Principles of Mathematics in Operations Research

238 Solutions => Hi = 16 - V208 M2 = 16 + V208 Then, ||yl(2)|| ||^(2)_1|| = 1.2676(15.2111) = 19.2815 = c[A(2)}. We know, fi\ ...

Solutions 239 -4(4) = Q 0 Ro = -0.83812 0.52265 -0.15397 -0.02631 -0.41906 -0.44171 0.72775 0.31568 -0.27937 -0.52882 -0.13951 - ...

240 Solutions 4(4) 3 = Q3-R3 -1.0000000 0.0009338 -1.7566 x 10~^8 8.8905 x 10~^15 -0.0009338 -1.0000000 7.2955 x 10~^5 -8.7996 x ...

Solutions 241 Problems of Chapter 7 7.1 a) A zero dimensional polytope is a point. b) One dimensional polytopes are line segment ...

Solutions X (0.1.0) ts Fig. S.8. 3-dimensional pyramid 7.4 See Figure S.8 for a drawing of Pn+\. Let a^1 be the normal to face F ...

Solutions 243 Pn+\ is not a union of a cone at XQ and a poly tope. Pn+i is a direct sum of a cone at xo and Cn. P„+i is an inter ...

244 Solutions 7 j- / ~%i.i.-i) i / /v,-',-D j f(0,f',-« (-•,0,-f') (t,0,-f'> f*7^ RS^—™~™™™™™-T«™™| ^ / It/ r 1 * / #7 ' * * ...

Solutions 245 (•"',•.0) Fig. S.ll. The extreme points of the dodecahedron, cj>: golden ratio ...

246 Solutions Problems of Chapter 8 8.1 a) We have six variables and three constraints, therefore we have ( 3 ) candidate bases. ...

Solutions 247 D=(0,8,10) x3 A ,V,Y, V 72> Fig. S.12. Exercise 8.1: Primal and dual polyhedra Thus, S3 enters. B-iNss = 10-1 0 ...

248 Solutions xB = B~lb = [10-1] 01 0 10 0 [10" 8 2 = [ 6" 10 8 We are on point D. z = cTBxB = [0,2,2] 6 10 = 36. cl-clB^N =[1,0 ...

Solutions 249 w=[0,2,2]Ei^1 B-^1 = [2,2,0]. w = cTBB~l = [0,2,2] rXl = cXl - wNxi = 1 - [2,2,0] rS2 = cS2 - wNS2 = 0 - [2,2,0] r ...

250 Solutions revised simplex with B = QR decomposition: Let XB = (si,x 3 ,x 2 )T, XN = (xi,s 2 ,s 3 )T. Then, B 101 010 001 i ...

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 en ...

252 Solutions So, they are rays. Since every pair of the above vectors have zeros in different places, we cannot express one ray ...

Solutions 253 6' 1 } 4 i i ~ 2 2 0 1 0 0 1 + 2 4 0 0 5 0 + 0 4 5 1 3 0 0 + 0 1 5 0 2 5 0 1 + 1 2 1 0 3 0 + 1 1 0 0 2 1 convex co ...

254 Solutions 0 0 0 < Si «2 S3 = 20 12 11 1 1 0 a < 6 — 1 = 5(bounds of x\) =>a = min{20,12,5} = 5. xi leaves immed ...

Solutions 255 where Af = {xi,si,X3,X4}. Then, Bland's rule (lexicographical order) marks the first variable. Since the reduced c ...

256 Solutions 3x 3 + x 4 <20 (2/3) -xi < -1 (2/4) x\ < 6 (1/5) £2 < 10 (l/e) -x 3 < -3 (2/T) X3 < 9 (j/ 8 ) x ...

8.4 Solutions 257 Fig. S.13. A multi-commodity flow instance Let us take the instance given in Figure S.13, where K = 3 and V = ...