Question

Sketch the constraint set for each noncanonical linear programming problem below. On the basis of this constraint set, formulate a conjecture as to whether or not the solution of the given problem is the same as the solution of the associated canonical linear programming problem where all independent variables are constrained to be nonnegative. Verify your conjecture by solving both linear programming problems.

c. Maximize f(x, y)= - x + 2y subject to -x+y-1 2x - y = -2

Answer 1

Given LP prophen is Maximine tears) = -x + 2 subject to atuant 2a-4a-2. at can be written as het za ten,g). sal ect to Max zi -x+2y subre - cons tocant axtsa al- anay s-2 / on the first constrain, b. =-120 I so multiply with ii, we get n x-4 & ③ similarly multiply with is in the second constraint, we get -2xty ? 2- 47. wig > 0. ence the transformed LPpio max zanatay subject to anyal -2xty > 2 and a yo

To solve the Lpp byuring simplex matter of (Big-m.nothes) - first we convert the LPP into Canonical form by adding slack, surplus and artificial Viables as approfint. * as the constraint (a-y=1) is of type ">", we should subhat subtract surplus racible S, and add artificial Varithe An * As the constraint (2x+4), 2) in of type "s" we shood fubtract surplus variable Sy and add entiticed vaille Az. After introdusing the surplus and artitiaid Variables, the team the resultant LPP in Max z = -x +2y + oos, tosat 19 A-MA subject to x-y-sitt, al - 2x + y - S₂t Az =9 x, y, si, S2, A,, Az>o and . I turation - Min Ratio z=-30 Z; -cj Mtl M tl -21 key column] inden is the Negative minimum z; -(; is "& and its column so, the entering variable is the Minimum ratio in " and it yow inden is 's'. so the leaving Variable is "A.

:. The pirot element is "1" . Entering Deporting - Az, key clement al. By using the formale, new valus a old walese al xey rowelement to find the second iteration. key element, c; la Heration 2 Min Ratio z=-3m44 zj M-4 M-2 m - 3 M-2 Here alt z; -c, 30 Hence the optimality and reached. Thas the optimal solution is arrived with vake of Variables as a o and y =2. Max zinxto 24 = -(0)+ 2(2) = 0 + de 4 Hence the optimal solution of the app is Maxz= 4 at x 20 and y=23 But this solution is not final feasible, because the solution ver violates the first constocent xayall also the artificial variable A, appears in the basis with positive Valee 3;

GRAPHICAL METHOD Consider the LPP max a = -x +2y to subject -*+ S-1 21-4 -2 *),y >o. To solve this Lpp by using graphical, Mathed, equating the constraints as equations. -x+y and 2x-YE-2 put a co I yari put u zo tyind put yo xil put o anal 1 * 10 Tyllo we polt the constraints in x-y plane solution in infeasible Now The objective A ad B Now we calculte the function vakes at A = (@, g) = -x+2y o +212) SUO (max z = 4 1 (B) fig? 1 mex z EX+24 -1 +200) Z(B) all Hence the LPP has no feasible solution Max z = 4