1. 3 points [You should study the self-study material on the standard forms of LP, avail- able in the Canvas/Files.] Put the following linear programming problem in standard form, that is, standard inequality form. (Do not solve it.) minimize 1 - 3x2 - x3 subject to - 2x2 + x3 = 3 -x1 + x 1 X1 0 2 unconstrained X3 1. 2 Reduction to standard form vbmda-1158 LPs in non-standard form can be reduced to standard form. For example: min 3x12x2 + x3 +1 Here are the reduction rules to use: Rule x=a min f = max(-f) max(f+const) max f f(x) a -f(x) -a a and --a x a I'=x-a0 xa x=a-x>0 No restriction on a I s.t.1+2 -3 2x1 + x 2 1 + 2 + 3 = 4 1 -2 2 3 The objective function then becomes: x,x+,x->0 The inequalities become: -3x1 + 2x2x3 = -3(-2) +2(3-2)-3+x3 =-3x2x2x3 + x3 +12 vilamponi burbusie Changes max-3x1 + 2x2-3-1 max -3x1 + 2x2 3 1- 3 x1 + x2 + x3 4,-21-22-23 -4 Replace z with a -2, then 12-2 x 20 Replace 22 with 3-2, then x2 3 Replace 23 with 2-3 and 3, 20 (the +12 can be dropped) T-T2 3 (1-2)-(3-x) 3 +2 8 2x1 + x2 22(x-2)+(3-x2) 2 2x1-x2 3 1 + x2 + x3 4(x-2)+(3-) + (x - 3) 4 x-x+x-x3 -21-22-23 -4-(-2) +-(3-2)-(-x-x3) -4 -x+x-x+x-3 20 The resulting standard form is: pianot villamponi od tamojt dopraful max - 3x12x2-x3+xz s.t. x + x2 8 2x1-x2 3 x-x+x-x3 3 - x + x-x + x3 -3 x1, x2, x3, 20 After finding an optimal solution to this LP problem, we get the optimal solution to the original problem by using the reverse transformations: x = x-2 rd & driw dairenosib of 'I sau sw tadt stol x2 = 3-2 19dow) to anot add bas egneds f'assob & axtos mort 18olo yllau ai boen ai ted x3 = x3 x3 odni sotilaupsal moet vos stated as sure will voitonut svitido sda jadi cela stol and am ni nattinw ad neds nao mat viileupas bisbaste od ddiw land of op v usod aktileupe notov oilT colderiny 19to 101 stan smise (eoldalov asala basqqs ew 190) bas min 3x12x2 + x3 + 1 = -max -3x1 + 2x2 - 3 - 1 = 1-max(-3x1 + 2x2x3)