Question

please explain to get standard form after adding artificial variable and also how to get the feasible dictionary

dont use tableau please

3:32 No SIM minimize subject to 224 -3i + 2 1.3 1. Illustrate the feasible area of problem (P) 2.) For the problem (P), use the nonnegative variable x3 for inequality constraint 1 and the nonnegative variable x4 for inequality constraint 2 and the nonnegative variable 5 for inequality 3 to Show the equation standard form of the problem (P). (3) Find all feasible basis solutions of the equation standard form of the problem (P) obtained in (2). (4) from the standard form matrix that obtain in number 2,Consider the artificial problem (the problem of the first phase) when applying the two-step method, the introduced artificial variable are v1 and v2. For the artificial problem, find disctionary for base variable v1,v2,v5 5) From problem 4, show the optimal dictionary 6. Use the feasible basis solution obtained from the optimal dictionary in (5) find first dictionary from the

Answer 1

Answer: - 1).

Solution Find solution using Simplex method (BigM method) MIN Z = x1 + x2 subject to x1 + 2x2 >= 4 2x1 + x2 >= 6 x1 x2 <1 and x1,x20 Solution Problem is Min Z = x1 + x2 subject to x2x24 2xx226 and x1,x220; The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate 1. As the constraint-1 is of type"2 'we should subtract surplus variable S, and add artificial variable A 2. As the constraint-2 is of type '2 ' we should subtract surplus variable s, and add artificial variable., 3. As the constraint-3 is of type''we should add slack variable S

After introducing slack,surplus,artificial variables subject to x1 - 2x2 - S Iteration-1 MinRatio 3.M 3.M 3M-1 ↑ 3M-1

Positive maximum -G is 3M-1 and its column index is 1 . So, the entering variable is x1. Minimum ratio is 3 and its row index is 2. So, the leaving basis variable is A2. . The pivot element is 2 Entering -x, Departing-4, Key Element-2 R,(new)-R(old) +Ri(new)-R(old) - R2(new) +R (new)(odR2(new)

Iteration-2 MinRatio -3-0.6667- 32.6667 3M 1 M 1 3A 1 M 1 3M 1 Positive maximum Z- C, is and its column index is 2. So, the entering variable is x2 Minimum ratio is 0.6667 and its row index is 1. So, the leaving basis variable is A

.. The pivot element is Entering -x Departing -41. Key Element - R,(new)-R,(old) +R2(new)- R2(old) 5R1(new) + Ry(new) - R;(old) - iRi(new)

Iteration-3 MinRatio S. 10 Since all Z,-Cs0 Hence, optimal solution is arrived with value of variables as 10 Min Z =-

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