LPP.(i) Minimize Z = 2x1 + 2x2

Subject to the constraints 2x1 + 4x2 1 ; -x1 2x2 -1 ; 2x1 + x2 1 & x1, x2 0.

Sol: Min Z = 4/3, x1 = 1/3, x2 = 1/3.

(ii) Maximize Z = 3x1 + 2x2Subject to the constraints X1 + x2 1 ; X1 + x2 7 ; X1 +2x2 10 ; X2 3 & x1, x2 0.

Sol: Max Z =21, x1 = 7, x2 = 0.

(iii)Minimize Z =x1x2+x3Subject to the constraints x1 x3 4 ; x1 x2 + 2x3 3 & x1, x2,x3 0.


15.Consider the LPP Max Z = 2x1 + x2 + 4x3x4Subject to the constraints X1 +2x2 + x3-3x4 8 ; -x2 + x3+2x4 0 ;

2x1 + 7x2 5x3 10x4 21 & x1, x2, x3, x4 0.

(a) Solve the LPP(b)Discuss the effect of change of b2 to 11.(c) Discuss the effect of change of b to [3,-2,4].Sol: (a) Max Z =16, x1 = 8, x2 = x3 = x4 = 0.

(b) Max Z = 87/2, x1 = 49/2, x2 = x3 = 0, x4 = 11/2.

(c) Infeasible solution.

16.Consider the LPP Max Z = 3x1 + 4x2 + x3 + 7x4Subject to the constraints 8x1 + 3x2 + 4x3 + x4 7 ; 2x1 + 6x2 + x3 + 5x3 3 ;

x1 + 4x2 + 5x3 + 2x4 8 & x1, x2, x3, x4 0.

(a) Solve the LPP.(b) If a new constraint 2x1 + 3x2 + x3 + 5x4 4 is added to the above LPP, discuss the effect of change in

the solution of the original problem.

(c) If a new constraint 2x1 + 3x2 + x3 + 5x4 2 is added (or) if the upper limit of the above constraint isreducd to 2, discuss the effect of change in the otimum solution of the original problem.

Sol: (a) Min Z = 83/19, x1 = 16/19, x2 = 0, x3 = 0, x4 = 5/19.

(b) Redundant.

(c) Max Z = 113/38, x1 = 33/38, x2 = x3 = 0, x4 = 1/19.

17.Consider the LPP Max Z = 2x1 + x2 + 4x3 x4Subject to the constraints X1 + 2x2 + x3 3x4 8 ; -x2 +x3 + 2x4 0 ;

2x1 + 7x2 5x3 10x4 21 & x1, x2, x3, x4 0.

(a) Solve the LPP.(b) Discuss the effect of change of c1 to 1.


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(c) Discuss the effect of change of (c3, c4) to (3, 4).(d) Discuss the effect6vcof change of (c1, c2, c3, c4) to (1, 2 , 3 , 4).

Sol:

(a) Max Z = 16, x1 = 8, x2 = x3 = x4 = 0.(b)Max Z 40/3, x1= x4 = 0, x2 = 8/3, x3 = 8/3.(c) Max Z = 163/5, x1 =0, x2 = 21/2, x3 = 11/10, x4 = 47/10.(d)Max Z = 431/10, x1= 0, x2 = 21/2, x3 = 11/10, x4 = 47/10.

18.Consider the LPP Max Z = 5x1 + 12x2 + 4x3Subject to the constraints X1 + 2x2 + x3 5 ; 2x1 x2 + 3x3 = 2 & x1, x2, x3 0.

(a) Discuss the effect of changing a3 to (2,5) from (1,3)(b)Discuss the effect of changing a3 to (-5,2) from (1,3).(c) Discuss the effect of changing a3 to (-1,2) from (1,3).

Sol:(a) Unbounded solution.(b)Max Z = 60, x1 = 0, x2 = 4, x3 = 3.

19.Consider the LPP Max Z = 3x1 + 5x2Subject to the constraints X1 4 ; 3x1 + 2x2 18 & x1, x2 0.

(a) Solve this LPP.(b) If a new variable x5 is added to this problem wih a column (1,2) and c5 = 7 find the change in the

optimal solution.

Sol:

(a) Max Z = 45, x1 = 0, x2 = 9.(b)Max Z = 53, x1 = 0 , x2= 5, x5 = 4.

20.Consider the optimal table of a maximization problem

Find the change in the optimal solution, when the basic variable x2 is deleted.

Sol: Max Z = 28, x1 = 0, x2 = 0 & x5 = 4.

Cj 3 5 0 0 7

CB YB XB x1 x2 x3 x4 x5

7 x5 4 1 0 1 0 1

5 x2 5 1/2 1 -1 1/2 0

Zj - Cj 53 13/2 0 2 5/2 0


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TRANSPORTATION AND ASSIGNMENT PROBLEM

1. What do you understand by Transportation problem?It is a special type of linear programming model in which the goods are shipped from various origins

to different destinations. The objective is to find the best possible allocations of goods from various

origins to different destinations such that the total transportation cost is minimum.

2. Write the mathematical form of Transportation Problem.m n

Min Z = Cij xij subject to the constraints

i=1 j=1

n

xij = ai i = 1,2, .., m

j=1

m

xij = bj j = 1,2, .., n Where

i=1

ai = Number of units available in ith

the origin

bj = Number of units required in jth

destination

Cij = transportation cost from ith

the origin to jth

destination

3.

What is an unbalanced transportation problem? How to solve it?A transportation problem is said to be unbalanced if the total supply is not equal to the total

demand ie

m n

ai bj

i=1 j=1

the unbalanced transportation problem is converted in to a balanced one by adding a dummy row or

dummy column whichever is necessary. The unit transportation cost for the dummy row or column

elements are assigned zero. Then the problem is solved by the usual procedure.

4. Define (i) Feasible solution (ii) Basic feasible solution (iii) Non-degenerate solution.Feasible solution: A set of non-negative values xij satisfies the constraint equation is called a feasible

solution.

Basic feasible solution: A basic feasible solution is said to be basic, if the number of positive allocations

are m+n-1. If the number of allocations is equal to m+n-1, it is called non-degenerate basic feasible

solution.


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5. What do you understand by degeneracy in a transportation problem?If the number of occupied cells in a m x n transportation problem, is less than (m+n-1), then the

problem is said to be degenerate.

6. When does a TP have a unique solution?While doing optimality test, if a empty cell evaluations ie ij = Cij (ui + vj) are positive, then the

problem is said to be have an unique solution.

7. What is the purpose of MODI method?MODI method is the test procedure for optimality involves examination of each unoccupied cell to

determine whether or not making an allocation in it reduce the total transportation cost and then

repeating this procedure until lowest possible transportation cost is obtained.

8. List any three approaches used with TP for determining the starting solution (or) the initial basicfeasible solution.

(i) NorthWest corner rule (ii) Least cost entry method (iii) Vogels approximations method.

9. How will you identify that a TP has got an alternate optimal solution?While doing optimality test, if any empty cell evaluation i.e., ij = Cij (ui + vj) = 0 then the problem

is said to have an alternate optimal solution.

10.When do you say that the occupied cell is in independent position?When it is not possible to draw a closed loop from the allocations, the occupied cell is in

independent position .11.What is maximization type transportation problem and how these problems are solved?

The main objective of transportation problem is to minimize the transportation cost. In

maximization type problems, the objective is to maximize the profit or maximize the total sales. To

solve these problems, we have to convert all the cell entries by multiplying 1. Then the problem is

solved by the usual method.

12.Write down the basic steps involved in solving a transportation problem.a. To find the initial basic feasible solutionb. To find an optimal solution by making successive improvements from the initial basic feasible

solution.

13.State the necessary and sufficient condition for the existence of a feasible solution to atransportation problem?

The necessary and sufficient condition for the existence of feasible solution is a solution that

satisfies all conditions of supply and demand.


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14.What is an assignment problem? Give two applications?It is a special type of transportation problem in which the number of jobs allocated for different

machines or operators. The objective is to maximize the overall profit or minimize the total cost for a

given assignment schedule.

a. It is used in production environment in which the number of jobs are assigned to number ofworkers or machines in such a way that the total time to complete all the jobs will be minimum.

b. It is used in traveling salesman.15.What do you mean by an unbalanced AP?

Since the assignment is one to one basis the problem have a square matrix. If the given problem is

not a square matrix ie the number of rows and columns are not same then it is called unbalanced

assignment problem. To make it a balanced assignment add a dummy row or dummy column and then

convert it into a balanced one. Assign zero cost values for the dummy row or column and solve it by

usual assignment method.

16.State the difference between TP & AP.Assignment Transportation

Allocations are made one to one basis.

Therefore only one occupied cell will

be Present in each row and each

column. Hence the table will be a

square matrix.

More than one allocation is possible in

each row and each column. Hence it neet

not be a square matrix.

It will always provide degeneracy. It will not provide degeneracy

The supply at any row and demand at

any column will be equal to 1.

The supply and demand may have any

positive quantity.

17.How do you convert the maximization problem in to a minimization one?To solve the maximization problem in to minimization assignment problem, first convert the given

maximization matrix in to an equivalent minimization matrix form by multiplying 1 in all the cost

elements. Then the problem is a maximization one and can be solved by the usual assignment method.

18.Give the linear programming form of A.P.Objective is to minimize the total cost involve.

n n

Min Z = Cij xij subject to the constraints

i=1 j=1

x11 +x12 + x13+ + x1n = 1

x21 +x22 + x23+ + x2n = 1

x31 +x32 + x33+ + x3n = 1

.

xn1 + xn2 + xn3+ . + xnn = 1


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19.What is the name of the method used in getting the optimum assignment?Hungarian method.

20.What is the indication of an alternate solution in an assignment problem?If the final cost matrix contains more than the required number of zero for assignment at

independent position then it indicated that the problem has an alternate optimal solution.

21.What do you inderstand by resticted assignments? Explain how should one overcome it?In assignment problems, it is assured that the performance of all the machines and operators are

same. Hence any machine can be assigned to any job. But in practical cases, a machine cannot do all

the operations of a job and operator cannot do all kinds of tasks. Therefore a high processing time is

assigned to the impossible cell (M or ) and then it will be solved by the usual assignment method. In

the final assignment the restricted cell will not be present.

22.Write two theorems that are used for solving assignment problems.Theorem 1: The optimum assignment schedule remains unaltered if we add or subtract a constant to /

from all the elements of the row or column of the assignment cost matrix.

Theorem 2: If for an assignment problem all Cij 0, then an assignment schedule (xij) which satisfies

Cij xij = 0 , must be optimal.

23.Write the mathematical formulation of an assignment problem.n n

Min Z = Cij xij subject to the constraintsi=1 j=1

n

xij = 1

i=1

n

xij = 1

j=1

xij = 1, if i th job is assigned to j th operator

0, otherwise

Cij = Cost of assigning the nth

job to mth

machine.

24.What is traveling salesman problem and what are its objectives?In this model a salesman has to visit n cities. He has to start from a particular city, visit each city

once and then return to his starting point. The main objective of a salesman is to select the best

sequence in which he visited all cities in order to minimize the total distance traveled or minimize the

total time.


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25.Why assignment problem will always provide degeneracy?In assignment problem, the allocation is one to one basis therefore, the number of occupied cells in

each row and each column will be exactly equal to 1. Hence assignment problem will always provide

degeneracy.

26.Why a transportation technique or the simplex method cannot used to solve an assignment problem.The transportation technique or simplex method cannot be used to solve the assignment problem

because of degeneracy.

PART B

1. Obtain the initial solution for the following TP using NWCR, LCM, VAM.

2. Solve the TP where the cell entries denote the unit transportation costs.

3. Solve the following TP.

Source

Destination

1 2 3 Capacity

1 2 2 3 10

2 4 1 2 15

3 1 3 1 40

Demand 20 15 30

Source

Destination

A B C Supply

1 2 7 4 5

2 3 3 1 8

3 5 4 7 7

4 1 6 2 14

Demand 7 9 18 34

Source

Destination

A B C D Supply

P 5 4 2 6 20

Q 8 3 5 7 30

R 5 9 4 6 50

Demand 10 40 20 30 100


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4. Find the minimum transportation cost.

Factory

Warehouse

D1 D2 D3 D4 Supply

F1 19 30 50 10 7

F2 70 30 40 60 9

F3 40 8 70 30 18

Demand 5 8 7 14

5. Find the optimal solution by using VAM.

Factory

Warehouse

A B C D E F Available

1 9 12 9 6 9 10 5

2 7 3 7 7 5 5 6

3 6 5 9 11 3 11 2

4 6 8 11 2 2 10 9

Requirement 4 4 6 2 4 2

6. Solve the TP.

7. Solve the following TP to maximize the profit.

Source

Destination

A B C D Supply

1 11 20 7 8 50

2 21 16 20 12 40

3 8 12 8 9 70

Demand 30 25 35 40

Source

Destination

A B C D Supply

1 40 25 22 33 100

2 44 35 30 30 30

3 38 38 28 30 70

Demand 40 20 60 30


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8. Solve the following unbalanced TP.

FROM

TO

1 2 3 Supply

1 5 1 7 10

2 6 4 6 80

3 3 2 5 15

Demand 75 20 50

9. Consider the following transshipment problem involving 4 sources and two destinations. The supplyvalues of the sources S1, S2, S3 and S4 are 200 units, 250 units, 200 units and 450 units respectively.

The demand values of the destinations D1 and D2 are 550 units and 550 units, respectively. The

transportation cost per unit between different sources and destinations are summarized in the

following table. Solve the transshipment problem.

Source

Destination

S1 S2 S3 S4 D1 D2

S1 0 6 24 7 24 10

S2 10 0 6 12 5 20

S3 15 20 0 8 45 7

S4 18 25 10 0 30 6

D1 15 20 60 15 0 10

D2 10 25 25 23 4 0

10.A firm having 2 sources S1 & S2 wishes to ship its products to 2 destinations D1 & D2. The number ofunits available at S1 & S2 are 5 & 25 resp. and the product demanded at D1 & D2 are 20 & 10 units

respectively. The firm instead of shipping directly decides to investigate the possibility of

transshipment. The unit transportation costs (in rupees) are given in the following table. Find the

optimal shipping schedule.

Source DestinationAvailable

S1 S2 D1 D2

SourceS1 0 2 3 4 5

S2 2 0 2 4 25

Destination

D1 3 2 0 1 -

D2 4 4 1 0 -

Demand - - 20 10


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11.Solve the AP:A B C D

I 1 4 6 3

II 9 7 10 9

III 4 5 11 7

IV 8 7 8 5

12.Solve the AP.A B C D

I 10 25 15 20

II 15 30 5 15

III 35 20 12 24

IV 17 25 24 20

13.Solve the Assignment problem:

Tasks

Men

A B C D E

I 1 3 2 8 8

II 2 4 3 1 5

III 5 6 3 4 6

IV 3 1 4 2 2

V 1 5 6 5 4

14.Solve the Assignment problem:

Jobs

Machine

1 2 3 4 5

A 11 17 8 16 20

B 9 7 12 6 15

C 13 16 15 12 16

D 21 24 17 28 26

E 14 10 12 11 15

15.A company is faced with the problem of assigning 4 machines to different jobs (one machine to one jobonly). The profits are estimated as follows.

Job

Machine

A B C D

1 3 6 2 6

2 7 1 4 4

3 3 8 5 8

4 6 4 3 7

5 5 2 4 3

6 5 7 6 4

Solve the problem to maximize the profit.


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16.Determine the optimum assignment schedule for the following assignment Problem. The cost matrix isgiven below.

Jo

b

Machine

1 2 3 4 5 6

A 11 17 8 16 20 15

B 9 7 12 6 15 13C 13 16 15 12 16 8

D 21 24 17 28 2 15

E 14 10 12 11 15 6

If the job C cannot be assigned to machine 6, will the optimum solution change?

17.A company has four machines to do three jobs. Each job can be assigned to one and only one machine.The cost of each job on each machine is given in the following table.

What are job assignments which will minimize the cost?

18.Write the algorithm for Hungarian method.19.There are four machines in a machine shop. On a particular day the shop got Orders for executing five

jobs (A, B, C, D & E). the expected profit for each job on each job on machine is as follows:

1 2 3 4

A 32 41 57 18

B 48 54 62 34

C 20 31 81 57

D 71 43 41 47

E 52 29 51 50

Find the optimal assignment of job to machines to maximize the profit. Which job should be

rejected?

20.A marketing manager has five salesmen working under his control to be assigned to five salesterritories. Taking into account the sales potential of the territories and the capabilities of the

salesman, the marketing manager estimated the sales per month (in thousands of Rs.) for eachcombination and is presented as below. Find the assignment of salesmen to sales Territories to

maximize the sales value per month.

1 2 3 4

A 18 24 28 32

B 8 13 17 19

C 10 15 19 22


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SalesmenSales Territories

A B C D E

1 32 38 40 28 40

2 40 24 28 21 36

3 41 27 33 30 37

4 22 38 41 36 36

5 29 33 40 35 39

21.Solve the following assignment problem using Hungarian method. The matrix entries are processingtimes in hours.

JobOperator

1 2 3 4 5

1 20 22 35 22 18

2 4 26 24 24 7

3 23 14 17 19 19

4 17 15 16 18 15

5 16 19 21 19 25

22.Five wagons are available at five stations 1, 2, 3, 4 and 5. These are required at five stations I, II, III, IVand V. the mileages between various stations are given by

From ToI II III IV V

1 10 5 9 18 11

2 13 29 6 12 14

3 3 2 4 4 5

4 18 9 12 17 15

5 11 6 14 19 10

How should the wagons be transported so as to minimize the total mileage covered?


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UNIT IVLINEAR PROGRAMMING

FormulationGraphical solution Simplex methodTwo phase methodTransportation and Assignment Problems

FORMULATION

1. Define a LPP.

The problem of finding the values of a set of variables satisfying a given set of constraints, non - negative restrictions and

maximizing the objective function is known as LPP.

2. Define Mathematical Form of LPP.

1 1 2 2 ... n nMax z c x c x c x ------------ (1)

Subject to constraints

11 1 12 2 1 1

11 1 12 2 1 2

1 1 2 2

...

...

.............................................

.............................................

.............................................

...

n n

n n

m m

a x a x a x b

a x a x a x b

a x a x a

mn n m

x b

------------ (2)

and 1 2, ,..., 0nx x x ------------ (3).

3. Basic assumptions (or) Conditions (or) Requirements (or) Limitations of LPP.

(a) LPP deals with only one objective function

(b) Objective function of LPP is linear

(c) Constraints of the LPP are linear

(d) Constraints and variables of a LPP are finite in number

(e) Variables are continuous taking nonnegative values

(f) R.H.S of each constraints of LPP should no negative values.

4. Canonical and Standard form of LPP.

The LPP with inequality constraints is said to be in canonical form.

The LPP in which all the constraints are expressed in terms of equations is said to be in standard form.

GRAPHICAL SOLUTION

1. What are the limitations (or) disadvantages (or) drawback of graphical method of solving LPP.

Graphical method can be applied for solving the LPP having two variables only. So if LPP has three (or) more variables

we cannot use graphical method.

SIMPLEX METHOD

1. What are Slack and Surplus variables?

The variables used for converting inequality constraints into an equation are called Slack variables.

Eg:- 1 23 4 5x x

1 2 13 4 5x x S where 1S are called slack variable.

The variables used for converting inequality constraints into an equation are called Surplus variables.

Eg:- 1 23 4 5x x

1 2 13 4 5x x S where 1S are called surplus variable.


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TRANSPORTATION PROBLEMS (T.P)

1. Define a T.P?

Consider m origins and n destinations with known capacities and demands. The problem of finding the number of units

of an item to be transported from various origins to various destinations minimizing the total transportation cost is known as

transportation problem.

2. Write down the matrix representation of T.P?

Consider a T.P of m origins 1 2, , ..., mO O O with capacities 1 2, , ..., ma a a and n destinations 1 2, , ..., nD D D with demands

1 2, ,..., nb b b . (i.e., ijC = Cost of transporting one unit of an item from iO to jD .

Then the matrix representation of T.P is given below

1D 2D 3D . . . nD Capacity

Origins

1O

11C 12C 13C . . . 11C 1a

2O

2a

3O

3a

. .

. .

mO 1mC 2mC 3mC . . . mnC ma

Demands 1b 2b 3b . . . nb

3. When do you say that a T.P is balanced?

A T.P is said to be balanced, if the sum of all origin capacities is equal to be sum of all the destination demands.

4. When do you say that a balanced T.P has initial Basic feasible solution (IBFS)?

A balanced T.P with m x n cost matrix is said to have IBFS, if it has 1m n allocations.

5. What do you mean by Degeneracy in a T.P?

A T.P m x n cost matrix is said to have degenerate solutions, if the number of allocation is less than 1m n .

The process of getting degenerate solutions is known as Degeneracy.

6. Name the method used for finding IBFS of a T.P.

(i) NorthWest Corner Rule

(ii) Matrix Minima Method (or) Least Cost Method

(iii) Vogels Approximation (VAM) Method

7. What do you mean by Optimality test in a T.P?

The process of checking whether the IBFS of a balanced T.P can be minimized further is known as optimality test. This

is carried out by a method known as UV Method (or) Modified Distribution (MODI) Method.

ASSIGNMENT PROBLEMS (A.P)

1. Define an A.P.

A problem whose objective is to assign a given number of origins into an equal number of destinations on a one one

basis that minimizes the total assignment cost is known as A.P.

2. When do you say that A.P is balanced?

An A.P is said to be balanced if it has a square cost matrix.

3. Name the method of solving an A.P?

Hungarian Method

unit 4 operations research

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