Constraint in an LP model restricts
A. value of the objective function
B. value of the decision variables
C. use of the available resources
D. all the options
ANSWER: D
Constraints in an LP model represents
A. limitations
B. requirements
C. balancing limitations and requirements
D. all the options
ANSWER: D
While solving a LP model graphically, the area bounded by the constraints is called
A. feasible region
B. infeasible region
C. unbounded solution
D. None of the options
ANSWER: A
While plotting constraints on a graph paper terminal points on both the axes are connected by a straight
line because
A. the resources are limited in supply.
B. the objective function is a linear function.
C. the constraints are linear equation or inequalities.
D. All the options.
ANSWER: C
Constraint in an LP model becomes redundant because
A. two iso profit lines may be parallel to each other.
B. the solution is unbounded
C. this constraint is not satisfied by the solution values.
D. None of the options.
ANSWER: D
If two constraints do not intersect in the positive quadrant of the graph then
A. the problem is infeasible.
B. the solution is unbounded.
C. One of the constraints is redundant.
D. None of the options
ANSWER: A
Alternative solutions exist of an LP model when
A. one of the constraints is redundant.
B. objective functions equation is parallel to one of the constraints.
C. two constraints are parallel.
D. All the options
ANSWER: B
If a non redundant constraint is removed from a LP problem then
A. feasible region will become larger.
B. feasible region will become smaller.
C. Solution will become infeasible.
D. None of the options.
ANSWER: A
For a maximization problem the objective function coefficient for an artificial variable is
A. +M
B. -M
C. Zero
D. None of the options
ANSWER: B
In the optimal simplex table cj-zj=0 value indicates
A. unbounded solution
B. cycling
C. Alternative solution
D. infeasible solution.
ANSWER: C
If for a given solution a slack variable is equal to zero then
A. the solution is optimal.
B. the solution is infeasible.
C. the entire amount of resource with the constraint in which the slack variable appears has been
consumed.
D. All the options
ANSWER: C
The dual of the primal maximization LP problem having m constraints and n non negative variables
should
A. have n constraints and m non negative variables.
B. be a minimization LP problem
C. Both the options
ANSWER: C
The dummy source or destination in a transportation problem is added to
A. satisfy rim condition
B. prevent solution from becoming degenerate
C. ensure that total cost does not exceed a limit
D. None of the options
ANSWER: A
The solution to a transportation problem with m rows and n columns is feasible if the number of positive
allocations are
A. mxn
B. m+n
C. m+n-1
D. m+n+1
ANSWER: C
While solving an assignment problem an activity is assigned to a resource through a square with zero
opportunity cost because the objective is to
A. Minimize the total cost of assignment.
B. Reduce the cost of assignment to zero.
C. Reduce the cost of that particular assignment to zero.
D. All the options
ANSWER: A
If all incoming variable column is negative then the solution is
A. Unbounded solution
B. Infeasible solution
C. Feasibel Solution
D. None of the options
ANSWER: A
The role of artificial variables in the simplex method is
A. To aid in finding the initial solution.
B. To find optimal dual prices in the final simplex table.
C. To start phases of simplex method.
D. All the options.
ANSWER: D
If an artificial variable is present in the basic variable column of optimal simplex table then the solution
is
A. Solution is unbounded
B. Solution is infeasible
C. Degenerate
D. None of the options
ANSWER: B
One disadvantage of using North-West corner rule to find initial solution to the transportation problem
is that
A. It is complicated to use.
B. It does not take into account cost of transportation.
C. It leads to a degenerate initial solution
D. All the options.
ANSWER: B
Which of the following method is used to verify the optimality of the current solution of the
transportation problem
A. Least cost method
B. Vogel approximation method
C. Modified distribution method
D. All the options
ANSWER: C
Problem 4 (15 Points) An oil refining company can purchase crude oil from 2 countries: Kuwait and
Canada. One barrel of crude oil yields useable gasoline, jet fuel, and lubricant, in barrels, according to
the following table (there is a 10% loss due to waste in the refining process): Product Yield (in barrels)
Kuwait Canada 0.3 0.4 0.2 0.4 0.2 0.3 Gasoline Jet Fuel Lubricant Kuwait can offer up to 9000 barrels of
crude oil at a cost of $20 per barrel, while Canada can offer up to 6000 barrels of crude oil at a cost of
S15 per barrel. To keep up with demand, this company must produce 2000 barrels of gasoline, 1500
barrels of jet fuel, and 500 barrels of lubricant. Answer using any approach you like, just make sure to
write a thorough mathematical formulation. 4.1 If the company purchases 100% of crude oil from
Kuwait to meet the minimum business requirements, what cost does the company incur? 4.2 If the
company wishes to minimize cost, how much crude oil should they purchase fronm each country?
(Blending & Refining LP optimization problem) American Oil of Middle East has a refinery in Kuwait.... HomeworkLib
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time
and 3 hours of craftmans time in its making while a cricket bat takes 3 hour of machine time and 1 hour
of craftmans time. In a day, the factory has the availability of not more than 42 hours of machine time
and 24 hours of craftsmans time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is Rs.20 and Rs.10 respectively, find the maximum profit of the
factory when it works at full capacity.
Medium
Solution
Verified by Toppr
Let number of tennis racket be made is X and number of cricket bat be made is Y
Since, tennis bat requires 1.5 hours and cricket bat requires 3 hours of machine time. Also, there is
maximum 42 hours of machine time available.
∴1.5X+3Y≤42
⇒X+2Y≤28 ...(1)
Since, tennis bat requires 3 hours and cricket bat requires 1 hours of craftmans time. Also, there is
maximum 24 hours of craftmans time available.
∴3X+Y≤24 ...(2)
Since, count of an object can't be negative.
∴X≥0,Y≥0 ...(3)
We have to maximize profit of the factory.
Here, profit on tennis racket is 20 Rs and on cricket bat is 10 Rs
So, objective function is Z=20X+10Y
Plotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the
image.
Corner points Value of Z=20X+10Y
A (0,14)
140
B (4,12)
200 (maximum)
C (8,0)
160
Hence,
(i) 4 tennis rackets and 12 cricket bats must be made so that factory will work at full capacity.
(ii) Maximum profit of factory will be 200 Rs
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the
time (in minutes) required for each toy on the machines is given below:
Types of Toys Machines Machines Machines
I
II
III
A
12
18
6
B
6
0
9
Each machine is available for a maximum of 6 hours per day. If the profit on each toy of
type A is Rs.7.50 and that on each toy of type B is Rs.5, show that 15 toys of type A and 30 of
type B should be manufactured in a day to get maximum profit.
Solution
Verified by Toppr
Let's assume that number of toys of type A be X and number of toys of type B be Y.
Since, toy A need 12 minutes and toy B need 6 minutes of machine I time. Also machine I is available for
maximum 6 hours ( 360 minutes ).
So, 12X+6Y≤360
⇒2X+Y≤60 ...(1)
Since, toy A need 18 minutes and toy B need 0 minutes of machine II time. Also machine I is available for
maximum 6 hours ( 360 minutes ).
So, 18X+0Y≤360
⇒X≤20 ...(2)
Since, toy A need 6 minutes and toy B need 9 minutes of machine III time. Also machine III is available
for maximum 6 hours ( 360 minutes ).
So, 6X+9Y≤360
⇒2X+3Y≤120 ...(3)
Since, count of toys can't be negative.
∴X≥0 and Y≥0 ...(4)
Now, profit on toy A is 7.50 Rs and profit on toy B is 5 Rs
So, total profit (Z)=7.5X+5Y
We have to maximize the total profit of manufacturer.
After plotting all the constraints give by equation (1), (2), (3) and (4), we get the feasible region as shown
in the image.
Corner points Value of Z=7.5X+5Y
A (0, 40)
200
B (15, 30)
262.50 (Maximum)
C (20, 20)
250
D (20, 0)
150
Hence maximum profit that manufacturer can earn is 262.50 Rs when 15 toys of type A and 30 toys of
type B manufactured.
A producer of helical gears manufactures two different quality rated versions of the same product. Both
the gears require three machines designated 1, 2, 3 for their fabrication. Gear A uses three hours, one
hour and one hour respectively of machining time. Gear A yields a profit of $50 per unit. Gear B uses one
hour, one hour and two hours respectively of machining time. Gear B yields a profit of $70 per unit.
Machine 1 is available 24 hours a day, but due to other product line usage, Machine 2 is only available
10 hours a day and Machine 3 is available 16 hours a day. As a senior engineer, it is your job to
determine the most profitable mix of production level of both gears A and B. a) Using the method of
Linear Programming, develop the merit function, constraints and bounds for this problem. b) Determine
the most profitable number of gears of each type to manufacture as well as the profit. (25 pts.)
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