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These are the important key points of lecture slides of Introduction to Operations Research are:Iteration, Extreme Point, Adjacent Extreme Point, Better Extreme Point, Two Hyperplane, One Hyperplane, Adjacent Extreme Point, Simplex Tableau, Functional Constraint, Decision Variable
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We are at an extreme point of the feasible region We want to move to an adjacent extreme point. We want to move to a better extreme point. Observation : A pair of basic feasible solutions which differ only in that a basic and non-basic variable are interchanged corresponds to adjacent feasible extreme points.
Two hyperplane (constraints) in common
Only one hyperplane in common
(adjacent)
(not adjacent)
It is convenient to describe how the Simplex Method works using a table (=tableau).
There are a number of different layouts for these tables.
All of us shall use the layout specified in the lecture notes.
It is convenient to incorporate the objective function into the formulation as a functional constraint.
We can do this by viewing z, the value of the objective function, as a decision variable , and introduce the additional constraint
z = Σj=1,...,n cjx (^) j
or equivalently
z - c 1 x 1 - c 2 x 2 - ... - cnx (^) n = 0
Terminology : We refer to the last row as the Z-row , and to the coefficient of x their as reduced costs. For example, the reduced cost of x 1 is −4.
BV Eq. # Z (^) x 1 x 2 x 3 x 4 x 5 RHS x 3 1 0 2 1 1 0 0 40 x 4 2 0 1 1 0 1 0 30 x 5 3 0 1 0 0 0 1 15 Z Z (^1) − 4 − 3 0 0 0 0
Issue :
Observation :
The Z-row tells us how the value of the objective function (Z) changes as we change the decision variables: z - c 1 x 1 - c 2 x 2 - ... - c (^) nxn = 0
Since we try to maximize the objective function, it would be better to select a non-basic variable with a large (positive) cost coefficient (large cj).
Thus, if we do the selection via the reduced costs, we will prefer a variable with a negative reduced cost.
If we maximize the objective function, to improve (increase) the value of the objective function we have to select a non-basic variable whose reduced cost is negative.
(Continued)
The most negative reduced cost in the Z-row is −4, corresponding to j=1. Thus, we select x 1 as the new basic variable.
BV Eq. # Z (^) x 1 x 2 x 3 x 4 x 5 RHS x 3 1 0 2 1 1 0 0 40 x 4 2 0 1 1 0 1 0 30 x 5 3 0 1 0 0 0 1 15 Z Z (^1) − 4 − 3 0 0 0 0
Step 2:
Suppose we decided to select x (^) j as a new basic variable.
Since the number of basic variables is fixed (m), we have to take one variable out of the basis.
Which one?
Suppose we select x 1 as the new basic variable.
Since x 2 is a nonbasic variable, its value is zero. Thus the above system can be simplified!
2 x 1 (^) + x 2 + x 3 = 40 x 1 + x 2 + x 4 = 30 x 1 + x 5 = 15
Each equation involves only two variables :
We can thus express the old basic variables in terms of the new one! x 3 = 40 - 2x 1 x 4 = 30 - x (^1) x 5 = 15 - x^1
thus the critical values are obtained from:
0 = 40 - 2x 1 (x 1 *=20) 0 = 30 - x 1 (x 1 *=30) 0 = 15 - x 1 (x 1 *=15)
Conclusions:
The critical value of x 1 is 15.
We take x 5 out of the basis.
If we select x (^) j as the new basic variable , then for each of the functional constraints we have
a (^) ij x (^) j + x (^) i = b (^) i (i=1,2,...,m)
where x (^) i is the old basic variable associated with constraint i.
