7490.lesson07simplexmethodctnd

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Unit 1

Lesson 7: Simplex Method Ctnd.

Learning outcomes:

Set up and solve LP problems with simplex tableau. Interpret the meaning of every number in a simplex tableau.

Dear students, today we discuss small cases or case-lets on the above topic. The

situations presented below are real business problems, modified slightly to suit ourpurpose.

We start now.

Case-let-1

A firm makes air coolers of three types and markets these under the brand name

Symphony.

The relevant details are as follows:-

Profit / unit

(Rs.)

300 Product A

(hrs. / unit)

700 Product B

(hrs. / unit)

900 Product C

(hrs,/ unit)

Totoal hrs.

available

Designing 0 10 20 320

Manufacturing 60 90 120 1600

Painting 30 40 60 1120

What Qty. of each product must be made to maximize the total profit of the firm?

Solution

Let X1, X2, X3 denote the Qty. of each product made

Then: Maximize: 300 X1+ 700 X2 +900 X3

Subject to: 0 X1 +10 X2 + 20 X3 320

60 X1 +90 X2 + 120 X3 1600


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30 X1 +40 X2 + 60 X3 1120

Introducing slacks:

Maximize: 300 X1 +700 X2 + 900 X3 +0S1+ 0S2+ 0S3

Subject to: 0 X1 +10 X2 + 20 X3 +S1+ 0S2+ 0S3 =32060 X1 +90 X2 + 120 X3 +0S1+ S2+ 0S3 =1600

30 X1 +40 X2 + 60 X3 +0S1+ 0S2+ S3 =1120

1st

Tableau

CJ(Rs.) 300 700 900 0 0 0

(Rs.) Basic

Act.

Qty. X1 X2 X3 S1 S2 S3

0 S1 320 0 10 20 1 0 00 S2 1600 60 90 120 0 1 0

0 S3 1120 30 40 60 0 0 1ZJ 0 0 0 0 0 0 0

CJ - ZJ 300 700 900 0 0 0

Pivot column = X3

Pivot row = S2

===

3

56

60

1120:,

3

40

120

1600:,16

20

320:S 321 SS

Pivot element = 120

Pivot row updating : 0,120

1,0,1,

4

3

120

90,

2

1

120

60,

3

40

120

1600===

Up-dating S1 row: Updating S3 row:

3

160

3

4020320 =

x 320

3

4060

20

1=

x

102

11200 =

X 0

2

16030 =

X

54

32020 =

x 54

36040 =

X

( ) 50201 = x ( ) 00600 = x

6

1

120

1200

=

x

2

1

120

1600

=

x

( ) 00201 = x ( ) 10601 = x


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IInd Tableau

Cj (Rs.)(Rs.)

Basic Act. Qty. 300x1

700x2

900X3

0S1

0S2

0S3

0

9000

S1

X3S3Zj (Rs.)Cj-Zj

(Rs.)

160/3

40/3320

2000

-10

0

450

-150

-5

-5

675

25

0

10

900

0

1

00

0

0

-1/6

1/120-1/2

15/2

-15/2

0

01

0

0

Since, There is still one + ve term in row, pivoting is requd.

Pivot Column = x2

Pivot row =

== .)(

5

320:,

9

160

4

3

3

40:),3(5

3

160: 3313 NotconsiSxdconsidereNotSx

Pivot element =3/4

Pivot-row updating:

0,90

1

4

3

120

1,0,

3

4

4

31,1,

3

2

4

3

2

1,

9

160

4

3

3

40====

Updating S1 row: Updating S3 row:

9

1280

9

1605

3

160 =

x

9

3680

9

1605320 =

x

3

20

3

2510 =

x

3

10

3

25 =

x0

( ) 0155 = x ( ) 0155 = x

3

20

3

450 =

x

3

20

3

45 =

x0

( ) 1051 = x ( ) 0050 = x

9

1

90

15

6

1 =

x

9

4

90

15

6

1 =

x

( ) 0050 = x ( ) 1051 = x


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IIIrd Tableau

Cj (Rs.)(Rs.) Basic Act. Qty. 300x1700x2

900x30S1

0S20S3

0700

0

x1S2

S3Zj (Rs.)

Cj-Zj(Rs.)

1280/9160/9

3680/9

112000

-20/302/3

10/3

1400/3

-500/3

0

10

700

0

20/3

4/320/3

2800/3

-100/3

1-

00

0

0

1/9

1-90-4/9

70/9

-70/9

0

01

0

0

Since Cj Zj row has no. + ve value left.

It is the optional soln.

Case-let-II

Two materials A & B are required to construct table & book cases. For one table, 12 units

of A & 16 units of B are required while for a book case 16 units of A and 8 units of B arerequired. The profit on book case is Rs 25 and Rs 20 on a table. 100 units of A & 80 units

of B are available. Formulate as a Linear programming problem & determine the optimal

number of book cases & tables to be produced so as to maximise the profits.

Solution

Let x1 = no. of unit of tables to be produced, and

X2 = no. of unit of book cases to be produced.

Formulating as a LP problem, we have:-

Maximise Z ( objective function )21 2520 xx +=

Subject to the constraints: 1001612 21 + xx

16 800

00

80

8 21+

xx, 21 xx

Introducing the slack variable, we have:

12 116 121 ++ Sxx

16 8 221 ++ Sxx


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Re-writing as:

=

+

+

+

80

100

100

1

0

1

8

16

16

122121 SSxx

or PoSPSPxPxP =+++ 24132211

our problem becomes MaximiseSubject to 20F 2121 0025 SSxx +++= (1)

Where P PoSPSPxPx =+++ (ii)24132211

=

=

+

+

8

16

16

13

0

1

0

1

80

10022420 PPPPP

element of Cj row are values of Po, P2,P4,P1, in (i) by comparing with (2) i.e. elements of

Cj are 0, 0, 20, 25.

Simplex Method

Cj 0 0 0 20 25 Ratio

Vectors Po P2 P4 P1 P30 P3 100 1 0 12 16 100/16=6.25

0 P4 80 0 1 16 0 80/8=10

Zj 0 0 0 0 0 6.25 is least ratio

Zj Cj

0 0 0 20 25 replaced vector is P2

R1 is

16

1R

R2 = R2 R1 x 8/16

Since 25 is the most postiveNo. in row C4 - Z4Replacing vector is P2

25 P2

4

25

1/16 0

4

3

1

4

25/

4

5=

3

25=8.33

0 P4 30

3

1

1 10 0 30/10=3

Zj

4

625

16

25

0

4

75

25

R1

R2Stage I

R1

R2

Stage I

Zj

Cj 4

625

16

25

0

4

5

0

3 is least ratio replaced vector is P4

4

5 is least ve no.

replaced vector is P1

10

1

2

R

isR

R1 = R R240

3

10

4/321 RR =

The elements of column Cj are values of P3 and P4 is (I) as compared with (2). The

column P0,P2,P4,P1,P3 are values of P0,P2,P1 etc. in (3)


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R1 25 P2 4 1/10 /3/40 0 1

R2 20 P4 3 -1/20 1 1 0

Stage I Zj 160 1.5 1/8 20 25

Zj Cj 160 1.5 1/8 0 0

Since in stage III all elements of row Zj Cj are +ve or zero, hence an optimal solutionhas been achieved.

The solution is given by column is Stage III.

210 43 PPP +=

compare it with 024132211 PSPSPxPxP =+++

0,0,4,3 2121 ==== SSxx

Note I. the elements of row Zj is the sum of product of corresponding elements of column

Cj with Po,P3,P4,P1,P2 respectively.For example in stage II, the elements of row Zj are

4

625300

4

25=+ xx25

16

25

2

10

16

1

=+ xx25

25x 0+0 x 1=0

4

75100

4

3=+ xx25

25 x 1 + 0x 0 =25.

Note 2. When we replace vector P2 in place of P3 in stage II, the elements to left of P3

under column Cj is also changed.Note. To check the results.

At x=3, x2 = 4

Constraints are 10010064361612 21 =+=+ xx

16 808032488 21 =+=+ xx

Case-let-III

A manufacturer produces childrens bicycles & scooters both of which one processed

through two machines. Machine 1 has a maximum of 120 hours available & machine 2has a maximum of 180 hours. Manufacturing a bicycle requires 6 hours on machine 1 x 4hours on machine 2. A scooter requires 3 hours on machine 1 & ten hours on machine 2.

If the profit is Rs 45 on a bicycle & Rs. 55 on a scooter determine the number of bicycles& scooters that should be produced in order to maximise profit.


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Solution

Let no. of bicycles and scooters produced by x and y units respectively.

L.P. problem is Maximize P = 45x + 55ySubject to 12036 + yx

4 0,0,18010 + yxyx

Introducing the slack variables, we get

12036 12 +++ Syx

180104 22 +++ Syx

writing it in vector form

(1)

=

+

+

+

180

120

1

0

0

1

4

3

4

621 SSyx

or P1x+P2y+p2S1+P4S2=P0

Our L.P.P is Max, P=45x+55y+OS1+OS2 Subject to P1x + P2y+P3S1 + P4S2=Po (2)Comparing (1) and (3), we have

=

=

=

1

0,

0

1,

10

3'

4

64321 PPPP

'180

120

=oP

The elements of row are value of

Po, P2,P4,P1,P3, in (1) by comparing it with (2)

i.e., element of row are o, o, 45,45.,55.Note. The element of row are sum of product of corresponding element of columns andcolumn vectors

Simplex Method

Cj 0 0 0 45 55

Vectors Po P3 P4 P1 P2

0 P3 120 1 0 6 3

0 P4 180 0 1 4 10

Zj 0 0 0 0 0

R1

R2

Stage I

Zj-Cj 0 0 0 -45 -55

Ratio

454/18041/40 == 20 is least ratio, so

replace vector is P3

- 55 is least no. in rowZj-Cj, replacing vector

is P1

Row6

1

31

11

R

a

RR =


8/10

Row6

412

31

41122 RR

a

aR=RR .

Further calculations are left to the students as an exercise.

Case-let-IV

.A firm has the following availabilities :

Type-available Amount-available (kg)

Wood

Plastic

Steel

240

370

180

The firm produces two products A & B having a selling price of Rs. 4 per unit & Rs. 6

per unit respectively. The requirements for the manufacture of A & B are as follows:

Requirements of (kg)Product

Wood Plastic Steel

A

B

1

3

3

4

2

1

Formulate as a LP problem & solve by using the simplex method to maximise the gross

income of the firm.

Case-let-V

Ace- advantage Ltd. faces the following situation:Media available - electronic (A) & print (B)Cost of available in - media A: Rs.1000

Media B: Rs.1500Annual advertising budget - Rs. 20000

The following constraints are applicable:

Electronic Media (A) can not have more than 12 advertisements in a yearand not less than 5 advertisement must be placed in the print media (B).

The estimated audience are as follows:Electronic media (A) - 40000Print media (B) - 55000

You are required to develop a mathematical model & solve it formaximizing the total effective audience.


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Case-let-VI

Khalifa & sons sells two different books B1 & B2 at a profit

margin of Rs. 7 and Rs. 5 per book respectively. B1 requires 5units of raw material & B2 requires 1 units of raw material. The

maximum availability of raw materials is limited to 15units. To

maintain the high quality of books, it is desired to follow the given

quality constraint: 3x1 + 7x2 21. FormulatingAs a LP model determine the optimal solution.

Dear students, we have now reached the end of our discussion scheduled for today.See you all in the next lecture.

Bye.


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7490.lesson07simplexmethodctnd

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