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Presented By: Saba Arif Rao M Nasir Bilal Ahmed Toor Akif Jamal
8/8/2019 Linear Programming (2)
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Presented By:
Saba Arif
Rao M Nasir
Bilal Ahmed Toor
Akif Jamal
8/8/2019 Linear Programming (2)
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What Is Linear Programming?
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Why it is important?
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Properties of Linear Programming solution:
Feasible Solution
Optimal solution
Alternate Optimal solution
Unbounded solution
Infeasible solution
Degenerate solution
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Linear Programming Methods:
Simplex Method.
Big M Method.
Dual Simplex Method.
Two-phase Method.
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SITUATION
Mr. Faraz Ahmed has Rs.10,000,000 to invest in
several alternative;
National Saving Certificate with an (9%)
Defense Saving with an (11.5%) Advertising Agency with an (13%)
Sports Goods with an (8%)
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GUIDELINES:
(1) No more than 25% of the total investment should
be in advertising agency.
(2) At least 30% of the investment should be inNational Saving Certificate & Defense Saving .
(3) The amount invested in advertising agency
should not exceed the amount invested in the
other three alternatives.
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DECISION VARIABLES:
There are four decision variables in this situation
representing the monitory amount invested in each
invested alternatives. X1= the amount invested in national saving
certificate.
X2= the amount invested in defense saving.
X3= the amount invested in advertising agency. X4= the amount invested in sports goods.
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MODEL CONSTRAINTS:
(1) No more than 25% of the total investment should
be in advertising agency.
X3 < 2,500,000
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(2) At least 30% of the investment should be in
National Saving Certificate & Defense Saving .
X1+ X2 > 3,000,000
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(3) The amount invested in advertising agency
should not exceed the amount invested in the
other three alternatives.
X3 < X1 + X2 + X4
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Finally Mr. Faraz Ahmed desire to invest all Rs.
10,000,000 in the four alternatives.
X1 + X2 + X3 + X4
= 10,000,000
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The complete LP model for this situation can be
summarized as;
Maximize Z= (0.09)(X1) + (0.115)(X2) + (0.13)(X3) + (0.08)(X4)
Subject to;X3 < 2,500,000
X1+ X2 > 3,000,000
X3 < X1 + X2 + X4
X1 + X2 + X3 + X4 = 10,000,000X1 , X2 , X3 , X4 >0
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After putting all required values in linear programming
software then,
X1 X2 X3 X4 R.H.S Dual
Maximize 0.09 0.115 0.13 0.08
Constraint 1 0 0 1 0 <= 2,500,000 0.015
Constrain 2 1 1 0 0 >= 3,000,000 0
Constrain 3 -1 -1 1 -1 <= 0 0
Constrain 4 1 1 1 1 = 10,000,000 0.115
solution 0 7,500,000 2,500,000 0 1,187,500
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CONCLUSION:
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THANK YOU
