Tag: linear programming problem

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Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

In transportation models designed in linear programming, points of demand is classified as

  1. ordination

  2. transportation

  3. destinations

  4. origins
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In linear programming, transportation modeltransportation model are applied to problems related to the study of efficient transportation routes. i.e., how effectively the available resources are transported to different destinations with minimum cost.

Therefore, the points of demand is classified as destinations.

Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

Consider the following linear programming problem:

Maximize $12X + 10Y$
Subject to: $4X + 3Y ≤ 480$
  $2X + 3Y ≤ 360$
  all variables $ ≥0$

Which of the following points $(X,Y)$ could be a feasible corner point?

  1. $(40,48)$
  2. $(120,0)$
  3. $(180,120)$
  4. $(30,36)$

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given constraints, $4x+3y\leq 480$ and $2x+3y\leq 360$


first, draw the graph for equations $4x+3y= 480$ and $2x+3y= 360$

for $4x+3y= 480$
substitute y=0 we get, $4x=480 \implies x=120$
substitute x=0 we get, $3y=480 \implies y=160$
therefore, $4x+3y= 480$ line passes through (120,0) and (0,160) as shown in fig.
Hence, $4x+3y\leq 480$ includes the region below the line.


for $2x+3y= 360$
substitute y=0 we get, $2x=360 \implies x=180$
substitute x=0 we get, $3y=360 \implies y=120$
therefore, $4x+3y= 480$ line passes through $(180,0)$ and $(0,120)$ as shown in fig.
Hence, $2x+3y\leq 360$ includes the region below the line.

the shaded region as shown in figure is intersection region. From the figure, $(0,120)$ and $(120,0)$ are feasible corner points.

Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

Consider the following linear programming problem:

Maximize $12X + 10Y$
Subject to: $4X + 3Y ≤ 480$
  $2X + 3Y ≤ 360$
all variables $ ≥0$

Which of the following points $(X,Y)$ is feasible?

  1. $(10,120)$
  2. $(120,10)$
  3. $(30,100)$
  4. $(60,90)$

  5. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider point $(10,120)$ , It doesnt satisfy the inequality $2X+3Y \le 360$

Consider point $(120,10)$ , it doesnt satisfy the inequality $4X+3Y \le 480$
Consider point $(30,100)$ , it satisfies all the inequalities . So this point is feasible
Consider point $(60,90)$ , it doesnt satisfies the inequalities $2X+3Y \le 360$ and $4X+3Y \le 480$
Therefore the correct option is $C$

Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

Unboundedness is usually a sign that the LP problem.

  1. has finite multiple solutions.

  2. is degenerate.

  3. contains too many redundant constraints.

  4. has been formulated improperly.

  5. none of the above.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A linear programming problem is said to have unbounded solution if it has infinite number of solutions. I.e., the problem has been formulated improperly

Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

The first step in formulating an LP problem is

  1. graph the problem.

  2. perform a sensitivity analysis.

  3. identify the objective and the constraints.

  4. define the decision variables.

  5. understand the managerial problem being faced.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

The first step in formulating an linear programming problem is to understand the managerial problem being faced i.e., determine the quantities that are needed to solve the problem.

Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

Consider the following linear programming problem:

Maximize $5X + 6Y$
Subject to: $4X + 2Y ≤ 420$
  $1X + 2Y ≤ 120$
  all variables  $≥0$

Which of the following points $(X,Y)$ is in the feasible region?

  1. $(30,60)$
  2. $(105,0)$
  3. $(0,210)$
  4. $(100,10)$

  5. None of the above
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation
Feasible points are the points that satisfy the constraints.
Therefore, substitute the options in the constraint equations and verify. 

$\mathrm A.$ substituting  $(30,60)$ in $4x+2y\leq 420$
we get $4\times 30+2\times 60\leq 420$
$\implies 120+120 \leq 420 \implies 240\leq 420$ True
substituting  $(30,60)$ in $1x+2y\leq 120$
we get $1\times 30+2\times 60\leq 120$
$\implies 30+120 \leq 120 \implies 150\leq 120$ False

$\mathrm B.$ substituting  $(105,0)$ in $4x+2y\leq 420$
we get $4\times 105+2\times 0\leq 420$
$\implies 420+0 \leq 420 \implies 420\leq 420$ True
substituting  $(105,0)$ in $1x+2y\leq 120$
we get $1\times 105+2\times 0\leq 120$
$\implies 105+0 \leq 120 \implies 105\leq 120$ True

$\mathrm C.$ substituting  $(0,210)$ in $4x+2y\leq 420$
we get $4\times 0+2\times 210\leq 420$
$\implies 0+420 \leq 420 \implies 420\leq 420$ True
substituting  $(0,210)$ in $1x+2y\leq 120$
we get $1\times 0+2\times 210\leq 120$
$\implies 0+240 \leq 120 \implies 240\leq 120$ False


$\mathrm D.$ substituting  $(100,10)$ in $4x+2y\leq 420$
we get $4\times 100+2\times 10\leq 420$
$\implies 400+20 \leq 420 \implies 420\leq 420$ True
substituting  $(100,10)$ in $1x+2y\leq 120$
we get $1\times 100+2\times 10\leq 120$
$\implies 100+20 \leq 120 \implies 120\leq 120$ True

Therefore option B and D are the points in the feasible region.

Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

In order for a linear programming problem to have a unique solution, the solution must exist

  1. at the intersection of the nonnegativity constraints.

  2. at the intersection of a nonnegativity constraint and a resource constraint.

  3. at the intersection of the objective function and a constraint.

  4. at the intersection of two or more constraints.

  5. none of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In order for a linear programming problem to have a unique solution, the solution must exist at the intersection of two or more constraints. Then the problem becomes convex and has a single optimum(maximum or minimum) solution. 

Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

Consider the following linear programming problem:

Maximize $5X + 6Y$
Subject to: $4X + 2Y ≤ 420$
  $1X + 2Y ≤ 120$
  all variables $≥ 0$

Which of the following points $(X,Y)$ is feasible?

  1. $(50,40)$
  2. $(30,50)$
  3. $(60,30)$
  4. $(90,20)$

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Feasible points are the points that satisfy the constraints.
Therefore, substitute the options in the constraint equations and verify. 

$\mathrm A.$ substituting  $(50,40)$ in $4x+2y\leq 420$
we get $4\times 50+2\times 40\leq 420$
$\implies 200+80 \leq 420 \implies 280\leq 420$ True
substituting  $(50,40)$ in $1x+2y\leq 120$
we get $1\times 50+2\times 40\leq 120$
$\implies 50+80 \leq 120 \implies 130\leq 120$ False

$\mathrm B.$ substituting  $(30,50)$ in $4x+2y\leq 420$
we get $4\times 30+2\times 50\leq 420$
$\implies 120+100 \leq 420 \implies 220\leq 420$ True
substituting  $(30,50)$ in $1x+2y\leq 120$
we get $1\times 30+2\times 50\leq 120$
$\implies 30+100 \leq 120 \implies 130\leq 120$ False

$\mathrm C.$ substituting  $(60,30)$ in $4x+2y\leq 420$
we get $4\times 60+2\times 30\leq 420$
$\implies 240+60 \leq 420 \implies 300\leq 420$ True
substituting  $(60,30)$ in $1x+2y\leq 120$
we get $1\times 60+2\times 30\leq 120$
$\implies 60+60 \leq 120 \implies 120\leq 120$ True

$\mathrm A.$ substituting  $(90,20)$ in $4x+2y\leq 420$
we get $4\times 90+2\times 20\leq 420$
$\implies 360+40 \leq 420 \implies 400\leq 420$ True
substituting  $(90,20)$ in $1x+2y\leq 120$
we get $1\times 90+2\times 20\leq 120$
$\implies 90+40 \leq 120 \implies 130\leq 120$ False

Therefore option C (60,30) is the feasible point

Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

Which of the following statements about an LP problem and its dual is false?

  1. If the primal and the dual both have optimal solutions, the objective function values for both problems are equal at the optimum

  2. If one of the variables in the primal has unrestricted sign, the corresponding constraint in the dual is satisfied with equality

  3. If the primal has an optimal solution, so has the dual

  4. The dual problem might have an optimal solution, even though the primal has no (bounded) optimum
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

if one of the problems(primal, dual) is infeasible then the other problem is infeasible. Hence, the option D is the false statement.

Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

Mark the wrong statement:

  1. The primal and dual have equal number of variables.

  2. The shadow price indicates the change in the value of the objective function, per unit increase in the value of the RHS.

  3. The shadow price of a non-binding constraint is always equal to zero.

  4. The information about shadow price of a constraint is important since it may be possible to purchase or, otherwise, acquire additional units of the concerned resource.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The number of variables in dual is equal to the number of constraints in the primal and the  number of variables in primal is equal to the number of constraints in the dual.


Therefore, the primal and dual doesn't have equal number of variables.