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Unformatted text preview: reasing x1 to this bound
3
gives the solution of the next step:
5
1
x1 = , x2 = 0, x3 = 0, x4 = 0, x5 = 1, x6 =
(9.9)
2
2
which gives a result z = 25 improving the last value z = 0 of (9.8).
2
Now, we have to ﬁnd a new feasible solution that is better than (9.9). However, this task
is not as simple as before. Why? As a matter of fact, we had at disposal the feasible solution
of (9.8), but also the system of linear equations (9.6) which led us to a better feasible solution.
Thus, we should build a new system of linear equations related to (9.9) in the same way as (9.6)
is related to (9.8).
Which properties should have this new system? Note ﬁrst that (9.6) express the strictly
positive variables of (9.8) in function of the null variables. Similarly, the new system has to
express the strictly positive variables of (9.9) in function of the null variables of (9.9): x1 , x5 , x6
(and z) in function of x2 , x3 and x4 . In particular, the variable x1 , whose value just increased 9.2. THE SIMPLEX METHOD 133 from zero to a strictly positive value, has to go to the left side of the new system. The variable
x4 , which is now null, has to take the opposite move.
To build this new system, we start by putting x1 on the left side. Using the ﬁrst equation of
(9.6), we write x1 in function of x2 , x3 , x4 :
x1 = 53
1
1
− x2 − x3 − x4
22
2
2 (9.10) Then, we express x5 , x6 and z in function of x2 , x3 , x4 by substituting the expression of x1
given by (9.10) in the corresponding lines of (9.6).
53
1
1
x5 = 11 − 4
− x2 − x3 − x4 − x2 − 2x3
22
2
2
= 1 + 5x2 + 2x4 ,
53
1
1
x6 = 8 − 3
− x2 − x3 − x4 − 4x2 − 2x3
22
2
2
11
1
3
=
+ x2 − x3 + x4 ,
2 2
2
2
53
1
1
z=5
− x2 − x3 − x4 + 4x2 + 3x3
22
2
2
25 7
1
5
=
− x2 + x3 − x4 .
2
2
2
2
So the new system is
x1
x5
x6
z =
=
=
= 5
2 − 3 x2 −
2
1 + 5 x2
1
1
2 + 2 x2 −
25
7
2 − 2 x2 + 1
2
1
2
1
2 x3 − 1 x4
2
+ 2 x4
x3 + 3 x4
2
x3 − 5 x4 .
2 (9.11) As done at the ﬁrst iteration, we now try to increase the value of z by increasing a right
variable of the new system, while keeping the other right variables at zero. Note that raising x2
or x4 would lower the value of z, against our objective. So we try to increase x3 . How much?
The answer is given by (9.11) : with x2 = x4 = 0, the constraint x1 ≥ 0 implies x3 ≤ 5, x5 ≥ 0
impose no restriction and x6 ≥ 0 implies that x3 ≤ 1. To conclude x3 = 1 is the best we can do,
and the new solution is
x1 = 2, x2 = 0, x3 = 1, x4 = 0, x5 = 1, x6 = 0 (9.12) and the value of z increases from 12.5 to 13. As stated, we try to obtain a better solution but
also a system of linear equations associated to (9.12). In this new system, the (strictly) positive
variables x2 , x4 , x6 have to appear on the right. To build this new system, we start by handling
the new left variable, x3 . Thanks to the third equation of (9.11) we rewrite x3 and by substitution 134 CHAPTER 9. LINEAR PROGRAMMING in the remaining equations of (9....
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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
 smith
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