Question

Solve using the M’s method. Formulate the standard model and once defined the solution express your answer with the values ​​of Z and the variables.

Ejercicio Il Considere el siguiente problema, Minimizar Z = 2000X, + 500X, Sujeto a 7= 7000 ES=18 X2 = 14 X = E3-A4 =A6=0 2X, + 3X, 42 3X, + 6X, 266 X, X, zo

Not sure if the answers are correct

Answer 1

Problem is Max Z 2000 1 5002 subject to 2 3 2 42 3 1 62 z 66 and 1, 20; The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate 1. As the constraint-1 is of type' 'we should subtract surplus variable S1 and add artificial variable A1 2. As the constraint-2 is of type'> 'we should subtract surplus variable S2 and add artificial variable A2 After introducing surplus,artificial variables МА1 Max Z 2000 500 2 0 S1 0 S2 М А2 subject to A S2 21 3 2 S1 = 42 A2 66 3 1 6 2 and 1, 2, Si, S2, A1, A2 0 Cj 00M-M 2000 500 Iteration-1 MinRatio Хв Si S2 A Хв Св A2 В 1 42 0 1 A1 42 -M 3 -1 0 3 = 14 66 (6) Ар 3 -1 1 - M 6 = 11 Z = - M-M М м -9M -5M - 108M Zj - С, — 2000 9M -5M М М 0 500 79 66

Negative minimum z-c, is -9M-500 and its column index is 2. So, the entering variable is x2. Minimum ratio is 11 and its row index is 2. So, the leaving basis variable is A2 . The pivot element is 6. Entering x2, Departing A2, Key Element 6 R2(new) R2(old) 6 +R,(new) R,(old)- 3R2(new) 500 0 2000 0 -M Iteration-2 MinRatio Хв Св Хв x2 S1 S2 A1 X1 Х1 9 = 18- 0.5 А -M (0.5) 0 -1 0.5 1 11 = 22 0.5 500 0.5 0 -0.1667 0 X2 11 1 500 M -0.5M -83.3333-M Z 9M 5500 -0.5M 250 Z- C-0.5M-1750 М -0.5M - 83.3333 0 Negative minimum z, - C, is -0.5M - 1750 and its column index is 1. So, the entering variable is x. Minimum ratio is 18 and its row index is 1. So, the leaving basis variable is A1. :. The pivot element is o.5. Entering x,, Departing A, Key Element 0.5 +R,(new) R,(old) 0.5 R2(new) R(old) - 0.5R, (new) Privacy- Terms

CJ Iteration-3 2000 500 0 0 MinRatio Хв Хв S2 Св S1 В X2 X1 S1 2000 18 1 0 -2 1 X1 2 2 1 (1) 500 2 0 1 -0.6667 X2 Z 37000 3500 2000 500 1666.6667 -3500 0 0 1666.6667 Negative minimum z,-c, is -3500 and its column index is 3. So, the entering variable is s1. Minimum ratio is 2 and its row index is 2. So, the leaving basis variable is X2- The pivot element is 1. Entering S1, Departing x2, Key Element 1 R2(new) R(old) +R,(new) R,(old) 2R2(new) 2000 500 0 0 Iteration-4 MinRatio Хв Св Хв В Х1 X2 S2 22 2000 1 2 0 -0.3333 X1 S1 0 2 0 1 1 -0.6667 Z 44000 2000 4000 0 -666.6667 Z- C 3500 0 -666.6667 च।-

Variable S2 should enter into the basis, but all the coefficients in the S2 column are negative or zero. So S2 can not be entered into the basis Hence, the problem has an infeasible solution