6/11/2014 Question: XYZ Furniture Ltd. is involved in producing Chairs and Tables. The firm makes a profit of Rs. 200 per chair and Rs. 300 per table.

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Question: XYZ Furniture Ltd. is involved in producing Chairs and Tables. The firm makes a profit of Rs. 200 per chair and Rs. 300 per table. Each of these items is produced using three machines M1, M2 and M3. The labour hours required on each of these machines are as follows:

Machine

Hours Required

Available Hour/Week

Chairs Table

M1 6 6 72

M2 10 4 100

M3 4 12 120

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Formulate the above problem into Linear Programming Problem.

Use graphical as well as Simplex Techniques to arrive at Optimum Solution

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Solution:

a) Formulation Of Linear Programming Problem:

Decision Variable: let there be two decision variables x1 and x2, such thatx1 = number of chairs produced andx2 = number of tables produced.

Objective Function: the objective of the producer is to determine the the number of chairs and tables produced, so as to maximize the total profit, ie.,Z = 200x1 + 300x2

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(iii) Constraints:

First Constraint: On machine M1, each chair requires 6 hours and each table requires 6 hours. The total of working hours is given by 6x1 + 6x2.

Second Constraint: On machine M2, each chair requires 10 hours and each table requires 4 hours. The total of working hours is given by 10x1 + 4x2.

Third constraint: On machine M1, each chair requires 4 hours and each table requires 12 hours. The total of working hours is given by 4x1 + 12x2.

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Further, since the manufacturer does not have more then 72 hours of working on machine M1, does not have more then 100 hours of working on machine M2 and does not have more then 120 hours of working on machine M3. So that the constraint are as follows:

06x1 + 06x2 ≤ 7210x1 + 04x2 ≤ 10004x1 + 12x2 ≤ 120

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ivNon - Negativity Constraint:Since the production process can never be negative, we must have x1 ≥ 0 and x2 ≥ 0.

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Hence the manufacturer’s allocation problem can be put in the following mathematical form:Find two real numbers x1 and x2, such that to maximize the expression (Objective Function)

Z = 200x1 + 300x2Subject to the Constraints,06x1 + 06x2 ≤ 7210x1 + 04x2 ≤ 10004x1 + 12x2 ≤ 120and x1; x2 ≥ 0

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SIMPLEX ALGORITHIM:Introducing the three slack variables s1, s2, and s3 to the left hand side of the three constraint inequalities to convert them into equality and assign a zero coefficient to these in the objective function:Z = 200x1 + 300x2 + 0s1 + 0s2 + 0s3

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Subject to the constraints,06x1 + 06x2 + s1 = 72

10x1 + 04x2 + s2 = 100

04x1 + 12x2 + s3 = 120

And x1,x2, s1, s2 ands3 ≥ 0.

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Design the initial Feasible Solution: An initial basic feasible solution is obtained by putting

x1 = x2 = 0Thus we get,

s1 = 72, s2 = 100, s3 = 120.

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Setup the initial simplex table: For computational efficiency and simplicity, initial basic feasible solution, the constraints of the standard Linear programming problem as well as the objective function can be displayed in the tabular form, called Simplex Table.

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Cj ( contribution/unit) 200 300 0 0 0

Minimum ratioContribut

ion of basic

variable/unit

Basic Varia

ble

Value of the Basic

VariableCoefficient Matrix Identify Matrix

CB B b (= XB) x 1 x 2 s1 s2 s3 bj / xj

0 s1 b1 = 72 6 6 1 0 0 72/6 = 12

0 s2 b2 =100 10 4 0 1 0 100/4 = 25

0 s3 b3 = 120 4 12 * 0 0 1 120/12 = 10 →**

Contribution Loss per unit

Zj =

Σ CBj aij = 0 0 0 0 0 0

Net Contribution per unit

Cj-Zj 200300

↑***0 0 0

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*** Incoming Column** Outgoing Row* Key Element

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Compute the new key row values by using the formulaNew Table Key row values = Old Table Key row values / Key ElementNew Table Key row values

= 1/12 (120 4 12 0 0 1)= (120/12 4/12 12/12 0/12 0/12 1/12)= (10 1/3 1 0 0 1/12)

Computing all other row values using the formulaNew Table Row values=Old row values-Corresponding Coefficient in Key ColumnxCorresponding new table key row valueNew First Row Values = Old First Row Values -

6 x Corresponding New Table Key row values.= (72 6 6 1 0 0) - (60 2 6 0 0 1/2)= (12 4 0 1 0 -1/2)

New Second Row Values = Old Second Row Values- 4 x Corresponding New Table Key row values.

= (100 10 4 0 1 0) - (40 4/3 4 0 0 1/3)= (60 26/3 0 0 1 -1/3)

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Cj ( contribution/unit) 200 300 0 0 0

Minimum ratio

Contribution

of basic

variable/unit

Basic Variabl

e

Value of the Basic

VariableCoefficient Matrix Identify Matrix

CB B b (= XB) x 1 x 2 s1 s2 s3 b j / xj

0 s1 b1 = 12 4 * 0 1 0 -1/2 12/4 = 3 **

0 s2 b2 =60 26/3 0 0 1 -1/3 60/(26/3) = 90/13

300 x2 b3 = 10 1/3 1 0 0 1/12 10/(1/3) = 30

Contribution Loss per unit

Zj =

Σ C Bj aij =

3000

100 300 0 0 25

Net Contribution per unit

Cj-Zj100

↑***0 0 0 25

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Compute the new key row values by using the formulaNew Table Key row values

= Old Table Key row values / Key ElementNew Table Key row values

= 1/4(12 4 0 1 0 -1/2)= (12/4 4/4 0/4 1/4 0/4 -1/2 x 1/4)= (3 1 0 1/4 0 -1/8)

Computing all other row values using the formulaNew Table Row values=Old row values-Corresponding Coefficient in Key Column x Corresponding new table key row value

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New Second Row Values = Old Second Row Values

- 26/3 x Corresponding New Table Key row values.

= (60 26/3 0 0 1 -1/3)- (26 26/3 0 13/6 0 -13/12)= (34 0 0 -13/6 1 3/4)

New Third Row Values =Old Third Row Values – 1/3 x Corresponding

New Table Key row values= (10 1/3 1 0 0 1/12)- (1 1/3 0 1/12 0 -1/24)= (9 0 1 -1/12 0 1/8)

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Cj ( contribution/unit) 200 300 0 0 0

Contribution of basic

variable/unit

Basic Variable

Value of the Basic Variable

Coefficient Matrix Identify Matrix

CB B b (= XB) x 1 x 2 s1 s2 s3

200 x1 b1 = 3 1 0 ¼ 0 -1/8

0 s2 b2 =34 0 0 -13/6 1 3/4

300 x2 b3 = 9 0 1 -1/12 0 1/8

Contribution Loss per unitZj =

Σ CBj aij = 3300 200 300 25 0

Net Contribution per unit Zj - Cj 0 0 25 0

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The above simplex table – 3 yields the optimum solution and it is x1 = 3 and x2 = 9 with maximumZ = c1 x 1 + c2 x2

= 200 x 3 + 300 x 9 = 3300