business maths..linear programming

Linear Programming

Group Members

Names Roll Nos.

Syed Sherjeel Hasan 25Naveed Ahmed 20Sajid Fayyaz 17Zakria Fawad 41Zeeshan Mehmood 34Tayyaba Noreen 28

Question No. 1Question No. 1MaximizingMaximizing Z=14x1+20x2Z=14x1+20x2Subject toSubject to 2x1+6x2>=502x1+6x2>=50

x1+4x2 >=40x1+4x2 >=40 x1+ x2 <=60x1+ x2 <=60

(x1, x2>=0)(x1, x2>=0)Solution:

l1 2x1+6x2 =50l2 x1+4x2 =40l3 x1+ x2 =60

x-intercept and y-intercept of “l1”(25, 0) (0, 16.67)x-intercept and y-intercept of “l2”(40, 0) (0, 10)x-intercept and y-intercept of “l3”

(60, 0) (0, 60)

Corner points:

A (0, 10)B (0, 60)C (60, 0)D (40, 0)

Z=14x1+20x2

For point A Z=14(0) +20(10) =200For point B Z=14(0) +20(60) =1200For point C Z=14(60) +20(0) =840For point D Z=14(40) +20(0) =560

Z maximizes at point B (0, 60) = 1200 (Ans.)

Question No. 2Question No. 2MinimizingMinimizing Z=18x1+20x2Z=18x1+20x2Subject toSubject to 2x1+3x2 =202x1+3x2 =20

4x1+6x2 >=304x1+6x2 >=305x1+10x2<=505x1+10x2<=50(x1, x2>=0)(x1, x2>=0)

Solution:l1 2x1+3x2 =20l2 4x1+6x2 =30l3 5x1+10x2=50

x-intercept and y-intercept of “l1”(10, 0) (0, 6.67)x-intercept and y-intercept of “l2”(7.5, 0) (0, 5)x-intercept and y-intercept of “l3”(10, 0) (0, 5)

No feasible solution exists. (Ans.)

Question No. 3Question No. 3MaximizingMaximizing Z=4x1+3x2Z=4x1+3x2Subject toSubject to x1+ x2 >=4 x1+ x2 >=4

8x1+6x2 <=488x1+6x2 <=48x1x1 <=3 <=3 x2 <=6 x2 <=6(x1, x2 >=0)(x1, x2 >=0)Solution:

l1 x1+ x2 =4l2 8x1+ 6x2 =48l3 x1+ 0 =3l4 0+ x2 =6x-intercept and y-intercept of “l1”(4, 0) (0, 4)x-intercept and y-intercept of “l2”(6, 0) (0, 8)x-intercept and y-intercept of “l3”(3, 0) (0, 0)x-intercept and y-intercept of “l4”(0, 0) (0, 6)

Corner points:

A (4, 0) For point ‘C’ For point ‘D’ For point ‘E’B (6, 0) l2 – 6l4 l2 – 8l3 l1 – l3

C (1.5, 6) (1.5, 6) (3, 4) (3, 1)D (3, 4)E (3, 1)

Z=4x1+3x2

For point A Z=4(4) +3(0) = 16For point B Z=4(6) +3(0) = 24For point C Z=4(1.5) +3(6) = 24For point D Z=4(3) +3(4) = 24For point E Z=4(3) +3(1) = 15

Z maximizes at 3 points B (6, 0), C (1.5, 6) & D (3, 4) = 24 (Ans.)

Question No. 4Question No. 4 MinimizingMinimizing Z=5x1+7x2Z=5x1+7x2 Subject toSubject to 2x1+5x2 <=202x1+5x2 <=20

2x1+2x2 >=102x1+2x2 >=10 x1 x1 <=8 <=8 x1+2x2 >=6 x1+2x2 >=6(x1, x2 >=0)(x1, x2 >=0)Solution:

l1 2x1+ 5x2 =20l2 2x1+ 2x2 =10l3 x1+ 0 =2l4 x1+ 2x2 =6x-intercept and y-intercept of “l1”(10, 0) (0, 4)x-intercept and y-intercept of “l2”(5, 0) (0, 5)x-intercept and y-intercept of “l3”(8, 0) (0, 0)x-intercept and y-intercept of “l4”(6, 0) (0, 3)

Corner points:

A (1.67, 3.33) For point ‘A’ For point ‘B’ For point ‘C’B (8, 0.8) l1 – l2 l1 - 2l3 l2 – l4

C (4, 1) (1.67, 3.33) (8, 0.8) (4, 1)D (6.67, 0)E (8, 0)

For point A Z=5(1.67) +7(3.33) =31.67For point B Z=5(8) +7(0.8) =45.6For point C Z=5(4) +7(1) =27For point D Z=5(6.67) +7(0) =33.35For point E Z=5(8) +7(0) =40

Z minimizes at point C (4, 1) = 27 (Ans.)

Question No. 5Question No. 5MaximizingMaximizing Z=9x1+6x2Z=9x1+6x2Subject toSubject to x1+2x2 >=6 x1+2x2 >=6

3x1+2x2 <=303x1+2x2 <=302x1+x2 >=52x1+x2 >=5 x1 x1 <=8 <=8(x1, x2>=0(x1, x2>=0))Solution:

l1 x1+ 2x2 =6l2 3x1+ 2x2 =30l3 2x1+ x2 =5l4 x1+ 0 =8x-intercept and y-intercept of “l1”(6, 0) (0, 3)x-intercept and y-intercept of “l2”(10, 0) (0, 15)x-intercept and y-intercept of “l3”(2.5, 0) (0, 5)x-intercept and y-intercept of “l4”(8, 0) (0, 0)

Corner points:

A (0, 5) For point ‘C’ For point ‘D’B (0, 15) l2 – 3l4 2l1 – l3

C (8, 3) (8, 3) (1.33, 2.33)D (1.33, 2.33)E (6, 0)F (8, 0)

Z=9x1+6x2

For point A Z=9(0)+ 6(5) =30For point B Z=9(0)+ 6(15) =90For point C Z=9(8)+ 6(3) =90For point D Z=9(1.33)+ 6(2.33) =25.95For point E Z=9(6)+ 6(0) =54For point F Z=9(8)+ 6(0) =72

Z maximizes at points B (0, 15), C (8, 3) = 90 (Ans.)

Q No. 6: Subject to 3x1 + 4x2 <=124x1 + 3x2 >=12(x1, x2 >=0)

Maximize z = 3x1 + x2 Solution:

3x1 - 4x2 + S1 =124x1 - 3x2 - E2 + A2 =12(x1, x2, S1, E2, A2 >=0)

Std. form z = 3x1 + x2 + 0S1 + 0E2 - MA2m= 2 n= 5n-m= 5-2 = 3{Non Basic Var.}

2{Basic Var.}Non Basic Variables = {x1, x2, E2}Basic Variables = {S1, A2}

Basic Variables

Z x1 x2 S1 E2 A2 Solution Ratio

Z 1 -3 -1 0 0 M 0

S1 0 3 4 1 0 0 12

A2 0 4 3 0 -1 1 12

* Key Col.

Basic Variables

Z x1 x2 S1 E2 A2 Solution Ratio

Z 1 -3-4M -1-3M 0 M 0 -12M

S1 0 3 4 1 0 0 12 4

A2 0 4 3 0 -1 1 12 3

Ro’ = Ro - MR2

R1’ = R1

R2’ = R2

*Key Col.

Basic Variables

Z x1 x2 S1 E2 A2 Solution Ratio

Z 1 0 5/4 0 -¾ ¾+M 9

S1 0 0 7/4 1 ¾ -¾ 3 4

x1 0 1 ¾ 0 ¼ ¼ 3

R2” = ¼R2’

R1” = R1’ - 3R2”

Ro” = Ro’ + (3+4M) R2”

Basic Variables

Z x1 x2 S1 E2 A2 Solution Ratio

Z 1 0 3 1 0 M 12

E2 0 0 7/3 4/3 1 -1 4

x1 0 1 4/3 1/3 0 0 4

R1”’ = 4/3 (R1”)

Ro”’ = Ro” + ¾(R1”’)

R2”’ = R2” + ¼(R1”’)

Z maximizes at 12 where x1 = 4, x2 = 0, S1 = 0, E2 = 4, A2 = 0 (Ans.)

Q No. 7: Subject to x1 - x2 >=123x1 - x2 + 3x3 <=12(x1, x2, x3 >=0)

Maximize z = 14x1 + 2x2 + x3 Solution:

x1 - x2 -E1 + A1 =23x1 - x2 + 3x3 +S2 =10(x1, x2, x3, E1, A1, S2 >=0)

Std. form z = 14x1 + 2x2 + x3 + 0E1 - MA1 + 0S2m= 2n= 6n-m = 6 – 2 = 4{Non Basic Var.}

2{Basic Var.}Non Basic Variables = {x1, x2, x3, E1}Basic Variables = {A1, S2}

Basic Variables

Z x1 x2 x3 E1 A1 S2 Solution Ratio

Z 1 -14 -2 -1 0 M 0 0

A1 0 1 -1 0 -1 1 0 2

S2 0 3 -1 3 0 0 1 10

*Key Col.

Basic Variables

Z x1 x2 x3 E1 A1 S2 Solution Ratio

Z 1 -14-M -2+M -1 M 0 0 -2M

A1 0 1 -1 0 -1 1 0 2 2

S2 0 3 -1 3 0 0 1 10 3.33

Ro’ = Ro – MR1

R1’ = R1

R2’ = R2

*Key Col.

Basic Variables

Z x1 x2 x3 E1 A1 S2 Solution Ratio

Z 0 0 -16 -1 -14 14+M

0 28

x1 0 1 -1 0 -1 1 0 2

S2 0 0 2 3 3 -3 1 4 2Ro” = Ro’ + (14+M) R1’

R1” = R1’

R2” = R2’ – 3 R1”

Basic Variables

Z x1 x2 x3 E1 A1 S2 Solution Ratio

Z 0 0 0 23 10 -10+M 8 60

x1 0 1 0 3/2 ½ -½ ½ 4

x2 0 0 1 3/2 3/2 -3/2 ½ 2

R2”’ = ½ R2”

R1”’ = R1” + R2”’

Ro”’ = Ro” + 16 R2”’

{Z maximizes at 60 where x1 = 4, x2 = 2, E1 = 0, A1 = 4, S2 = 0}

(Ans.)

Q No. 8: Subject to x1 - 3x2 >=6x1 + x2 <=6(x1, x2 >=0)

Maximize z = 10x1 + 5x2 Solution:

x1 - 3x2 -E1 + A1 =6x1 + x2 +S2 =10(x1, x2, E1, A1, S2 >=0)

Std. form z = 10x1 + 5x2 + 0E1 - MA1 + 0S2m= 2n= 5n-m = 5 – 2 = 3{Non Basic Var.}

2{Basic Var.}Non Basic Variables = {x1, x2, E1}Basic Variables = {A1, S2}

Basic Variables

Z x1 x2 E1 A1 S2 Solution Ratio

Z 1 -10 -5 0 M 0 0

A1 0 1 -3 -1 1 0 6

S2 0 1 1 0 0 1 6

*Key Col.

Basic Variables

Z x1 x2 E1 A1 S2 Solution Ratio

Z 1 -10-M -5+3M M 0 0 -6M

A1 0 1 -3 -1 1 0 6

S2 0 1 1 0 0 1 6 6

Ro’ = Ro – MR1

R1’ = R1

R2’ = R2

*Key Col.

Basic Variables

Z x1 x2 E1 A1 S2 Solution Ratio

Z 1 -5-4M 0 M 0 5-3M 30-24M

A1 0 4 0 -1 1 3 24 6

x2 0 1 1 0 0 1 6 6Ro” = Ro’ + (5-

3M) R2”

R1” = R1’ +3 R2’

R2” = R2’

Basic Variables

Z x1 x2 E1 A1 S2 Solution Ratio

Z 0 0 0 -5/4 5/4 + M 35/4 60

x1 0 1 0 - ¼ ¼ ¾ 6

x2 0 0 -1 - ¼ ¼ -¼ 0

R1”’ = (¼) R1”

R2”’ = – R2”+ R1”’

Ro”’ = Ro” + (5+4M) R1”’

{Negative Ratios so we can’t go further}No feasible solution exists.{Z maximizes at 60 where x1 = 6, x2 = 0, E1 = 0, A1 = 0, S2 = 0} (Ans.)

Q No. 9:Subject to 20x1 + 30x2 >= 100

20x1 + 10x2 >= 80(x1, x2 >=0)

Maximize z = 15x1 + 10x2

Solution:20x1 + 30x2 - E1 + A2 = 10020x1 + 10x2 - E2 + A2= 80

(x1, x2, E1, E2, A1, A2 >=0)Std. form z = 15x1 + 10x2 + 0E1 + 0E2 – MA1 – MA2

m= 2n= 6n-m= 6-2 = 4{Non Basic Var.}

2{Basic Var.}Non Basic Variables = {x1, x2, E1, E2}Basic Variables = {A1, A2}

Basic Variables

Z x1 x2 E1 E2 A1 A2 Solution Ratio

Z 1 -15 -10 0 0 M M 0

A1 0 20 30 -1 0 1 0 100

A2 0 20 10 0 -1 0 1 80

*Key Col.

Basic Variables

Z x1 x2 E1 E2 A1 A2 Solution Ratio

Z 1-M -15-M -10-M -M -M 0 0 -M

A1 0 20 30 -1 0 1 0 100 5

A2 0 20 10 0 -1 0 1 80 4

Ro’ = Ro – MR1’ = R1

R2’ = R2

*Key Col.

Basic Variables

Z x1 x2 E1 E2 A1 A2 Solution Ratio

Z 1-M 0 -(5+M)/2 -M -(15+21M)/20 0 (15+M)/20 60+3M

A1 0 0 -20 1 -1 -1 1 -20 20

x1 0 1 1/2 0 -1/20 0 1/20 4

R2” = (1/20) R2’

R1” = – R1’ + 20

R2”

Ro” = Ro’ + (15+M) R2”Basic

Variables

Z x1 x2 E1 E2 A1 A2 Solution Ratio

Z 1-M 0 (25+41M)/2 -(15+41M)/20 0 (15+21M)/20 -M 75+24M

E2 0 0 20 -1 1 1 -1 20

x1 0 1 3/2 -1/20 0 1/20 0 5

R1”’ = - (R1”)

R2”’ = R2” + (1/20) R1”’

Ro”’ = Ro” + {(15+21M)/20} R1”’

{Negative Ratios so we can’t go further}If we put M = 100 thenZ maximizes at 2475 where

x1 = 5, x2 = 0, E1 = 0, E2 = 20, A1 = 0, A2 = 0 (Ans.)

Q No. 10: Subject to 2x1 + x2 + x3 <=23x1 + 4x2 + 2x3 >=8(x1, x2, x3 >=0)

Maximize z = 3x1 + 2x2 + 3x3Solution:

2x1 + x2 + x3 + S1 =23x1 + 4x2 + 2x3 - E2 + A2 =10(x1, x2, x3, S1, E2, A2 >=0)

Std. form z = 3x1 + 2x2 + 3x3 + 0S1 + 0E2 - MA2 m= 2n= 6n-m = 6 – 2 = 4{Non Basic Var.}

2{Basic Var.}Non Basic Variables = {x1, x2, x3, E2}Basic Variables = {S1, A2}

Basic Variables

Z x1 x2 x3 S1 E2 A2 Solution Ratio

Z 1 -3 -2 -3 0 0 M 0

S1 0 2 1 1 1 0 0 2

A2 0 3 4 2 0 -1 1 8

*Key Col.

Basic Variables

Z x1 x2 x3 S1 E2 A2 Solution Ratio

Z 1 -3M-3 -4M-2 -2M-3 0 M 0 -8M

S1 0 2 1 1 1 0 0 2 2

A2 0 3 4 2 0 -1 1 8 2

Ro’ = Ro – MR2

R1’ = R1

R2’ = R2

Basic Variables

Z x1 x2 x3 S1 E2 A2 Solution Ratio

Z 1 5M+1 0 2M-1 4M+2 M 0 4

x2 0 2 1 1 1 0 0 2

A2 0 -5 0 -2 -4 -1 1 0

R1” = R1’

R2” = R2’ – 4R1”

Ro” = Ro’ + (4M+2) R1”

Z maximizes at 4 where

x1 = 0, x2 = 2, x3 = 0, S1 = 0, E2 = 0, A2 = 0 (Ans.)