Question

Use the two-phase method to find the optimal solution to the following LP:

Min z = 3x₁ + 2x₂

s.t.:      3x₁ + x₂ ≥ 3

4x₁ + 3x₂ ≥ 6

x₁ + 2x₂ ≤ 3

x₁, x₂ ≥ 0

Answer: z = 4.2, x₁ = 0.6, x₂ = 1.2.

Answer 1

Solution:

Please find the images attached for a detailed solution:

Find solution using Two-Phase method MIN z= 3x1 + 2x2 subject to 3x1 + x2 >= 3 4x1 3x26 x1 2x23 x10 x2 0 and x1,x2 >= 0 Solution Problem is Min z 3x1 + 2x2 subject to 3x1+ x2 23 4x1 + 3x26 x1 2x2S3 x220 andx1,x2 2 0; ->Phase-1<- 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 A 2. As the constraint-2 is of type' e'we should subtract surplus variable S2 and add artificial variable A2 3. As the constraint-3 is of type' s'we should add slack variable S3 4. As the constraint-4 is of type'2 'we should subtract surplus variable s4 and add artificial variable A 5. As the constraint-5 is of type'2' we should subtract surplus variable S5 and add artificial variable A4

After introducing slack,surplus,artificial variables Min 2- subject to 3x12- S, 1 +2* Iteration-1 MinRatio S. Positive maximum 2,- C, is 8 and its column index is 1. So, the entering variable is x Minimum ratio is 0 and its row index is 4. So, the leaving basis variable is.A3 . The pivot element is 1 Entering,DepartingA, Key Element -1

+R4(new) R4(od) + R,(new)R,(old) - 3R4(new) + R (new)-R2(old) 4R,(new) + R (new) R(old) - R4(new) + R (new) R,(old) Iteration-2 MinRatio Z.-C ositive maximum Z - C, is 7 and its column index is 6. So, the entering variable is S4 Minimum ratio is 1 and its row index is 1. So, the leaving basis variable is.A

The pivot elament is 3 Entering-34. Departing = Al-Kay Element = 3 +Rnew) -Rg(old) -R1(new) Iteration-3 MinRatio S3 -12

Positive maximum, ls and its column indlex is 2 So, the antering vaniable is Minimum ratio is 0 and hs row index is 5. So, the leaving basis varlable sA. The plvot element is 1 EnteriDeparting-A Key Element -1 tR,lnew-R(old) tne-,new) S1 -2 -1 チS

Posiltive maxkimum Z-C, lsnd scomn index is 7. So, the entering varlable is s, Minimurn ratio is 12 and its row index is 2 So, tha leaving basis variabla is A The pivot element is 3 Entering - S,, Departing -4, Key Element - +23(new) _ R101d-782(new) + Rs(new) = Rsfold) + R2(new) Iteration-5 ci Cs S1 MinRatio r- 0

Since all -QSO Hence, optimal solution is arrived with value of variables as Min:-0 >Phase 2 we eliminate the artificial variables and change the objective function for the original Iteration-1 Si S3 MinRat S5 21 ろ 弓-G Since all 2,-Cso Hence, optimal solution is arrtved with value of varlables as 21 Min