LINEAR PROGRAMMING is the technique of optimum allocation of limitedresources

Linear is used to describe the proportionate relationship of two or morevariable

Programming is used to specify sort of planning that involves economicallocation of resources

STRUCTURE OF L.P.P.

Decision variable - They are under the control of decision maker

All decision variables are continuous, controllable, non negative

x1, x2. . . .xn ≥ 0

Objective function - Objective function of LP program is mathematicalrepresentation of objective in terms of measurable quantity such as profit, cost,revenue

Optimise (Max. or Min.) Z = c1x1+c2x2+. . . .cnxn

Constraints

There are certain limitation on the use of limited resources e.g. labour, capital

Assumption of Linear Programming

- Certainity

- Divisibility

- Additivity

- Linearity

Special cases in LP

- Infeasible Soln

Max Z = 4x1 + 2x2

2x1 + 3x2 ≤ 18

x1 + x2 ≥ 10

x1x2 ≥ 0

(0, 6) (9, 0)

(0, 10) (10, 0)

Multiple Soln

Max Z = 4x2 + 3x2

3x1 + 4x2 ≤ 24

8x1 + 6x2 ≤ 48

x1 ≤ 5 (0, 6) (8, 0)

x2 ≤ 6 (0, 8) (6, 0)

10

8

6

4

2

02 4 6 8 10

0 (0, 0) A (5, 0)

B (5, 4/5)

C (24/7, 24/7)

D (0, 6)

ROC - 24

Onbounded Solution

Max Z = 3x1 + 2x2

x1 - x2 ≤ 1

x1 - x2 ≥ 3

x1, x2 ≥ 0

(x, 0) (0, -1) (1, 0)

x1 = 0 x2 = 3 x2 = 0 x1 = 3

Redundent constraint

x1 ≥ 20

x1 ≥ 10

The standard weight of a special purposebrick is 5 kg. and it contains two basicingredient 1 & B2, B1 cost Rs. 5 per Kg.and B2 costs Rs. 8 per kg. Strengthconsideration dictate that brick containnot more than 4 kg. of B1 & Min. of 2 kg.of B2 kind out graphically they minimumcost of brick satisfying above condition

Min Z = 5x1 + 8x2

s.t. x1 ≤ 4

x2 ≥ 2

x1 + x2 = 5

8

6

4

2

0

B

C

x1 ≥ 5

x2 ≥ 6

3

2

1

-1

1 2 3

Use graphical method to solve the following problem

Max Z = 15x1+10x2

s.t. 4x1 + 6x2 ≤ 360

3x1 + 0x2 ≤ 180

0x1 + 5x2 ≤ 200

x1, x2 ≥ 0

2x1 + 5x2 ≤ 180

(0, 60) (90, 0)

3x1 = 180 x1 = 60

5x2 < 200 x2 = 40

5

4

3

2

1

1 2 3 4 5

x1 = 4

x2 = 2

0 (0, 0) = 0

A (60, 0) = 900

B (60, 20) = 1100

C (30, 40) = 850

The manager of an oil refinary must decide on the optimal mix of two possibleblending process of which the inputs & output per production run are asfollows :

Input Output

Process Grade A Grade B Gasoline X Gasoline Y

1 5 3 5 8

2 4 5 4 4

Max. amount of available crude A & B are 200 units & 150 units respectively.Market requirement show that atleast 100 units of Gasoline X & 80 units ofCasol & must be produced. The profits per production run for process 1 &process 2 are Rs. 300 & Rs. 400 respective solve LP by graphical method

Soln Max Z = 300x1 + 400x2

Input constraint

5x1 + 4x2 ≤ 200

3x1 + 5x2 ≤ 150

Output constraint

5x1 + 4x2 ≥ 100

8x1 + 4x2 ≥ 80

100

80

60

40

20

020 40 60 80 100

B (60, 20)

x2 = 40

x1 = 60

C (30, 40)

(0, 40)

A (20, 0) = 6,000

B (40, 0) = 12,000

C (400/13, 50/23) = 1,80,000/13

D (0, 30) = 12,000

E (0, 25) = 10,000

e.g. A company makes two kinds of leather belts, Belt A is a high quality belt &belt B is of lower quality. The respective profits are Rs. 4 & Rs. 3 per belt. Eachof type A require twice as much time as a belt of type B. The company couldmake 1000 per day. The supply of leather is sufficient for only 800 belts per day(both A & B combined). Belt A requires fancy buckle & only 400 per days areavailable. There are only 700 buckles a day available for B.

What should be daily production of each type of belt. Formula (PP & solve it bysimplex method

Max z = 4x1 + 3x2

s.t. 2x1 + x2 ≤ 1000

x1 + x2 ≤ 800

x1 ≤ 400

x2 ≤ 700

x1, x2 ≥ 0

2x1 + x2 + s1 = 1000

x1 + x2 + s2 = 800

x1 + s3 = 400

x2 + s4 = 700

50

40

30

20

10

010 20 30 40 50

C

or

Max z = 4x1 + 3x2 + 0s1 + 0s2 + 0s4

2x1 + x2 + 1s1 + 0s2 + 0s3 + 0s4 = 1000

x1 + x2 + 0s1 + 1s2 + 0s3 + 0s4 = 800

x1 + 0x2 + 0s2 + 0s2 + 1s3 + 0s4 = 400

0x1 + x2 + 0s1 + 0s2 + 0s3 + 1s4 = 700

Cj 4 3 0 0 0 0

CB Xb b x1 x2 s1 s2 s3 s4 Ratio Q

0 s1 1000 2 1 1 0 0 0 500

0 s2 800 1 1 0 1 0 0 800

0 s3 400 1 0 0 0 1 0 400-

0 s4 700 0 1 0 0 0 1 00-

Zj 0 0 0 0 0 0

Cj - Zj 4 3 0 0 0 0

↑

Second Simplex

Cj 4 3 0 0 0 0

CB Xb b x1 x2 s1 s2 s3 s4 Ratio Q

0 s1 200 0 1 1 0 -2 0 200 →

0 s2 400 0 1 0 1 -1 0 400

4 x1 400 1 0 0 0 1 0 -

0 s4 700 0 1 0 0 0 1 700

Zj 4 0 0 0 4 0

Cj - Zj 0 3 0 0 -4 0

↑

Third Simplex

Cj 4 3 0 0 0 0

CB Xb b x1 x2 s1 s2 s3 s4 Ratio Q

3 x2 200 0 1 1 0 -2 0 -

0 s2 200 0 0 -1 1 1 0 200 →

4 x1 400 1 0 0 0 1 0 400

0 s4 500 0 0 -1 0 2 1 250

Zj 4 3 3 0 -2 0

TABLE IV

Cj 4 3 0 0 0 0

CB Xb b x1 x2 s1 s2 s3 s4

3 x1 600 0 1 -1 2 0 0

0 s3 200 0 0 -1 1 1 0

4 x1 200 1 0 1 -1 0 0

0 s4 100 0 0 -3 -2 0 1

Zj 4 3 1 2 0 0

Zj - Cj 0 0 -1 -2 0 0

x1 = 200 x2 = 600

Max z = 4 x 200 + 3 x 600 = 2600

Max z = 15x1 + 10x2

s.t. 4x1 + 6x2 < 3603x1 + 0x2 < 1800x1 + 5x2 < 200x1 x2 > 0

Max z = 15x1 + 10x2

4x1 + 6x2 + 1s1 + 0s2 + 0s3 = 3603x1 + 0x2 + 0s1 + 1s2 + 0s3 = 1800x1 + 5x2 + 0s1 + 0s2 + 1s3 = 200

Cj 15 10 0 0 0

CB Xb b x1 x2 s1 s2 s3 Ratio θ

0 s1 360 4 6 1 0 0 90

0 s2 180 3 0 0 1 0 60 →

0 s3 200 0 5 0 0 1 —

Zj 0 0 0 0 0

Cj - Zj 15 10 0 0 0

↑

SECOND SIMPLEX TABLE

Cj 15 10 0 0 0

CB Xb b x1 x2 s1 s2 s3 Ratio θ

0 s1 120 0 6 1 -4/3 0 20 →

15 x1 60 1 0 0 1/3 0 -

0 s3 200 0 5 0 0 1 40

Zj 15 0 0 5 0

Cj - Zj 0 10 0 -5 0

↑

THIRD SIMPLEX TABLE

Cj 15 10 0 0 0

CB Xb b x1 x2 s1 s2 s3 Ratio θ

10 x2 20 0 1 1/6 -2/9 0

15 x1 60 1 0 0 1/3 0

0 s3 100 0 0 -5/6 10/9 1

Zj 15 10 5/3 25/9 0

Cj - Zj 0 0 5/3 25/9 0

Use Penalty (Big M) method to solve the following LPP

Min z = 5x1 + 3x2

s.t. 2x1 + 4x2 < 122x1 + 2x2 = 105x1 + 2x2 > 10

or 2x1 + 4x1 + s1 = 122x1 + 2x2 + A1 = 105x1 + 2x2 - s2 + A2 = 10

Cj 5 3 0 0 M M Ratio θ

CB Xb b x1 x2 s1 s2 A1 A2 (Min.)

0 s1 12 2 4 1 0 0 0 6

M A1 10 2 2 0 0 1 0 5

M A2 10 5 2 0 -1 0 1 2 →

Zj 5M + 2M 4M 0 -M M M

Cj - Zj 5-7M 3-4M 0 M 0 0

↑

Revised R3 2 1 2/5 0 -1/5 0 1/5

SECOND SIMPLEX TABLE

Cj 5 3 0 0 M

CB XB b x1 x2 s1 s2 A1 Ratio θ

0 s1 8 0 16/15 1 2/5 0 5/2 →

M A1 6 0 6/5 0 2/5 1 5S x1 2 1 2/5 0 -1/5 0 5

Zj 5 6M/5+2 0 2M/5-1 MCj - Zj 0 -6M/5+1 0 -2M/5+1 0

↑

THIRD SIMPLEX TABLE

5 3 0 0 M

CB XB b x1 x2 s1 s2 A1 Ratio θ

0 s1 5/2 0 1 5/16 1/8 0 20

M A1 3 0 0 -3/8 1/4 1 12 →S x1 1 1 0 -1/8 -1/4 0 —

Zj 5 3 85

83

−−M

45

4−

MM

Cj - Zj 0 0 85

83

+M

45

4+

M0

FOURTH SIMPLEX TABLE

CJ → 5 3 0 0

CB Xb b X1 X2 S1 S2 Ratio θ

3 X2 1 0 1 ½ 0

0 S2 12 0 0 -3/2 1

5 X1 4 1 0 -½ 0

CJ 5 3 -1 0

CJ - ZJ 0 0 1 0

Point to be RememberedWhen objective fn is max Z 8

Constraints are of ≤ type introduceSlack variable (S1, S2 .....)

In case of Min Zconstraints are of ≥ type introduce(- S1, - S2 ..) surplus & corresponding A1, A2constraints are of = introduce A1, A2

Give the dual of the following

Min Z = x1 + x2 + x3

st x1 - 3x2 + 4x3 = 52x1 - 2x2 ≤ 32x2 - x3 ≥ 5

Min Z = x1 + x2 + x3

st x1 - 3x2 + 4x3 ≥ 5- x1 + 3x2 - 4x3 ≥ - 5- 2x1 + 2x2 + 0x3 ≥ - 30x1 + 2x2 - x3 ≥ 5

Max ZD = 5W1 - 52 - 3W3 + 5W4W1 - W2 - 2W3 + 0W4 ≤ 1- 3W1 + 3W2 + 2W3 + 2W4 ≤ 14W1 - 4W2 + 0W3 - W4 ≤ 1

Max Z = 3x1 + 2x2 + 4x3

st 4x1 + 3x2 + 5x3 ≤ 2000

3x1 + 2x2 + 4x3 ≤ 2500

x1 ≥ 100 - x1 ≤ - 100

x2 ≥ 200 - x2 ≤ - 200

x3 ≥ 500 - x3 ≤ - 50

x1 < - 50

or

Min ZD = 2000W1 + 2500W2 - 100W3 - 200W4 - 50W4 + 150W64W1 + 3W2 - W3 + 0W4 + 0W5 + W6 ≥ 33W1 + 2W2 + 0W3 - W4 + 0W5 + 0W6 ≥ 2SW1 + 4W2 + 0W1 + 0W2 + 0W3 + 0W4 - W5 + 0W6 ≥ 4

eg

Max Z = 2x1 + 4x2 + 3x3st 3x1 + 4x2 + 3x3 ≤ 3600

2x1 + x2 + 3x3 ≤ 2400x1 + 3x2 + 3x3 ≤ 4800

Dual

Min ZD = 3600W1 + 2400W2 + 4800W3st 3W1 + 2W2 + W3 ≤ 2

4W1 + W2 + 3W3 ≤ 43W1 + 3W2 + 3W3 ≥ 3

Min Z = x1 + x2st 3x1 + 2x2 ≥ 4

- x1 + 3x2 ≥ 54x1 + 2x2 ≥ 52x1 + x2 ≥ 1

Max ZD = 4W1 + 5W2 + 5W3 + 5W4st 3W1 - W2 + 4W3 + 2W4 ≤ 1

2W1 + 3W2 + 2W3 + W4 ≤ 1

or

3W1 - W2 + 4W3 + 2W4 + S1 = 1

2W1 + 3W2 + 2W3 + W4 + S2 = 1

CJ 4 5 5 1 0 0

CB WB b W1 W2 W3 W4 S1 S2 Ratio θ(Min)

0 S1 1 3 - 1 4 2 1 0 -

0 S2 1 2 3 2 1 0 1 (1/3)Key element →

ZJ 0 0 0 0 0 0

CJ - ZJ 4 5 5 1 0 0

↑

CJ 4 5 5 1 0 0

CB WB b W1 W2 W3 W4 S1 S2 Ratio θ(Min)

0 S1 4/3 11/3 0 14/3 7/3 1 1/3 2/7→

5 W2 1/3 2/3 1 2/3 1/3 0 1/3 1/2

CJ 10/3 5 10/3 5/3 0 5/3

CJ - ZJ 2/3 0 5/3 - 2/3 0 - 5/3

↑

4 5 5 1 0 0

CB WB b W1 W2 W3 W4 S1 S2

5 W3 2/7 11/14 0 1 1/2 3/14 1/14

5 W2 1/7 1 0 0 - 1/7 2/7

ZJ 65/14 5 5 5/2 5/14 25/14

CJ - ZJ - 9/14 0 0 - 3/2 - 5/14 - 25/14

W1 = 0 W2 = 1/7 W3 = 2/7 W4 ZD = 15/7

Soln of prima x1 = 5/14 x2 = 25/14

Zp = 5/14 + 25/14 = 15/7

NO PHASE METHOD

Min Z = 600x1 + 400x2 Min Z = 0x1 + 0x2 + 0S1 + 0S2 + A1 + A2 + A3st 3x1 + 3x2 ≥ 40 3x1 + 3x2 - S1 + A1 = 40

3x1 + x2 ≥ 40 3x1 + x2 - S2 + A2 = 402x1 + 5x2 ≥ 44 2x1 + 5x2 - S3 + A3 = 44

SOURCE : - V.K. KAPOOR

CJ→ 0 0 0 0 0 1 1 1

CB XB b x1 x2 S1 S2 S3 A1 A2 A3 Min ratio

1 A1 40 3 3 - 1 0 0 1 0 0 40/3

1 A2 40 3 1 0 - 1 0 0 1 0 40

1 A3 44 2 5 0 0 - 1 0 0 1 44/5→

ZJ 8 9 - 1 - 1 - 1 1 1 1

CJ - ZJ - 8 - 9 1 1 1 0 0 0

↑

0 0 0 0 0 1 1 Ratio

CB XB b x1 x2 S1 S2 S3 A1 A2 Q

1 A1 68/5 9/5 0 - 1 0 3/5 1 0 68/9

1 A2 156/5 13/5 0 0 - 1 1/5 0 1 12

0 x2 44/5 2/5 1 0 0 - 1/5 0 0 22

ZJ 22/5 0 - 1 - 1 4/5 1 1

CJ - ZJ - 22/5 0 1 1 - 4/5 0 0

↑

CJ 0 0 0 0 0 1

CB XB b x1 x2 S1 S2 S3 A2 Ratio Q

0 x1 340/45 1 0 - 5/9 0 1/3 0 -

1 A2 520/45 0 0 13/9 - 1 - 2/3 1 8→

0 x2 260/45 0 1 2/9 0 - 1/3 0 26

ZJ 0 0 13/9 - 1 - 2/3 1

CJ - ZJ 0 0 - 13/9 1 2/3 0

↑

FOURTH SIMPLEX

CJ → 0 0 0 0 0

CB XB b x1 x2 S1 S2 S3

0 x1 12 1 0 0 - 5/13 + 1/13

0 S1 8 0 0 1 - 9/13 - 6/13

0 x2 4 0 1 0 2/13 - 3/13

ZJ 0 0 0 0 0

CJ - ZJ 0 0 0 0 0

PHASE

CJ 600 400 0 0 0

CB xb b x1 x2 S1 S2 S3

600 x1 12 1 0 0 - 5/13 1/13

0 S1 8 0 0 1 - 9/13 - 6/13

400 x2 4 0 1 0 2/13 - 3/13

ZJ 600 400 0 - 2200/13 - 600/13