Question

Question 1 (24 marks]. (a) Rewrite the following linear program in standard inequality form: minimise 78 - 12 +503 - 1034 subject to -11x1 - 12.12 - 13 + 14 > 1, 11-472-873 = 12 21 +672 +31, 57. 11,1220, 13 50 I, unrestricted (b) Consider the following linear program in standard equation form: maximise + 212-383 + 70s subject to I +212 +2.13 + 14 = 3 I + 2x2 + 713 + IS 2x + 4x2 + 7x's + 16 = 6, 11, 12, 13, 14, 15, 1620 For each of the following values of x'=(11, 12, 13, 14, 15, 16) say whether or not this value is a basic feasible solution of this linear program. Justify your answers. (i) x = (1,1,0,0,0,0) (ii) x = (1,0,0, 2, 2, 4) (iii) x'=(0,0,0,3,3,6) (c) Which, if any, of (i), (ii), and (iii) from question 1(b) are extreme point solutions to the given linear program? Justify your answer. You may use, without proving, any theorems discussed in the lectures. (d) Consider an arbitrary linear program in standard equation form: maximise cx subject to Ax = b. x>0 Suppose that x is an optimal solution to this linear program. Show that if x is not an extreme point solution then we can express x as x = ly + (1 - A)z where 1 € (0.1) and y and z are two different optimal solutions of this program.

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Answer 1

กว่าห์? ΟΥ and from min 1-10 na subject to 74,- H₂ + 5813-1004 subject to 0 -113, -1222 - +3+7147,1 ay - 422 -883 =12 224 +62 +304 < 7 26, 27,0,-8350, na unrestricted. let -X₂ = 13 then from ④ we have 24 - AM2-873 = 12 is 34 - 47, +87=12 re 34 - 42 + 8x3 =12 x4 - Ash +84 712 and from the 24-4^2 +847,12 and 24+401-877-12 from 24 + 6 m+30 27 »-244-63-3747-7 - 11 - 12 - xy + x47,1 =) - 1104-12 2 +3 +247, the standard form 74-32-54 -114-12 % + 2y +47) 2 - 4 + 8u to na 7,12 - † Ally-821 -0.1434-12 -2x62 toon - 3 na> - 7 41, 92, 970, pa umrestorited ☺ 377 (1,1,0,0,0,0), 9f we chooxe 14, & bave variables sh, 23, Na, H, N. all are Satisfying non negative restriction. and satisfies the equality constrainty we putting M = 45 26-0 get & +234 + 20+ xq = 1+2=3. + 2x + 7 + = 1 + 2 = 3 284 +4+ 733 + 86 = 2+4 = 6 satisfies constrainty. fence it inget = (1,1, -0,0,0,0) 14 hand feasible solution set. (1,0,0,2,3,4) can not choute 6-3=3 basis vanables and basic variables are not zero in this care Hence and also it not a banix feasible solution af at Co,o,,3,3,6), 2,5%=3=0 as banc Variables and it satisfier all contraint and nonegative sestriction lane fenstelle solution () 40 given z = 4+2 22 32 + 715 60) 7 1+2 = 3 fii= (0,9, 9, 3,3,6) 7x3 = 21 2,4) 7 = 1 + 7 x2 = 15 Here nt=C0,0,0,3,3, 6) is extreme point. Since the extreme is point one of the barie feasible Saluting since ge = (0,0,0,3,3,6) is banie feasible solution and 2 = 21 & extreme point solution. If y & z be two extreme point solutions and Ay=b & Azab 1 (Ay) + (-1)Az) – 1b +(-x)b = b 외 Ay) + (l_) A2 A ( ny + (1-1)2) = Ax = b became x is a bare feamble solution CS Scanned with Camscanner we tene for (i) at C1,1,0,0, Z / (9,0,0,2, 'cy then