← 2013 Paper 2
UPSC 2013 Maths Optional Paper 2 Q4c — Step-by-Step Solution
20 marks · Section A
Simplex method (basic) · Linear Programming · asked 9× in 13 yrs · Read the full method →
Question
Minimize z=5x1−4x2+6x3−8x4 subject to the constraints
x1+2x2−2x3+4x4≤40,2x1−x2+x3+2x4≤8,4x1−2x2+x3−x4≤10,xi≥0.
Technique
Standard simplex with slack variables; two pivots reach optimum.
Solution
Strategy. Standard simplex with slack variables. Since all constraints are ≤ with non-negative RHS, the slacks form a feasible starting basis.
Introduce slacks s1,s2,s3≥0:
x1+2x2−2x3+4x4+s12x1−x2+x3+2x4+s24x1−2x2+x3−x4+s3=40,=8,=10.
Min z=5x1−4x2+6x3−8x4 (zero objective contribution from slacks).
Initial BFS: xj=0, (s1,s2,s3)=(40,8,10), z=0.
Step 2 — Iteration 1: x4 enters, s2 leaves
For minimisation, entering variable = one with most positive zj−cj. Compute zj−cj=−cj (initial cB=0):
| Var | zj−cj |
|---|
| x1 | −5 |
| x2 | +4 |
| x3 | −6 |
| x4 | +8 ← largest |
Min-ratio test on x4 column: 40/4=10, 8/2=4, 10/(−1) skip (negative coefficient). Min = 4 at s2.
Pivot on entry (s2,x4)=2. Resulting tableau (after row operations):
| Basis | x1 | x2 | x3 | x4 | s1 | s2 | s3 | RHS |
|---|
| s1 | −3 | 4 | −4 | 0 | 1 | −2 | 0 | 24 |
| x4 | 1 | −1/2 | 1/2 | 1 | 0 | 1/2 | 0 | 4 |
| s3 | 5 | −5/2 | 3/2 | 0 | 0 | 1/2 | 1 | 14 |
cB=(0,−8,0), z=−32.
Step 3 — Iteration 2: x2 enters, s1 leaves
Recompute zj−cj:
| Var | zj | cj | zj−cj |
|---|
| x1 | −8 | 5 | −13 |
| x2 | 4 | −4 | +8 ← largest |
| x3 | −4 | 6 | −10 |
| s2 | −4 | 0 | −4 |
Largest positive: x2 (with +8). Enter x2.
Min-ratio on x2 column: s1: 24/4=6; x4: −1/2 (skip); s3: −5/2 (skip). Min = 6 at s1.
Pivot on (s1,x2)=4. Resulting tableau:
| Basis | x1 | x2 | x3 | x4 | s1 | s2 | s3 | RHS |
|---|
| x2 | −3/4 | 1 | −1 | 0 | 1/4 | −1/2 | 0 | 6 |
| x4 | 5/8 | 0 | 0 | 1 | 1/8 | 1/4 | 0 | 7 |
| s3 | 25/8 | 0 | −1 | 0 | 5/8 | −3/4 | 1 | 29 |
cB=(−4,−8,0), z=−4(6)+(−8)(7)=−24−56=−80.
Step 4 — Optimality check
Compute zj−cj for non-basic variables:
| Var | zj | cj | zj−cj |
|---|
| x1 | −4(−3/4)+(−8)(5/8)+0=3−5=−2 | 5 | −7 |
| x3 | −4(−1)+(−8)(0)+0(−1)=4 | 6 | −2 |
| s1 | −4(1/4)+(−8)(1/8)+0=−1−1=−2 | 0 | −2 |
| s2 | −4(−1/2)+(−8)(1/4)+0=2−2=0 | 0 | 0 |
All zj−cj≤0 — optimal. The value zs2−cs2=0 indicates alternative optima exist along the edge into s2.
Step 5 — Optimum
Answer
x1=0,x2=6,x3=0,x4=7;zmin=−80.