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Consider  the following LPP:

$\begin{array}{ll} \text{Min.} Z= & x_{1}+x_{2}+x_{3} \\ \text{Subject to } & 3x_{1}+4x_{3}\leq 5 \\ & 5x_{1}+x_{2}+6x_{3}=7 \\ & 8x_{1}+9x_{3}\geq 2, \\ &x_{1},x_{2},x_{3} \geq 0 \end{array}$

The standard form of this LPP shall be:

1.  $\begin{array}{ll} \text{Min.}Z= & x_{1}+x_{2}+x_{3}+0x_{4}+0x_{5}\\ \text{Subject to} & 3x_{1}+4x_{3}+x_{4}=5;\\ & 5x_{1}+x_{2}+6x_{3}=7;\\ & 8x_{1}+9x_{3}-x_{5} = 2; \\ & x_{1},x_{2},x_{3},x_{4},x_{5}\geq 0 \end{array}\\$
2. $\begin{array}{ll} \text{Min.}Z= & x_{1}+x_{2}+x_{3}+0x_{4}+0x_{5}-1(x_{6})-1(x_{7})\\ \text{Subject to} & 3x_{1}+4x_{3}+x_{4}= 5;\\ & 5x_{1}+x_{2}+6x_{3}+x_{6}=7;\\ & 8x_{1}+9x_{3}-x_{5}+x_{7}= 2;\\ & x_{1}\ \text{to} \ x_{7}\geq 0 \end{array}$
3. $\begin{array}{ll} \text{Min.}Z= & x_{1}+x_{2}+x_{3}+0x_{4}+0x_{5}+0x_{6} \\ \text{Subject to} & 3x_{1}+4x_{3}+x_{4}=5;\\ & 5x_{1}+x_{2}+6x_{3}=7;\\ &8x_{1}+9x_{3}-x_{5}+x_{6} =2;\\ &x_{1}\ \text{to}\ x_{6}\geq 0 \end {array}\\$
4. $\begin{array}{ll} \text{Min.}Z= & x_{1}+x_{2}+x_{3}+ 0x_{4}+ 0x_{5}+ 0x_{6}+ 0x_{7}\\ \text{Subject to} & 3x_{1}+4x_{3}+x_{4}=5;\\ & 5x_{1}+x_{2}+6x_{3}+x_{6}=7\\ & 8x_{1}+9x_{3}-x_{5}+x_{7}= 2; \\ & x_{1}\ \text{to} \ x_{7} \geq 0 \end{array}$

To convert into standard for  we need to convert  <= and >=  equations into = form  for this either some extra variables (slack variable or surplus variable ) are added or subtracted from given equation and that variable with 0 coefficient is included in objective function

so 3x1 + 4x3 + x4 = 5   ( slack variable x4 is added)

5x1+x2+6x3 =7    (already in = form)

8x1 + 9x3 - x5 =2  (surplus x5 is subtarcted)

now objective function becomes

Minimize x1+x2+x3+0x4+0x5

so ans is A

https://ocw.mit.edu/courses/sloan-school-of-management/15-053-optimization-methods-in-management-science-spring-2013/tutorials/MIT15_053S13_tut06.pdf

A is right

Use kill by options method

Option B C D does not satisfy the criteria of basic slack and surplus variable for converting an inequality to equality.