← 2015 Paper 2

UPSC 2015 Maths Optional Paper 2 Q3c-ii — Step-by-Step Solution

10 marks · Section A

LPP: standard form; basic, basic feasible, optimal solutions · Linear Programming · asked 6× in 13 yrs · Read the full method →

Question

Without solving the problem, show that it has an optimal solution. Which of the basic feasible solution(s) is/are optimal?

Technique

Substitute equality constraints to reduce ZZ to a single variable x3x_3; observe Z=207x3Z=20-7x_3 is bounded above (since x30x_3\ge 0); by LPP fundamental theorem, optimum is attained; check each BFS.

Solution

LPP (recap):

Maximize Z=x1+2x23x3+4x4,x1+x2+2x3+3x4=12,  x2+2x3+x4=8,  xi0.\text{Maximize } Z=x_1+2x_2-3x_3+4x_4,\quad x_1+x_2+2x_3+3x_4=12,\;x_2+2x_3+x_4=8,\;x_i\ge 0.

Step 1 — Existence of optimal solution

Theorem (LPP fundamental). A linear program with non-empty feasible region and bounded objective on the feasible region attains its optimum at an extreme point (= basic feasible solution).

Non-empty feasible region: From Q3(c)(i), we have at least one BFS — e.g., (4,8,0,0)(4,8,0,0) — so the feasible region is non-empty.

Boundedness of ZZ above: Let’s show ZZ is bounded above on the feasible region.

From the second constraint x2+2x3+x4=8x_2+2x_3+x_4=8, we have x4=8x22x3x_4=8-x_2-2x_3.

Substitute into the first: x1+x2+2x3+3(8x22x3)=12x_1+x_2+2x_3+3(8-x_2-2x_3)=12, so x1=12x22x324+3x2+6x3=12+2x2+4x3x_1=12-x_2-2x_3-24+3x_2+6x_3=-12+2x_2+4x_3.

So x1=2x2+4x312x_1=2x_2+4x_3-12 and x4=8x2x4x_4=8-x_2-x_4 \dots wait, x4=8x22x3x_4=8-x_2-2x_3.

Substitute into ZZ:

Z=x1+2x23x3+4x4=(2x2+4x312)+2x23x3+4(8x22x3)Z=x_1+2x_2-3x_3+4x_4=(2x_2+4x_3-12)+2x_2-3x_3+4(8-x_2-2x_3) =2x2+4x312+2x23x3+324x28x3=2x_2+4x_3-12+2x_2-3x_3+32-4x_2-8x_3 =(2+24)x2+(438)x3+(3212)=(2+2-4)x_2+(4-3-8)x_3+(32-12) =0x27x3+20.=0\cdot x_2-7x_3+20.

So Z=207x3Z=20-7x_3 on the feasible region (with x10,x40x_1\ge 0,x_4\ge 0 as additional constraints).

Since x30x_3\ge 0, Z=207x320Z=20-7x_3\le 20, with equality iff x3=0x_3=0. So ZZ is bounded above by 20.

By the fundamental theorem, the LPP has an optimal BFS, attaining Z=20Z=20.

Step 2 — Identify the optimal BFS

Z=207x3Z=20-7x_3 is maximized when x3=0x_3=0. Looking at the 4 BFS from Q3(c)(i):

BFSx3x_3Z=207x3Z=20-7x_3
(4,8,0,0)(4,8,0,0)020
(4,0,4,0)(4,0,4,0)42028=820-28=-8
(0,6,0,2)(0,6,0,2)020
(0,0,3,2)(0,0,3,2)32021=120-21=-1

Two BFS attain the maximum Z=20Z=20: (4,8,0,0)(4,8,0,0) and (0,6,0,2)(0,6,0,2).

Answer

  Zmax=20, attained at BFS (4,8,0,0) and (0,6,0,2).  \boxed{\;Z_{\max}=20\text{, attained at BFS }(4,8,0,0)\text{ and }(0,6,0,2).\;}
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