MTH601-11 Kal

Muhammad Rizwan Ashraf
BS(CS)
Virtual University of Pakistan
MTH 601 Final term Data
Note:- Please confirm the answers by your own too
I made it with a lot of hard work as there was not much data on web for this subject
Please duaon main yad rakhiay ga
While solving an LP problem by the Simplex method, in the standard table, the element at the
intersection of key column and key row is called ------------- element.
Entering
Leaving
Slack
Pivot
In the initial table of Simplex method, the objective function should not contain the terms
involving-----------variables
Artificial
Degenerate
Basic
Non basic
If a balanced Transportation problem with ‘m’ sources and ‘n’ sinks then degeneracy will arise
only if there are less than ---------- independent allocations in the solution.
m–n–1
m+n-1
(In this question all options are in (-) negative range like m–n–1 , m–n–2 m–n–3, m–n–4 )
and there was no one true. correct was m+n–1 .however i select m–n–1 ).
Which of the following type of Elementary matrix operations are performed while solving an
Assignment problem by Hungarian’s method?
Row operations
Column operations
Both of the above
None of the above
Check Not sure
While solving an Assignment problem by Hungarian’s method, in the modified cost matrix if the
minimum number of horizontal and vertical lines to cover zeros are not equal to the number of
rows (or columns), then which of the following operation is done?
Subtract smallest element of uncovered rows from all other elements of uncovered cells
Which of the following is a characteristic of critical path scheduling techniques?

The jobs or tasks are dependent on each other
Interruptions may be permitted if they are documented
The jobs or tasks must be well defined, and their completion marks the end of the project
The jobs or tasks do not necessarily have to be in a predetermined sequence
Search and cutting methods are used to solve following types of problems
Transportation
Assignment
Integer Programming
Queuing problems
A feasible solution is called a _______________ solution if the number of non negative

allocation is equal to “m + n – 1” (Here m is the number of rows, n is the number of columns in
transportation problem)
basic feasible
non basic feasible
basic infeasible
optimal
Consider the following cost table

Destinations
D1 D2 D3 Supply
Source
S1 5 1 2 25
S2 7 2 4 10
S3 3 3 5 15
Demand 10 20 20
Using Vogel’s Approximation Method, which one is the starting cell?
S1 à D1
S3 à D1
S2 à D2
S2 à D1
Using Vogel’s Approximation method:

These below are the three cells which can be next cell:
Which one is the correct option?
I, II and III
I only
II only
III only
I and II only
This is the initial basic feasible solution of a question by North-West Corner Method.
Finding optimal basic feasible solution: Consider U1=0 and V1=5, what is the value of V2?
10
15
25
5
Transportation problem is a special class of linear programming problem in which transport a

commodity from the source to a destination in such a way that the total transportation cost is
______.
Zero
Maximum
Minimum
Average
We solve a assignment problem by subtracting least number in each row and we get the table
below:
After doing next step, which table should be next?
►
►
Selected answer
If the arrival rate is 5 per hour and the service rate is 10 per hour, then the average system time
is:
1/5
5
1
2

If the arrival rate is 5 per hour and the service rate is 10 per hour, then (traffic intensity or
system utilization) =?
If the arrival rate is 5 per hour and the service rate is 10 per hour, then the expected number in
the queue or average queue length is:
Formula for the expected number in the system L is:


L
 

L
 

L


L
 
Total cost per period = Item cost + Order cost + Holding cost + _____________.
Shortage cost
Optimum Shortage (S*)
Economic Oreder Quantity. (Q*)
Maximum Inventory. (I max.)
In Manufacturing Model with no shortage, the replacement rate is finite and ___________ the
demand rate
greater than
less than
equal to
Formula of EOQ in manufacturing model with shortages is
Q*  2C2 D / C3 (1  D / R). C3  C4 / C4
Q*  2C2 D / C3 . R / R  D
Q*  2C2 D / C3
None of them
To identify and maintain the proper precedence relationship between activities those are not
connected by event, we introduce
Parallel Activity
Dummy Activity
Sequential Activity
None of these
PERT is based on
Deterministic times
Probabilistic Times
Dummy Times
Estimated times
Non-critical activities in the following network are -------
3
6
5
1 3
Activity (1,2)
Activity (2,3)
Dummy Actvity
Activity (1,3)
If t0 = 10, tm = 10 and tp = 16, then S.D = _________.

Answer will be 1
Which of the following is an essential characteristic of a linear programming problem?

The relationship between variables and constraints must be linear
The model must have an objective function
The model must have structural constraints
All are equivalent
In the simplex table for a linear programming problem, we select the leaving basic variable
corresponds to --------------
Maximum non-negative ratio
Maximum negative ratio
Minimum non-negative ratio
Minimum negative ratio
A technique for solving a linear programming problems in which artificial variables are included
with coefficients of very large number say equal 10 times of any cost coefficient of decision
variables is known as ---------
Big M – Method.
Least Cost Method
Hungarian Method
Branch and Bound Method
Check
If the __________ variables appear in the final basic set, then the LP (Linear Program) has no
solution.
slack
surplus
non – basic
artificial
In Two – Phase method, if all the _____________ variables become zero, we stop the phase I
and proceed to phase-II.
artificial
basic
non basic
positive slack
negative slack
In two phase method, for the phase-I, a new objective function is expressed as -----.
sum of slack variables
difference of slack variables
difference of artificial variables
sum of artificial variables
Page # 133 – My Ok
If “MaxZ=x–y, subject to x>3, x<2, x,y>0” is solved by two phase method, then which of the
following would be the objective function of 1st phase?
Max Z =A
In two phase method, for the phase-I, the problem has infeasible solution if the minimum value
of objective function
zero
greater than zero
less than zero
In two phase method, for the phase-I, a new objective function in terms of artificial is to be
minimized
Maximized
In two phase method, which of the following will be taken as starting solution for 2nd phase?
Basic solution of 1st phase containing non-zero artificial variables
Non-basic solution of 1st phase containing non-negative slacks
Optimum feasible solution of 1st phase
Infeasible solution of 1st phase
Which of the following order pair would minimize the objective function of the linear
programming problem; z = x + y subject to x≥3, y≥0 ?
(3,3)
(3,0)
(0,3)
(0,0)
In a Transportation Problem, the objective function ’Z’ gives ----------.

Total Cost of transportation
Total Time of transportation
Total Profit of transportation
Total inventory be supplied in transportation
For North West Corner method, in the first row and first column, resource and sink contain ‘5’
and ‘7’ units respectively; then after allocating the appropriate amount ‘x11’ in the cell (1,1), we
will move towards which of the following cell?
missed
To convert an Assignment problem into a maximization problem, which of the following

operations would have to apply?
Deduct all elements of the row from highest element in that row
If the total demand is equal to total supply as per requirement of a balanced transportation
problem i-e “a1+a2+---+an = b1+b2+---+bn” then which of the following is true?
“a1=b1, a2=b2,---, an=bn” is necessarily implied (not sure)
In case the cost elements of one or two cells are not given in the problem, it means that ---------
The routes connected by those cells are not available
If the cost matrix in an Assignment problem is not square then which the following modification
will be made to balance the given problem?
Add a dummy row(column) with negative cost elements (not sure)
The cost coefficient of artificial variable in Objective function is -------------.
M (sure)
While solving a Linear Programming problem by Simplex Method, an inequality saysμ‘2x-3y<5’

is transformed into strict equality of the form---------, where ‘s>=0’.
2x-3y +s =5
In the Simplex method to solve an LP problem, Gauss Jordan Elimination method demands that
all the key column's entries should be --------- except key row(pivot) entry.
strictly negative
While solving an LP problem by Simplex method, the inclusion of slacks in the constraints’
inequalities helps in finding ------------- variables
Check for options
In a feasible region, if the end points of the line segment are basic solutions, then all the points
between these two will ------------.
also be the basic
In a feasible region, if the end points of a line segment are basic solutions which BOTH also give
the optimal solution, then ------------ will also serve for optimal.
the point which divides segment in ratio 1:2
After converting constraints into the respective Standard equalities, we have an LP problem of
‘4’ equations in ‘6’ variables, if the initial basic feasible solution is say;(2,4,0,2), then it is --------
-- solution.
degenerate feasible
The positive slack variables indicate -----------------.

excess resource available (sure)
If in a LP problem, the objective is to maximize: z = 2x+3y, with all associated constraints of

‘=<’ type, then in the 1st iteration, z = ---------.
0
In M-method, which of the following is the standard equality corresponds to the

constraintμ“3x+5y=2”?
3x+2y +A=2
3x+2y–A=2
3x+2y–MA=2
3x+2y+MA=2
In the final iteration of M-method to obtain the optimal solution, the artificial variable must-------
the basis.
add in each variable of
subtract from each variable of
leave
include in
If Primal Problem is a maximization problem, then the dual will be -------

minimization Problem sure
Graphically, in case of optimal degenerate solution,feasible solution always exists at corner

points -----------.
lie on any axis
The number of variables in the Primal will be the number of -------- in Dual and vice versa.
constraints sure
When the total allocations in a transportation model of m*n size is not equals to “m + n – 1”, the
situation is known as:
Unbalanced situation
In North West Corner method, the first step after choosing the appropriate cell in 1st row, we
allocate -------------so that the capacity of first row or first column is exhausted.
as much as possible
The row, which is introduced in the matrix to balance an unbalnced Transportation problem, is
known as ----------.
dummy row (conform)
The solution of a transportation problem with m rows (supplies) and n (destinations) is feasible if
numbers of positive allocations are
m+n-1 (conform)
The amounts shipped from a dummy source represent shortages at the receiving destinations
True (conform)
False
In the Vogel’s approximation Method for solving a Transportation problem, Penalty measure for
any row or column, is given by which of the following?
Difference between the smallest unit cost to the next smallest cost in the same row(column)
While solving a LP problem by Simplex method in a given iteration, the new basic variable is ---
------- and the variable remove from the basis is called ---------
Entering Variable, Leaving variable
Page # 113 - My Ok
Due to which of the following reason, Simplex method is not preferred to solve a Transportation
problem
Since it contains large number of decision variables ‘xij’s’ so that it becomes complicated
Since Transportation problem contains constraints of ‘=’ type
Since Transportation problem does not contain positive slacks
Since Transportation problem does not contain negative slacks
If a variable in the Primal is unrestricted in sign, then the corresponding constraint in the dual
will be of -------- type and vice versa
>=
<=
=
None
In M-method, which of the following is true about the coefficient (M) of artificial variable(A) in
the objective function
M ->+infinitie
M->-infinite
M-> zero
M-> optimal solution
For finding the maximum profit in an enterprise of selling two products such that ‘freezing’ the
sale of one product and keep selling the other
Degeneracy
Duality
In Hungarian method of solving assignment problem, the cost matrix is obtained by----------.
Dividing each row by the elements of the row above it
Subtracting the elements of the row from the elements of the row above it
Subtracting the smallest element from all other elements of the row
Subtracting all the elements of the row from the highest element in the matrix
Page # 204 Or Click Here – My Ok
Which of the following statement applies to both transportation model and assignment model?
The inequalities of both problems are related to one type of resource.
Both have objective function, structural constraint and non-negativity constraints
Both use Volgel’s approximation for grtting basic feasible solution
Both are tested by Steping Stone for optimality
Click Here – My Ok
The similarity between Assignment problem and Transportation problem is _______

Both are rectangular matrices
Both are square matrices
Both can be graphical method
Both have objective function and non negativity constraints
Click Here – My Ok
For the project of a firm, if ‘5’ sales persons are assigned ‘5’ different sales territories, then in
how many ways a single territory can be assigned to a single sale person?
120
25
10
5
In which of the following stage, the machine operates at highest efficiency and its production
rate will be high and hence no need of replacement?
Infant stage
Youth stage
Old stage
None of the above
When money value changes with time at Ten percent(10%), then Power Worth Factor(PWT) for
first year is.
1
0.909
0.852
0.9
Which of the following is the correct assumption for replacement policy when money value does
not change with time
No Capital cost
No scrap value
Constant scrap value
Zero maintenance cost
My Ok - google Books
Which of the following department is more responsible for the development of queuing theory.
Railway station
Municipal office
Telephone department
Health department
My Ok - google books
This department is responsible for the development of queuing theory:

Railway station
Municipal office
Health department
If the number of arrivals during a given time period is independent of the number of arrivals that
have already occurred prior to the beginning of time interval, then the new arrivals follow --------
distribution.
Erlang
Poisson
Exponential
Normal
The arrival rate of customers in a Queue is an example of ______ variable

Continuous
Complex
Stochastic
Discrete
Primal of a Primal is-------.

Primal
Dual
Primal Primal
Primal dual
My Ok
Primal of a Dual is-------.

Primal
Dual
Primal Primal
Primal dual
My Ok
The dual of a dual problem yields the original primal.

True
False
My Ok - Note
Dual of a dual is
Primal
Dual
Primal Primal
Primal dual
My Ok
1. Primal of a Prima is Primal

2. Dual of a Dual is Primal.
3. Primal of a Dual is Primal.
4. Dual of a Primal is Dual.
5. Dual of a Dual of a Dual is Primal.
An unrestricted primal variable will result in an equality dual constraint. Conversely a primal
equation produces an unrestricted dual variable.
True (I think)
False
If the primal (either) problem has an unbound solution, then the dual has no solution
Optimal solution
Infeasible solution
Bounded solution
No solution
My Ok – Google books
If the primal has an unbound solution, then the dual has no solution
Optimal solution
Infeasible solution
Bounded solution
No solution
In two phase method, for the Phase-I, if the objective function has zero value with all vanishing
artificial variables then we _________
Necessarily have optimal solution
Infeasible solution
Can’t proceed for 2nd phase
Proceed for 2nd phase
In M-method, if the given LP problem has the feasible solution then the optimization algorithm
enforces artificial variable to
+infinitie
-infinite
zero
optimal solution
Check
Which of the following is the standard form of objective function corresponding to,MinZ=2x–
11y, subject to x=5 and y=7? Where As are artificial
Z=2x-11y-MA1-MA2
Z=2x-11y+MA1+MA2
Z=2x-11y-MA1+MA2
Z=2x-11y+MA1-MA2
Not confirm – check
The insensitivity of the solution relative to the original decision variables in an LP problem
which is solved by M-method is overcome by ----------
Simplex method
Graphical method
Two phase method
Duality principle
In Simplex method, which of the following is the standard equality corresponding to the
constraint“3x+5y>=2”?
3x+2y+S+A=2
3x+2y–S+A=2
3x+2y–S–A=2
3x+2y+S–A=2
In the initial iteration of Big M-method, the artificial variables appear in ---------.
Basis
Non-basic variables’ set
In two phase method, for the phase-I, if the given problem has feasible solution then----------
both objective function and artificial are zero
objective function is zero but artificial may arbitrary
objective function is arbitrary but artificial is zero
both objective and artificial can have arbitrary values
Under which of the following condition to solve an LP by using two phase method, we can’t
proceed for 2nd phase?
Objective function of 1st phase has zero value
Objective function of 1st phase has positive value
In replacement analysis, the maintenance cost is a function of _______ My Quiz

Time
Final investment
Initial investment
Resale value
Replacement of an item will become necessary when_________

The company has surplus funds to spend
Old item becomes too expensive to operate or maintain
Your opponent changes his machine in his unit
Your operator desires to work on a new machine
Which of the following type of failure that usually occurs in old age of the machine and hence
become a reason of replacement? My Quiz
Random failure
Early failure
Wear-out failure
Chance failure
My Ok – Google Books
The type of failure that usually occurs in old age of the machine is
(a) Random failure
(b) Early failure
(c) Chance failure
(d) Wear - out failure
Group replacement policy applies to _______

Irreparable items
Reparable items
Items that fails partially
Items that fail completely
Which of the following Replacement policy is imposed on an item irrespective of its failure?
Group replacement
Individual replacement
Repair spare replacement
Successive replacement
The replacement policy that is imposed on an item irrespective of its failure is

(a) Group replacement
(b) Individual replacement,
(c) Repair spare replacement,
(d) Successive replacement. ( )
Which of the following cost is irrelevant to replacement analysis?

Purchase cost of the machine
Operating cost of the machine
Maintenance cost of the machine
Machine hour rate of the machine
Which cost of the following is irrelevant to replacement analysis?

(a) Purchase cost of the machine,
(b) Operating cost of the machine,
(c) Maintenance cost of the machine,
(d) Machine hour rate of the machine. ( )
In a queue, the arrival pattern can be expressed in terms of --------.
Probabilities
Exact continuous variables
Customer behavior
Number of servers
If there is infinite number of servers then all the customers are served ______ on arrival.
Randomly
Instantaneously
To convert the transportation problem into a maximization model we have to------------

write the inverse of the coefficient matrix
multiply the feasibility condition by –1
multiply the coefficient matrix by –1
We can’t convert the transportation problem into a maximization problem, as it is basically
a minimization problem.
Transportations models consist of ------- like the production centers and --------- which may be
the sales centers.
(sinks, sources)
(sources, sinks) Most
(origins, sources)
(sinks, destinations)
In which of the following models, Simplex algorithm is not preferred to use due to laborious
computations?
Transportations models
Degenerate Linear models (Waqar Sidhu)
Non-degenerate Linear models
Dual or unbounded linear models
2 x  3 y 18
The inequality is equivalent to
2 x  3 y   18
2 x  3 y 18
2 x  3 y  18
2 x  3 y   18
The inequality 2 x  3 y 18 is equivalent to

2 x  3 y   18
2 x  3 y 18
2 x  3 y  18
2 x  3 y   18
MaxZ=2x+3y
Subject to
x  2  x  s1  A  2 
 
y  3   y  s2  3 
 x, y, s , s , A  0
x, y  0 1 2 
Which of the following is associated objective function of the1st phase ?
Which of the following is associated objective function of the 1st phase?
MazZ=2x+3y+A
MinZ=2x+3y+A
MaxZ=A
MinZ=A
By using two phase method to solve a linear programming problem, in Phase I, a new objective
function is formed by assigning on left hand side, zero to every original variable (including slack
and surplus variables) and ______ to each of the artificial variables.
M
-M
+1
-1
Zero valued artificial variables may appear as ________ variables in the final solution, when one
or more of the original constraints equations is reduendant.
Non basic
Basic
Slacks
Surplus
Artificial
In the big-M method, if the introduced ______ variables do not leave the basis in final iteration,
then this indicates that the give linear programming problem can’t be optimized.
Entering
Positive slack
Negative slack
Artificial
While solving a linear programming problem by using M-Method, traditionally the ________
variables are chosen in the initial basic feasible solution.
Negative slack
Positive slack
Entering
Artificial
The _________ variable is chosen by examining the cost coefficients in the objective function.
Entering
Leaving
Positive slack
Negative slack
If a company manufacture ‘x’ units of product ‘A’ and ‘y’ units of ‘B’ with associated profits of
Rs.5 and Rs.3 then which of the following is the objective function to maximize is the profit?
Z=15xy
Z=5x-3y
Z=3x-5y
Z=5x+3y
In a linear Programming problem (LPP), which of the following must be hold?

Only objective function is linear
Both objective function and constraints are linear
Only constraints needs to be linear
At least one of objective function or constraint should be linear
Solution region for constraint y  0 is the

Half plane below y-axis
Half plane below the line y=0
Set of all those points where ordinates are non-positive
All are equivalent
In PERT, activity time estimates are distributed according to

Beta distribution
Normal Distribution
Poisson Distribution
Binomial Distribution
Page # 35, 36 – My Ok
Best possible time estimate that a given activity would take under normal conditions which often
exist, is called
Most likely time estimate
Pessimistic time estimate
Smallest time estimate
Activity time estimate
Page # 35 – My Ok
For any activity backward pass computations provide its

Earliest start times
Latest start time
Moderate start times
Completion time
Page # 32 – My Ok
The following network is an example of
Redundancy
Dangling
Cycling
Dummy
MAD=_________ S.D
2


2
2
3
3
2
Standard Deviation (S.D) = ________ MAD


2
2

2
3
3
2
Product of ‘item cost’ and ‘ordered item’ is

Crash cost
Cost period
Set up cost
Shortage cost
Page # 51 – My Ok
Formula for geometric series

Formula for geometric series “x+vx+v2x+… is
1  vn
P ( n)  x
1 v
1  vn
P ( n) 
1 v
1 v
P ( n)  x
1  vn
1 v
P ( n) 
1  vn
Page # 255 - My Ok
The present worth of a rupee to be spent after a year is denoted by v and given by
v= (1 + r)
v= (1 / r)
v = (1 + r) / 10
v = 1/ (1 + r)
If “Ni” be the Number of replacement made at the end of the ith week and “Pj” be the probability
of failure during the ith week, then N1 = ------------------.
N0P1
N1P1
N0P0
N1P2
If “Ni” be the Number of replacement made at the end of the ith week and “Pj” be the probability
of failure during the ith week, then N2 = ------------------.
N1P2 + N1P1
N0P2 + N1P1
N0P1 + N1P2
N0P2 + N1P2
Check
A repairman services three machines. For each machine the time between service requirements is
8 hours following exponential distribution. The time of repair also has the same distribution with
a mean of 2 hours. Then the mean service time “  ” is
½=0.5
4
¼=0.25
2
A repairman services three machines. For each machine the time between service requirements is
8 hours following exponential distribution. The time of repair also has the same distribution with
a mean of 2 hours. Then the average rate “  ” is
1/8=0.125
8
¼=0.25
4
A duplicating machine maintained for office use is used and operated by people in the office who
need to make copies. Since the work to be copied varies in length (number of pages of the
original) and copies required, the service rate is randomly distributed, the arrival rate is 5 per
hour and the service rate is 10 per hour then the equipment utilization “  ” is equal to
0.50
0.20
5
2
In a bank, every 15 minutes one customer arrives for cashing the cheque. The staff in the
payment counter takes only 10 minutes for serving a customer on an average, then the service
rate “  ” =__________
6 per hour
4 per hour
10 per hour
1/6 per hour
If the mean arrival and mean service rates are 4 and 7 respectively in a queue then expected
waiting time in system is
1/3
3
28
7/4
We go in probabilistic replacement model when period between installation and failure is

Varying exponentially
Varying linearly
Constant
Is not constant (varying arbitrarily)
If a basic feasible solution contains less than “m + n – 1” (Here m is the number of rows, n is the
number of columns in transportation problem) non negative allocation, then it is said to be
Degenerate (Waqar Sidhu)
Multiple Solutions
Non degenerate
Alternative Optima
Any set of non negative allocation (Xij>0) which satisfy the row and column sum is called a
_____ solution
Feasible (Waqar Siddhu)
Non basic feasible
Basic infeasible
Optimal
Degeneracy in a 5x6 transportation problem occurs when the number of occupied cell is less than
20 but greater than 10
10
Zero
Infinity
Check
In sequencing problem, the Johnson’s algorithm in finding the optimal ordering of n jobs through
3 machines can be applied, if the problem is converted into following number of machine
problems
3n
3n!
2*3=6
2
During a replacement if the value of money decreases at the rate of 3% then the present worth
factor of unit amount to be spent after one year is given by
0.25
0.333
0.9708
4
If the money carries a rate of interest of 12% per year, the present worth factor of one rupee
due in one year is
0.08333
0.89285 (Waqar Sidhu)
0.0769
13
Which of the following binary operation in assignment problem among all the elements in the
given profit matrix from the highest element in the matrix
Subtraction
Division
Multiplication
Addition
The cost matrix in assignment problem is always diagonal matrix
identity matrix
zero matrix
diagonal matrix
square matrix (Waqar Sidhu)
The cost matrix in assignment problem is always

identity matrix
zero matrix
diagonal matrix
square matrix (Waqar Sidhu)
The important characteristic of Cost matrix associated with Assignment problem, while solving
it by Hungarian’s method is ______
It will have zero as element of one diagonal
It will have zero as the element of both diagonals
It will have at least one zero in each column and each row
It will not have zero as its element
In which of the following age, the Replacement decision is very much common?
Infant stage
Old age
Youth
In all the above
The assignment problem is unbalanced if the cost matrix is not a ____ matrix.
Square
Rectangle
Non-singular
Singular
In the assignment problem, the cost matrix is ______

Rectangular
Square
Non-Singular
Singular
Which of the following replacement policy is considered to be dynamic in nature? My Quiz

Time is continuous variable and the money value does not change with time.
When money value does not changes with time and time is a discrete variable.
When money value changes with time.
When money value remains constant for some time and then goes on changing with time.
To balance Assignment matrix we have to:

(a) Open a Dummy row,
(b) Open a Dummy column,
(c) Open either a dummy row or column depending on the situation,
(d) You cannot balance the assignment matrix
To balance the assignment matrix we have to:

(a) Open a Dummy row,
(b) Open a Dummy column,
(c) Open either a dummy row or column depending on the situation,
(d) You cannot balance the assignment matrix.
Note google book waly sary answers dobara sy confirm krny hain lazmi
In an assignment problem, while applying Hungarian’s method, if in the final modified matrix
any row or column does not have single zero, then which of the following is true?
No assignment can be made
Unique assignment will be made
Multiple assignments can be made
None of the above
For a Transportation Problem, if it’s initial feasible solution is evaluated by Least cost method,
the quality of this initial solution is better than________
North West Corner Method
Vogal’s approximation Method
For an unbalanced Transportation problem, if the total demand is MORE than total supply then
which of the following is true in order to balance the problem?
One constraint will have evacuate
One constraint will have to add
A dummy sink would have to include with demand equal to surplus
A dummy source would have to include with supply equal to shortage
Check
In the assignment problem, the decision variable ‘xij’ can attain which of the following value?
Only ‘1’
Only zero
Any arbitrary non-negative
Zero or 1
In Least cost Method, for any cell both demand and supply are satisfied then which of the
following will be crossed out?
Only row
Only column
Both row and column
Either of row or column
Which of the following method is used to find the basic feasible solution of a transportation
problem?
North West Corner Method
Least Cost’s Method
Vogel’s approximation Method
All above methods are applicable
If shortage cost is infinity, then _________

No shortages are allowed
No inventory carrying cost is allowed
Ordering cost is zero
Purchase cost = Carrying cost
Many decision making problems involve a process that takes place in multiple stage in such a
way that at each stage, the process is dependent on the strategy chosen. Such types or problems
are called.
Linear Programming problems
Integer programming problems
Dynamic programming problems
Assignment problems
A primal’s constraint equation produces _____ dual variable

Non basic
An unrestricted
Artificial
Basic
In the Simplex table, if the coefficient of non-basic variable says x1 in the Z-row is zero, then it
indicates that value of z________
Increase
Decrease
Does not change
Becomes infinity
Consider the following simplex table corresponding to maximization problem

Basic x1 x2 x3 x4 A1 A2 R.H.S
Z 0 0 M+1/2 M 16M-11 0 -0.6M-5.7
x1 1 0 5 0 -1 0 3
x2 0 1 -5 0 3 0 9
A2 0 0 -1 -1 -0.6 1 0.6
This is the indication that problem has ___________.
Infeasible Solution
Temporarily Degenerate Solution
Unbounded Solution
Degenerate Solution
Consider the following simplex table corresponding to maximization problem
Basic x1 x2 x3 S1 S2 S3 R.H.S
Z -4 -3 -2 0 0 0 0
x1 0 0 1 1 0 0 8
x2 -2 1 3 0 1 0 9
A2 0 2 1 0 0 1 3
This is the indication that problem has ___________.
Problem has Unbounded Solutions
Problem has Infeasible Solutions
Problem has temporarily degenerate
Problem has Multiple Optimal Solutions
For a linear programming problem, the Unboundedness is related to which of the following?
Solution space only
Objective value only
Neither solution space nor to objective value
Both solution space and objective value
If the degeneracy arises in the initial stage, then which of the following is true?
One of the basic variables is zero
More than one variable is eligible to leave the baiss
Tie the ratio of each row, by taking right hand side of each row and dividing by the
corresponding element of the key column.
Iterative Cycling of basic solution exists without reaching the optimality
If the problem has a ________ solution such that the minimum value of the objective function is
zero along with zero values of artificial variables. Then proceed to phase II method.
Unbounded
Feasible
Non feasible
De-generate feasible solution
Min. Z = 0 and no artificial variable appears in the basic variables. __________ solution to
original problem has been found and we proceed to phase II method.
A basic feasible
An optimal
An infeasible
If no ___________ variable appears in the basis and the optimality conditions are satisfied, then
the current solution will be optimal one.
artificial
non basic
slack
surplus
The constrained 3x1+2x2  18 in standard form can be as:
(Where R is artificial variable and S1 is slack variable)
3x1+2x2+S1+R=18
3x1+2x2+S1-R=18
3x1+2x2-S1+R=18
3x1+2x2+S1 =18
If the objective function of a linear programming problem needs further improvement then which
of the following will have to proceed?
A new decision variable to be entered and other decision variable to leave the basis
A new s;ack variable to be entered and other slack to leave the basis
A new non-degenerate variable to be entered and other no-degenerate to leave the basis
A new basic variable to be included and other basic variable to leave the basis
It is a property of simplex method that there always exist infinite number of basic feasible
solution.
True
False
Early finish time of the event is ‘3’ in the following is _______
14 days
2 days
8 days
6 days
Which of the following give the excess of available time over the activity time when all jobs start
as early as possible?
Total float
Free float
Independent float
Early finish time
Let FS = Free Slack, TS = Total Slack, INDS = Independent Slack, then which relation is true
I. TS ≤ FS
II. INDS ≤ FS
III. FS ≤ TS
Then which relation is true
Only I
Only II
Only III
Only I & II
Only II & III
In replacement Models, Capital Cost = (Item / Machine Cots) – (--------------------)

Resale Cost
Running Cost
Maintenance Cost
Operating Cost
In assignment problem, each job requires exactly one resource.

True
False
May not be true
The name of the method in getting the optimum assignment is

Simplex method
North-West corner method
Vogel’s approximation method
Hungarian method
The Vogel Approximation method is an iterative procedure for computing a _______ solution of
the transportation problem
Basic feasible
Non basic feasible
Basic infeasible
CPM can help you figure out EXCEPT:

How long your complex project will take to complete
Which activities are “critical” meaning that they have to be done on time or else the whole
project will take longer
You can try to speed up the project
What is the most costly way to speed up the project
Transportation technique or the simplex method cannot be used to solve the assignment problem
because of
Degeneracy
Involvement of dummy activities
Looping among the entries of assignment matrix
Inequilibrium between demand and supply
The representation of reality in some physical form or in some form of mathematical equation
may be discussed under the topic entitled as
Transportation problems
Game theory
Simulation
Replacement problems
Under the individual replacement policy an item is replaced on its

Success
Failure
Maximum utilization
Minimum utilization
Note this is a very important link from google books jahan sy is subject ky past quizzes hain
https://books.google.com.pk/books?id=u-
eRHBbJW7AC&pg=PA263&lpg=PA263&dq=transportation+model+and+assignment+model+
The+inequalities+of+both+problems+are+related+to+one+type+of+resource.&source=bl&ots=9
3F4IylWQB&sig=e84VsaivWDsZYmMAVJO0T1ZvBeE&hl=en&sa=X&ved=0ahUKEwjC9qv
Du5fcAhUHOJoKHRfrB0oQ6AEILjAB#v=onepage&q=transportation%20model%20and%20as
signment%20model%20The%20inequalities%20of%20both%20problems%20are%20related%2
0to%20one%20type%20of%20resource.&f=false
Note Yeh MCQs Quiz k hain lakin in k options nhi hain
Which of the following represents “Arrival-->Service-->Service-->Service-->Out-->”?

(a) Single Channel Single Phase system
(b) Multi channel single-phase system
(c) Single channel multi phase system
(d) Multi channel multi phase system. ( )
4. Distribution of service time in Queuing Model is ------------exponential distribution.

5. Which of the following replacement model is said to be probabilistic model?
6. Queuing models measure the effect of-----------.
In Simplex standard table to solve an LP problem of Minimization, we choose the candidate for
entering variable in ________
Objective function with most positive coefficient
Objective function with most negative coefficient
Constraint with most positive coefficient
Constraint with most negative coefficient
In Simplex standard table to solve an LP problem of Maximization, we choose the candidate for
entering variable in ________
Objective function with most positive coefficient
Objective function with most negative coefficient
Constraint with most positive coefficient
Constraint with most negative coefficient
If an artificial variable is not included while converting a constraint of type’>=’ into equation,
then we will have _______ solution
Feasible
Infeasible
Degenerate
Non-degenerate
By Simplex method, to minimize ‘Z=2x+λy’ of an LP problem, if ‘z=A>0’ for the initial

iteration then for its next improved solution (0<A<100), which of the following would be the
next entering variable
X<0
Y<0
x>0
y>0
Transportation problem is basically a
Maximization model
Minimization model
Transshipment problem
Iconic model
The column, which is introduced in the matrix to balance the rim requirements, is known as:
Key column
Idle column
Slack column
Dummy Column
The row, which is introduced in the matrix to balance the rim requirement, is known as:
Key row
Idle row
Dummy row
Slack row
One of the differences between the Resource allocation model and Transportation Model is:
The coefficients of problem variables in Resource allocation model may be any number and
in transportation model it must be either zeros or ones. (I Think)
The coefficients of problem variable in Resource allocation model must be either zeros or ones
and in Transportation model they may be any number
In both models they must be either zeros or ones only
In both models they may be any number
Check
To convert the transportation problem into a maximization model we have to

To write the inverse of the matrix
To multiply the rim requirements by –1
To multiply the matrix by –1 (I think)
We cannot convert the transportation problem in to a maximization problem, as it is basically a
minimization problem
Check
In a transportation problem where the demand or requirement is equals to the available resource
is known as
Balanced transportation problem (I think)
Regular transportation problem,
Resource allocation transportation problem
Simple transportation model
Check
The total number of allocation in a basic feasible solution of transportation problem of m × n size
is equal to:
m×n
(m / n ) – 1
m + n +1
m+n–1
When the total allocations in a transportation model of m × n size is not equals to m + n – 1 the
situation is known as:
Unbalanced situation
Tie situation
Degeneracy (I Think)
None of the above
Check
The opportunity cost of a row in a transportation problem is obtained by:

Deducting the smallest element in the row from all other elements of the row,
Adding the smallest element in the row to all other elements of the row,
Deducting the smallest element in the row from the next highest element of the row
Deducting the smallest element in the row from the highest element in that row.
In Northwest corner method the allocations are made

Starting from the left hand side top corner,
Starting from the right hand side top corner
Starting from the lowest cost cell
Starting from the lowest requirement and satisfying first
VAM stands for:

Value added method
Value assessment method
Vogel Adam method
Vogel’s approximation method
MODI stands for

Modern distribution
Mendel's distribution method
Modified distribution method
Model index method
In the optimal solution, more than one empty cell have their opportunity cost as zero, it indicates
The solution is not optimal
The problem has alternate solution
Something wrong in the solution
The problem will cycle. ( )
In case the cost elements of one or two cells are not given in the problem, it means:
The given problem is wrong
We can allocate zeros to those cells
Allocate very high cost element to those cells
To assume that the route connected by those cells are not available
To solve degeneracy in the transportation problem we have to:

Put allocation in one of the empty cell as zero
Put a small element epsilon in any one of the empty cell
Allocate the smallest element epsilon in such a cell, which will not form a closed loop with
other loaded cells.
Allocate the smallest element epsilon in such a cell, which will form a closed loop with other
loaded cells
A problem where the produce of a factory is stored in warehouses and then they are transported
to various demand point as and when the demand arises is known as:
Transshipment problem
Warehouse problem
Storing and transport problem
None of the above ( )
Implied Cost in transportation problem sets (in the existing program):

The lowest limit for the empty cell beyond which it is not advisable to include in the programme,
The highest limit for the empty cell beyond which it is not advisable to include in the
programme,
The opportunity cost of the empty cell,
None of the above
In transportation model, the opportunity cost is given by

Implied cost + Actual cost of the cell
Actual cost of the cell – Implied cost,
Implied cost – Actual cost of the cell (I think)
Implied cost × Actual cost of the cell
Check
If ui and vj are row and column numbers respectively, then the implied cost is given by:
ui + vj
ui – vj
ui × vj
ui /vj ( )
If a transportation problem has an alternate solution, then the other alternate solutions are derived
by:
(Given that the two matricides of alternate solutions are A and B, and d is any positive fraction
number)
A + (1 – d) × B
A ( 1 – d) + B
dA + dB
dA + (1 – d) × B
Check
Assignment Problem is basically a
Maximization Problem
Minimization Problem
Transportation Problem
Primal problem
The Assignment Problem is solved by

Simplex method
Graphical method
Vector method
Hungarian method
In Index method of solving assignment problem

The whole matrix is divided by smallest element
The smallest element is subtracted from whole matrix
Each row or column is divided by smallest element in that particular row or column
The whole matrix is multiplied by – 1.
In Hungarian method of solving assignment problem, the row opportunity cost matrix is obtained
by:
Dividing each row by the elements of the row above it,
By subtracting the elements of the row from the elements of the row above it
By subtracting the smallest element from all other elements of the row
By subtracting all the elements of the row from the highest element in the matrix
In Flood's technique of solving assignment problem the column opportunity cost matrix is
obtained by:
Dividing each column by the elements of a column which is right side of the column
By subtracting the elements of a column from the elements of the column which is right side of
the column
By subtracting the elements of the column from the highest element of the matrix
By subtracting the smallest elements in the column from all other elements of the column
The property of total opportunity cost matrix is

It will have zero as elements of one diagonal
It will have zero as the elements of both diagonals
It will have at least one zero in each column and each row
It will not have zeros as its elements
The horizontal and vertical lines drawn to cover all zeros of total opportunity matrix must be:
Equal to each other,
Must be equal to m × n (where m and n are number of rows and columns)
m + n ( m and n are number of rows and columns)
Number of rows or columns
The assignment matrix is always is a
Rectangular matrix
Square matrix
Identity matrix
None of the above.
To balance the assignment matrix we have to:

Open a Dummy row
Open a Dummy column
Open either a dummy row or column depending on the situation
You cannot balance the assignment matrix
In cyclic traveling salesman problem the elements of diagonal from left top to right bottom are
Zeros
All negative elements
All are infinity
all are ones
In cyclic traveling salesman problem the elements of diagonal from left top to right bottom are
Zeros
All negative elements
All infinity
All ones
To convert the assignment problem into a maximization problem

Deduct smallest element in the matrix from all other elements.
All elements of the matrix are deducted from the highest element in the matrix.
Deduct smallest element in any row form all other elements of the row.
Deduct all elements of the row from highest element in that row. ( )
The similarity between Assignment Problem and Transportation problem is:

Both are rectangular matrices
Both are square matrices,
Both can be solved by graphical method
Both have objective function and nonnegativity constraints
The following statement applies to both transportation model and assignment model
The inequalities of both problems are related to one type of resource.
Both use VAM for getting basic feasible solution
Both are tested by MODI method for optimality
Both have objective function, structural constraint and non-negativity constraints
To test whether allocations can be made or not (in assignment problem), minimum number of
horizontal and vertical lines are drawn. In case the lines drawn is not equal to the number of rows
(or columns), to get additional zeros, the following operation is done:
Add smallest element of the uncovered cells to the elements to the line
Subtract smallest element of uncovered rows from all other elements of uncovered cells
Subtract the smallest element from the next highest number in the element.
Subtract the smallest element from the element at the intersection of horizontal and vertical line
The total opportunity cost matrix is obtained by doing:

Row operation on row opportunity cost matrix,
By doing column operation on row opportunity cost matrix,
By doing column operation on column opportunity cost matrix
None of the above
The total opportunity cost matrix is obtained by doing:

Row operation on row opportunity cost matrix,
Column operation on row opportunity cost matrix,
Column operation on column opportunity cost matrix,
None of the above
Flood’s technique is a method used for solving

Transportation problem
Resource allocation model
Assignment mode.
Sequencing model
The assignment problem will have alternate solutions when total opportunity cost matrix has
At least one zero in each row and column,
When all rows have two zeros,
When there is a tie between zero opportunity cost cells,
If two diagonal elements are zeros
The following character dictates that assignment matrix is a square matrix:

The allocations in assignment problem are one to one
Because we find row opportunity cost matrix
Because we find column opportunity matrix
Because one to make allocations, one has to draw horizontal and veridical lines. ( )
When we try to solve assignment problem by transportation algorithm the following difficulty
arises:
There will be a tie while making allocations
The problem will get alternate solutions,
The problem degenerate and we have to use epsilon to solve degeneracy
We cannot solve the assignment problem by transportation algorithm
The objective of sequencing problem is:

To find the order in which jobs are to be made
To find the time required for completing all the jobs on hand
To find the sequence in which jobs on hand are to be processed to minimize the total time
required for processing the jobs
To maximize the effectiveness
The time required for printing of four books A, B, C and D is 5, 8, 10 and 7 hours. While its data
entry requires 7, 4, 3 and 6 hours respectively. The sequence time that minimizes total elapsed
time is
ACBD
ABCD
ADCB
CBDA
If there are ‘n’ jobs and ‘m‘ machines, there will be ---------------sequences of doing the jobs.
n×m
m×n
nm
( n !) m ( )
In general sequencing problem will be solved by using ………….

Hungarian Method
Simplex method
Johnson and Bellman method
Flood‘s technique
In solving 2 machine and ‘n‘ jobs, the following assumption is wrongμ

No passing is allowed
Processing times are known
Handling time is negligible
The time of processing depends on the order of machining
The following is the assumption made in processing of ‘n’ jobs on 2 machines

The processing time of jobs is exactly known and is independent of order of processing
The processing times are known and they depend on the order of processing the job
The processing time of a job is unknown and it is to be worked out after finding the
sequence
The sequence of doing jobs and processing times are inversely proportional
The following is one of the assumptions made while sequencing ‘n’ jobs on 2 machines
Two jobs must be loaded at a time on any machine
Jobs are to be done alternatively on each machine
The order of completing the jobs has high significance
Each job once started on a machine is to be performed up to completion on that machine
This is not allowed in sequencing of ‘n’ jobs on two machinesμ

Passing
Loading
Repeating the job
Once loaded on the machine it should be completed before removing from the machine.
Write the sequence of performing the jobs for the problem given below:
Jobs: A B C D E
Time of machining
On Machine X: 6 8 5 9 1
They can be processed in any order.
As there is only one machine, sequencing cannot be done
This is not a sequencing problem.
None of the above
Johnson Bellman rule states that:

If smallest processing time occurs under first machine, do that job first
If the smallest processing time occurs under the second machine, do that job first
If the smallest processing time occurs under first machine, do that job last
If the smallest processing time occurs under second machine keep the processing pending
To convert ‘n’ jobs and 3-machine problem into ‘n’ jobs and 2-machine problem, the following
rule must be satisfied.
All the processing time on second machine must be same.
The maximum processing time of 2nd machine must be ≤ to minimum processing times of
first and third machine.
The maximum processing time of 1st machine must be ≤ to minimum processing time of other
two machines.
The minimum processing time of 2nd machine must be ≤ to minimum processing times of first
and third machine
If two jobs J1 and J2 have same minimum process time under first machine but processing time
of J1 is less than that of J2 under second machine, then J1 occupies:
First available place from the left
Second available place from left
First available place from right
Second available place from right
If Job A and B have same processing times under machine I and Machine II, then prefer
Job A
Job B
Both A and B
Either A or B
The given sequencing problem will have multiple optimal solutions when the two jobs have
same processing times under:
First Machine,
Under both machines,
Under second machine
None of the above
The given sequencing problem will have multiple optimal solutions when the two jobs have
same processing times under:
First machine
Both machines,
Second machine
None of the above
If a job is having minimum processing time under both the machines, then the job is placed in:
Any one (first or last) position,
Available last position,
Available first position,
Both first and last position
FIFO is most applicable to sequencing of

One machine and ‘n’ jobs,
2 machines and ‘n’ jobs,
3 machine ‘n’ jobs,
‘n’ machines and 2 jobs. ( )
At a petrol Bunk, when ‘n’ vehicle are waiting for service then this service rule is usedμ
LIFO
FIFO
Service in Random Order
Service by highest profit rule
Consider the following sequencing problem, and write the optimal sequence:
Jobs: 1 2 3 4 5
Processing M/C X 1 5 3 10 7
Time in Hrs.
M/C Y 6 2 8 4 9
12345
13542
54321
1 4 3 5 2 (I think)
Check
In a 3 machine and 5 jobs problem, the least of processing times on machine A, B and C are 5, 1,
and 3 hours and the highest processing times are 9, 5, and 7 respectively, then Johnson and
Bellman rule is applicable if order of the machine is:
B-A-C
A-B-C
C-B–A
Any order
In maximization case of sequencing problem of 2 machines and ‘n’ jobs, the job is placed at
available left first position if it has ---------------- process time under machine --------------.
Least, first
highest, first
least, second
highest, second
The fundamental assumption of Johnson‘s method of sequencing isμ

No Passing rule
Passing rule
Same type of machines are to be used
Non zero process time
If a job has zero process time for any machine, the job must be
Possess first position only
Possess last position only
Possess extreme position
Be deleted from the sequencing
The assumption made in sequencing problems i.e. No passing rule means:

A job once loaded on a machine should not be removed until it is completed
A job cannot be processed on second machine unless it is processed on first machine
A machine should not be started unless the other is ready to start
No job should be processed unless all other are kept ready to start
The technological order of machine to be operated is fixed in a problem having:

1 machine and ‘n’ jobs
2 machines and ‘n’ jobs
‘n’ machines and 2 jobs
A sequencing problem is infeasible in case of:

2 jobs and ‘n’ machines
In a 2 jobs and ‘n’ machine problem a lie at 45° representsμ

Job 2 is idle
Job 1 is idle
Both jobs are idle
both jobs are under processing
In a 2 jobs and ‘n’ machine problem, the elapsed time for job 1 is calculated as (Job 1 is
represented on X -axis).
Process time for Job 1 + Total length of vertical line on graph
Process time for Job 2 + Idle time for Job 1
Process time for job 1 + Total length of horizontal line on graph
Process time for job 2 – Idle time for job 1
In a 2 jobs and ‘n’ machine-sequencing problem the horizontal line on a graph indicates:
Processing time of Job I
Idle time of Job I,
Idle time of both jobs,
Processing time of both jobs
In a 2 job, ‘n’ machine sequencing problem, the vertical line on the graph indicatesμ
Processing time of Job 1
Processing time of Job 2
Idle time of Job 2
Idle time of both jobs
In a 2 job and ‘n’ machine sequencing problem we find thatμ

Sum of processing time of both the jobs is same,
Sum of idle time of both the jobs is same,
Sum of processing time and idle time of both the jobs is same,
Sum of processing time and idle time of both the jobs is different
One of the important basic objective of Inventory management is

To calculate EOQ for all materials in the organization
To go in person to the market and purchase the materials
To employ the available capital efficiently so as to yield maximum results
Once materials are issued to the departments, personally check how they are used
The best way of improving the productivity of capital is

Purchase automatic machines
Effective labour control
To use good financial management
Productivity of capital is to be increased through effective materials management
Materials management is a body of knowledge, which helps manager to

Study the properties of materials
Search for needed material
Increase the productivity of capital by reducing the cost of material
None of the above
The stock of materials kept in the stores in anticipation of future demand is known as
Storage of materials
Stock of materials
Inventory
Raw materials
The stock of animals reared in anticipation of future demand is known as:

Live stock Inventory
Animal inventory
Flesh inventory
None of the above
The working class of human beings is a class of inventor known as:

Live stock
Human inventory
Population
Human resource inventory
In general the percentage of materials cost in product is approximately equals to:

40 to 50 %
5 to 10 %
2 to 3 %
90 to 95%
Materials management bring about increased productivity of capital by:

Very strict control over use of materials
Increasing the efficiency of workers
Preventing large amounts of capital locked up for long periods in the form of inventory
To apply the principles of capital management
We can reduce the materials cost by

Using systematic inventory control techniques,
Using the cheap material,
Reducing the use of materials,
Making hand to mouth purchase
The basis for ABC analysis is

Interests of Materials manager
Interests of the top management
Pareto’s 80-20 rule
None of the above
ABC analysis depends on the

Quality of materials
Cost of materials
Quantity of materials used
Annual consumption value of materials
‘A’ class materials consumesμ

10 % of total annual inventory cost
30% of total annual inventory cost
70 to 75% of total inventory cost
90 % of total annual inventory cost
‘B’ class of materials consumes ---------% of annual inventory cost.
60 to 70%
20 to 25%
90 to 95%
5 to 8%
‘C’ class materials consume -------% of annual inventory cost.

5 to 10 %
20 to 30%
40 to 50%
70 to 80%
The rent for the stores where materials are stored falls under:
Inventory carrying cost
Ordering cost,
Procurement cost
Stocking cost
Insurance charges of materials cost falls under:

Ordering cost
Stock out cost
Procurement cost
As the volume of inventory increases, the following cost will increase:

Stock out cost
Ordering cost
Procuring cost
As the order quantity increases, this cost will reduce:

Ordering cost
Insurance cost
Stock out cost
Procurement cost may be clubbed with:

Inventory carrying charges
Stock out cost
Loss due to deterioration
Ordering cost
The penalty for not having materials when needed is:

Loss of materials cost
Loss of order cost,
Stock out cost
General losses
Losses due to deterioration, theft and pilferage comes under,

Inventory Carrying charges
Losses due to theft,
Does not come under any cost
Consumption cost
Economic Batch Quantity is given by: (where, C1 = Inventory carrying cost, C3 = Ordering cost,
r = Demand for the product)
(2C1 / C3)1/2
(2 C3 / C1r )1/2
2C3r / C1
(2C3r / C1)1/2
If is the annual demand, C1 = Inventory carrying cost, i = rate of inventory carrying charges, p
= unit cost of material in Rs., then EOQ =
(2C ip)1/2
2C ip,
(2 C3 / ip
C3 ip)1/2, ( )
If C1 = Carrying cost, C3 is the ordering cost, r = demand for the product, then the optimal
period for placing an order is given by:
(2 C3/C1r)1/2
(2C1 C3/r )1/2
( 2C3r/C1)1/2
(2C1C3r)1/2
When C1 = Inventory carrying cost, C3 = ordering cost, r = demand for the product, the total
cost of inventory is given by:
(2C1C3r)
(2C1C3)1/2
(2C3r/C1)1/2
(2C1C3r)1/2
When is the annual demand for the material, p = unit price of the material in Rs., C3 is the
ordering cost, q = order quantity, then the total cost including the material cost is given by:
(q/2) ip qC p
2C ip p
(q/2) ip p
(2C3q ip)1/2
In VED analyses, the letter V stands for:

Very important material
Viscous material
Weighty materials
Vital materials
In VED analysis, the letter D strands for:

Dead stock
Delayed material
Deserved materials
Diluted materials
The VED analysis depends on:

Annual consumption cost of materials
Unit price of materials
Time of arrival of materials
Criticality of materials
In FSN analysis the letter S stands for:

Slack materials
Stocked materials,
Slow moving materials
Standard materials
In FSN analyses, the letter N stands for:

Non moving materials
Next issuing materials,
No materials
None of the above
FSN analysis depends on:

Weight of the material
Volume of material,
Consumption pattern
Method of moving materials
MRP stands for:

Material Requirement Planning
Material Reordering Planning,
Material Requisition Procedure
Material Recording Procedure
A system where the period of placing the order is fixed is known as:
q – system
Fixed order system
p – system
Fixed quantity system
A system in which quantity for which order is placed is constant is known as:
q – System
p - system,
Period system
Bin system
LOB stands for:

Lot of Bills
Line of Batches
Lot of Batches
Line of Balance
High reliability spare parts in inventory are known as:

Reliable spares
Insurance spares
Capital spares
Highly reliable spares
The property of capital spares is:

They have very low reliability;
These can be purchased in large quantities, as the price is low,
These spares have relatively higher purchase cost than the maintenance spares.
They are very much similar to breakdown spares.
Re-usable spares are known as:

Multi use spares
Repeated useable stores,
Scrap materials
Rotable spares
JIT stands for:

Just in time Purchase
Just in time production,
Just in time use of materials
Just in time order the material
The cycle time, selected in balancing a line must be:

Must be greater than the smallest time element given in the problem,
Must be less than the highest time element given in the problem,
Must be slightly greater than the highest time element given in the problem,
Left to the choice of the problem solver
The cycle time, selected in balancing a line must be:

Greater than the smallest time element given in the problem,
Less than the highest time element given in the problem,
Slightly greater than the highest time element given in the problem,
Left to the choice of the problem solver
The lead-time is the time:
To place orders for materials,
Time of receiving materials,
Time between receipt of material and using materials
Time between placing the order and receiving the materials
The lead-time is the time:

To place orders for materials,
Of receiving materials,
Between receipt of material and using materials,
Between placing the order and receiving the materials
The PQR classification of inventory depends on:

Unit price of the material
Annual consumption value of material,
Criticality of material
Shelf life of the materials
The PQR classification of inventory depends on:

Unit price of the material
Annual consumption value of the material,
Criticality of the material
Shelf life of the materials
The classification made on the weight of the materials is known as:

PQR analysis
VED analysis
XYZ analysis
FSN analysis
At EOQ
Annual purchase cost = Annual ordering cost
Annual ordering cost = Annual carrying cost
Annual carrying cost = annual shortage cost
Annual shortage cost = Annual purchase cost
If shortage cost is infinity,

No shortages are allowed
No inventory carrying cost is allowed,
Ordering cost is zero
Purchase cost = Carrying cost
The most suitable system for a retail shop is

FSN Analysis
ABC analysis,
VED analysis
GOLF analysis
The inventory maintained to meet unknown demand changes is known as

Pipeline inventory
Anticipatory inventory
De coupling inventory
Fluctuatory inventory
The most suitable inventory system for a Petrol bunk is

P- System
2 Bin system,
Q- System
Probabilistic model
The water consumption from a water tank follows

P – system
PQ -system,
Q – System
EOQ System
Which of the following inventories is maintained to meet expected demand fluctuations?

Fluctuatory Inventory
Buffer stock,
De-coupling inventory
Which of the following inventory is maintained to meet expected demand fluctuations?

Fluctuatory Inventory
Buffer stock
De- coupling inventory
Which of the following increases with quantity ordered per order

Carrying cost
Ordering cost,
Purchase cost
Demand
The ordering cost per order and average unit carrying cost are constant, and demand suddenly
falls by 75 % then EOQ will:
Decreases by 50 %
Does not change
Increases by 50 %
Decreases by 40%
In JIT system, the following is assumed to be zero.
Ordering cost
Transportation cost
Carrying cost
Purchase cost
Which of the following analyses neither considers cost nor value?

ABC
XYZ
HML
VED
Which of the following analysis neither considers cost nor value:

ABC
XYZ
HML
VED
As per queue discipline the following is not a negative behaviour of a customer:

Balking
Reneging
Boarding
Collusion
The expediting or follow up function in production control is an example of

LIFO
FIFO
SIRO
Pre emptive
The expediting or follow up function in production control is an example of

LIFO
FIFO
SIRO
Preemptive
In M/M/S: N/FIFO the following does not apply

Poisson arrival
Limited service
Exponential service
Single server
The dead bodies coming to a burial ground is an example of:

Pure Birth Process
Pure death Process
Birth and Death Process
Constant rate of arrival
The system of loading and unloading of goods usually follows:

LIFO
FIFO
SIRO
SBP
A steady state exist in a queue if:
≤µ
≥µ
Note this question font should be symbol
If the operating characteristics of a queue are dependent on time, then is said to be:
Transient state
Busy state
Steady state
Explosive state
A person who leaves the queue by losing his patience to wait is said to be: ( )
Reneging
Balking
Jockeying
Collusion
The characteristics of a queuing model is independent of:

Number of service stations
Limit of length of queue
Service Pattern
Queue discipline
The unit of traffic intensity is:

Poisson
Markow
Erlang
Kendall
In (M /M /1) : (∞ / FCFS) model, the length of the system Ls is given by:

p2/1/p
p/1–
–
–
Note this question font should be symbol
In (M / M
Ls, Length of the system
Lq length of the queue
Wq Waiting time in queue
Ws Waiting time in system. ( )
The queue discipline in stack of plates is:

SIRO
Non-Pre-Emptive
FIFO
LIFO
Office filing system follows:

LIFO
FIFO
SIRO
SBP
SIRO discipline is generally found in:

Loading and unloading
Office filing
Lottery draw
Train arrivals at platform
The designation of Poisson arrival, Exponential service, single server and limited queue selected
randomly is represented by:
(M / M / S) : (N / SIRO)
(M / M / 1) : ( N / SIRO)
For a simple queue (M / M

Poisson busy period
Random factor,
Traffic intensity
Exponential service factor
With respect to simple queuing model which on of the given below is wrong:
Lq Wq
β
Ws = Wq + μ
Ls = Lq + β
When a doctor attends to an emergency case leaving his regular service is called:
Reneging
Balking
Pre-emptive queue discipline
Non-Pre-Emptive queue discipline ( )
A service system, where customer is stationary and server is moving is found with:
Buffet Meals
Out patient at a clinic
Person attending the breakdowns of heavy machines
Vehicle at Petrol bunk
In a simple queuing model the waiting time in the system is given by:
(Lq
Wq + µ ( )
This department is responsible for the development of queuing theory:

Railway station
Municipal office
Health department.
If the number of arrivals during a given time period is independent of the number of arrivals that
have already occurred prior to the beginning of time interval, then the new arrivals follow --------
---distribution.
Erlang
Poisson
Exponential
Normal
The figure given represents:

Single Channel Single Phase system
Multi channel single-phase system
Single channel multi phase system
Multi channel multi phase system
In queue designation A/B/S : (d/f), what does S represents:

Arrival Pattern
Service Pattern
Number of service channels
Capacity of the system
When the operating characteristics of the queue system dependent on time, then it is said to be:
Steady state
Explosive state
Transient state
Any one of the above
The distribution of arrivals in a queuing system can be considered as a:

Death Process
Pure Birth Process
Pure live process
Sick process
Queuing models measure the effect of:

Random arrivals
Random service
Effect of uncertainty on the behaviour of the queuing system
Length of queue
Traffic intensity is given by:

Mean arrival rate/Mean service rate,
,
Number present in the queue / Number served
Variance of queue length is :
–
–
–
If the value of the game is zero, then the game is known as:
Fair strategy
Pure strategy
Pure game
Mixed strategy
The games with saddle points are:

Probabilistic in nature
Normative in nature
Stochastic in nature
Deterministic in nature
When Minimax and Maximin criteria matches, then

Fair game is exists
Unfair game is exists,
Mixed strategy exists
Saddle point exists
When the game is played on a predetermined course of action, which does not change throughout
game, then the game is said to be
Pure strategy game
Fair strategy game
Mixed strategy game
Unsteady game
If the losses of player A are the gins of the player B, then the game is known as:
Fair game
Unfair game
Non- zero sum game
Zero sum game
Identify the wrong statement:

Game without saddle point is probabilistic
Game with saddle point will have pure strategies
Game with saddle point cannot be solved by dominance rule.
Game without saddle point uses mixed strategies
In a two person zero sum game, the following does not hold correct:
Row player is always a loser
Column Player is always a winner.
Column player always minimizes losses
If one loses, the other gains
If a two person zero sum game is converted to a Linear Programming Problem,

Number of variables must be two only
There will be no objective function
If row player represents Primal problem, Column player represent Dual problem
Number of constraints is two only
In case, there is no saddle point in a game then the game is

Deterministic game
Fair game,
Mixed strategy game
Multi player game
When there is dominance in a game then

Least of the row ≥ highest of another row
Least of the row ≤ highest of another row
Every element of a row ≥ corresponding element of another row.
Every element of the row ≤ corresponding element of another row. ( )
When the game is not having a saddle point, then the following method is used to solve the
game:
Linear Programming method
Minimax and maximin criteria
Algebraic method
Graphical method
Consider the matrix given, which is a pay off matrix of a game. Identify the dominance in it.
B
XYZ
P173
AQ564
R720
P dominates Q
Y dominates Z
Q dominates R
Z dominates Y
Identify the unfair game:

CDCD
(a) A 0 0 (b) A 1 –1
B 0 0 B –1 1
CDCD
(c) A –5 +5 (d) A 1 0
B + 10 – 10 B 0 1
If there are more than two persons in a game then the game is known as:
Non zero sum game
Open game
Multiplayer game
Big game
For the pay of matrix the player A always uses:

B
I II
I –5 –2
A
II 10 5
First strategy
Mixed strategy of both II and I
Does not play game
Second strategy
For the pay off matrix the player prefers to play

B
I II
I –7 6
A
II –10 8
Second strategy
First strategy
Keep quite
Mixed strategy
For the game given the value is:

B
I II
123
A
II –5 5
3
–5
5
2
In the game given the saddle point is:

B
I II III
I 2 –4 6
A II 0 –3 –2
III 3 –5 4
–2
0
–3
2
A competitive situation is known as:

Competition
Marketing
Game
None of the above
One of the assumptions in the game theory is:

All players act rationally and intelligently
Winner alone acts rationally
Loser acts intelligently,
Both the players believe luck
A play is played when:

The manager gives green signal
Each player chooses one of his courses of action simultaneously
The player who comes to the place first says that he will start the game
When the latecomer says that he starts the game. ( )
The list of courses of action with each player ……………

Is finite
Number of strategies with each player must be same
Number of strategies with each player need not be same
None of the above
A game involving ‘n’ persons is known asμ

Multi member game
Multi player game
n - person game
not a game
Theory of games and economic behavior is published by:

John Von Neumann and Morgenstern
John Flood
Bellman and Neumann
Mr. Erlang
In the matrix of a game given below the negative entries are:

B
I II
I 1 –1
A
II –1 1
Payments from A to B
Payments from B to A
Payment from players to organizers
Payment to players from organizers
In Dynamic Programming Problems, the decisions are made in

Single stage
2-stages
Multi-stages
No decision making process
In dynamic programming problems, the main problem is divided into subproblems. Each sub-
problem is known as:
Part
Stage
State
Mini problem
The technique of Dynamic Programming problem is developed by:

Taylor
Gilberth
Richard Bellman
Bellman and Clarke
Another name used to Dynamic Programming is:

Multistage problem
Recursive optimization
State problems
No second name
If the outcome at any decision stage is unique and known for the problem, then the Dynamic
programming problem is known as:
Probabilistic dynamic programming problem
Stochastic dynamic programming problem
Static dynamic programming problem
Deterministic dynamic programming problem
The possible decisions at any stage are known as:

States
Steps
Parts
None
The rule which determines the decision at each stage is known as

State
Stage
Policy
Decision
The conclusion of a process designed to weigh the relative utilities of a set of available
alternatives to select most preferred one is known as:
Concluding session
Conclusion
End of the process
Decision
The body of knowledge that deals with the analysis of making of decisions is known as
Decisions
Knowledge base
Decision theory
Decision analysis
Decisions that are meant to solve repetitive and well structured problems are known as:
Repetitive decisions
Structured decisions,
Programmed decisions
Linear programming
Decisions that handle non-routine, novel, and ill structured problems are known as:
Non-programmed decisions
Programmed decisions,
Ill-structured decisions
Non-linear programming
Operations Research is the outcome of

National emergency
Political problems
Combined efforts of talents of all fields
Economics and Engineering
O.R. came into existence during

World War I
India and Pakistan War,
World War II
None of the above
The name of the subject Operations Research is due to the fact that
Problems can be solved by war approach
The researchers do the operations
The war problems are generally known as operations and inventing a new way of solving such
problems.
Mathematical operations are used in solving the problems
The first country to use Operations Research method to solve problems is

India
China,
U.K
U.S.A
The name Operations Research is first coined in the year

1945
1935,
1940
1950
The person who coined the name Operations Research is:

Bellman
Newman,
McClosky and Trefrhen
None of the above
O.R. Society of India is founded in the year

1965
1970,
1959
1972
The objective of Operations Research is:
To find new methods of solving Problems,
To derive formulas
Optimal utilization of existing resources
To utilize the services of scientists
Operations Research is the art of giving

Good answers for war problems,
Bad answers to war problems,
Bad answers to problems where otherwise worse answers are given.
Good answers to problems where otherwise bad answers are given
Operations Research is
Independent thinking approach
Group thinking approach
Inter-disciplinary team approach
None of the above
The first step in solving Operations Research problem is

Model building
Obtain alternate solutions,
Obtain basic feasible solutions
Formulation of the problem
The model, which gives physical or visual representation of the problem, is

Analogue model
Static model,
Iconic model
Symbolic model
One of the properties of O.R. model is

Model should be complicated,
Model is structured to suit O.R. techniques,
Model should be structured to suit all the problems we come across,
Model should be easy to derive
The problem, which is used to disburse the available limited resources to activities, is known as
O.R. Model
Resources Model
Allocation Model
Activities model
A wide class of allocation models can be solved by a mathematical technique know as:
Classical model
Mathematical Model,
Descriptive model
Linear Programming model
One of the properties of Linear Programming Model is

It will not have constraints,
It should be easy to solve,
It must be able to adopt to solve any type of problem,
The relationship between problem variables and constraints must be linear
The constraints of Maximisation problem are of

Greater than or equal type
Less than or equal type
Less than type
Greater than type
The slack variables indicate

Excess resource available
Shortage of resource available,
Nil resources
Idle resource
In graphical solution of solving Linear Programming problem to convert inequalities into

equations, we
Use Slack variables,
Use Surplus variables
Use Artificial surplus variables,
Simply assume them to be equations
To convert ≤ type of inequality into equations, we have to

Assume them to be equations
Add surplus variables,
Subtract slack variables
Add slack variables
To convert ≥ type of inequality into equations, we have to

Add slack variable
Subtract slack Variable
Subtract surplus variable
Add surplus variable
In Graphical solution of maximisation problem, the line, which we move from origin to the
extreme point of the polygon is :
Any one side of the polygon
Iso cost line,
Iso profit line
An imaginary line
The key row indicates
Incoming variable
outgoing variable
Slack variable
Surplus variable
The key column indicates

Outgoing variable
Incoming variable
Independent variable
Dependent variable
The penalty for not taking correct decision is known as

Fine
Loss,
Cost
Opportunity cost.
To transfer the key row in simplex table we have to

Add the elements of key row to key number
Subtract the elements of key row from topmost no key row
Divide the elements of key row by key number
None of the above.
The solution of the Linear programming problem in graphical solution lies in

First quadrant
Second quadrant,
Third quadrant
Fourth quadrant
When we solve maximization problem by simplex method the elements of net evaluation row of
optimal solution must be (when we use opportunity cost concept
Either zeros or positive numbers
Either zeros or negative numbers
All are negative numbers
All are zeros
When all the elements of replacement ratio column are equal, the situation is known as
Tie
Degeneracy
Break
None of the above
When the elements of net evaluation row of simplex table are equal, the situation is known as
Tie
Degeneracy
Break
Shadow price
The number at the intersection of key row and key column is known as
Column number
Row number,
Key number
Cross number
Dual of a Duel is
Primal
Dual
Prima dual
None of the above
Primal of a Primal is :
Primal
Dual
Prima primal
duo primal
Dual of a Dual of Dual is

Dual
Primal
Double dual
Single dual
Primal of a dual is
Primal
Dual,
Prime dual
Prime primal
If Dual has a solution, then the primal will

Not have a solution
Have only basic feasible solution,
Have a solution
None of the above
If Primal Problem is a maximisation problem, then the dual will be

Maximisation Problem
Minimisation Problem
Mixed Problem
None of the above
To get the Replacement ration column elements we have to

Divide Profit column elements by key number,
The first column elements of identity is divided by key number
Divide the capacity column elements by key number
Answer mentioned there is c option but there is no C option
The cost coefficient of slack variable is

Zero
One,
> than one
< than one, ( )
The cost coefficient of artificial surplus variable is

0
1
M
> than 1
If the primal has an unbounded solution, then the dual has

Optimal solution
No solution
Bound solution
None of the above
One of the important basic objectives of Inventory management is:

To calculate EOQ for all materials in the organisation,
To go in person to the market and purchase the materials,
To employ the available capital efficiently so as to yield maximum results
Once materials are issued to the departments, personally check how they are used
The best way of improving the productivity of capital is:

Purchase automatic machines
Effective labour control
To use good financial management
Productivity of capital is to be increased through effective materials management
Materials management is a body of knowledge, which helps manager to:

Study the properties of materials
Search for needed material
Increase the productivity of capital by reducing the cost of material
None of the above
The stock of materials kept in the stores in anticipation of future demand is known as:
Storage of materials
Stock of materials,
Inventory
Raw materials
The stock of animals reared in anticipation of future demand is known as:
Live stock inventory
Animal inventory,
Flesh inventory
None of the above
The working class of human beings is a class of inventor known as:

Live stock
Human inventory,
Population
Human resource inventory
In general, the percentage of materials cost in product is approximately equal to:

40 to 50 %
5 to 10 %
2 to 3 %
90 to 95%
Materials management brings about increased productivity of capital by:

Very strict control over use of materials,
Increasing the efficiency of workers,
Preventing large amounts of capital locked up for long periods in the form of inventory
To apply the principles of capital management
We can reduce the materials cost by:

Using systematic inventory control techniques
Using the cheap material,
Reducing the use of materials,
Making hand to mouth purchase, ( )
The basis for ABC analysis is

Interests of Materials manager
Interests of the top management,
Pareto’s 80-20 rule
None of the above
ABC analysis depends on the:

Quality of materials,
Cost of materials,
Quantity of materials used,
Annual consumption value of materials
‘A’ class materials consumeμ

10% of total annual inventory cost
30% of total annual inventory cost,
70 to 75% of total inventory cost
90% of total annual inventory cost. ( )
‘B’ class of materials consumes ............% of annual inventory cost.

60 to 70%
20 to 25%
90 to 95%
5 to 8% ( )
‘C’ class of materials consume ..............% of annual inventory cost.
5 to 10 %
20 to 30%
40 to 50%
70 to 80% ( )
The rent for the stores where materials are stored falls under:
Ordering cost,
Procurement cost
Stocking cost
Insurance charges of materials cost fall under:

Ordering cost
Stock out cost
Procurement cost
As the volume of inventory increases, the following cost will increase:

Stock out cost
Ordering cost,
Procuring cost
As the order quantity increases, this cost will reduce:

Ordering cost
Insurance cost,
Stock out cost
Procurement cost may be clubbed with:

Inventory carrying charges
Stock out cost
Loss due to deterioration
Ordering cost
The penalty for not having materials when needed is:

Loss of materials cost
Loss of ordering cost,
Stock out cost
General losses
When load is the annual demand for the material, p = unit price of the material in Rs., C3 is the
ordering cost, q = order quantity, then the total cost including the martial cost is given by:
(q/2) ip qC p
2C ip p
(q/2) ip p
( 2C3 q ip) 1/2 ( )
Contractual maintenance or agreement maintenance with manufacturer is suitable for equipment,

which is
In its infant state
When machine is old one,
Scrapped
None of the above
When money value changes with time at 10 %, then PWF for first year is :
1
0.909
0.852
0.9
When money value changes with time at 20%, the discount factor for 2nd year is:
1
0.833
0
0.6955 ( )
Which of the following maintenance policy is not used in old age stage of a machine? My Quiz
Operate up to failure and do corrective maintenance
Reconditioning
Replacement
Scheduled preventive maintenance
Which of the following maintenance policies is not used in old age stage of a machine?
Operate up to failure and do corrective maintenance,
Reconditioning,
Replacement,
Scheduled preventive maintenance
Which of the following replacement policies is considered to be dynamic in nature?

Time is continuous variable and the money value does not change with time,
When money value does not change with time and time is a discrete variable,
When money value changes with time
When money value remains constant for some time and then goes on changing with time
Which of the following replacement policy is considered to be dynamic in nature?

Time is continuous variable and the money value does not change with time.
When money value does not changes with time and time is a discrete variable.
When money value remains constant for some time and then goes on changing with time
When the probability of failure reduces gradually, the failure mode is said to be:
Regressive
Retrogressive
Progressive
Recursive
The following replacement model is said to be probabilistic model:

When money value does not change with time and time is a continuous variable
When money value does not change with time and time is a discrete variable
Preventive maintenance policy (I think)
Check
A machine is replaced with average running cost___________ My Quiz

Is not equal to current running cost,
Till current period is greater than that of next period,
If current period is greater than that of next period,
If current period is less than that of next period
The curve used to interpret machine life cycle is

Bath tub curve
Time curve,
Product life cycle
Ogive curve
Decreasing failure rate is usually observed in ………………. stage of the machine

Infant
Youth
Old age
Any time in its life
Which cost of the following is irrelevant to replacement analysis?

Purchase cost of the machine,
Operating cost of the machine,
Maintenance cost of the machine,
Machine hour rate of the machine
The type of failure that usually occurs in old age of the machine is
Random failure
Early failure,
Chance failure
Wear-out failure
Group replacement policy is most suitable for: My Quiz

Trucks
Infant machines,
Street light bulbs
New cars
The chance failure that occur on a machine are commonly found on a graph of time Vs
Failure rate (on X and Y axis respectively as
Parabolic
Hyperbolic
Line nearly parallel to X axis
Line nearly parallel to Y-axis
The chance failure that occurs on a machine is commonly found on a graph of time Vs failure
rate (on X and Y axes respectively as
Parabolic
Hyperbolic,
Line nearly parallel to X-axis
Line nearly parallel to Y-axis
Replacement of an item will become necessary when

Old item becomes too expensive to operate or maintain
When your operator desires to work on a new machine
When your opponent changes his machine in his unit
When company has surplus funds to spend
The production manager will not recommend group replacement policy My Quiz
When large number of identical items are to be replaced
In case Low cost items are to be replaced, where record keeping is a problem
For items that fail completely
For Reparable items
Which of the following is the correct assumption for replacement policy when money value does
not change with time? My Quiz
No Capital cost
No scrap value,
Constant scrap value
Zero maintenance cost
Which one of the following does not match the group?

Present Worth Factor (PWF)
Discounted rate (DR),
Depreciation value (DV)
Mortality Tables (MT)
Reliability of an item is
Failure Probability
1 / Failure probability,
1 - failure probability
Life period / Failure rate
The following is not discussed in group replacement policy: My Quiz

Failure Probability
Cost of individual replacement,
Loss due to failure
Present worth factor series
It is assumed that maintenance cost mostly depends on:

Calendar age
Manufacturing date
Running age
User’s age
Group replacement policy applies to:

Irreparable items
Reparable items
Items that fail partially
Items that fail completely
If a machine becomes old, then the failure rate expected will be:
Constant
Increasing
Decreasing
We Cannot be said
Replacement is said to be necessary if

Failure rate is increasing
Failure cost is increasing,
Failure probability is increasing
Any of the above
In this stage, the machine operates at highest efficiency and its production rate will be high.
Infant stage
Youth stage
Old age
None of the above
Replacement decision is very much common in this stage:
Infant stage
Old age
Youth
In all the above
The replacement policy that is imposed on an item irrespective of its failure is My Quiz
Group replacement
Individual replacement
Repair spare replacement
Successive replacement
When certain symptoms indicate that a machine is going to fail and to avoid failure if
maintenance is done it is known as:
Symptoms maintenance
Predictive maintenance
Repair maintenance
Scheduled maintenance
In retrogressive failures, the failure probability ------------------ with time

Increases
Remains constant,
Decreases
None of the above

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What are some characteristics of critical path scheduling techniques?

What are some characteristics of critical path scheduling techniques?

Muhammad Rizwan Ashraf

Which of the following is a characteristic of critical path scheduling techniques?

A feasible solution is called a _______________ solution if the number of non negative

Consider the following cost table

Using Vogel’s Approximation method:

Transportation problem is a special class of linear programming problem in which transport a

After doing next step, which table should be next?

Formula for the expected number in the system L is:

Non-critical activities in the following network are -------

If t0 = 10, tm = 10 and tp = 16, then S.D = _________.

Which of the following is an essential characteristic of a linear programming problem?

In a Transportation Problem, the objective function ’Z’ gives ----------.

To convert an Assignment problem into a maximization problem, which of the following

While solving a Linear Programming problem by Simplex Method, an inequality saysμ‘2x-3y<5’

The positive slack variables indicate -----------------.

If in a LP problem, the objective is to maximize: z = 2x+3y, with all associated constraints of

In M-method, which of the following is the standard equality corresponds to the

If Primal Problem is a maximization problem, then the dual will be -------

Graphically, in case of optimal degenerate solution,feasible solution always exists at corner

The similarity between Assignment problem and Transportation problem is _______

This department is responsible for the development of queuing theory:

The arrival rate of customers in a Queue is an example of ______ variable

Primal of a Primal is-------.

Primal of a Dual is-------.

The dual of a dual problem yields the original primal.

1. Primal of a Prima is Primal

In replacement analysis, the maintenance cost is a function of _______ My Quiz

Replacement of an item will become necessary when_________

Group replacement policy applies to _______

The replacement policy that is imposed on an item irrespective of its failure is

Which of the following cost is irrelevant to replacement analysis?

Which cost of the following is irrelevant to replacement analysis?

To convert the transportation problem into a maximization model we have to------------

The inequality 2 x  3 y 18 is equivalent to

In a linear Programming problem (LPP), which of the following must be hold?

Solution region for constraint y  0 is the

In PERT, activity time estimates are distributed according to

For any activity backward pass computations provide its

The following network is an example of

Standard Deviation (S.D) = ________ MAD

Product of ‘item cost’ and ‘ordered item’ is

Formula for geometric series

We go in probabilistic replacement model when period between installation and failure is

The cost matrix in assignment problem is always

In the assignment problem, the cost matrix is ______

Which of the following replacement policy is considered to be dynamic in nature? My Quiz

To balance Assignment matrix we have to:

To balance the assignment matrix we have to:

If shortage cost is infinity, then _________

A primal’s constraint equation produces _____ dual variable

Consider the following simplex table corresponding to maximization problem

Early finish time of the event is ‘3’ in the following is _______

In replacement Models, Capital Cost = (Item / Machine Cots) – (--------------------)

In assignment problem, each job requires exactly one resource.