Principles of Mathematics in Operations Research

(Rick Simeone) #1

252 Solutions


So, they are rays. Since every pair of the above vectors have zeros in different
places, we cannot express one ray as a linear combination of the others, they
are extreme rays.


c)



  1. Xi + X 2 +£3:
    c^1 = (l,l,l,0,0)T=>
    ( (cl)Trl = f + l + f+0 + 0=^>0<-)- unbounded


(c^1 1\T„2 ) f + o + f + o- 0 I > 0 "-> unbounded

(ci)Tr (^3) = 2 + l + 0 + 0 + 0 = 3>0--» unbounded
_ (ci)Tr (^4) = 1 + 0 + o + 0 + 0 = l>0<-> unbounded
Thus, there is no finite solution.



  1. —2xi — X2 — 3x3:
    c^ = (-2,-l,-3,0,0)J =•
    ( (c2)Tri :=:| 1 _9+o + 0=-f^O-> bounded


(c,2\T2 ^2 ) -l-o-i + o- 0= -| ^ 0 abounded

(c2)Tr (^3) = _ 4 _1 + 0 + o + 0 -5 i> 0 °-> bounded
(c^2 )Tr^4 = -2 + 0 + 0 + 0 + 0 = -2 ^ 0 -->• bounded
Thus, there is finite solution.



  1. -x\ - 2^2 + 2a;3:
    c^3 = (-l,-2,2,0,0)T^
    ( {cz)Trl = -§-2+§+0 + 0 = -§^0<-+ bounded
    (c^3 )Tr^2 = -|+0+|+0 + 0=|>0^ unbounded


(c3JTr (^3) = _ 2 2 + 0 + 0 + 0 = -4^0'-> bounded
[ (c^3 )Tr^4 = -1 + 0 + 0 + 0 + 0=-1^0^ unbounded
Thus, there is no finite solution,
d) xi = 6, x 2 = 1, x 3 = \ Si = ±5
*i — 2 '
«3 = 1
r 6 n
1
1
J
2
i

= a
~2
0
1
0
0
4
0
0
5
0



  • A«i
    4 5 1 3 5 0 0

  • M2
    i 5 0 2 5 0 1

  • /^3
    2
    1
    0
    3
    0

  • M4
    1
    0
    0
    2
    1
    (l-«)
    a,Mi)M2.M3,A«4 > 0
    We have 5 unknowns and 5 equations. The solution is

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