Principles of Mathematics in Operations Research

Solutions 251

b)

Method 1:

At (2,0,1,0,0)T, we have

B~]N =

x 2 si .S3

_i —2 _i

5 5 5

_3 1 _2

5 5 5

, B-^:

X\ - 5^2 = 2

If x 2 enters 2 3 - |x 2 = 1

x 2 >0

,T 2 = (9 => r = (2 + §0,0,1 + f 0,0,0)T is

feasible for 9 > 0. Thus, r^1 = (|,l,f,0,0)T is an unboundedness direction
and hence an extreme ray.
xi - ±s 3 = 2 \
If s 3 enters X3 - |,S;3 = i \ -» s 3 = 0 =» r = (2 + ±0,0,1 + |0,O,0)r
s 3 > 0 J
is feasible for 0 > 0. Thus, r^2 = (±,0, f, 0,1)T is another unboundedness
direction and hence an extreme ray.
At (4,0,0,5,0)^7 \ we have

B-lN

x 2 x 3 s 3

-3 2 -1

-3 5 -2

B~lb-

x'l - 2x 2 = 4 I
If x 2 enters .si - 3x 2 --= 5 > <-+ x 2 = 0 => r = (4 + 2(9,0, 0,5 + 30, O)^7 ' is
x 2 > 0 I
feasible for 0 > 0. Thus. (2,1,0,3,0) is an unboundedness direction
and hence an extreme ray.
Xi - S 3 = 4 ~)
If .s 3 enters sx - 2s 3 = 5 \ <-> s 3 = 0 => r = (4 + 0,0,0, 2(2, 0)r is feasible
•S3 > 0 J
for 0 > 0. Thus, r^4 = (1,0,0,2,1)T is another unboundedness direction and
hence an extreme ray.
Method 2:
Try to find some nonnegative vectors in N{A).

r

0<rl =(l,l,l,0,o\ eAf(A).

0 <r^2 Q,0,^,0,1) eAf(A).

6><r^3 = (2,l,0,3,0)rGAA(yl).

0<r^4 = (1,0,0,2, l)Te//(>*)•