MAP 5236 Feb 12, 2005
HW 2 Remarks Prof. S. Hudson
I graded HW2 based mainly on problems 4.3.1 and 4.6.13a and 5.1.1b.
There was a question about Problem 5.2.2, which asks you to work though a problem
using the revised simplex method. Make sure you can do this kind of calculation for Exam
I. Matrices are awkward to type - but maybe you can check your work from this. Please
forgive the organization. The initial tableaux is
−5−8−7−4−6 0 0 0
2 3 3 2 2 1 0 20
3 5 4 2 4 0 1 30
We see that x2enters and x7leaves. From column 2 we get η= [−3/5,1/5]T(this
is easier to type than a column vector) and we put it into B−1(see below). Then v=
[0,8]B−1= [0,8/5] and vA = (8/5)[3,5,4,2,4] and vA −c= [−1/5,0,−3/5,−4/5,2/5]
which still has negative numbers in it.
So, x4enters. Column 4 is B−1[2,2]T= [4/5,2/5]T. The RHS is B−1[20,30]T=
[2,6]T. The minimal ratio is 2 over 4/5, so x6leaves. We get η= [5/4,−1/2]Tand the new
B−1is computed from this and from the old one (see below). Then v= [4,8]B−1= [1,1]
and by coincidence we get vA −c= [0,0,0,0,0]. Since there are no negatives, we can
stop. An optimal solution is [x4, x2]T=B−1b= [2.5,5]T(other variables are 0) with
Z=cx = 50.
However, the large number of zeroes in vA −cis a warning sign. Notice that x1is
non-basic, but has a coefficient of 0 in row (0). This means it can enter, and produce a
different optimal solution [10,0,0,0,0] which also gives Z= 50. Likewise, we could let x3
or x5enter and collect more optimal CPF solutions. Unfortunately, the textbook’s answer
key only mentions the first one and I won’t try to list them all.
Here are the B−1matrices for this problem, not including the first one, which is always
I, and not including any used to find the additional optimal solutions.
B−1=1−3/5
0 1/5B−1=5/4 0
−1/2 1 1−3/5
0 1/5=5/4−3/4
−1/2 1/2
1