Duality LPP min:Z=3x₁+2x₂

Step 1: Write the dual. For the primal (minimisation, all constraints ≥), the dual is:

Primal: Minimize Z = 3x₁ + 2x₂
  subject to  2x₁ + 4x₂ ≥ 10
              4x₁ + 2x₂ ≥ 10
              x₂ ≥ 4
              x₁, x₂ ≥ 0

Dual:   Maximize W = 10y₁ + 10y₂ + 4y₃
  subject to  2y₁ + 4y₂ ≤ 3
              4y₁ + 2y₂ + y₃ ≤ 2
              y₁, y₂, y₃ ≥ 0

Here y₁, y₂, y₃ are the dual variables, one for each primal constraint.

Step 2: Standard form of the dual. Add slack variables s₁, s₂ (one per dual constraint). Since all right sides are positive, the slack variables give the starting basis.

Maximize W = 10y₁ + 10y₂ + 4y₃
subject to
  2y₁ + 4y₂ + s₁ = 3
  4y₁ + 2y₂ + y₃ + s₂ = 2

Step 3: Simplex tables for the dual

Table 1

          Cj 10 10  4  0  0        
Basis     CB y₁ y₂ y₃ s₁ s₂ b Ratio
-----------------------------------
s₁         0  2  4  0  1  0 3   3/2
s₂         0  4  2  1  0  1 2   1/2
-----------------------------------
          Zj  0  0  0  0  0 0      
       Cj-Zj 10 10  4  0  0        
Entering: y₁   Leaving: s₂   Pivot = 4

Table 2

          Cj 10  10    4  0    0          
Basis     CB y₁  y₂   y₃ s₁   s₂   b Ratio
------------------------------------------
s₁         0  0   3 -1/2  1 -1/2   2   2/3
y₁        10  1 1/2  1/4  0  1/4 1/2     1
------------------------------------------
          Zj 10   5  5/2  0  5/2   5      
       Cj-Zj  0   5  3/2  0 -5/2          
Entering: y₂   Leaving: s₁   Pivot = 3

Table 3

          Cj 10 10    4    0    0           
Basis     CB y₁ y₂   y₃   s₁   s₂    b Ratio
--------------------------------------------
y₂        10  0  1 -1/6  1/3 -1/6  2/3     -
y₁        10  1  0  1/3 -1/6  1/3  1/6   1/2
--------------------------------------------
          Zj 10 10  5/3  5/3  5/3 25/3      
       Cj-Zj  0  0  7/3 -5/3 -5/3           
Entering: y₃   Leaving: y₁   Pivot = 1/3

Table 4

          Cj  10 10  4    0  0     
Basis     CB  y₁ y₂ y₃   s₁ s₂    b
-----------------------------------
y₂        10 1/2  1  0  1/4  0  3/4
y₃         4   3  0  1 -1/2  1  1/2
-----------------------------------
          Zj  17 10  4  1/2  4 19/2
       Cj-Zj  -7  0  0 -1/2 -4     

Step 4: Read the answer. All Cj − Zj ≤ 0, so the dual is solved: y₁ = 0, y₂ = 3/4, y₃ = 1/2, Max W = 19/2.

By the duality theorem, Min Z = Max W. The values of the primal variables are the numbers (without sign) in the Cj − Zj row under the slack columns: x₁ = 1/2 (under s₁), x₂ = 4 (under s₂).

Check: Z = 3(1/2) + 2(4) = 19/2.

Answer: x₁ = 1/2, x₂ = 4, Minimum Z = 19/2.