MCQ
The __________ is the method available for solving an L.P.P
  • A
    Graphical method
  • B
    Least cost method
  • C
    MODI method
  • D
    Hungarian method

Answer

  1. Graphical method

Solution:

There are different methods to solve an linear programming problem.

Such as Graphical method, Simplex method, Ellipsoid method, Interior point methods. 

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

$\int_{}^{} {{e^x}\frac{{({x^2} + 1)}}{{{{(x + 1)}^2}}}dx = } $
The angle between lines $3x + 2y + z = 0 = x + y -2z$ and $2x -y -z = 0 = 7x + 10y -8z$ is
The value of $\sin ^{ -1 }{ \left( \cos { \frac { 53\pi }{ 5 } } \right) }=\sin ^{ -1 }{ \left( \cos { \frac { 50\pi+3\pi }{ 5 } } \right) }:$
  1. $ \frac {- \pi }{ 1 }$
  2. $ \frac {- \pi }{ 7 }$
  3. $ \frac { \pi }{ 10 }$
  4. $ \frac {- \pi }{ 10 }$
Choose the correct answer from the given four options:
let $\text{P}(\text{A})=\frac{7}{13},\text{P}(\text{B})=\frac{9}{13}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{13}.$ Then $\text{P}\Big(\frac{\text{A'}}{\text{B}}\Big)$ is equal to:
The equation x2 - x - 2 = 0 in three dimensional space is represented by:
  1. A pair of parallel planes
  2. A pair of straight lines
  3. A pair of perpendicular plane
  4. None of these
A particle moves in a straight line with a velocity given by $\frac{{dx}}{{dt}} = x + 1$($x$ is the distance described). The time taken by a particle to traverse a distance of $99$ metre is
If A and B are two independent events with $\text{P(A)}=\frac{3}{5}$ and $\text{P(B)}=\frac{4}{9},$ then $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$ equals,
  1. $\frac{4}{15}$
  2. $\frac{8}{45}$
  3. $\frac{1}{3}$
  4. $\frac{2}{9}$
The orthogonal projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is:
  1. $\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{a}}}{|\vec{\text{a}}|^2}$
  2. $\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{b}}}{\big|\vec{\text{b}}\big|^2}$
  3. $\frac{\vec{\text{a}}}{|\vec{\text{a}}|}$
  4. $\frac{\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
If $tan^{-1 } \frac{1}{1+2}+tan^{-1 } \frac{1}{1+(2)(3)}+ tan^{-1 }\frac{1}{1+(3)(4)}+$ ......$tan^{-1}\frac{1}{{1 + n(n + 1)}}=tan^{-1} \theta,$ then $\theta$ =
$\int {\frac{{2\left( {{x^3} - 1} \right)}}{{x\left( {2{x^3} + 1} \right)}}} \,dx$ is equal to (where $C$ is the constant of integration)