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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Linear Programming

BITSAT 2015 · Mathematics · Linear Programming

Minimise Z=∑_ j=1 ^ n ∑_ i=1 ^ m c_ i j x_ i j Subject to ∑_ i=1 ^...

Minimise Z=∑j=1n∑i=1mcijxijZ=\sum_{j=1}^{n} \sum_{i=1}^{m} c_{i j} x_{i j} Subject to ∑i=1mxij=bj,j=1,2,…,n∑j=1nxij=bj,i=1,2,…,m\sum_{i=1}^{m} x_{i j}=b_{j}, j=1,2, \ldots, n\sum_{j=1}^{n} x_{i j}=b_{j}, i=1,2, \ldots, m is a LPPL P P with number of constr

Options

  1. Am−nm-n
  2. Bmm
  3. Cm+n\mathrm{m}+\mathrm{n}
  4. Dmn\frac{\mathrm{m}}{\mathrm{n}}

Correct answer

C. m+n\mathrm{m}+\mathrm{n}

Step-by-step solution

Constraints will be x11+x21+….+xml=b1x12+x22+……xm2=b2xln⁡+x2n+….+xmn=bnx11+x12+…+x1n=b1x21+x22+…+x2n=b2xml+xm2+…+xmn=bn\mathrm{x}_{11}+\mathrm{x}_{21}+\ldots.+\mathrm{x}_{\mathrm{ml}}=\mathrm{b}_{1} \mathrm{x}_{12}+\mathrm{x}_{22}+\ldots \ldots \mathrm{x}_{\mathrm{m} 2}=\mathrm{b}_{2}\mathrm{x}_{\ln }+\mathrm{x}_{2 \mathrm{n}}+\ldots.+\mathrm{x}_{\mathrm{mn}}=\mathrm{b}_{\mathrm{n}}\mathrm{x}_{11}+\mathrm{x}_{12}+\ldots+\mathrm{x}_{1 \mathrm{n}}=\mathrm{b}_{1}\mathrm{x}_{21}+\mathrm{x}_{22}+\ldots+\mathrm{x}_{2 \mathrm{n}}=\mathrm{b}_{2}\mathrm{x}_{\mathrm{ml}}+\mathrm{x}_{\mathrm{m} 2}+\ldots+\mathrm{x}_{\mathrm{mn}}=\mathrm{b}_{\mathrm{n}} So, total number of constraints =m+n=\mathrm{m}+\mathrm{n}
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