Question

Problem 3. (a) Solve the following LP problem using the Simplex Method. Use the smallest- subscript rule to choose entering and leaving variables. Show all steps. maximize xi+ 5.02 + 5x3 + 524 subject to X1+ 412 + 3x3 + 3x4 < 17 12 + x3 + x4 <4 Xit 202 + 2x3 + 3x4 < 10 X1, ..., 84>0. (b) Is the optimal solution you found the only one? Explain.

Answer 1

(a) The Problem is Max Z - X1 + 5X2 + 5x3 + 5x4 subject to X1 + 4x2 + 3x3 + 3x4 517 .. . x2 + x3 + x454 ..........(2) x1 + 2x2 + 2x3 + 3x4 5 10 ....... and X1, X2, X3, X4 2 0; .. (1) As the constraint-1 is of type's' we should add slack variable S1 As the constraint-2 is of type' S'we should add slack variable S2 As the constraint-3 is of type's' we should add slack variable S; After introducing slack variables the prolem becomes Max Z = X1 + 5x2 + 5x3 + 5x4 + 0S + 0 S2 + 0 S3 subject to X1 + 4x2 + 3x3 + 3x4 + s = 17 x2 + x3 + x4 + s = 4 xy + 2x2 + 2x3 + 3x4 + S3 = 10 and X1, X2, X3, X4, S1, S2, S320 Iteration-1 ci 5 5 500 MinRatio 4 - 425 0 (1) 1 22 3 O Z-C, -1 -515 -5 0 0 0 Iteration 2 1 5 55 Xasi MinRatio X S3 1 0 Z; Z;-C; 05 -11 0000

Iteration-3 15 5 5000 MinRatio - o 1 1 0 Z-21 0 Iteration-4 C; 155 sooo X3 X4 S S2 S3 MinRatio z = 22 3 01 -1 -1 510012- 121 2; 15 56 031 2;- C000031 Since all Z; -C;20 Hence, optimal solution is arrived with value of variables as : Xı = 2, X2 = 3, x3 = 1, x4 = 0 Max Z = 22 (b) The solution X, = 2, x2 = 3, x3 = 1, x4 = 0 otained in (a) is the only one (unique) as in the final Simplex Table 24-C4> 0 there is no scope for x4 to enter into the optimal basis