Duality LPP min:Z=20x+50y

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

Primal: Minimize Z = 20x + 50y
  subject to  2x + 5y ≥ 12
              3x + 7y ≥ 17
              x, y ≥ 0

Dual:   Maximize W = 12y₁ + 17y₂
  subject to  2y₁ + 3y₂ ≤ 20
              5y₁ + 7y₂ ≤ 50
              y₁, y₂ ≥ 0

Here y₁, y₂ 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 = 12y₁ + 17y₂
subject to
  2y₁ + 3y₂ + s₁ = 20
  5y₁ + 7y₂ + s₂ = 50

Step 3: Simplex tables for the dual

Table 1

          Cj 12 17  0  0         
Basis     CB y₁ y₂ s₁ s₂  b Ratio
---------------------------------
s₁         0  2  3  1  0 20  20/3
s₂         0  5  7  0  1 50  50/7
---------------------------------
          Zj  0  0  0  0  0      
       Cj-Zj 12 17  0  0         
Entering: y₂   Leaving: s₁   Pivot = 3

Table 2

          Cj   12 17     0  0            
Basis     CB   y₁ y₂    s₁ s₂     b Ratio
-----------------------------------------
y₂        17  2/3  1   1/3  0  20/3    10
s₂         0  1/3  0  -7/3  1  10/3    10
-----------------------------------------
          Zj 34/3 17  17/3  0 340/3      
       Cj-Zj  2/3  0 -17/3  0            
Entering: y₁   Leaving: y₂   Pivot = 2/3

Table 3

          Cj 12   17    0  0    
Basis     CB y₁   y₂   s₁ s₂   b
--------------------------------
y₁        12  1  3/2  1/2  0  10
s₂         0  0 -1/2 -5/2  1   0
--------------------------------
          Zj 12   18    6  0 120
       Cj-Zj  0   -1   -6  0    

Step 4: Read the answer. All Cj − Zj ≤ 0, so the dual is solved: y₁ = 10, y₂ = 0, Max W = 120.

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: x = 6 (under s₁), y = 0 (under s₂).

Check: Z = 20(6) + 50(0) = 120.

Answer: x = 6, y = 0, Minimum Z = 120.

Note: in Table 2 the two ratios tied (10 and 10), so the next solution is degenerate (s₂ = 0 in Table 3).

Remark (alternative optimum). The cost line 20x + 50y = 10(2x + 5y) is parallel to the first constraint 2x + 5y = 12. So the point (1, 2), which lies on that constraint and also satisfies 3x + 7y ≥ 17, gives the same Z = 20 + 100 = 120. Every point on the segment from (6, 0) to (1, 2) is optimal; the simplex method reports the corner (6, 0).