UNIT-1 INTRODUCTION TO LINEAR PROGRAMMING

Short Answers

1. Define Operations Research? Operations Research is the application of scientific methods, techniques and tools to operations

of systems to obtain optimal solution to the problems; it provides a quantitative technique to the

managers for making better decisions for operations under control.

2. What is an Iconic or physical model in O.R.? This is a physical or a pictorial representation of various aspects of a system. Properties of

the real system are represented by the properties themselves with a change of scale. (e.g).

Model of solar system, Scaled up model of a cell in biology.

3. Write applications of O.R.

* Production, blending, product mix.

* Inventory control, demand forecast, sale and purchase.

*Transportation, repair and maintenance, scheduling and sequencing.

*Planning, scheduling and controlling of projects.

*Optimal allocation of men, machines, materials, time and money.

*Location and size of warehouses, distributions centres, retail depots, etc.

*Cash management so that all sections are departments receive adequate supply of funds.

4. 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.

5. 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 + + a 1nxn b1

a21x1 +a22x2 + a23x3 + + a 2nxn b2

a31x1 +a32x2 + a33x3 + + a 3nxn b3

am1x1 + am2x2 + am3x3 + . + a mn xn bm

and non-negativity restrictions x1 , x2, x3 x n 0

6. Specify the components of a LPP (OR) Specify the basic assumptions of LPP

(i) Proportionality (ii) Additively (iii) Divisibility (iv) Certainty (or) Deterministic (v) Finiteness (vi) Optimality.

7. 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.

8. What is the use of artificial variable in LPP

Non-negative variables that are added to the constraints of (>=) or (=) type. The purpose of

introducing artificial variables is just to obtain an initial basic feasible solution.

9. 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 ..x n 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.

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 optimal simplex 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 changes in 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 ie 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

PART-B (1) A firm produces 3 products. These products are processed on 3 different machines.

The time required to manufacture one unit of each of the 3 products and the daily

capacity of the 3 machines are given below:

Machine Time per unit (minutes) Machine Capacity Product1 Product2 Product3 (Minutes/day) M1 2 3 2 440 M

2 4 - 3 470 M

3 2 5 - 430 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.

Solution: 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.

Property Raw material

A B C

Specific

gravity 0.92 0.97 1.04

Chromium 7% 13% 16%

Melting 440C 490C 480C

point

Cost of the various raw materials per unit ton 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. Solution: 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 a cutting machine and a finishing machine relevant data are

Product

P1

P2

Cutting hrs (per unit) 2 1

Finishing hrs (perunit) 3 3

Profit (perunit) Rs.6 Rs.4

Max. sales (per week) - 210

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 ? Nov/Dec 2003

Solution: 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 and the 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 of more than Rs.6 per week, assuming that he

cannot house more than 20 hens. Formulate this as a L.P.P.

Solution: Max Z = 0.5 x2 0.1x 1 Sub to

2x1 + 5x2 80

x1 + x2 20

0.5x2 0.1x 1 6 & x1, x2 0 (5) A television company operates 2 assembly sections, section A and section B. Each

section is used to assemble 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 B

Colour 3 1

Standard 1 1

Economy 2 6 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. Solution : Minimize Z = 6000x1 + 4000x2

Sub to 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 purchases costing 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

(6) Solve Graphically: Maximize Z = 3x1 + 9x2

sub to x1 + 4x2 8

x1 + 2x2 4 & x1, x2 0

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

(7)Solve graphically : Maximize Z = 2x1 + 4x2

sub to x1 + 2x2 5

Solution : Alternate solution x1 + x2 4 & x1, x2 0

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

(8)Solve Graphically: Maximize Z = 2x1 + x2

sub to x1 x 2 10

2x1 40 & x1,x2 0. Solution : Unbounded solution

(9) Solve graphically : Maximize Z = 3x1 + 2x2

sub to 2x1 + x2 2

3x1 + 4x2 12 & x1,x2 0.

Solution : Infeasible solution.

(10)Solve graphically : Maximize Z = 100x1 +

40x2 5x1 +2 x2 1000

3 x1 +2 x2 900

x1 + 2x2 500 & x1 , x2 0

(11)Solve by Simplex Method: Maximize Z = 3x1 + 9x2

sub to x1 + 4x2 8

x1 + 2x2 4 & x1,x2 0

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

(12)Solve by Simplex Method : Maximize Z = 2x1 + 4x2

sub to x1 + 2x2 5

Solution : Alternate solution x1 + x2 4 & x1,x2 0

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

(13) Solve by Simplex Method : Maximize Z = 2x1+ x2

sub to x1 x 2 10

Solution : Unbounded solution 2x1 40 & x1,x2 0.

(14) Solve by Simplex Method: Maximize Z = 3x1 + 2x2

sub to 2x1 + x2 2

Solution : Infeasible solution. 3x1 + 4x2 12 & x1,x2 0.

(15) Solve by Simplex Method : Maximize Z = 20 x1 + 30x2

Sub to 2x1 + 3x2 120

x1 + x2 35

2x1 + 1.5x2 90 & x1, x2 0. Apr/May 2004

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

(16)Maximize Z = 15x1 +6 x2 +9 x3+2x4

Sub to 2x1 +x2 + 5x3 +6x4 20

3x1 +x2 + 3x3 +25x4 24

7x1 + x4 70 & x1, x2,x3, x4 0.

(16) Solve by Simplex Method : Minimize Z = 8x1 2x 2

Sub to -4x1 + 2x2 1

5x1 4x 2 3 & x1, x2 0.

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

(17)Use Big-M OR Penalty Method to solve Maximize Z = 2x1 + x2 + x3

Sub to 4x1 + 6x2 + 3x3 8 3x1 6x 2 4x 3 1 2x1 + 3x2 5x 3 4. & x1, x2,x3 0. Solution: Max Z = 64/21, x1 = 9/7, x 2 =

10/21 , x3 = 0

(18)Use Big-M OR Penalty Method to solve Minimize Z = 4x1 + 3x2

Sub to 2x1 + x2 10 -3x1 + 2x 2 6 x1 + x2 6 & x1,x2 0. Solution : Min Z = 22, x1 = 4, x2 = 2.

(19)Use Two-phase method to solve Maximize Z = 5x1 + 8x2 Sub to

3x1 + 2x2 3 x1 + 4x2 4

x1 + x2 5 & x1,x2 0. Solution : Max Z = 40, x1 = 0, x2 =5.

(20) Use Two-phase method to solve Minimize Z = -2x1 x 2 Sub to

X1 + x2 2 X1 + x2 4 & x1,x2 0.

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

Variants of the Simplex Method:

(21`) Solve the L.P.P by Simplex Method : Maximize Z = x1 + 2x2 + x3 Sub to

2x1 + x2 x 3 2 -2x1 + x2 5x 3 -6 4x1 + x2+ x3 6 & x1,x2,x3 0.

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

(22) Solve the L.P.P by Simplex Method : Maximize Z = 3x1 x 2 Sub to

X1 x 2 10

X1 20 & x1,x2 0. Solution : Unbounded solution.

(23) Solve the L.P.P by Simplex Method : Maximize Z

= 6x1 + 4x2 Sub to

2x1 + 3x2 30

3x1 + 2x2 24

x1 + x2 3 & x1,x2 0. Solution : Max Z = 48, x1 = 8, x2 = 0 & Max Z = 48, x1 = 12/5, x2 = 42/5

(24) Solve the L.P.P by Simplex Method : Maximize Z

= 3x1 + 2x2 Sub to

2x1 + x2 2 3x1 + 4x2 12 & x1,x2 0. Solution : Infeasible solution.

(25) Solve the L.P.P by Simplex Method : Maximize Z

= 2x1 + 3x2 Sub to

-x1 + 2x2 4 x1 + x2 6 x1 + 3x2 9 & x1, x2 are unrestricted.

Solution : Max Z = 27/2, x1 = 9/2, x2 = 3/2. (26)Solve by Dual Simplex Method: Minimize Z

= 3x1 + x2 Sub to

3x1 + x2 3

4x1 + 3x2 6 x1 + x2 3 & x1,x2 0.

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

(27) Solve by Dual Simplex Method:

Minimize Z = 2x1 + 3x2 Sub to

2x1 + 2x2 30

x1 + 2x2 10 & x1, x2 0. (28)Write down the dual of the following LPP and solve it.

Max Z = 4x1 + 2x2 Sub to

-x1 x 2 -3 -x1 + x2 -2 & x1, x2 0.

Solution : Infeasible solution

(29)Use duality to solve the following LPP Minimize Z = 2x1 + 2x2 Sub to

2x1 + 4x2 1 -x1 2x 2 -1 2x1 + x2 1 & x1, x2 0. Solution : Min Z = 4/3, x1 = 1/3, x 2

= 1/3.

(30)Use duality to solve the following LPP Maximize Z = 3x1 + 2x2 Sub to

X1 + x2 1 X1 + x2 7 X1 +2x2 10 X2 3 & x1, x2 0. Solution: Max Z =21, x1 = 7, x2 = 0.

(31)Use duality to solve the following LPP Minimize Z =x1 x 2+x3 Sub to

x1 x 3 4

x1 x 2 + 2x3 3 & x1, x2,x3 0. Solution: Infeasible solution.

(32) Consider the LPP

Max Z = 2x1 + x2 + 4x3 x 4 Sub to X1 +2x2 + x3-3x4 8 -x2 + x3+2x4 0

2x1 + 7x2 5x 3 10x 4 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].

Solution: (a) Max Z =16, x1 = 8, x2 = x3 = x4 = 0.

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

(33)Consider the LPP

Max Z = 3x1 + 4x2 + x3 + 7x4 Sub to 8x1 + 3x2 + 4x3 + x4 7 2x1 + 6x2 + x3 + 5x3 3

x1 + 4x2 + 5x3 + 2x4 8 & x1, x2, x3, x4 0. (a) Solve the LPP.

Solution : (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.

(34)Consider the LPP

Max Z = 2x1 + x2 + 4x3 x

4 Sub to

X1 + 2x2 + x3 3x 4

8 -x2 +x3 + 2x4 0

2x1 + 7x2 5x 3 10x 4 21 & x1, x2, x3, x4 0. (a) Solve the LPP.

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

(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) Solution :

(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.

(35)Consider the LPP

Max Z = 5x1 + 12x2 +

4x3 Sub to

X1 + 2x2 + x3 5

2x1 x 2 + 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). Solution : (b) Unbounded solution.

(d) Max Z = 60, x1 = 0, x2 = 4, x3 = 3. (36)Consider the LPP

Max Z = 3x1 +

5x2 Sub to

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.

Solution : (a) Max Z = 45, x1 = 0, x2 = 9.

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