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- 2082 Baishakh · 7 marks
- 2081 Baishakh · 8 marks
Duality LPP min:Z=8x+9y
Short answer
Construct dual: Maximize W=4y₁+5y₂ subject to y₁+2y₂≤8, 3y₁+y₂≤9, y₁,y₂≥0. Solve by simplex: y₁=2, y₂=3, W=23. From final Cj-Zj row, primal x=11/5, y=3/5, Min Z=23.
Solve the following Linear Programming Problem by constructing the dual: Minimize Z = 8x + 9y subject to x + 3y ≥ 4, 2x + y ≥ 5 and x, y ≥ 0.
Step 1: Write the dual. For the primal (minimisation, all constraints ≥), the dual is:
Primal: Minimize Z = 8x + 9y
subject to x + 3y ≥ 4
2x + y ≥ 5
x, y ≥ 0
Dual: Maximize W = 4y₁ + 5y₂
subject to y₁ + 2y₂ ≤ 8
3y₁ + y₂ ≤ 9
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 = 4y₁ + 5y₂
subject to
y₁ + 2y₂ + s₁ = 8
3y₁ + y₂ + s₂ = 9
Step 3: Simplex tables for the dual
Table 1
Cj 4 5 0 0
Basis CB y₁ y₂ s₁ s₂ b Ratio
--------------------------------
s₁ 0 1 2 1 0 8 4
s₂ 0 3 1 0 1 9 9
--------------------------------
Zj 0 0 0 0 0
Cj-Zj 4 5 0 0
Entering: y₂ Leaving: s₁ Pivot = 2
Table 2
Cj 4 5 0 0
Basis CB y₁ y₂ s₁ s₂ b Ratio
------------------------------------
y₂ 5 1/2 1 1/2 0 4 8
s₂ 0 5/2 0 -1/2 1 5 2
------------------------------------
Zj 5/2 5 5/2 0 20
Cj-Zj 3/2 0 -5/2 0
Entering: y₁ Leaving: s₂ Pivot = 5/2
Table 3
Cj 4 5 0 0
Basis CB y₁ y₂ s₁ s₂ b
--------------------------------
y₂ 5 0 1 3/5 -1/5 3
y₁ 4 1 0 -1/5 2/5 2
--------------------------------
Zj 4 5 11/5 3/5 23
Cj-Zj 0 0 -11/5 -3/5
Step 4: Read the answer. All Cj − Zj ≤ 0, so the dual is solved: y₁ = 2, y₂ = 3, Max W = 23.
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 = 11/5 (under s₁), y = 3/5 (under s₂).
Check: Z = 8(11/5) + 9(3/5) = 23.
Answer: x = 11/5, y = 3/5, Minimum Z = 23.
✓ Answer checked by Claude Opus 5.5 (high effort)
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