Jordan

SOUTHERN UNIVERSITY BANGLADESH

An Assignment on Managerial Economics

Topic: Linear Programming

Submitted To:

M. A. Bari

Professor

Business Administration

Submitted By:

1. Marzina Zaman Tania 027-32-49

2. Md. Shiduzzaman 136-31-16(M)

3. Md. Main Uddin 027-32-38

4. Shamar kanti Barua 027-32-15

5.

Date of Submission: 12th August, 2011

Assignment Problem:

Maximize π = 3x+ 6y

Subject to,

20x + 50y ≤ 3300

4x + 3y ≤ 380

With x ≥ 0 y ≥ 0

Solution:

We are given,

Maximize π = 3x+ 6y

Subject to,

20x + 50y ≤ 3300

4x + 3y ≤ 380

With x ≥ 0 y ≥ 0

Addition of slack Variable (si) to the inequality constraint leads to the following system of equation.

20x + 50y + S1 = 3300

4x + 3y + S2 = 380

If x=0 and y=0 then S1=3300, S2=380 and π=o.

Further we can write as

π = 0 + 3x+ 6y

S1 = 3300 – 20x – 50y

S2 = 380- 4x- 3y

Now the initial basic feasible solution can be expressed in matrix form.

Basic Variable | Solution Variable | Non- basic Variable | Minimum Ratio |

| | x | y | |

x | 0 | 3 | 6 | 0 |

S1 | 3300 | -20 | -50 | 66 |

S2 | 380 | -4 | -3 | 95 |

Considering the column of non –basic variable element 6 in the firsy row under column y is the largest number.

Hence, y column is the pivot column dividing each number of the solution column by the absolute value of the corresponding negative number of pivot column. The minimum ratio indicates the pivot row. In this matrix s1 is the pivot row. The variable s1 will leave the basis and be replaces by x. The reciprocal of the pivot element will be the entry in new matrix.

The reciprocal of -50 is 1-50 . The entire new row will be like this

[3300|-50|...