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## General Linear Programming Problem GLPP and Standard LPP

General Linear Programming Problem (GLPP)

Maximize / Minimize Z = c

_{1}x_{1}+ c_{2}x_{2}+ c_{3}x_{3}+.................+ c_{n}x_{n}Subject to constraints

a

_{11}x_{1}+ a_{12}x_{2}+ .............+a_{1n}x_{n}(≤ or ≥) b_{1}a

_{21}x_{1}+ a_{22}x_{2}+ ...........+a_{2n}x_{n}(≤ or ≥) b_{2}.

.

.

a

_{m1}x_{1}+ a_{m2}x_{2}+ ..........+a_{mn}x_{n}(≤ or ≥) b_{m}&

x

_{1}≥ 0, x_{2 }≥ 0,..., x_{n }≥ 0Where 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 x_{1}+2x_{2}≤ 12, we can write asx

_{1}+2x_{2}+ s_{1}= 12.o If the constraint is of '≥' type, we deduct a variable known as

Surplus variableand change it to an equation. For instance2x

_{1}+x_{2}≥ 152x

_{1}+x_{2}- s_{2}= 15Step 3- After that, the right side element of every constraint should be made non-negative2x

_{1}+x_{2}- s_{2}= -15-2x

_{1}- x_{2}+ s_{2}= 15 (Multiplying throughout by -1)Step 4- All variables should have non-negative values.For instance: x

_{1}+x_{2}≤ 3x

_{1 }> 0, x_{2 }is unrestricted in sign_{ }Then x

_{2}can be written as x_{2}= x_{2}? - x_{2}?? where x_{2}?, x_{2}??^{ }≥ 0Thus the inequality takes the shape of equation as x

_{1}+ (x_{2}? - x_{2}??) + s_{1}= 3Using all the above steps, we can change or write the GLPP in the form of SLPP.

Write the Standard LPP (SLPP) of the followingExample 1Maximize Z = 3x

_{1}+ x_{2}Subject to

2 x

_{1}+ x_{2 }≤ 23 x

_{1}+ 4 x_{2}≥ 12& x

_{1 }≥ 0, x_{2 }≥ 0SLPPMaximize Z = 3x

_{1}+ x_{2}Subject to

2 x

_{1}+ x_{2}+ s_{1}= 23 x

_{1}+ 4 x_{2}- s_{2}= 12x

_{1 }≥ 0, x_{2 }≥ 0, s_{1 }≥ 0, s_{2 }≥ 0Example 2Minimize Z = 4x

_{1}+ 2 x_{2}Subject to

3x

_{1}+ x_{2 }≥ 2x

_{1}+ x_{2}≥ 21x

_{1}+ 2x_{2}≥ 30& x

_{1 }≥ 0, x_{2 }≥ 0SLPPMaximize Z

^{?}= - 4x_{1}- 2 x_{2 }Subject to

3x

_{1}+ x_{2 }- s_{1}= 2x

_{1}+ x_{2}- s_{2}= 21x

_{1}+ 2x_{2}- s_{3}= 30x

_{1 }≥ 0, x_{2 }≥ 0, s_{1 }≥ 0, s_{2 }≥ 0, s_{3 }≥ 0Example 3Minimize Z = x

_{1}+ 2 x_{2 }+ 3x_{3}Subject to

2x

_{1}+ 3x_{2 }+ 3x_{3}≥ - 43x

_{1}+ 5x_{2}+ 2x_{3}≤ 7& x

_{1 }≥ 0, x_{2 }≥ 0, x_{3}is unrestricted in signSLPPMaximize Z

^{?}= - x_{1}- 2 x_{2 }- 3(x_{3}^{?}- x_{3}^{??})Subject to

-2x

_{1}- 3x_{2 }- 3(x_{3}^{?}- x_{3}^{??}) + s_{1}= 43x

_{1}+ 5x_{2}+ 2(x_{3}^{?}- x_{3}^{??}) + s_{2 }= 7x

_{1 }≥ 0, x_{2 }≥ 0, x_{3}^{?}≥ 0, x_{3}^{??}≥ 0, s_{1 }≥ 0, s_{2 }≥ 0Some fundamental DefinitionsSolution of LPPSolution of LPP is any set of variable (x

_{1}, x_{2}... x_{n}) which satisfies or fulfills the given constraint.Basic solutionIt is a solution achieved through setting any 'n' variable equivalent to zero and solving remaining 'm' variables. These 'm' variables are known as

Basic variablesand 'n' variables are known asNon-basic variables.Basic feasible solutionA 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 solutionIf 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 solutionIt is a basic feasible solution which has precisely 'm' positive x

_{i}, where i=1, 2, ... m. In other terms all 'm' basic variables are positive and left over 'n' variables are zero.Optimum basic feasible solutionA basic feasible solution is referred to be optimum when it optimizes (max / min) the objective function.

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