# HW-6 - d is a direction of unboundedness then both Ad =0...

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Solution for HW6 Section 4.4 3. Basic Variables Basic Feasible Solution Corner Point x 1 , x 2 x 1 =150 x 2 =100 s 1 =s 2 =0(150, 100) x 1 , s 1 x 1 =200, s 1 =150, x 2 =s 2 =0(200, 0) x 1 , s 2 x 1 =350, s 2 =-300, x 2 =s 1 =0 Infeasible x 2 , s 1 x 2 =400, s 1 =-450, x 1 =s 2 =0 Infeasible x 2 , s 2 x 2 =175, s 2 = 225, x 1 =s 1 =0(0, 175) s 1 , s 2 s 1 =350, s 2 =400, x 1 =x 2 =0 (0, 0) 4. Let σ A = 1/6, σ C = 1/3 and σ B = ½. Then ). 20 , 10 ( ) 40 , 0 ( 2 1 ) 0 , 30 ( 3 1 ) 0 , 0 ( 6 1 = + + 5. + 2 . 33 2 . 12 6 . 2 1 10 144 0 14 0 will do the trick. 6. Suppose we are given that A d = 0 and d>=0 and that x is feasible . Then for all c>=0, x+ c d>=0. Also A( x +c d)=Ax=b. Therefore for all c>=0 x +c d is feasible. We now show that if Ad =0 and d>=0 are not both true, then d is not a direction of unboundedness. This is equivalent to showing that if
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Unformatted text preview: d is a direction of unboundedness then both Ad =0 and d>=0 must be true. Clearly if Ad =0 is not true then x +c d is not feasible so d is not a direction of unboundedness. If d>=0 is not true then d must have an I’th component that is negative. Then for large enough c>=0 the I’th component of x +c d will be negative, so x +c d will not be feasible and d will not be a direction of unboundedness. 7. ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 3 1 1 is a direction of unboundedness. As we move in this direction z increases and we stay feasible....
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## This note was uploaded on 09/06/2009 for the course IEOR 162 taught by Professor Zhang during the Spring '07 term at Berkeley.

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