Linear Programming Problem - Management Science - Lecture Slides, Slides of Business Management and Analysis

This lecture is from Management Science. Key important points are: Linear Programming Problem, Hours Required, Linear Programming, Simplex Techniques, Optimum Solution, Decision Variable, Chairs Produced, Tables Produced, Objective Function, First Constraint

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2012/2013

Uploaded on 01/29/2013

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Linear Programming Problem
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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Linear Programming Problem

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

M 1 6 6 72

M 2 10 4 100

M 3 4 12 120

Formulate the above problem into

Linear Programming Problem.

Use graphical as well as Simplex

Techniques to arrive at Optimum

Solution

(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.

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 ≤ 72 10x1 + 04x2 ≤ 100 04x1 + 12x2 ≤ 120

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 + 300x

Subject to the Constraints,

06x1 + 06x2 ≤ 72

10x1 + 04x2 ≤ 100

04x1 + 12x2 ≤ 120

and x1; x2 ≥ 0

SIMPLEX ALGORITHIM:

Introducing the three slack

variables s 1 , s 2 , and s 3 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 + 0s

Design the initial Feasible

Solution : An initial basic feasible

solution is obtained by putting

x1 = x2 = 0

Thus we get,

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

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.

*** Incoming Column

** Outgoing Row

* Key Element

Compute the new key row values by using the formula

New Table Key row values

= Old Table Key row values / Key Element

New Table Key row values

Computing all other row values using the formula

New 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.

  • (60 2 6 0 0 1/2)

New Second Row Values = Old Second Row Values

- 4 x Corresponding New Table Key row values.

  • (40 4/3 4 0 0 1/3)

Compute the new key row values by using the formula

New Table Key row values

= Old Table Key row values / Key Element

New 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 formula

New Table Row values=Old row values-Corresponding

Coefficient in Key Column x Corresponding new table key

row value

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)

The above simplex table – 3 yields

the optimum solution and it is x1 =

3 and x2 = 9 with maximum

Z = c1 x1 + c2 x

= 200 x 3 + 300 x 9