Class 5.pptx

7/31/2019 Class 5.pptx

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Linear Programming (Graphical Method)

Linear programming problems with two decision variables can beeasily solved by graphical method.

Feasible Region

It is the collection of all feasible solutions. In the following

figure, the shaded area represents the feasible region.


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Convex Set

It is a collection of points such that for any two points onthe set, the line joining the points belongs to the set. In the

following figure, the line joining P and Q belongs entirely in R.

Thus, the collection of feasible solutions in a linear

programming problem form a convex set.


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Extreme Point

Extreme points are referred to as vertices or corner points.

In the following figure, P, Q, R and S are extreme points.


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Problem

A factory produces two types of raw mortar i.e. lean mix mortar

and rich mix mortar. Two basic materials, Cement and Sand are

used to produce the mixes. The maximum availability of cement is

800 cu.ft a day; that of sand is 3000 cu.ft a day.

The requirement of cement and sand per cu.ft of rich and lean

mix is given as under:

A market survey has established that the daily demand for the

lean mix does not exceed that of rich mix by more than 1000

cu.ft. The maximum demand for lean mix is limited to 1200 cu.ft

Rich Mix Lean Mix

Price in Rs. /

(cu.ft)

500 300

Cement (cu.

ft)

0.3 0.2

Sand (cu.ft) 1.0 1.0


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Decision Variables

1. Rich Mix produced daily = x1

1. Lean Mix produced daily = x2

. Objective Function

Z = 500 x1 + 300 x2

.

Constraints

. 0.3 x1 + 0.2x2 800 (Cement). x1 + x2 3000 (Sand). x2 x1 1000 ( relative diff. of lean and

rich).


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Constraint Equations

0 1000 2000 3000 4000 5000

0

1000

2000

3000

4000

5000

0.3x1 +0.2 x2

800

x

1

x

2

0 1000 2000 3000 4000 5000

0

1000

2000

3000

4000

5000

0 1000 2000 3000 4000 5000

0

1000

2000

3000

4000

5000

0 1000 2000 3000 4000 5000

0

1000

2000

3000

4000

5000

x

1


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Constraint Equations

x2

x1


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x

2

x

1

Z = 500 x1 +

300 x2

Lines for different values of Z are drawn parallel to Z

line passing through origin O which has beeni

-300

500

Z-Line

O

AB C

D

E


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For different values of decission variables, the values

obtained for Z are given in following table.

Z = 500 x1 + 300 x2 = 0, giving x1/x2 = -300

/ 500Cornerx1

x2 Z

Origin 0 0 0

A 0 1000 300000

B 150 1200 435000

C 1800 1200 1260000

D 2000 1000 1300000

E 2667 0 1333500

Class 5.pptx

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