General Linear Programming Problem (GLPP)
Maximize / Minimize Z = c1x1 + c2x2 + c3x3 +.................+ cnxn
Subject to constraints
a11x1 + a12x2 + .............+a1nxn (≤ or ≥) b1
a21x1 + a22x2 + ...........+a2nxn (≤ or ≥) b2
.
am1x1 + am2x2 + ..........+amnxn (≤ or ≥) bm
&
x1 ≥ 0, x2 ≥ 0,..., xn ≥ 0
Where constraints may be in the terms of any inequality (≤ or ≥) or even in terms of an equation (=) and at the end fulfill the non-negativity restrictions.
Steps to change GLPP to SLPP (Standard LPP)
Step 1 - First of all, write the objective function in the maximization form. If the known objective function is in minimization form then multiply all through by -1 and write Max z? = Min (-z)
Step 2 - Then, convert all the inequalities as equations.
o If an equality of '≤' appears then add a variable called Slack variable. We can change it to an equation. For instance x1 +2x2 ≤ 12, we can write as
x1 +2x2 + s1 = 12.
o If the constraint is of '≥' type, we deduct a variable known as Surplus variable and change it to an equation. For instance
2x1 +x2 ≥ 15
2x1 +x2 - s2 = 15
Step 3 - After that, the right side element of every constraint should be made non-negative
2x1 +x2 - s2 = -15
-2x1 - x2 + s2 = 15 (Multiplying throughout by -1)
Step 4 - All variables should have non-negative values.
For instance: x1 +x2 ≤ 3
x1 > 0, x2 is unrestricted in sign
Then x2 can be written as x2 = x2? - x2?? where x2?, x2?? ≥ 0
Thus the inequality takes the shape of equation as x1 + (x2? - x2??) + s1 = 3
Using all the above steps, we can change or write the GLPP in the form of SLPP.
Write the Standard LPP (SLPP) of the following
Example 1
Maximize Z = 3x1 + x2
Subject to
2 x1 + x2 ≤ 2
3 x1 + 4 x2 ≥ 12
& x1 ≥ 0, x2 ≥ 0
SLPP
2 x1 + x2 + s1 = 2
3 x1 + 4 x2 - s2 = 12
x1 ≥ 0, x2 ≥ 0, s1 ≥ 0, s2 ≥ 0
Example 2
Minimize Z = 4x1 + 2 x2
3x1 + x2 ≥ 2
x1 + x2 ≥ 21
x1 + 2x2 ≥ 30
Maximize Z? = - 4x1 - 2 x2
3x1 + x2 - s1 = 2
x1 + x2 - s2 = 21
x1 + 2x2 - s3 = 30
x1 ≥ 0, x2 ≥ 0, s1 ≥ 0, s2 ≥ 0, s3 ≥ 0
Example 3
Minimize Z = x1 + 2 x2 + 3x3
2x1 + 3x2 + 3x3 ≥ - 4
3x1 + 5x2 + 2x3 ≤ 7
& x1 ≥ 0, x2 ≥ 0, x3 is unrestricted in sign
Maximize Z? = - x1 - 2 x2 - 3(x3? - x3??)
-2x1 - 3x2 - 3(x3? - x3??) + s1= 4
3x1 + 5x2 + 2(x3? - x3??) + s2 = 7
x1 ≥ 0, x2 ≥ 0, x3? ≥ 0, x3?? ≥ 0, s1 ≥ 0, s2 ≥ 0
Some fundamental Definitions
Solution of LPP
Solution of LPP is any set of variable (x1, x2... xn) which satisfies or fulfills the given constraint.
Basic solution
It is a solution achieved through setting any 'n' variable equivalent to zero and solving remaining 'm' variables. These 'm' variables are known as Basic variables and 'n' variables are known as Non-basic variables.
Basic feasible solution
A basic solution that is feasible or possible (all basic variables are non negative) is called basic feasible solution. There are two kinds of basic feasible solution.
1. Degenerate basic feasible solution
If any of the basic variable of a basic feasible solution is zero (0) than it is referred to as degenerate basic feasible solution.
2. Non-degenerate basic feasible solution
It is a basic feasible solution which has precisely 'm' positive xi, where i=1, 2, ... m. In other terms all 'm' basic variables are positive and left over 'n' variables are zero.
Optimum basic feasible solution
A basic feasible solution is referred to be optimum when it optimizes (max / min) the objective function.
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