IM + OR Page
2-9
IM + OR 10-22
23-29
CONTENTS 30-38
39-43
Unit Chapters 44-45
1 Linear Programming 46-51
2 PERT CPM 52-58
3 Queuing Theory 59
4 Inventory Control 60-61
5 Transportation 62
6 Scheduling 63-64
7 Break Even Analysis 65-81
8 Forecasting
9 Line Balancing
10 Sequencing
11 Assignment
12 Material Requirement Planning
13 Industrial Engineering
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MECHANICAL ENGINEERING
1 LINEAR PROGRAMMING
1. If at the optimum in a linear programming problem, a dual variable corresponding to a particualr primal
constraint is zero, then if means that (GATE - ME - 96)
a) Right hand side of the primal constrraint will disturb the optimum Solution
b) Changing the right hand side of the primal constraint will disturb the optimum Solution.
c) The objective function is unbounded
d) The problem is degenerate
2. Solve the following linear programming problem bysimplex method (GATE - ME - 00)
Maximize 4x1 + 6x2 + x3
Subject to 2x1 - x2 + 3x3 < 5
x2, x2, x3 > 0
a) What is the solution to the above problem ?
b)Add the constraint x2 < 2 to the simplex table of part (a) and find the solution.
3. A furniture manufacturer produces chairs and tables. The wood-working department is capable of
producting 200 chairs or 100 tables or any proportionate combinations of these per week. The profit
from a chiar is Rs. 100 and that from a table is Rs. 300. (GATE - ME - 02)
a) Set up the problem as a linear program
b) Deternine the optimum product mix for maximizing the profit.
c) What is the maximum profit ?
d) If the profit of each table drops to Rs. 200 per unit, what is the optimal and profit ?
4. A manufacturer produces two types of products, 1 and 2, at production levels of x1 and x2
respectively. The profit is given is 2x1 + 5x2. The production constraints are (GATE - ME - 03)
x1 + 3x2 < 40
3x1 + x2 < 24
x1 + x2 < 10
x1 > 0, x2 < 0
The maximum profit which can meet the constraints is
a) 29 b) 38
c) 44 d) 75
5. A company produces two types of toys : P and Q. Production time of Q is twice that of P and the
company has a maximum of 2000 time units per day. The supply of raw material is just sufficient to
produce 1500 toys (of any type) per day. Toy type Q requires an electric switch which is available
@ 600 pieces per day only. The company make a profit of Rs. 3 and Rs. 5 on type P and Q
respectively. For maximization of profit, the daily production quantities of P and Q toys should
respectively be (GATE - ME - 04)
a) 100, 500 b) 500, 1000
c) 800, 600 d) 1000, 1000
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6 & 7 common data questions
Consider a linear programming problem with two variables and two constraints. The objective
function is : Maximize X1 + X2. The corner point of the feasible region are (0, 0), (0, 2), (2, 0) and
(4/3, 4/3). (GATE - ME - 05)
6. If an addition constrints X1 + X2 < 5 is added, the optimal solution is
a) 55 b) 44
, ,
33
33
c) 55 d) (5, 0)
,
22
7. Let Y and Y be the decision vaialbes of the dual and v and v be the slack variables of the dual of the
12 12
given linear programming problem. The optimum dual variables are
a) Y1 and Y2 b) Y1 and v1 c) Y1 and v2 d) v1 and v2
8 & 9 common data questions
Consider the Linear Programme (I, P) (GATE - ME - 08)
Max 4x + 6y
Subject to
3x + 2y < 6
2x + 3y < 6
x, y > 0
8. After introducting slack variables s and t, the initial basicfeasible solution is represented by the table
below (basic variables are s = 6 and t = 6, and the objective function value is 0).
-4 -6 0 0 0
832106
123010
x y s t RHS
After some simplex iterations, the following table is obtained
0 0 0 2 12 Copyright : Ascent Gate Academy 3
s 5/3 0 1 -1/3 2
y 2/3 1 0 1/3 2
x y s t RHS
From this, one can conclude that
a) the LP has a unique optimal solution
b) the LP has an optimal solution that is not unique
c) the LP is infeasible
d) the LP is unbounded
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9. The dual for the LP in Q 74 is (GATE - ME - 08)
a) Min 6u + 6v b) Max 6u + 6v
Subject to Subject to
3u + 2v > 4 3u + 2v < 4
2u + 3v > 6 2u + 3v < 6
u, v > 0 u, v > 0
c) Max 4u + 6v c) Max 6u + 6v
Subject to Subject to
3u + 2v > 6 3u + 2v < 6
2u + 3v > 6 2u + 3v < 6
u, v > 0 u, v > 0
10. Consider the following Linear Programming problem (LPP) (GATE - ME - 09)
Subject to x1 < 4
x2 < 6
3x1 + 2x2 < 18
x1 > 0, x2 > 0
a) The LPP has a unique optimal solution
b) The LPP is infeasible
c) The LPP is unbounded
d) The LPP has multiple otpimal solutions
11. Simplex method of solving linear programming problem uses (GATE - ME - 10)
a) all the points in the feasible region
b) only the corner points of the feasible region
c) intermediate points within the infeasible region
d) only the interior points in the feasible region
12 & 13 common data questions
One unit of product P1 requires 3 kg of resource R1 and 1 kg of resource R2. One unit of product P2
requires 2 kg of resource R1 and 2 kg of resource R1 and 2 kg of resource R2. The profits per unit by
selling product P1 and P2 are Rs. 2000 and Rs. 3000 respectively. The manufacturer has 90 kg of
resource R1 and 100 kg of resource R2. (GATE - ME - 11)
12. The unit worth of resource R2, i.e. dual price of resource R2 in per kg is
a) 0 b) 1350 c) 1500 d) 2000
13. The manufacturer can make a maximum profit of Rs.
a) 60000 b) 135000 c) 150000 d) 200000
**************************
“Everything is perfect in the universe,
even your desire to improve it”
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IM + OR
1 LINEAR PROGRAMMING
Previous IES Questions :
1. Solution to Z = 4x + 6x (ESE-92)
max 1 2
x1 + x2 < 4
3x1 + x2 < 12
x1 x2 > 0 is
(a) Unique (b) Unbounded (c) Degenerate (d) Infinite
2. Which of the following conditions are necessary of applying linear programming (ESE-92)
1. There must be a well defined objective function
2. The decision variables should be interrelated and non-negative
3. The resources must be in limited supply
(a) 1 and 2 only (b) 1 and 3 only (c) 2 and 3 only (d) 1, 2 and 3
3. Consider the following statements : (ESE-93)
Linear programming model can be applied to
1. Line balancing problem
2. Transportion problem
3. Project management
Of these statements
(a) 1, 2 and 3 are correct
(b) 1 and 2 are correct
(c) 2 and 3 are correct
(d) 1 and 3 are correct
4. A simple table for a linear programming problem is given below : (ESE-94)
5 2 3 00 0
x1 x2 x3 x4 x5 x6 B0
x4 1 2 2 1 0 0 8
x5 3 4 1 0 1 0 7
x2 3 4 00 1 10
6
Which one of the following correctly indicates the combination of entering and leaving variable
(a) x1 and x4 (b) x2 and x6 (c) x1 and x3 (d) x3 and x4
5. A feasible solution to the linear programming problem should (ESE-94)
(a) Satisfy the problem constraints
(b) Optimize the objective function
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MECHANICAL ENGINEERING
(c) Satisfy the problem constraints and non-negatively restrictions
(d) Satisfythe non-negativity restrictions
6. Which of the following subroutines does a computer implementations linear programming by the
simplex method use ? (ESE-96)
(a) Finding a root of a polynomial
(b) Finding the determinant of matrix
(c) Finding the eigen values of a matrix
(d) Solving a system of linear equations
7. Consider the following statements : (ESE-00)
1.Alinear programming problem with three variables and two constraints can be solved by graphical
method
2. For solutions of a linear programming problem with mixed constraints, Big-M-method can be
employed.
3. In the solution process of a linear programming problem using Big-M-method, when an artificial
variable leaves the basis, the column of the artificial variable can be removed from all subsequenctt
tables.
Which of these statements are correct ?
(a) 1, 2 and 3 (b) 1 and 2 (c) 1 and 3 (d) 2 and 3
8. The primal of a LP problem is maximization of objective function with 6 variables an 2 constraints.
(ESE-02)
Which of the following correspond to the dual of the problem stated ?
1. It has 2 variables and 6 constraints
2. It has 6 variables and 2 constraints
3. Maximization of objective function.
4. Miniation of objective function.
Select the correct answer using the codes given below :
Codes :
(a) 1 and 3 (b) 1 and 4 (c) 2 and 3 (d) 2 and 4
9. In case of solution of a two variable linear programming problem by graphical method one constraint
line comes parallel to the objective function line. Which one of the following is correct ? (ESE-04)
The problem will have
(a) infeasible solution
(b) unbounded solution
(c) Degenerate solution
(d) infinite number of optimal solutions
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IM + OR
10. Which of the following are correct in respect of graphically solved linear programming problems ?
(ESE-04)
1. The region of feasible solution has concavity property.
2. The boundarie of the region are lines or planes.
3. There are corners or extreme points on the boundary.
Select the correct answer using the codes given below :
(a) 1 and 2 (b) 2 and 3 (c) 1 and 3 (d) 1, 2 and 3
11. A tie for leaving variables in simplex procedure implies (ESE-05)
(d) Degeneracy
(a) Optimality (b) Cycling (c) No solution
12. In a linear programming problem, if a basic solution has no more than m positive xj (j = 1,2,......n), it is
called (ESE-06)
(a) Basic feasible solution (b) Unbounded solution
(c) Non-degenerate basic feasible solution (d) None of the above
13. In case of solution of linear programming problem using graphical method, if the constraint line of one
of the non-redundant constraints is parallel to the objective function line, then it indicates (ESE-06)
(a) an infeasible solution (b) a degenerate solution
(c) an unbound solution (d) a multiple number of optimal solution
14. If m is the number of constraints in a linear programming with two variables x and yand non-nregativity
constraints x >0, y > 0; the feasible region in the graphical solution will be surrounded by how many
lines ? (ESE-07)
(a) m (b) m + 1 (c) m + 2 (d) m + 4
15. Which of the following statements is not correct ? (ESE-08)
(a) A linear programming problem with 2 variables and 3 constraints can be solved by Graphical
Method
(b) In big-M method if the artificial variable can not be driven out it depiets an optimal solution.
(c) Dual of a dual is the primal problem
(d) For mixed constraints either big-M method or two phase method can be employed
16. In a linear programming problem, which one of the following is correct for graphical method ?
(ESE-09)
(a)Apoint in the feasible region is not a solution to the problem
(b) One of the corner points of the feasible region is not the optimum solution
(c)Any point in the positive quadrant does not satisfy the non negativity constraint
(d) The lines corresponding to different values of objective functions are parallel.
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MECHANICAL ENGINEERING
17. In a linear programming problem with mixed constraints (some) constrains of < type and some of >
type) can be solved by which of the following methods ? (ESE-09)
(a) Big-M method (b) Hungarian method
(c) Branch and bound technique (d) Least cost method
18. While solving a linear programming problem by simplex method, if all ratios of the right hand side (bi)
to the cofficient in the key row (aij) become negative, then the problem has which of the following type
of soltuion ? (ESE-09)
(a)An unbound solution (b) Multiple solutions
(c)Aunique solution (d) No solution
19. Which one of the following is true in case of simplex method of linear programming ? (ESE-09)
(a) The constants of constraints equation may be positive or negative
(b) Inequalities are not converted into equations
(c) It cannot be used for two variable problems
(d) The simplex algorithm is an iterative procedure
20. The linear programming is used for optimizartion problem which satisfythe following conditions :
(ESE-10)
1. Objective function expressed as a linear function of variables
2. Resoucres are unlimited
3. The decision variable are inter-related and non-regative
Which of the above statements is/are correct ?
(a) 1, 2 and 3 (b) 2 and 3 only (c) 1 only (d) 1 and 3
21. The leaning baisc variable in simplex method is the basic variable that (ESE-10)
(a) has the lowest value
(b) has the largest coefficient in the key row
(c) has the smallest cofficient in the key row
(d) goes to zero first as the entering basic variable is increased
22. A basic feasible solution of an m x n transportation problem is said to be non degenerate if the
allocations are in independent positions and starting basic feasible solution contains exactly following
number of individual allocations : (ESE-11)
(a) m + n (b) m x n (c) m + n - 1 (d) m + n + 1
23. An unbound solution of a linear programming problem is reflected in the simplex method, when :
(ESE-11)
(a)All the ratio of ‘right hand sides’to coefficients in key column become negative
(b)All the ratio of right hand sides to coefficients in key column become zero
(c)All right hand sides become negative
(d)All right hand sides become zero
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IM + OR
24. In simplex method, the variables which have not been assigned the value zero during the iteration, are
called (ESE-12)
(a) Basic variables (b)Actual variables (c)Artificial variables (d) None of the above
25. Assuming X and Y are the two control variables, the following are the constraints laid out for
maximizingtheprofit : (ESE-15)
Maximize profit (P) = 8X + 5Y subject to
Constraint-1 : 2X + Y < 1000
Constraint-2 : 3X + 4Y < 2400
Constraint-3 : X + Y < 800
Constraint-4 : X - Y < 350
Constraint-5 : X > 0
Constraint-6 : Y > 0
Which of the above constraints is a redundant one and does not have any effect on the solution ?
(a) Constraint - 1 (b) Constraint - 3
(c) Constraint - 4 (d) Constraint - 5 and Constraint - 6
26. Maximize Z = 2x1 + 3x2
subject to
2x1 + x2 < 6 (ESE-15)
x1, x2 > 0 (d) degenerate
The solution to the above LPP is
(a) optimal (b) infeasible (c) unbounded
27. Objective function
Z = 5X1 + 4X2 (Maximize) (ESE-15)
Subject to
0 < X1 < 12
0 < X2 < 9
3X1 + 6X2 < 66
X, X >0
12
What is the optimum value ?
(a) 6, 9 (b) 12, 5 (c) 4, 10 (d) 0, 9
****************
1. Linear Programming (Ans.) (IES)
1-a, 2-d, 3-c, 4 - ,5-c, 6-d, 7-a, 8-b, 9-d, 10-b, 11-d, 12-c, 13-d, 14-c, 15-b, 16-d, 17-a, 18-a, 19-d,
20-d, 21-d, 22-c, 23-a, 24-a, 25-b, 26-d, 27-b.
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MECHANICAL ENGINEERING
2 PERT CPM
1. In the construction of networks, dummy activities are introduced in order to : (GATE - ME - 90)
a) Compute the slack on all events
b) Transfer resources, if necessary during monitoring
c) Clearly designate a precedence relationship
d) Simplifythe crashing plan
2. For a small project with five jobs, the following data is given (GATE - ME - 93)
Job Immediate Duration (days)
predecessors mean Std. deviation
A- 10 2
B- 51
CA 16 2
DA 12 2
E B,C 15 1
(i) Draw the proejct network in activity on are mode
(ii) Under PERT assumptions, determine the distribution of proejct duration
3. In PERT, the distribution of activity times is assumed to be (GATE - ME - 95)
a) Normal b) Gamma (GATE - ME - 95)
Cost
c) Beta d) Exponential
50
4. A project with the following data is to be implemented 50
40
Activity Precede Duration 100
or (s) (days) (Rs/day) 100
A - 2 60
B - 4
C A 1
D B 2
E A,B 3
F E 2
a) What is the minimum duration of the proejct ?
b) Draw a Grant chart for the early schedule
c) Determine the peak requirement of money and the day on which it occurs in above schedule
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IM + OR
5. Adummy activity is used in PERT network to describe (GATE - ME - 97)
a) Precedence relationship b) necessary time delay
c) Resource restriction d) rresource idleness
6. A project plan is given below : (GATE - ME - 97)
Activity Time duration in weeks Predecessor
A2 None
B2 None
C7 A
D 12 A
E 10 B
F3 D, E
G4 C, F
a) Construct a PERT net work
b) Find the critical path and estimate the project duration
7. Given below are the data for a project network (GATE - ME - 00)
Activity Immediately Duration
Preceding
A3
B --- 5
C A, B 5
DA 4
E C, D 4
FA 4
G A, B 6
a) Draw the network for the above project capturing the precedence relationships.
b) Find the critical path and its duration
8. A proejct consists of three parallel paths with duration and variances of (10, 4), (12, 4) and (12, 9)
respectively.According to the standard PERT assumptions, the distribution of the project duration is
(GATE - ME - 02)
9. The precedence relations and durations of jobs in a project are given below : (GATE - ME - 02)
Job Predecessors (s) Duration (in days
A- 2
B- 4
CA6
DA 8
E B, C 6
F B,C 4
GF 2
HF 8
I D,E,G 6
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MECHANICAL ENGINEERING
a) Draw the activity -on - are project network
b) Determine all critical path(s) and their duration(s).
c) What is the total float for job B and D
10. Aproject consists of activitesAto M shown in the net in the following figure with the durition of the
activities marked in days [GATE - 2003]
The project can be completed
a) between 18, 19 days b) between 20, 22 days
c) between 24, 26 days d) between 60, 70 days
11. In PERT analysis a critical activity has (GATE - ME - 04)
a) Maximum Float b) Zero Float
c) Maximum Cost d) minimum Cost
12. A project has six activities (A to F) with respective activity durations 7, 5, 6, 6, 8, 4 days. The network
has three paths A-B, C-D and E-F.All the activities can be crashed with the same crash cost per day.
The number of activities that need to be carshed to reduced the project duration by 1 day is
(GATE - ME - 05)
a) 1 b) 2 c) 3 d) 6
Statement for linked answer questions 13 & 14 :
Consider a PERT network for a project involving six tasks ( a to f)
Taks Predecessor Expected taks time Variance of the taks
(in days) time (in days2)
a- 30 25
ba 40 64
ca 60 81
db 25 9
e b, c 45 36
f d, e 20 9
13. The expected completion time of the proejcts is (GATE - ME - 06)
d) 155 days
a) 238 days b) 224 days c) 171 days
14. The standard deviation of the critical path of the project is (GATE - ME - 06)
d) 238 days
a) 151 days b) 155 days c) 200 days
15. The expected time (te) of a PERT activity in terms of optimistic time (t0), pessimistic (tp) and most likely
time (t1) is given by (GATE - ME - 09)
t 4t +t t 4t +t L
a) te = 0 L p b) te = 0 P
6 6
t 4t +t t0 4 l + tL
c) te = 0 L P d) te =
P
3
3
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IM + OR
Common data question 16 & 17 6
Consider the following network
3
>
>
> >
1 >2 > 5 7
> >
4
The optimistic time, most likelytime and pessimistic time of ll the activities are given in the table below
: (GATE - ME - 09)
Activity T0 T m T p T0+4Tm+Tp = Tp-T0 =
Te = 6
6 Tp-T0
6
1-2 1 2 3 2 11
1-3 5 6 7 6 39
1-4 3 5 7 5 11
2-5 5 7 9 7 39
3-5 2 4 6 4 24
5-6 4 5 6 5 39
4-7 4 6 8 6 24
6-7 2 3 4 3 39
24
39
11
39
24
39
11
39
16. The critical path duration of the network (in days) is
a) 11 b) 14
c) 17 d) 18
17. The standard deviation of the critical path is
a) 0.33 b) 0.55
c) 0.77 d) 1.66
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MECHANICAL ENGINEERING
18. The project activities, prededence relationships and durations are described in the table. The critical
path of the proejct is (GATE - ME - 10)
Activity Precedence Duration
(in days)
___P_________________________-___________________________3__________
___Q_________________________-___________________________4__________
___R_________________________P___________________________5__________
___S_________________________Q___________________________5__________
___T_________________________R_,__S________________________7__________
___U_________________________R_,__S________________________7__________
___V_________________________T___________________________2__________
WU 10
a) P-R-T-V b) Q-S-T-V c) P-R-U-W d) Q-S-U-W
19. For the netwrok below, the objective is to find the length of the shortest path from node P to node G.
Let dij be the length of directed are from node I to node j. Let sj be the length of the shortest path from
P to node j. Which of the following equation can be used to find s ? (GATE - ME - 10)
G
a) sG = Min {sQ, sR} > >O
b) sG = Min {sQ - dQG, SR - dRG} >
c) sG = Min {sQ + dQG, SR + dRG} P >G
d) sG = Min {dQG, dRG}
> >R
***************
“If you cannot do great things, do small things in a great way”
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IM + OR
2 PERT & CPM
Previous IES Questions : (ESE-93)
1. Earliest finish time can be regarded as
(a) EST + duration of activity
(b) EST - duration of activity
(c) LFT + duration of activity
(d) LFT - duration of activity
2. Consider an activity having a duration time of Tij. E is the earliest occurrence time and L the latest
occurence time (see figure given) (ESE-93)
Consider the following statements in thie regard : (ESE-93)
1. Total float = Lj - Ei - Tij
1. Free float = Ej - Ei - Tij
1. Slack of the tail event = Lj - Ei
Of these statements
(a) 1, 2 and 3 are correct
(b) 1 and 2 are correct
(c) 1 and 3 are correct
(d) 2 and 3 are correct
3. Which one of the following networks is correctely drawn ?
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MECHANICAL ENGINEERING
4. APERT network has three activities on critical path with mean time 3, 8 and 6, and standard deviation
1, 2 and 2 respectively. The probability that the porject will be completed in 20 days is (ESE-93)
(a) 0.50 (b) 0.66 (c) 0.84 (d) 0.95
5. In a connected network of ‘n’ arcs (roads) joining ‘m’ vertices (town), a selection of roads is taken up
for resurfacing based on a minimum spanning tree of the network as being the least cost solution. This
spanning tree will contain. (ESE-94)
(a) m arcs (b) (m+1) arcs (c) (m-1) arcs (d) (m+n-1) arcs
6. For the PERT network shown in the given figure the probability of completing the project in
27 days is (ESE-94)
(a) 0.841 (b) 0.919 (c) 0.964 (d) 0.977
7. In PCM, the cost slope is determined by (ESE-94)
(a) Crash cost
Normal cost
(b) Crash cost - Normal cost
Normal time - crash time
(c) Normal cost
Crash cost
(d) Normal cost - crash cost
Normal time - crash time
8. Assertion (A) : Generally PERT is preferred over CPM for the purpose of project evaluation.
Reason (B) : PERT is based on the approach of multiple time estimates for each activity. (ESE-96)
9. Which of the following are the guidelines for the construction of a network diagram ? (ESE-96)
1. Each activity is represented by one and only one arrow in the network
2. Two activities can be identified by the same beginning and end events
3. Dangling must be avoided in a network diagram
4. Dummy activity consumes no time or resource
Select the correct answer using the codes given below (ESE-96)
(a) 1, 2 and 3 (b) 1, 3 and 4 (c) 1, 2 and 4 (d) 2, 3 and 4
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IM + OR (ESE-96)
10. In the network shown below the critical path is along
(a) 1-2-3-4-8-9 (b) 1-2-3-5-6-7-8-9
(c) 1-2-3-4-7-8-9 (d) 1-2-5-6-7-8-9
11. In a PERT network, expected proejct duration is found to be 36 days from the start of the project.
The variance is four days. The probability that the project will be completed in 36 days is (ESE-97)
(a) zero (b) 34% (c) 50% (d) 84%
12. The variance (V1) for critical path
a b = 4 time units, b c = 16 time units, c d = 4 time units, d e = 1 time unit.
The standard deviation of the critical path is a e is (ESE-97)
(a) 3 (b) 4 (c) 5 (d) 6
13. Estimated time Te and variance of the activities ‘V’on the critical path in a PERT net work are given in
the following table : (ESE-98)
The probability of completing the project in 43 days is
(a) 15.6 % (b) 50.0% (c) 84.13% (d) 90.0%
14. The earliest occurence time for event ‘1’ is 8 weeks and the latest occurrence time for event ‘1’ is
26 weeks. The earliest occurence time for event ‘2’ is 32 weeks and the latest occurrence time for
event ‘2’is 37 weeks. If the activity time is 11 weeks, then the total float will be (ESE-98)
(a) 11 (b) 13 (c) 18 (d) 24
15. Time estimates of an activity in a PERT network are :
Optimistic time t0 = 9 days ; pessimistic time tp = 21 days and most likely time tm = 15 days. The
approximates probability of completion of this activity in 13 days is (ESE-99)
(a) 16% (b) 34% (c) 50% (d) 84%
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MECHANICAL ENGINEERING
16. Dummy activities are used in a network to (ESE-00)
(a) Facilitate computation of slacks
(b) Satisfy precedence requirements
(c) Determine project completion time
(d) Avoid use of resources
17. If the earliest starting time for an activity is 8 weeks, the latest finish time is 37 weeks and the duration
time of the activity is 11 weeks, then the total float is equal to (ESE-00)
(a) 18 weeks (b) 14 weeks (c) 56 weeks (d) 40 weeks
18. Latest start time of an activity in CPM is the (ESE-01)
(a) Latest occurence time of the successor event minus the duration of the activity
(b) Earliest occurrence time for the precessor event plus the duration of the activity
(c) Latest occurence time of the successor event
(d) Earliet occurence time for the predecessor event
19. For the network shown in the given figure, the earliest expected complection time of the proejct is
(ESE-01)
(a) 26 days (b) 27 days (c) 30 days (d) indeterminable
20. Consider the following statements regarding updating of the network :
1. For short duration project, updating is done frequently
2. For large duration porject, frequency of updating is decreased as the project is nearing completion
3. Updating is caused by overstimated or underestimated times of activities
4. The outbreak of natural calamity necessitates updating
Which of the above statements are correct ? (ESE-02)
(a) 1, 2 and 3 (b) 2, 3 and 4 (c) 1, 3 and 4 (d) 1, 2 and 4
21. Which one of the following is assumed for timing the activities in PERT network ? (ESE-02)
(a) distribution
(b) distribution
(c) Binomial distribution
(d) Erlangian distribution
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IM + OR
22. The three time estimates of a PERTactivityare : optimistic time = 8 min, most likelytime = 10 min and
pessimistic time = 14 min. The expected time of the activity would be (ESE-02)
(a) 10.00 min (b) 10.33 min (c) 10.66 min (d) 11.00 min
23. For the network shown in the figure, the variance along the critical path is 4. (ESE-02)
The probability of completion of the project in 24 days is
(a) 68.2% (b) 84.1% (c) 95.4% (c) 97.7%
24. The variance of the completion time for a project is the sum of variances of (ESE-03)
(a)All activitytimes
(b) Non-critical activity times
(c) Critical activitytimes
(d)Activity times of first and last activities of the proejct
25. The earliest time of the completion of the last event in the above network in week is (ESE-04)
(a) 41
(b) 42
(c) 43
(d) 46
26. Consider the following statements with respect to PERT : (ESE-04)
(d) 1, 2 and 4
1. It consists of activities with uncertain time phases
2. This is evolved from Gantt chart
3. Total slack along the critical path is not zero
4. There can be more than one critical path in PERT network
5. It is similar to electrical network
Which of the statements given above are correct ?
(a) 1, 2 and 5 (b) 1, 3 and 5 (c) 2, 4 and 5
27. In a small engineering project, for an activity, the optimistic time is 2 minutes, the most likely time is
5 min and the pessimistic time is 8 minutes. What is the expected time of the activity ? (ESE-05)
(a) 1 minute (b) 5 minute (c) 8 minutes (d) 18 minutes
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MECHANICAL ENGINEERING
28. The critical path of a network is the path that (ESE-05)
(a) Takes the shortes time (b) Takes the longest time
(c) Has the minimum variance (d) Has the maximum variance
29. In a network what is total float equal to ? (ESE-06)
(a) LFTj - ESTi + ti-j
(b) EFT - LET + t
j i i-j
(c) EST - LFT - t
j i i-j
(d) LFT - EST - t
j i i-j
LFT = Latest finish time of an activity
ESET = Earliest start time of an activity
tij = time of activity i-j
30. Consider the following statements in respect of PERT and CPM (ESE-06)
1. PERT is event-oriented while CPM is activity-oriented
2. PERT is probabilistic which CPM is deterministic
3. Levelling and smoothing are the techniques related to resource schedule in CPM
What of the statements given above are correct ?
(a) 1, 2 and 3 (b) only 1 and 2 (c) only 2 and 3 (d) only 1 and 3
31. Match List-I with List-II and select the correct answer using the code given below the lists : (ESE-07)
List I (Term) List II (Characteristics)
A. Dummyactivity 1. Follows distribution
B. Critical path 2. It is built on activitly-oriented diagram
C. PERT activity 3. Constructed only to establish sequence
D. Criticl path method 4. Has zero total slack
Code A B C D
(a) 3 4 1 2
(b) 4 2 3 1
(c) 3 4 2 1
(d) 4 2 1 3
32. Consider the following statements : (ESE-07)
PERT considers the following time estimates
1. Optimistic time
2. Pessimistic time
3. Most likely time
Which of the statements given above are correct ?
(a) 1, 2 and 3 (b) 1 and 2 only (c) 3 only (d) 1 and 3 only
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IM + OR
33. Which of the following statements in not correct ? (ESE-07)
(a) PERT is probabilistic and CPM is deterministic
(b) In PERT, events are used and in CPM activities are used
(c) In CPM, the probability to complete the project in a given time-duration is calculated
(d) In CPM crashing is carried out
34. APERT activity has an optimistic time estimate of 3 days, most likely time estimate of 8 days, most
likely time estimate of 8 days, and a pessimistic time estimate of 10 days. What is the expected time of
this activity? (ESE-08)
(a) 5.0 days (b) 7.5 days (c) 8.0 days (d) 8.5 days
35. Which one of the following statements is not correct ? (ESE-08)
(a) PERT is activity oriented and CPM is event oriented
(b) In PERT, three time estimates are made, whereas in CPM only one time estimate is made
(c) In PERT slack is calculated whereas in CPM floats are calculated
(d) Both PERT and CPM are used for project situations
36. What is additional time available for the performance of an activity in PERT and CPM calculated on
the basis that all activities will start at their earliest start time, called ? (ESE-08)
(a) Slack (b) Total float (c) Free float (d) Independent float
37. Consider the below network. Activitly times are given in number of days. The earliest expected
occurrence time (TE) for event 50 is (ESE-08)
(a) 22
(b) 23
(c) 24
(d) 25
38. If critical path of a project is 20 months with a standard deviation 4 months, what is the probability that
the project will be completed in 24 months (ESE-08)
(a) 15.85% (b) 68.3% (c) 84.2% (d) 95.50%
39. In CPM, the project duration can be reduced by crashing (ESE-10)
(a) One or more non-critical activities (b) One or more critical activities only
(c) One or more dummy activities only (d)Activities having independent float
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MECHANICAL ENGINEERING
40. In project scheduling, networks have an advantages over Gantt charts since networks (ESE-10)
(a) slow start and finish of an activity clearly
(b) show activityand activity times clearly
(c) show inter-relationship among activities clearly
(d) are easy to draw
41.
Consider the 3 activities of a CPM network as shown above. Earliest and latest occurence times of
these events are given on the nodes. The float on activity 20-30 is (ESE-10)
(a) 0 (b) 2 (c) 3 (d) 5
42. In PERT and CPM network the dummy activity (ESE-10)
(a) consumes time (b) consumes resources
(c) is used to preserve the logic (d) is a real activity
43. Which one of the following statements is not correct ? (ESE-10)
(a) PERT is probabilistic whereas CPM is deterministic
(b) In PERT slack on various events is calculated whereas in CPM floats are calculated
(c) Critical path in a network is the path on which events have no slack
(d) More than four dummy activities cannot be used in a PERT network.
44. In PERT, the distribution of activties times, is assumed to be (ESE-12)
(a) Normal (b) Gamma (c) Beta (d) Exponential
45. Which of the following is true with respect to a PERT network ? (ESE-12)
(a)Activity duration is beta - distributed and project duration is normally distributed
(b)Activity duration is deterministic and project duration is beta distributed
(c)Activity duration is deterministic and hence porject duration is also deterministic
(d) Four time estimates are used for determining average duration of an activity
46. Adummy activity is used in PERT network for : (ESE-13)
(a) Precedence relationship (b) Necessary time delay
(c) Rersource restriction (d) Resource idleness
41. Statement (I) : The change in critical path requires rescheduling in a PERT network.
Statement (II) : Some of the activities cannot be completed in time due to unexpected breakdown of
equipment or no availability of raw materials. (ESE-15)
*******************
2. PET & CPM (Ans.) (IES)
1-a, 2-b, 3-a,b 4-c, 5-d, 6-a, 7-b, 8-a, 9-b, 10-b, 11-c, 12-c, 13-c, 14-c, 15-a, 16-b, 17-a, 18-a, 19-c,
20-a, 21-b, 22-b, 23-d, 24-c, 25-d, 26-d, 27-b, 28-b, 29-d, 30-a, 31-a, 32-a, 33-c, 34-b, 35-a, 36-c,
37-b, 38-c, 39-b, 40-c, 41-d, 42-c, 43-d, 44-c, 45-a, 46-a, 47-b.
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IM + OR
3 QUEUING THEORY
1. On the average 100 customers arrive at a palce each hour, and on the average the server can process
120 customers per hour. What is the proportion of time the server is idel ? (GATE - ME - 10)
2. The cost of providing service in a queing system increases with (GATE - ME - 97)
a) Increased mean time in the queue
b) Increased arrivle rate
c) Decreased mean time in the queue
d) Decreased arrival rate
3. People arrive at a hotel in a Poissson distributed arrival rate of 8 per hour. Service time distribution is
closely approximated by the negative exponential. The average service time is 5 minutes. Calculate
(a) the mean number in the waiting line ;
(b) the mean number in the system;
(c) the waiting time in the queue;
(d) the mean time in the system and
(e) the utiliztion factor. (GATE - ME - 97)
4. At a production machine, parts arrive according to a Poisson process at the rate of 0.35 parts per
minute. Processing time for parts have exponential distribution with mean of 2 minutes. What is the
probability that a random part arrival finds that there are already 8 parts in the system (in machine + in
queue) ? (GATE - ME - 99)
a) 0.0247 b) 0.0576
c) 0.0173 d) 0.082
5. In a single serve infinite population queuing model, arrivals follow a Poisson distribution with mean
= 4 per hour. The service times are exponential with mean service time equal to 12 minutes. The
expected length of the queue will be (GATE - ME - 00)
a) 4 b) 3.2
c) 1.25 d) 24.3
6. Arrival at a telephone booth are considered to be Poisson, with an average time of 10 minutes
between successive arrivals. The length of a phone call is distributed exponentially with mean
3 minutes. The probability that an arrival does not have to wait before service is (GATE - ME - 02)
a) 0.3 b) 0.5 c) 0.7 d) 0.9
7. A maintenance service, facility has Poisson, arrival rates, negative exponential service time and
operates on ‘first come first served’ queue discipline. Breakdowns occur on an average of 3 per day
with a range of zero to eight. The maintenance crew can service an average of 6 machines per day with
a range of zero to seven. The mean waiting time for an item to be serviced would be
(GATE - ME - 04)
a) 1 day 1 c) 1 day d) 3 days
6 b) day
3
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MECHANICAL ENGINEERING
8. Consider a single server queuing model with Poisson arrivals = 4 / hour) and exponential service
( = 4/hour). The number in the system is restricted to a maximum of 10. The probability that a person
who comes in leaves without joining the queue is (GATE - ME - 05)
a) 1 b) 1 c) 1 1
11 10 9 d)
2
9. The number of customers arriving at a railway reservation counter is Poisson distributed with an arrival
rate of eight customers per hour. The reservation clerk at this counter takes six minutes per customer
on an average with an exponentially distributed service time. The average number of the customerse in
the queue will be (GATE - ME - 06)
a) 3 b) 3.2 c) 4 d) 4.2
10. In an M/M/1 queuing system, the number of arrivals in an interval of length T is a Poission random
variable (i.e. the probability of there being n arrivals in an interval of length T is e-T (T)n ).
n!
The probabilitydensity function f(t) of the inter-arrival time is given by (GATE - ME - 08)
a) (e-t) b) e-t
c) e-t
e-t
d)
11. Little’s law is a relationship between (GATE - ME - 10)
a ) stock level and lead time in an inventory system
b) waiting time and length of the queue in a queuing system
c) number of machines and job due dates in a scheduling problem
d) uncertainly in the activity time and project completion time
12. Cars arrive at a service station according to Poisson’s distribution with a mean rate of 5 per hour. The
service time per car is exponential with a mean of 10 minutes.At state, the average waiting time in the
queue is (GATE - ME - 11)
a) 10 min b) 20 min c) 25 min d) 50 min
**************
“Just try to be the best you can be and
hope that is the best ever”
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IM + OR
3 QUEUING THEORY
Previous IES Questions : (ESE-92)
1. Service time in queuing theory is usuallyassumed to follow
(a) Normal distribution
(b) Poissons distribution
(c) Erlang distribution
(d) Exponential law
2. For a M/M/1 : /FCFS queue, the probability of the queue size being greater than N is given by
= mean arrival rate and = mean service rate (ESE-93)
N N (c) ()N (d) N
(a) (b)
3. Assertion (A) : In waiting line model, it is assumed that arrival rate is described by a poisson
probability distribution.
Reason (R) : The arrival rate is a probabilistic variable in queue disclipline is first come first served.
(ESE-93)
4. If the arrival takes place every10 minutes with a service time 4 minutes per unit then the mean arrival
rate, mean service rate and the probability that one would have to wait will be respectively(ESE-94)
(a) 10, 4 and 0.25 (b) 0.1, 0.25 and 0.4 (c) 10, 0.4 and 0.25 (d) 0.1, 0.25 and 0.1
5. In a M/M/I queuing system, expected waiting time of a unit that actually waist is given by (ESE-94)
(c) (d)
(a) (b)
6. Assertopm (A) : In a queuing model, the assumptin of exponential distribution with onlyone parameter
family for service times is found to be unduly restrictive.
Reason (R) : This is partly because the exponential distribution has the property that smaller service
times are inherently less probable than larger service time. (ESE-95)
(a) Both Aand R are true and R is the correct explanation of A
(b) Both A and R are true but R is not a correct explanation of A
(c) Ais true but R is false
(d) Ais false but R is true
7. If the arrival rate of units is and the service rate is for a waiting line system having ‘m’ number of
service stations, then the probability of a service stations, then the probability of a service unit being
turned out in time interal (t, t + t) is (ESE-96)
(a) Zero (b) t (c) m.t (d) n.t
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MECHANICAL ENGINEERING
8. Consider two queuing disciplines in a single server queue. Case 1 has first come fir server queue. Case
I has first come first served discipline and case 2 has a last come first served discipline. If the average
waiting times in the two cases are W1 and W2 respectively, then which one of the following inferences
would be true ? (ESE-97)
(a) W1 > W2 (b) W1< W2 (c) W1 = W2
(d) Data insufficient to draw anytangible inference
9. In a single server queue customers are served at a rate of . If W and Wq represent the mean waiting
time in the system and mean waiting time in the queue respectively, then W will be equal to(ESE-97)
(a) Wq - (b) Wq + (c) Wq + (d) Wq -
10. For a M/M/1 : /FCFS queue, the mean arrival rate is equal to 10 per hour and the mean service rate
is 15 per hour. The expected queue length is (ESE-98)
(a) 1.33 (b) 1.53 (c) 2.75 (d) 3.20
11. Assertion (A) : The number of vehicles arriving per unit time at a vehicle maintenance shop at any
instant of time can be assessed if the probability of that number and the average rate of arrival is know.
Reason (R) : The vehicles, ‘k’ arriving at a vehicle maintenance shop with probability (k) is given by
lk e-l (ESE-98)
p(k) = k!
12. The average time between two arrivals of customers at a counter in a ready-made garment store is
4 minutes. The average time of the counter clerk to serve the customer is 3 minutes. The arrivals are
distributed as per Poisson distribution and the services are as per the exponential distribution. The
probability that a customer arriving at the counter will have to wait, is (ESE-99)
(a) Zero (b) 0.25 (c) 0.50 (d) 0.75
13. Arrivals at a telephone booth are considered to be Poisson with an average time of 10 minutes
between one arrival and the next. The length of a phone call is assumed to be distributed exponentially
with a mean of 3 minutes. The probability that a person arriving at the booth will have to wait, is
(ESE-00)
(a) 0.043 (b) 0.300 (c) 0.429 (d) 0.700
14. Arrivals at a telephone booth are considered to be according to Poisson distrubion, with an average
time of 10 minutes between one arrival and the next. The length of a phone call is assumed to be
distributed exponentially with mean of 3 minutes. The probability that a person arriving at the booth
will have to wait, is (ESE-01)
(a) 3 7 (c) 7 13
10 (b) 10 30 (d) 10
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IM + OR
15. The reason for diversifiation is to (ESE-02)
(a) Reduce production cost (b) Balance low demand high capacity situation
(c) Satisfy more customers (d) Improve capacityutiliztion
16. Customers arrive at a counter randomly at the rate of 6 customers per hour. The service is provided at
the counter by a server. The mean time of the service is 4 minutes per customer. The services are
exponentially distributed. What is the probability that a newly arrived customer has to wait ?
(ESE-04)
(a) 0.4 (b) 0.6 (c) 0.66 (d) 0.8
17. Which one of the following statements is correct ? (ESE-04)
Queuing theoryis applied best in situations where
(a) arrival rate of customers equal to service rate
(b) average service time is greater than averaeg arrival time
(c) there is only one channel of arrival at random and the service time is constant
(d) the arrival and service rate cannot be analyzed through any standard statistical distribution
18. The arrivals are completely random, then what is the probability distribution of number of arrivals in a
given time (ESE-05)
(a) Negative exponential (b) Binomial
(c) Normal (d) Poisson
19. In single server queuing model if arrival rate is and service rate is when what is the probability of
the system being idle ? (ESE-05)
(d) 1 -
(a) (b) (c) 1 -
20. In a queuing problem, if the arrivals are completelyrandom, then the provabilitydistribution of number
of arrivals in a given time follows (ESE-06)
(a) Poisson distribution (b) Normal distribution
(c) Binomial distribution (d) Exponential distribution
21. In the number of arrivals in queue follows the Poisson distribution, then the inter arrival time obeys
which one of the following distributions ? (ESE-07)
(a) Poisson’s distributions (b) Negative distribution
(c) Normal distribution (d) Binomial
22. If average arrival rate in a queue is 6/hr, which one of the following is the average number of customers
in the line, including the customer being served ? (ESE-07)
(a) 0.3 (b) 0.6 (c) 1.2 (d) 1.5
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MECHANICAL ENGINEERING
23. In a single server queuing system with arrival rate of ‘’ and mean service time of ‘’ the expected
number of customers in the system is . What is the expected waiting time per customer in
the system ? (b) (ESE-08)
(c) -
(a) (d)
24. The inter-arrival times at a tool crib are exponential with an average time of 10 minutes and the length
of the service time is assumed to be exponential with mean 6 minute. The probability that a person
arriving at the booth will have to wait is equal to : (ESE-08)
(a) 0.15 (b) 0.40 (c) 0.42 (d) 0.6
25. Which of the following distributions is followed by the number of arrivals in a given time in a single
server queuing model ? (ESE-09)
(a) Negative exponential distribution (b) Poisson distribution
(c) Normal distribution (d) Beta distribution
26. If the arrivals at a service facility are distributed as per the Poisson distribution with a mean rate of 10
per hour and the services are exponentially distributed with a mean service time of 4 minutes, what is
the probability that a customer may have to wait to be served ? (ESE-09)
(a) 0.40 (b) 0.50 (c) 0.67 (d) 1.00
27. A single-bay car wash with a Poisson arrival rate and exponential service time has cars arriving at an
average rate of 10 minutes a part and an average service time of 4 minutes. What is the system
utilization ? (ESE-09)
(a) 1.00 (b) 0.67 (c) 0.40 (d) 0.24
28. In the Kendall’s notation for representing queuing models the first position represents (ESE-10)
(a) Probability law for the arrival (b) Probability law for the service
(c) Number of channels (d) Capacity of the system
29. For a single server queue, the mean arrival rate is equal to 8/hr and the mean service rate is 12/hr. The
expected number in waiting line is equal to : (ESE-11)
(a) 0.5 (b) 1.33 (c) 2 (d) 3
30. In the queuing theory, if the arrivals in a single server model follow Poisson distribution, the time
between arrivals will follow a : (ESE-11)
(a) Gamma distribution (b) Exponential distribution
(c) Binomial distribution (d)Weibull distribution
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IM + OR
31. In queuing theory, the number of arrivals per unit time is estimated by (ESE-12)
(a) Binomial distribution (b) Poisson distribution
(c) Normal distribution (d) Bath tub analogy
32. Consider the follwing statements :
In a single-server queuing model
1. the arrivals is a memoryless process
2. the arrivals is discribed as a Poisson distribution
3. uncetrainty concerning the demand for service exists
Which of the above statements are correct ? (ESE-15)
(d) 2 and 3 only
(a) 1 and 2 only (b) 1 and 3 only (c) 1, 2 and 3
33. Acompanyhas a store which is manned by 1 attendant who can attend to 8 technicians in an hour. The
technicians wait in the queuea and they are attended on first - come - first - served basic. The techni
cians arrive at the store on an average 6 per hour.Assuming the arrivals to follow poisson and servicing
to follow exponential distribution, what is the expected time spent by a technicians in the system, what
is the expected time spent by a technician in the queue and what is the expected time spent by a
technician in the queue and what is the expected number of technicians in the queue ? (ESE-15)
(a) 22.5 minutes, 30 minutes and 2.75 technicians
(b) 30 minutes, 22.5 minutes and 2.25 technicians
(c) 22.5 minutes, 22.5 minutes and 2.75 technicians
(d) 30 minutes, 30 minutes and 2.25 technicians
***********
3. Queuing Theory (Ans.) (IES)
1-d, 2-a, 3-a, 4-b, 5-a, 6-c, 7-c, 8-c, 9-c, 10-a, 11-a, 12-d, 13-b, 14-a, 15-c, 16-a, 17-c, 18-d, 19-c,
20-a, 21-a, 22-d, 23-c, 24-d, 25-b, 26-c, 27-c, 28-a, 29-b, 30-b, 31-b, 32-a, 33-b.
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MECHANICAL ENGINEERING
4 INVENTORY CONTROL
1. In an ideal inventory control system, the economic lot size for a part is 1000. If the annual demand for
the part is doubled, the new economic lot size required will be : (GATE-ME-89)
a) 500 b) 2000 c) 10002 d) 10002
2. When the annual demand of a product is 24000 units, the EOQ (Economic Order Quantity) is
2000 units. If the annual demand is 48000 units the most appropirate EOQ will be
(GATE-ME-91)
a) 1000 units b) 2000 units c) 2800 units d) 4000 units
3. If the demand for an item is doubled and the ordering cost halved, the economic order quantity
(GATE-ME-95)
a) remains unchanged
b) increases by factor of 2
c) is doubled
d) is halved
4. Consider the following data for a product : (GATE-ME-95)
Demand = 1000 unis /year
Order cost = Rs. 40/order
Holding cost = 10% of the unit cost/unit-yr
Unit cost = Rs. 500
a) What is the economic order quantity ?
b) Under the EOQ what is the number of annual orders ?
c) With a policy of ordering every month what would be the total actual cost as a percentage of the
cost at EOQ ?etup costs do not include (GATE-ME-97)
a) Labour cost of setting up machines b) Ordering cost of raw material
c) Maintenance cost of the machiens d) Cost of processing the work piece
5. Determine the number of production runs and also the total the incremental cost in a factory for the
data given below : (GATE-ME-97)
Annual requirement = 15,000 units
Preparation cost per order = Rs. 25
Inventoryholding cost = Rs. 5/unit/year
Production rate = 100 units/day
Number of working days = 250/year
6. One of the following statements about PRS (Periodic Recording System) is not true. Identify
(GATE-ME-98)
a) PRS requires continuous monitoring of inventory levels
b) PRS is useful in control of perishable items
c) PRS provides basis for adjustments to account for variations in demand
d) In PRS, inventory holding costs are higher than in Fixed Recorder Quantity System
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7. In inventoryplanning, extra inventory is unneccessarily carried to the end of the planning period when
using one of the following lot size decision polices : (GATE-ME-98)
a) Lot - for - lot production b) Economic Order Quantity (EOQ) lot size
c) Period Order Quantity (POQ) lot size d) Part Period total cost balancing
8. In computing Wilson’s economic lot size for an item, by mistake the demand rate estimate used was
40% higher than the tree demand rate. Due to this error in the lot size computation, the total cost of
setup plus inventory holding per unit time. Would rise above the true optimum by approximately
(GATE-ME-99)
a) 1.4% b) 6.3% c) 18.3% d) 8.7%
9. A company places orders for supply of two items A and B. The order cost for each of the items is
Rs. 300/- order. The inventory carrying cost is 18% of the unit price per year per unit. The unit prices
of the items are Rs. 40 and Rs. 50 respectively. The annual demands are 10,000 and 20,000
respectively,
(a) Find the economic order quantities and the minimum total cost ?
(b)Asupplier is willing to give a 1% discount on price, if both the items are ordered from him and if the
order quantity for each item is 1000 units or more. Is if profitable to avail the discount ?
(GATE-ME-00)
10. A company is offered the following price breaks for order uantity (GATE-ME-01)
Order quantity Price (Rs)
0 - 100 150
101 and above 100
Ordering cost is Rs. 60 per order while the holding costs is 10% of the purchases price. Determine the
economic order quantity (EOQ) if the annual requirement is 1000 Units.
11. An item can be purchased for Rs. 100. The ordering cost is Rs. 200 and the inventroy carrying cost is
10% of the item cost per annum. If the annual demand is 4000 units, the economic order quantity
(in units) is (GATE-ME-02)
a) 50 b) 100 c) 200 d) 400
12. Market demand for springs is 8,00,000 per annum. A company purchase these springs in lost and
sells them. The cost of making a purchase order is Rs. 1,200. The cost of storage of springs is Rs. 120
per stored piece per annum. The economic order quantity is (GATE-ME-03)
a) 400 b) 2,828 c) 4,000 d) 8,000
13. There are two products P and Q with the following characteristics (GATE-ME-04)
Product Demenad Order Cost Hoding Cost
(Units) (Rs./order) (Rs./unit/year)
P 100 50 4
Q 400 50 1
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The economic order quantity (EOQ) of products P and Q will be in the ratio
a) 1 : 1 b) 1 : 2 c) 1 : 4 d) 1 : 8
14. A company has an annual demand of 1000 units, ordering cost of Rs. 100 per order and carrying cost
of Rs. 1000 per unit-year. If the stock-out costs are estimated to be nearly Rs. 400 each time the
company runs out-of-stock, then safety stock ustified by the carrying cost will be (GATE-ME-04)
a) 4 b) 20 c) 40 d) 100
15. The distribution of lead time demand for an item is as follows : (GATE-ME-05)
Lead time demand Probability
80 0.20
100 0.25
120 0.30
140 0.25
The reorder level is 1.25 times the expected value of the lead demand. The service level is
a) 25% b) 50% c) 75% d) 100
16. Consider the following data for an itme.
Annual demand : 2500 units per year Ordering cost : Rs. 100 per order, Inventory holding rate :
25% of unit price (GATE-ME-06)
Price quoted by a supplier
Order quantity (units)Unit Price (Rs.)
< 500 10
> 500 9
17. A stockiest wishes to optimize the number of perishable items he needs to stock in any month in his
store. The demand distribution for this perishable item is (GATE-ME-06)
Demand (in units) 2345
Probability 0.10 0.35 0.35 0.20
The stockiest pays Rs. 70 for each item and he sells each at Rs. 90. If the stock is left unsold in any
month, he can sell the item at Rs. 500 each. There is no penaltyfor unfulilled demand. To miximize the
expected profit, the optimal stock level is
a) 5 units b) 4 units c) 3 units d) 2 units
18. The maximum level of inventoryof an item is 100 and it is achieved with infinite replenishment rate. The
inventory becomes zero over one and half month due to consumption at a uniform rate. This cycle
continues thoughtout the year. Ordering cost is Rs. 100 per order and inventory carrying
cost is Rs. 10 per item per month.Annual cost (in Rs.) of the plan, neglecting material cost, is
(GATE-ME-07)
a) 800 b) 2800 c) 7800 d) 6800
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19. In a mchine shop, pins of 15 mm diameter are produced at a rate of 1000 per month and same is
consumed at a rate of 500 per month. The production and consumption continue simulaneouslytill the
maximum inventoryis reached. Then inventoryis allowed to reduced to zero due to consumption. The
lot size of production is 1000. If backlog is not allowed, the maximum inventory levels is
(GATE-ME-07)
a) 400 b) 500 c) 600 d) 700
20. Acompanyuses 2555 units of an item annually. Deliverylead time is 8 time is days. The recorder point
(in number of units) to achieve optimum inventory is (GATE-ME-09)
a) 7 b) 8 c) 56 d) 60
21. Annual demand for window frames is 10000. Each frame costs Rs. 200 and ordering cost is Rs. 300
per order. Inventory holding cost is Rs. 40 per frame per year. The supplier is willing to offer 2%
discount if the order quantity is 1000 or more, and 4% if order quantity is 2000 or more. If the total
cost is to be minimized, the retailer should (GATE-ME-10)
a) order 200 frames every time b) accept 2% discount
c) accept 4% discount d) order Economic Order Quantity
22. The word kanban is most appropriately associated with (GATE-ME-11)
a) economic order quantity b) just-in-time production
c) capacity planning d) product design
**********
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4 INVENTORY CONTROL
Previous IES Questions :
1. Out of the following item listed below, which to items would you consider under category (C) under
ABC analysis. (ESE-92)
Item No Annual usage of items Unit cost Rs.
Annual usage x 1000
______A_________________________3__0__________________________0__._1_0_______
______B________________________3__0_0__________________________0__._1_5_______
______C__________________________2__________________________2_0__0_._0__0_____
____________D__________________________________________________6___0______________________________________________________0___.__1___0_____________
____________E_______________________________________________5__________________________________________________0___.__3__0____________
_________F_________________________________3__0__0____________________________________0__._1___0_________
______G_________________________1__0__________________________0__._0_5_______
______H__________________________7___________________________0__._1_0_______
_______I________________________2__0__________________________0__._1_0_______
J5 0.20
(a) Band F (b) C and E (c) E and J (d) G and H
(ESE-94)
2. There are two products Aand B with the following characteristics :
Product Demand Order cost Holding cost
(in Rs/order) (in Rs/unit)
A 100
B 400 100 4
100 1
The economic order quantities (EOQ) of product Aand B will be in the ratio of
(a) 1:4 (b) 1:2 (c) 1:4 (d) 1:8
3. In inventory control theory, the economic order quantity (EOQ) is (ESE-95)
(a) average level of inventory
(b) optimum lot size
(c) lot size corresponding to break-even analysis
(d) capacity of a warehouse
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4. Given tht = procurement cost per order, D = number of units demanded per year, H = holding cost
paer unit per year, i = rate of interest, p = purchase price per unit. The procurement quantity per order
(Q) is given by (ESE-96)
2..D 2..D 2..D 2.
(a) Q = (b) Q = iH + p (c) Q = H + ip (d) Q = D(H + ip)
H + ip
5. Annual demand for a product costing Rs. 100 per piece is Rs. 900. Ordering cost per order is Rs. 100
inventory holding cost is Rs. 2 per unit per year. The economic lost size is (ESE-97)
(a) 200 (b) 300 (c) 400 (d) 500
6. Which of the following cost elements are considerd while determining the Economic Lost size for
purchase ? (ESE-98)
1. Inventory carrying cost
2. Procurement cost
3. Set up cost
Select the correct answer using the codes given below :
(a0 1, 2 and 3 (b) 1 and 2 (c) 2 and 3 (d) 1 and 3
7. Consider the following costs :
1. Cost of inspection and return of goods
2. Cost of obsolescence
3. Cost of scrap
4. Cost of negotiation with suppliers
Which of these costs are related to inventory carrying cost ? (ESE-99)
(d) 2, 4 and 5
(a) 1, 2 and 3 (b) 1, 3 and 4 (c) 2, 3 and 4
8. A dealer sells a radio set at Rs. 900 and makes 80% profit on his investment. If he can sell it at Rs. 200
more, his profit as percentage of investment will be (ESE-99)
(a) 160 (b) 180 (c) 100 (d) 120
9. Match List I (Limits in normal distribution) with List II (Population covered) and select the correct
answer using the codes given below the lists : (ESE-02)
List - I List - II
A. + 3 1. 0.3413
B. + 2 2. 0.6826
C. + 1 3. 0.9973
4. 0.9545
Codes : ABC ABC
(a) 3 4 2 (b) 3 2 4
(c) 4 2 3 (d) 4 3 2
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MECHANICAL ENGINEERING
10. AShop owner with an annual constant demand of ‘A’units has ordering costs of Rs. ‘P’per oder and
carrying costs Rs. ‘I’per unit per year. The economic order quantity for a purchasing model having no
shortage may be determined from (ESE-02)
(a) 24P (b) 24AP (c) 2AP 2AI
AI AI AI (d) P
11. ABC analysis in materials management is a method of classifying the inventories based on (ESE-03)
(a) The value of annual usage of the itmes
(b) Economic order quantity
(c) Volume of material consumption
(d) Quantity of materials used
12. InABC analysis,Aitems requrie (ESE-05)
(a) No safety stokc (b) Low safety stock
(c) Moderate safety stock (d) High safety stock
13. Which one of the following is an inventory system that keeps a running record of the amount in storage
and replenishes the stock when it drops to a certain level by ordering a fixed quantity ? (ESE-06)
(a) EOQ (b) Periodic (c) Peripheral (d) ABC
14. In theABC method of inventory control, GroupAconstitues costly items. What is the usal percentage
of such items of the total iterms ? (ESE-06)
(a) 10 to 20% (b) 20 to 30% (c) 30 to 40% (d) 40 to 50%
15. If the annual demand of an item becomes half, ordering cost double, holding cost one-fourth and the
unit cost twice, then what is the ratio of the new EOQ and the earlier EOQ ? (ESE-06)
1 1 (c) 2 (d) 2
(a) 2 (b) 2
16. Which of the following are the benefits of inventory control ?
1. Improvement is customer’s relationship
2. Economyis purchasing
3. Elimination of the possibilityof duplicate ordering
Select the correct answer using the code given below : (ESE-07)
(d) 1 and 3 only
(a) 1, 2 and 3 (b) 1 and 2 only (c) 2 and 3 only
17. In the EOQ model, if the unit ordering cost is doubled, the EOQ (ESE-07)
(a) I halved (b) I doubled
(c) Increase 1.414 times (d) Decreaes 1.414 times
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18. In the basic EOQ model, demand is 60 per month, ordering cost is Rs. 12 per oder, holding cost is
RS. 10 per unit month, what is the EOQ ? (ESE-08)
(a) 12 (b) 144 (c) 24 (d) 28
19. Which one of the following is correct
In the basic EOQ model, if lead time increases from 5 to 10 days, the EOQ will (ESE-08)
(a) double (b) decrease by a factor of two
(c) remains the same (d) the data is insufficient to find EOQ
20. If demand is doubled and ordering cost, unit cost and inventory carrying cost are halved, then what
will be the EOQ ? (ESE-09)
(a) Half (b) Same (c) Twice (d) Four times
21. If, D = annual demand for a material (units per year)
Q = quantity of material ordered at each order point (units per order)
C = cost of carrying one unit in inventory for one year (rupees per unit per year)
S = average cost of completing an order for a material (rupees per order)
TSC = total anuual stocking costs for a material (rupees per year)
Then, the economic order quantity (EOQ) is (ESE-10)
(a) 2DS (b) 2DC (c) 2DC 2DS
C S S (d) C
22. In the inventorycontrol if the yearlydemand for a certain material is fixed, the economic order quantity
gives minimum of (ESE-10)
(a) Inventory carrying cost per year (b)Acquisition cost per year
(c) Total cost per year (d) Number of roders per year
23. EOQ is taken at the point where the cost of carrying equals the cost of (ESE-10)
(a) Ordering the materials (b) The mateial
(c) The safety stock (d) Both the material and the safety stock
24. The following is the general policy forAclass items inABC analysis :
1. Very strict control
2. Frequency review of their consumption
3. Safety stock kept
Which of the above statements is/are correct ? (ESE-10)
(d) 1, 2 and 3
(a) 1 only (b) 1 and 2 only (c) 2 only
25. When the ordering cost is increased to 4 times, the EOQ will be increased to : (ESE-11)
(a) 2 times (b) 3 times (c) 8 times (d) Reamin same
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26. In an economic order quantity based inventory control, when re-order level is greater than order
quantity, the number of orders outstanding at any time is (ESE-12)
(a) Never more than one (b) At least one
(c) No order outstanding (d) One only
27. ABC analysis is useful because it (ESE-12)
1. Identifies vital few and trivial many
2. Classifies items into three classes
(a) Neither 1 nor 2 (b) Both 1 and 2 (c) 1 only (d) 2 only
28. In a quantity discount model of inventory control, the relevant costs are (ESE-12)
(a) Annual purchase costs are
(b)Annual order cost and annual carrying cost
(c)Annual purchase cost, annual order cost and annual carrying cost
(d) Annual order cost
29. At break-even point, inventory carrying cost is : (ESE-13)
(a) Four times the preparatory cost (b) Three times the preparatory cost
(c) Two times the preparatory cost (d) Equal to the preparatory cost
30. Purification of inventorymeans : (ESE-13)
(a) Cleaning of inventories (b) Disposing of inventories
(c) Procesing of inventories (d) Storing of inventories in bins
31. Large size of inventory is a sign of (ESE-13)
(a) Better planning (b) Inefficiency
(c) Reliable control of vendors (d) Better scheduling
32. What is the ratio of anuual order cost to annual carrying cost when the order size is determined using
eonomic order quantity (EOQ) model ? (ESE-14)
(a) 0.5 (b) 0.25 (c) 0.75 (d) 1
33. The correct sequences of increasing production volume is (ESE-14)
(a) batch, job, flow and mass (b) mass, flow, bath and job
(c) job, flow, mass and batch (d) job, batch, mass and flow
34. In ABC inventory control of spare parts, the items A, B and C respectively refer to (ESE-14)
(a) high stock-out cost, moderate stock-out cost and low stock-out cost
(b) low stock-out cost, moderate stock-out cost and high stock-out cost
(c) moderate stock-out cost, high stock-out cost and low stock-out cost
(d) stock-out costs whose sequences depends on other factors also
35. The annual demand of a commodity in a supermarket is 80000. The cost of placing an order is
Rs. 4,000 and the inventory cost of each item is Rs. 40. What is the economic order quantity ?
(ESE-15)
(a) 2000 (b) 4000 (c) 5656 (d) 6666
***************
4. Inventory Control (Ans.) (IES)
1-d, 2-c, 3-b, 4-c, 5-b, 6-b, 7-d, 8-d, 9-a, 10-c, 11-a, 12-a, 13-a, 14-a, 15-s, 16-c, 17-c, 18-a, 19-c,
20-c, 21-a, 22-c, 23-a, 24-a, 25-a, 26-b, 27-d, 28-c, 29-d, 30-b, 31-b, 32-d, 33-d, 34-d, 35-b.
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5 TRANSPORTATION
1. Transportation costs (in INR/unit) from factories to respective markets are given in the table below.
The market requirements and factory capacities are also given. Using the North-West Corner method,
the quantity (in units) to be transported from factory R to market II is (GATE-PI-16)
Factory Requirements
(in units)
PQ R S
Market I 33 2 1 50
II 42 5 9 20
30
III 1 2 1 4
Factory
capacity 20 40 30 10
(in
units)
(a) 30 (b) 20 (c) 10 (d) 0
2. Given below is a basic feasible solution to a transportation problem with three supply points (A, B, C)
and three demand points (P, Q, R) that minimizes cost of transportation in the standard tabular format.
A B C Demand (GATE-ME-00)
4 88
P 150
50 100
12 8 11
Q
10 100 100
R 69
Supply 50 250
50 200
150 300
(a) Comoyte the cost corresponding to the present solution
(b) it is optimal ?
(c) Does an alternte optimum exist ?
3. The supplyat three sources is 50, 40 and 60 units respectivelywhile the demand at the four distinations
is 20, 30, 10 and 50 units. In solving this transportation problem (GATE-ME-02)
a) a dummy source of capacity 40 units is needed
b) a dummy destination of capacity 40 units is needed
c) no solution exists as the problem is infeasible
d) none solution existsas the problem is degenerate
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4. A company has tow factories S1, S2 and two warehouses D1m D2. The supplies from S1 and S2 are
50 and 40 units respectively. Warehouse D1 requires a minimum of 20 units and a maximum of
40 units. Warehouse D2 requires a minimum of 20 units and, over and above, it can take as much as
can be supplied. A balanced transportation problem is to be formulated for the above situation. The
number of supply points, the number of demand points, and the total supply (or total demand) in the
balanced transportation problem respectively are) (GATE-ME-05)
a) 2, 4, 90 b) 2, 4, 110 c) 3, 4, 90 d) 3, 4, 110
5 For the standard transportation linear programme with m sources and n destinations and total supply
euqaling total demand, an otimal solution (lowest cost) with the smallest number of non-zero xij values
(amounts from source i to destination j) is desired. The best upper bound for this number is
(GATE-ME-08)
a) mn b) 2(m +n) c) m + n d) m + n -1
*************
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5 TRANSPORTATION
Previous IES Questions :
1. In a 6 x 6 tranportation problem, degeneracy would arise, if the number of filled slots were (ESE-93)
(a) Equal to thirty six (b) More than twelve (c) Equal to twelve (d) Less then eleven
2. Assertion (A) : In distribution problem, unit cost of production as well as transportation cost are
considered.
Reason (R) : The Vogel’s approximation method can reduce the number of iterations required to move
from the initial assignment to the optimal solution. (ESE-94)
3. The solution in a transportation model (of dimension m x n) is said to be degenerate if it has (ESE-95)
(a) exaclty (m+n-1) allocations
(b) fewer than (m+n-1) allocations
(c) more than (m+n-1) allocations
(d) (mxn) allocation
4. Assertion (A) : Transportation problem can solved by VAM heuristic much faster as compared to the
solution through linear programming method.
Reason (R) : VAM heuristic gives an approximate solution. It is checked for optimality test. If it is
optimal, the algorithm stops there. If it is not an optimal solution, then improved solutions are found out
through veryfew iterations till optimalityis reached. (ESE-96)
(a) Both Aand R are true and R is the correct explanation of A
(b) Both A and R true but R is not a correct explanation of A
(c) Ais true but R is false
(d) Ais false but R is true
5. A solution is not a basic feasible solution in a transportation problem if after allocations. (ESE-96)
(a) there is no closed loop
(b) there is a closed loop
(c) total number of allocation is one less than the sum of number of sources and destinations
(d) there is degeneracy
6. When there are ‘m’ rows and ‘n’ columns in a transportation problem, degeneracy is said to occur
when the number of allocations is (ESE-97)
(a) less than (m+n-1) (b) greater than (m+n-1) (c) equal to (m-n-1) (d) less than (m-n-1)
7. In transporation problem, the materials are transported from 3 plants to 5 warehouses. The basic
feasible solution must contain exaclty, which one of the following allocated cells ? (ESE-98)
(a) 3 (b) 5 (c) 7 (d) 8
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8. Consider the following statements on transportation problem : (ESE-03)
1. In Vegel’s approximatin method, priority allotment is made in the cell with lowest cost in the column
or row with least penalty
2. The North-West corner method ensures faster optimal solution
3. If the total demand is higher than the supply, transportation problem cannot be solved
4.Afeasible solution man not be an optimal solution
Which of these statements are correct ?
(a) 1 and 4 (b) 1 and 3 (c) 2 and 3 (d) 2 and 4
9. Assertion (A) : In the solution of transportation problem, for applications required is m+n-1 and these
should be in independent positions.
Reason (R) : If the number of allocations is not m+n-1, value of all oddments,i.e. uiand vjcannot be
found (ESE-03)
10. Which one of the following conditions should be satisfied for the application of optimality test on an
initial solution of transportation model ? (ESE-04)
(a) Number of allocations should be less then m+n-1
(b) Number of allocations should be equal then m+n-1
(c) Number of allocations should be equal to m+n
(d) Number of allocations should be more than m+n
11. Which one of the following is not the solution method of transportation problem ? (ESE-06)
(a) Hungarian method (b) Northwest corner method
(c) Least cost method (d) Vogel’s approximation method
12. Consider the follwoing statements :
The assignment proboem is seen to be the special case of the transportation problem in which (The
symbols have the usual meaning)
1. m = n 2. all a = 1 3. s = 1
i ij
Which of the statement given above are correct ? (ESE-07)
(a) 1,2 and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only
13. In order for a transportation matrix which has six rows and four columns not to be degenerate, how
much must be the number of allocated cells in the matrix ? (ESE-07)
(a) 6 (b) 9 (c) 15 (d) 24
14. When solving the problem by Big-M method, if the objective functions row (evaluation row) shows
optimality but one or more artifical variables are still in the basis, what type of solution does it show ?
(a) Optimal solution (b) Pseudo optimal solution
(c) Degenerate solution (d) Infeasible solution (ESE-09)
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15. In a transportation problem, how is an unacceptable transportation route handled ? (ESE-11)
(a) By giving it a cost of zero (b) By carring an extremely high cost
(c) By introducing a dummy destination (d) Byintroducing a dummuyorigin
16. Which of the following method is NOT used for obtaining the intial basic feasible solution in
transportation problems ? (ESE-11)
(a) North west corner method (b) Least cost entry method
(c) Vogel’s approximation method (d) MODI method
17. A trnsporation problem consists of 3 sources and 5 destinations with approximate rim conditons. The
number of possible solutions is (ESE-15)
(a) 15 (b) 225 (c) 6435 (d) 150
*******************
5. Transportation (Ans.) (IES)
1-d, 2-b, 3-b, 4-a, 5-b, 6-a, 7-c, 8-a, 9-a, 10-b, 11-a, 12-a, 13-b, 14-d, 15-b, 16-d, 17-c.
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6 SCHEDULING
1. Service workshop has four jobs on hand to be processed. The date and processing time for each of
the jobs are given below [GATE-ME-90]
Job No. Processing time, days Due date
1 3 7
2 6 9
3 5 4
4 9 14
Considering mean lateness and mean flow time, evaluate the shortest-time rule and least-slack rule
and recommend the desirable rule.
2. A job shop incurs a cost of Rs. 60/- per day for each day a job is in the shop. At the beginning of a
month there are five jobs in the shop with the following data. [GATE-ME-96]
Job 1234 5
Processing time (days) 5382 6
Due date (days) 10 12 20 9 8
Which schedule will minimize the total cost? What is the minimum total cost ? Which jobs (if any) fail
to meet their due dates ?
3. A job shop has 6 orders to be completed by a single turning centre. The processing times and due
dates are as follows : [GATE-ME-98]
Order 123 456
Processing time 329 424
Due date 17 21 5 12 15 24
Assume that all orders are ready for processing. Give a production schedule that minimizes the
average flow time. Compare this schedule with one that minizes tardiness (lateness)
4. A set of 5 jobs is to be processsed on a single machine. The processing time (in days) is given in the
table below. The hloding cost for each job is Rs. K per day [GATE-ME-08]
Job Processing time
P5
Q2
R3
S2
T1
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A schedule that minimizes the total inventorycost is
a) T-S-Q-R-P b) P-R-S-Q-T c) T-R-S-Q-P d) P-Q-R-S-T
[GATE-ME-09]
5. Six jobs arrived in a sequences as given below :
Job Completion time
I4
II 9
III 5
IV 10
V6
VI 8
Average flow time (in days) for the above jobs using shortest Processing time rule is
a) 20.83 b) 23.16 c) 125.00 d) 139.00
Common data for Q. 6 and 7
Four jobs are to be processed on a machine as per data listed in the table.
Job Processing Due
time in days date
1 46
2 79
3 2 19
4 8 17
6. If the Earliest Due Date (EDD) rule is used to sequence the jobs, the number of jobs delayed is
[GATE-ME-10]
a) 1 d) 2 c) 3 d) 4
7. Using the Shorteset Processing Time (SPT) rule, total tardiness is [GATE-ME-10]
a) 0 b) 2 c) 6 d) 8
ANSWERS OF : **********
01. LINEAR PROGRAMMING : 1 - a, 2 - NA, 3 - NA, 4 - a, 5 - c, 6 - b, 7 - a, 8 - b, 9 - a, 10 - d, 11 - b, 12 - a, 13 - b.
02. PERT CPM : 1 - c, 2 - NA, 3 - c, 4 - NA, 5 - a, 6 - NA, 7 - NA, 8 - d, 9 - NA, 10 - c, 11 - b, 12 - c, 13 - d, 14 - a, 15 - a,
16 - d, 17 - c, 18 - d, 19 - c
03. QUEUING THEORY : 1 - NA, 2 - c, 3- NA, 4- NA, 5 - NA, 6 - c, 7 - b, 8 - a, 9 - b, 10 - c, 11 - b- 12 - d.
04. INVENTORY CONTROL : 1 - d, 2 - c, 3 - a, 4 - NA, 5 - b, 6 - NA, 7 - a, 8 - b, 9 - NA, 10 - NA, 11 - NA, 12 - d, 13 - c,
14 - c, 15 - a, 16 - c, 17 - c, 18 - b, 19 -d, 20 - b, 21 - c, 22 - c, 23 - b
05. TRANSPORTATION : 1 - c, 2 - NA, 3 - b, 4 - a, 5 - d
06. SCHEDULING : 1 - NA, 2- NA, 3 - NA, 4 - a, 5 - a, 6 - b, 7 - d.
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7 BREAK EVEN ANALYSIS
1. In the production of a product the fixed costs are Rs. 6,000/- and the variable cost is Rs. 10/- per
product. If the sale price of the product is Rs. 12/-, the break even volume of products to be made will be:
[IES-2008]
a) 2000 b) 3000 c) 4000 d) 6000
2. A standard machine tool and an automatic machine tool are being compared for the prodution of a
component. Following data refers to the two machines. [GATE-2004
Standard Machine Automatic Machine
Tool Tool
Setup time 300 min. 2 hours
Machining time per piece 22 min. 5 min
Machine rate Rs. 200 per hour Rs. 800 per hour
The breakeven production batch size above which the automatic machine tool will be economical to
use, will be
a) 4 b) 5 c) 24 d) 225
3. A component can be produced by any of the four processes I, II, II and IV. Process I has a fixed cost
of Rs. 20 and variable cost of Rs. 3 per piece. Process II has a fixed cost Rs. 50 and variable cost of Re. 1
per piece. Process III has a fixed cost of Rs. 40 and variable cost of rs. 2 per piece. Proces IV has a fixed
cost of Rs. 10 and variable cost of Rs. 4 per piece. If the company wishes to produce 100 pieces of the
component, from economic point of view it should choose. [GATE-2005]
a) Process I b) Process II c) Process III d) Process IV
4. Match List-I (Element of cost) with List-II (Nature of cost) and select the correct answer using the
codes given below the lists: [IES-1994]
List-I List-II
A. Interest on capital 1. Variable
B. Direct labour 2. Semi-variable
C. Water and electricity 3. Fixed
Codes: ABC
a) 3 1 2
b) 2 1 3
c) 3 2 1
d) 2 3 1
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IM + OR
5. Fixed investments for manufacturing a product in a particular year is Rs. 80,000/- The estimated
sales for this period is 2, 00,000/-. The variable cost per unit for this product is Rs. 4/-. If each unit is sold
at Rs.20/-, then the break even point would be: [IAS-1994]
a) 4,000 b) 5,000
c) 10,000 d) 20,000
6. The variable cost per unit associated with automated assembly line (VA), cellular manufacturing (VB),
and job shop production (VC) will be such that [IAS-1995]
a) VA > VB > VC b) VB > VA > VC
c) VC > VB > VA d) VC > VA > VB
7. For a small scale industry, the fixed cost per month is Rs. 5000/-. The variable cost per product is
Rs. 20/- and sales price is Rs. 30/- per piece. The break-even production per month will be: [IES-1995]
a) 300 b) 460
c) 500 d) 10000
8. Two alternative methods can produce a product first method has a fixed cost of Rs. 2000/- and
variable cost of Rs. 20/- per piece. The second method has a fixed cost of Rs. 1500/- and a variable cost of
Rs. 30/-. The break even quantity between the two alternatives is:
a) 25 b) 50
c) 75 d) 100
9. Assertion (A): Marginal cost in linear break-even analysis provides the management with useful
information for price fixing. [IAS-1996]
Reason (R): The marginal cost is the maximum value at which theproduct selling price must be fixed to
recover all the costs.
a) BothA and R are individually true and R is the correct explanation ofA
b) BothAand R are individually true but R is not the correct explanation ofA
c) Ais true but R is false
d) Ais false but R is true
10. IAssertion (A):Alarger margin of safetyin break-even analysis is helpful for management decision.
[IES-1997]
Reason (R): If the margin of safety is large, it would indicate that there will be profit even when there is a
serious drop in production.
a) BothAand R are individually true and R is the correct explanation ofA
b) BothAand R are individually true but R is not the correct explanation ofA
c) Ais true but R is false
d) Ais false but R is true
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MECHANICAL ENGINEERING
11. Process I requires 20 units of fixed cost and 3 units of variable cost per piece, while Process II
required 50 units of fixed cost and 1 unit of variable cost per piece. For a company producing 10 piece per
day [IES-1997]
a) Process I should be chosen
b) Process II should be chosen
c) Either of the two processes could be chosen
d) Acombination of process I and process II should be chosen
12. M/s. ABC & Co. is planning to use the most competitive manufacturing process to produce an
ultramodern sports shoe. They can use a fully automatic robot-controlled plant with an investment of Rs.
100 million; alternately they can go in for a cellular manufacturing that has a fixed cost of Rs. 80 million.
There is yet another choice of traditional manufacture that needs in investment of Rs. 75 million only. The
fully automatic plant can turn out a shoe at a unit variable cost of Rs. 25 per unit, whereas the cellular and the
job shop layout would lead to a variable cost of Rs. 40 and Rs. 50 respectively. The break even analysis
shows that the break even quantities using automatic plant vs traditional plant are in the ratio of 1: 2. The per
unit revenue used in the break even calculation is: [IES-1997]
a) Rs. 75 b) Rs. 87 c) Rs. 57 d) Rs. 55
13. Based on the given
graph, the economic
range of batch sizes to
be preferred for
general purpose
machine (OP), NC
machine (NC) and
special purpose
machine (SP) will be:
Codes:
GP NC SP
a) 2 5 4
b) 1 4 5
c) 3 2 4
d) 1 4 2
[IES-1997]
14. The fixed costs for a year is Rs. 8 lakhs, variable cost per unit is Rs. 40/- and the selling price of each
unit is Rs. 200/-. If the annual estimated sales is Rs. 20,00,000/-, then the break-even volume is :
[IAS-1997]
a) 2000 b) 3000 c) 3333 d) 5000
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IM + OR
15. On a lathe, the actual machining time required per work piece is 30 minutes. Two types of carbide
tools are available, both having a tool life of 60 minutes. [IES-1998]
Type I : Brazed type of original cost Rs. 50/-.
Type II : Throwaway tip (square) of original cost Rs. 70/-
If the overall cost of grinding the cutting edge is Rs. 10/-, assuming all the costs are the same for both the
types, for break even costs, the appropriate batch size would be:
a) 2 pieces b) 4 pieces c) 6 pieces d) 8 pieces
16. Match List-I (Methods) with List-II (Applications) and select the correct answer using the codes
given below the lists: [IES-1998]
List-I List-II
A. Break even analysis 1. To provide different facility at different locations
B. Transportation problem 2. To take action from among the paths with uncertainty
C.Assignment problem 3. To choose between different methods of manufacture
D. Decision tree 4. To determine the location of the additional plant
Codes: ABCD
a) 4 3 1 2
b) 3 4 1 2
c) 3 4 2 1
d) 4 3 2 1
17. Two jigs are under consideration for a drilling operation to make a particular part. JigAcosts
Rs. 800 and has operating cost of Rs. 0.10 per part. Jig B costs Rs. 1200 and has operating cost of Rs. 0.08
per part. The quantity of parts to be manufactured at which either jig will prove equallycostly is: [IAS-1998]
a) 8000 b) 15000 c) 20000 d) 23000
18. A company sells 14,000 units of its product. It has a variable cost of Rs. 15 per unit.
Fixed cost is Rs. 47,000 and the required profit is Rs. 23,000 Per unit product price (in Rs.) will be:
a) 60 b) 40 c) 30 d) 20 [IAS-1999]
19. Assertion (A): It is possible to have more than one break-even point in break even charts.
[IES-1999]
Reason (R):All variable costs are directly variable with production.
a) BothA and R are individually true and R is the correct explanation ofA
b) BothA and R are individually true but R is not the correct explanation ofA
c) Ais true but R is false
d) Ais false but R is true
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MECHANICAL ENGINEERING
20. Last year, a manufacturer produced 15000 products which were sold for Rs. 300 each. At that
volume, the fixed costs were Rs. 15.2 lacs and total variable costs were Rs. 21 lacs. The break even
quantity of product would be: [IES-2000]
a) 4000 b) 7800 c) 8400 d) 9500
21. IAS-4. If Break-even point = Total fixed cost ÷1 - Variable cost per unit , then X is the
X
a) Overheads b) Price per unit c) Direct cost [IAS-2000]
d) Materials cost
22. Match List-I (Symbols)
with List-II (Meaning) and
select the correct answer
using the codes given
below the Lists; related to
P/V chart on Break-Even
Analysis as shown in the
above figure:
List-I List-II [IAS-2002]
A. OR 1. Profit
B. PQ 2. Break-Even Point
C. SS 3. Profit/Volume Ratio
D. RQ 4. Cost for new design
5. Fixed cost
Codes: ABCD ABCD
a) 5 4 2 3 b) 2 1 3 5
c) 5 1 2 3 d) 2 4 3 5
23. The indirect cost of a plant is Rs 4,00,000 per year. The direct cost is Rs 20 per product. If the
average revenue per product is Rs 60, the break-even point is: [IES-2003]
a) 10000 products b) 20000 products c) 40000 products d) 60000 products
24. Which one of the following information combinations has lowest break-even point? [IES-2004]
Fixed cost Variable cost Revenue/units
(in Rs.) / unit (in Rs.) (in Rs.)
a) 30,000 10 40
b) 40,000 15 40
c) 50,000 20 40
d) 60,000 30 40
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