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Unformatted text preview: ith this notation, Equation (9.5) can now be
written as x4 ≥ 0. Similarly, we introduce the variables x5 and x6 for the two other constraints of
Problem (9.4). Finally, we use the classic notation z for the objective function 5x1 + 4x2 + 3x3 .
To summarize, for all choices of x1 , x2 , x3 we deﬁne x4 , x5 , x6 and z by the formulas
x4
x5
x6
z = 5 − 2x1
= 11 − 4x1
= 8 − 3x1
=
5x1 − 3x2
− x2
− 4x2
+ 4x2 −
x3
− 2x3
− 2x3
+ 3x3 . (9.6) With these notations, the problem can be written as:
Maximize z subject to x1 , x2 , x3 , x4 , x5 , x6 ≥ 0. (9.7) The new variables that were introduced are referred as slack variables, when the initial
variables are usually called the decision variables. It is important to note that Equation (9.6)
deﬁne an equivalence between (9.4) and (9.7). More precisely:
• Any feasible solution (x1 , x2 , x3 ) of (9.4) can be uniquely extended by (9.6) into a feasible
solution (x1 , x2 , x3 , x4 , x5 , x6 ) of (9.7). 132 CHAPTER 9. LINEAR PROGRAMMING • Any feasible solution (x1 , x2 , x3 , x4 , x5 , x6 ) of (9.7) can be reduced by a simple removal of
the slack variables into a feasible solution (x1 , x2 , x3 ) of (9.4).
• This relationship between the feasible solutions of (9.4) and the feasible solutions of (9.7)
allows to produce the optimal solution of (9.4) from the optimal solutions of (9.7) and vice
versa.
The Simplex strategy consists in ﬁnding the optimal solution (if it exists) by successive
improvements. If we have found a feasible solution (x1 , x2 , x3 ) of (9.7), then we try to ﬁnd a
new solution (x¯1 , x¯2 , x¯3 ) which is better in the sense of the objective function:
5x¯1 + 4x¯2 + 3x¯3 ≥ 5x1 + 4x2 + 3x3 . By repeating this process, we obtain at the end an optimal solution.
To start, we ﬁrst need a feasible solution. To ﬁnd one in our example, it is enough to set
the decision variables x1 , x2 , x3 to zero and to evaluate the slack variables x4 , x5 , x6 using (9.6).
Hence, our initial solution,
x1 = 0, x2 = 0, x3 = 0, x4 = 5, x5 = 11, x6 = 8 (9.8) gives the result z = 0.
We now have to look for a new feasible solution which gives a larger value for z. Finding
such a solution is not hard. For example, if we keep x2 = x3 = 0 and increase the value of x1 ,
then we obtain z = 5x1 ≥ 0. Hence, if we keep x2 = x3 = 0 and if we set x1 = 1, then we obtain
z = 5 (and x4 = 3, x5 = 7, x6 = 5). A better solution is to keep x2 = x3 = 0 and to set x1 = 2;
we then obtain z = 10 (and x4 = 1, x5 = 3, x6 = 2). However, if we keep x2 = x3 = 0 and if
we set x1 = 3, then z = 15 and x4 = x5 = x6 = −1, breaking the constraint xi ≥ 0 for all i. The
conclusion is that one can not increase x1 as much as one wants. The question then is: how much
can x1 be raised (when keeping x2 = x3 = 0) while satisfying the constraints (x4 , x5 , x6 ≥ 0)?
The condition x4 = 5 − 2x1 − 3x2 − x3 ≥ 0 implies x1 ≤ 5 . Similarly, x5 ≥ 0 implies x1 ≤ 11
2
4
and x6 ≥ 0 implies x1 ≤ 8 . The ﬁrst bound is the strongest one. Inc...
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This note was uploaded on 11/20/2013 for the course CS 101 taught by Professor Smith during the Fall '13 term at Mitchell Technical Institute.
 Fall '13
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