212 Solutions
Fig. S.3. The fundamental cycle defined by edge 10 in Problem2.1
Ax = b, x > 0 using a standard simplex algorithm over GF(2), we will get
the network simplex method.
2.2 (a)
4(5,2) =
00
00
00
00
20 0 0
06 0 0
0 0 12 0
00 0 20
[N\B]
where
UB =
[B\N] = \UB\UN] -)• [h\VN]
= 04x2, VN =
20 0 0'
06 0 0
0 0 12 0
00 0 20
, uN =
"0 0"
00
00
00
00
00
00
00
= 04x2-
Then,
TZ(A) = Span {2ei,6e2,12e3,20e4} = Span {e±, e2, e%, e^} = R^4.
The rank of A(n, k) is r = 4.
1Z(AT) = Span {2e3,6e4,12e5,20ee} = Span{e3,e4,e$,eQ} = R^4.
Af(A) = Span <
if
1
)
0
(^0 0)
0
I Voy
1
1
(o)
1
0
0
0
WJ
— Span {ei,e2}
JV(4r) = {0} , dimAf (AT) = 0.
Thus, R^6 = Tl(AT)®Af(A) = R^4 ©R^2 and R^4 = 1l{A)®N{AT) = R^4 00 = R^4.