Linear Programmingis a mathematical Linear Programmingis a mathematical technique for optimum allocation of limited technique for optimum allocation of limited or scarce resources, such as labor, material, machine, money, energy and so on , to machine, money, energy and so on , to several competing activities such as products, services, jobs and so on, on the basis of a services, jobs and so on, on the basis of a given criteria of optimality.

Max/ min z = c x + c x + ... + cnxnMax/ min z = c x + c x + ... + cnxn

subject to:subject to:a x + a x + ... + a nxn ( , =, ) ba x + a x + ... + a nxn ( , =, ) b

:a x + a x + ... + a nxn ( , =, ) b

:am x + am x + ... + amnxn ( , =, ) bm

xj = decision variablesbi = constraint levelsbi = constraint levelscj = objective function coefficientsaij = constraint coefficientsij

1.1. Plot model constraint on a set of Plot model constraint on a set of 1.1. Plot model constraint on a set of Plot model constraint on a set of coordinates in a planecoordinates in a planecoordinates in a planecoordinates in a plane

2.2. Identify the feasible solution space on the Identify the feasible solution space on the graph where all constraints are satisfied graph where all constraints are satisfied graph where all constraints are satisfied graph where all constraints are satisfied simultaneouslysimultaneously

3.3. Plot objective function to find the point on Plot objective function to find the point on boundary of this space that maximizes (or boundary of this space that maximizes (or boundary of this space that maximizes (or boundary of this space that maximizes (or minimizes) value of objective functionminimizes) value of objective function

Solve the following LPP by graphical method

Maximize Z = X + XMaximize Z = X + X

Subject to constraints

X + X

X

X

X

X

X , X

The first constraint X + X

can be represented as The first constraint X + X

can be represented as

follows.

We set X + X = We set X + X =

When X = in the above constraint, we get,

x + X =

X = X =

Similarly when X = in the above constraint, we get,

X + = X + =

X = / =

The second constraint X can be

represented as follows,represented as follows,

We set X = We set X =

The third constraint X can be

represented as follows,

We set X = We set X =

The constraints are shown plotted in the above figure.

Point X1 X2 Z = 5X1 +3X2

0 0 0 0

A 0 700 Z = 5 x 0 + 3 x 700 = 2,100

B 150 700Z = 5 x 150 + 3 x 700 = 2,850* Maximum

B 150 700Maximum

C 400 200 Z = 5 x 400 + 3 x 200 = 2,600

D 400 0 Z = 5

x 400

+ 3

x 0

= 2,000D 400 0 Z = 5

x 400

+ 3

x 0

= 2,000

The Maximum profit is at point B

When X = and X = When X = and X =

Z = Z =

Solve the following LPP by graphical methodSolve the following LPP by graphical method

Maximize Z = X + X

Subject to constraintsSubject to constraints

X + X

X + X

X

X

X , X

Solution:Solution:

The first constraint X + X can be

represented as follows.represented as follows.

We set X + X =

When X = in the above constraint, we get,

x + X = x + X =

X = / = .

Similarly when X = in the above constraint, we get,Similarly when X = in the above constraint, we get,

X + x =

X = / = .

The second constraint X + X

can be represented The second constraint X + X

can be represented

as follows,

We set X + X = We set X + X =

When X = in the above constraint, we get,

x + X =

X = / = X = / =

X = / = .

Similarly when X = in the above constraint, we get,Similarly when X = in the above constraint, we get,

X + x =

The third constraint X

can be represented as follows,The third constraint X

can be represented as follows,

We set X =

Point X1 X2 Z = 400X1 + 200X2

0 0 0 00 0 0 0

A 0 150Z = 400 x 0+ 200 x 150 = 30,000* MaximumA 0 150 Maximum

B 31.11 80 Z = 400 x 31.1 + 200 x 80 = 28,444.4

C 44.44 0 Z = 400

x 44.44

+ 200

x 0

= 17,777.8C 44.44 0 Z = 400

x 44.44

+ 200

x 0

= 17,777.8

The Maximum profit is at point A

When X = and X = When X = and X =

Z = ,

Solve the following LPP by graphical methodSolve the following LPP by graphical method

Minimize Z = X + XMinimize Z = X + X

Subject to constraintsSubject to constraints

X + X

X + X

X + X

X + X

X , X

The first constraint X + X

can be The first constraint X + X

can be

represented as follows.represented as follows.

We set X + X = We set X + X =

When X = in the above constraint, we get,When X = in the above constraint, we get,

x + X =

X = / =

Similarly when X = in the above constraint, we get,Similarly when X = in the above constraint, we get,

X + x =

X = / =

The second constraint X + X

can beThe second constraint X + X

can be

represented as follows,represented as follows,

We set X + X =

When X = in the above constraint, we get,

x + X = x + X =

X = / = X = / =

Similarly when X = in the above constraint, we

get,

X + x = X + x =

X = / = X = / =

The third constraint X + X can be

represented as follows,

We set X + X = We set X + X =

When X = in the above constraint, we get,When X = in the above constraint, we get,

x + X = x + X =

X = / = X = / =

Similarly when X = in the above constraint,Similarly when X = in the above constraint,

we get,

X + x =

X = / =

Point X1 X2 Z = 20X1 + 40X2

0 0 0 00 0 0 0

A 0 18 Z = 20 x 0 + 40 x 18 = 720

B 2 6 Z = 20

x2

+ 40

x 6

= 280B 2 6 Z = 20

x2

+ 40

x 6

= 280

C 4 2Z = 20 x 4 + 40 x 2 = 160* Minimum

D 12 0 Z = 20

x 12

+ 40

x 0

= 240D 12 0 Z = 20

x 12

+ 40

x 0

= 240

The Minimum cost is at point C

When X = and X =

Z =Z =

Solve the following LPP by graphical methodSolve the following LPP by graphical method

Maximize Z = . X + . X

Subject to constraintsSubject to constraints

X

,X

,

X ,

. X + . X

. X + . X

X + X ,

X , X

The first constraint X , can be

represented as follows.

We set X = ,We set X = ,

The second constraint X

, can be The second constraint X

, can be

represented as follows,represented as follows,

We set X = ,We set X = ,

The third constraint . X + . X can be

represented as follows, We set . X + . X = represented as follows, We set . X + . X =

When X = in the above constraint, we get,When X = in the above constraint, we get,

. x + . X =

X = / . = ,

Similarly when X = in the above constraint, we get,Similarly when X = in the above constraint, we get,

. X + . x = . X + . x =

X = / . = ,

The fourth constraint X + X , can be

represented as follows, We set X + X = ,represented as follows, We set X + X = ,

When X = in the above constraint, we get,When X = in the above constraint, we get,

+ X = ,

X = ,

Similarly when X = in the above constraint, we get,Similarly when X = in the above constraint, we get,

X + = ,X + = ,

X = ,

Point X1 X2 Z = 2.80X1 + 2.20X2

0 0 0 00 0 0 0

A 0 40,000Z = 2.80 x 0 + 2.20 x 40,000 = 88,000

B 5,000 40,000Z = 2.80 x 5,000 + 2.20 x 40,000 = 1,02,000

Z = 2.80

x 10,500

+ 2.20

x 34,500

= C 10,500 34,500

Z = 2.80

x 10,500

+ 2.20

x 34,500

= 1,05,300* Maximum

D 20,000 6,000Z = 2.80

x 20,000

+ 2.20

x 6,000

= D 20,000 6,000

Z = 2.80

x 20,000

+ 2.20

x 6,000

= 69,200

E 20,000 0Z = 2.80 x 20,000 + 2.20 x 0 = 56,00056,000

The Maximum profit is at point C

When X = , and X = ,When X = , and X = ,

Z = , ,

Solve the following LPP by graphical method

Maximize Z = X + XMaximize Z = X + X

Subject to constraints

X + X

X + X

X + X

X + X

X - X -

X X

X X

The first constraint X + X can be

represented as follows.represented as follows.

We set X + X = We set X + X =

When X = in the above constraint, we get,When X = in the above constraint, we get,

x + X = x + X =

X =

Similarly when X = in the above constraint, we get,

X + =

X = / = X = / =

The second constraint X + X

can beThe second constraint X + X

can be

represented as follows,

We set X + X =

When X = in the above constraint, we get,When X = in the above constraint, we get,

+ X = + X =

X = / = X = / =

Similarly when X = in the above constraint, Similarly when X = in the above constraint,

we get,

X + x =

X =

The third constraint X - X

- can be The third constraint X - X

- can be

represented as follows,

We set X - X = -

When X = in the above constraint, we get,When X = in the above constraint, we get,

- X = -

X = - / = .X = - / = .

Similarly when X = in the above constraint, we get,Similarly when X = in the above constraint, we get,

X x = -

X = -X = -

Point X1 X2 Z = 10X1 + 8X2

0 0 0 00 0 0 0

A 0 7.5 Z = 10 x 0 + 8 x 7.5 = 60

B 3 9 Z = 10

x 3

+ 8

x 9

= 102B 3 9 Z = 10

x 3

+ 8

x 9

= 102

C 6 8 Z = 10 x 6 + 8 x 8 = 124* Minimum

D 10 0 Z = 10

x 10

+ 8

x 0

= 100D 10 0 Z = 10

x 10

+ 8

x 0

= 100

The Maximum profit is at point C

When X = and X =

Z = Z =

Product Product AVALIABLE

Machine <=

x+y<=

Machine <=

x+y<=x+ y<=

A=( , ) profit=A=( , ) profit=B=( , ) profit=B=( , ) profit=C=( , ) profit= *

Click on Tools to invoke Solver.

Objective function

=E -F

=E -F

=C *B +D *B

Decision variables bowls(x )=B ; mugs (x )=B

=C *B +D *B

Copyright John Wiley & Sons, Inc.

Supplement -

(x )=B ; mugs (x )=B

After all parameters and constraints have been input, click on Solve.

Objective function

Decision variablesDecision variables

C *B +D *B

C *B +D *B

Click on Add to insert constraints

Copyright John Wiley & Sons, Inc.

Supplement -

Copyright John Wiley & Sons, Inc.

Supplement -