DAYANANDA SAGAR COLLEGE OF ENGINEERING
Shavige Malleshwara Hills, Kumaraswamy Layout, Bengaluru-560078
Department of Mechanical Engineering
Course:
Operations Research
2024-2025 Even Semester
Course Code: 22ME645
Maximum marks: 50
th
Semester:
6
Question Bank
Module 4 :PERT & CPM
1
A project schedule has the following characteristics
Activities
1-2 1-3 2-4 3-4 3-5 4-9 5-6 5-7 6-8 7-8 8-10 9-10
Time (Days)
4
1
1
1
6
5
4
8
1
2
5
7
From the above information you are required to i. Construct a network diagram ii. Compute the
earliest event time and latest event time iii. Determine the critical path and total project duration
iv. Compute total float for each activity.
Solution:
Activity
Activity
Name
Normal
Time
(tij)
1-2
A
4
1-3
B
1
2-4
C
1
3-4
D
1
3-5
E
6
4-9
F
5
5-6
G
4
5-7
H
8
6-8
I
1
7-8
J
2
8-10
K
5
9-10
L
7
Critical Path – 1-3-5-7-8-10
Earliest Time (TE)
Latest Time (TL)
Start(ES) Finish(EF) Start
Finish(LF)
(LS)
0
4
5
9
0
1
0
1
4
5
9
10
1
2
9
10
1
7
1
7
5
10
10
15
7
11
12
16
7
15
7
15
11
12
16
17
15
17
15
17
17
22
17
22
10
17
15
22
Total
Float
5
0
5
8
0
5
5
0
5
0
0
5
1
2
Small maintenance project consists of the following jobs, whose precedence relationship are given
below:
Job
1-2 1-3 2-3 2-5 3-4 3-6 4-5 4-6 5-6 6-7
Duration (Days) 15 15 3
5
8
12 1
14 3
14
i. Draw an arrow diagram representing the project
ii. Find the total float for each activity
iii. Find the critical path and the total project duration.
Solution:
Activity Normal Earliest
time
Start
Finish
3
Latest
Start
Finish
Floats
TF
FF
IF
1-2
15
0
15
0
15
0
0
1-3
15
0
15
3
18
3
0
2-3
3
15
18
15
18
0
0
2-5
5
15
20
32
37
17
0
3-4
8
18
26
18
26
0
0
3-6
12
18
30
28
40
10
0
4-5
1
26
27
36
37
10
0
4-6
14
26
40
26
40
0
0
5-6
3
27
30
37
40
10
0
6-7
14
40
54
40
54
0
0
Critical path – 1-2-3-4-6-7
Total time taken for project completion is 54 days.
The following table shows the jobs of a project with their duration in days. Draw the network and
determine the critical path. Also calculate the total float
Jobs
1- 1- 1- 2- 3- 4- 5- 5- 6- 6- 78910112
3
4
5
7
6
7
8
7
9
10 10 10 11
12
Duration 10 8
9
8
16 7
7
7
8
5
12 10 15 8
5
Solution:
2
Activity
4
Normal
Earliest Time (TE)
Latest Time (TL)
Total
Time
Start(ES) Finish(EF) Start (LS) Finish(LF) Float
(tij)
1-2
10
0
10
0
10
0
1-3
8
0
8
1
9
1
1-4
9
0
9
1
10
1
2-5
8
10
18
10
18
0
3-7
16
8
24
9
25
1
4-6
7
9
16
10
17
1
5-7
7
18
25
18
25
0
5-8
7
18
25
20
27
2
6-7
8
16
24
17
25
1
6-9
5
16
21
17
22
1
7-10
12
25
37
25
37
0
8-10
10
25
35
27
37
2
9-10
15
21
36
22
37
1
10-11
8
37
45
37
45
0
11-12
5
45
50
45
50
0
Critical path – 1-2-5-7-10-11-12
The following table shows the jobs of a network along with their time estimates. The time
estimates are in days
Job 1-2 1-6 2-3 2-4 3-5 4-5 5-8 6-7 7-8
a
3
2
6
2
5
3
1
3
4
m
6
5
12 5
11 6
4
9
19
b
15 14 30 8
17 15 7
27 28
i.
Draw the project network
ii.
Find the critical path
iii.
Find the probability of the project being completed in 31 days
Solution:
Job
Optimistic Time
Most Likely Time
Pessimistic Time
te
=
(to
+4tm+tp)/6
σ2 = (tp-t0/6)2
1-2
3
6
15
7
1-6
2
5
14
6
2-3
6
12
30
14
2-4
2
5
8
5
3-5
5
11
17
11
4-5
3
6
15
7
5-8
1
4
7
11
6-7
3
9
27
4
7-8
4
19
28
18
4
4
16
1
4
4
16
1
16
3
Expected duration of the project = 36 days
Critical path = 1-2-3-5-8
Project length variance σ2 = 4+16+4+1 = 25
σ=5
The probability that the project is completed in 31 days is given by P (Z≤D)
D = (Ts – Te)/ σ = (31-36)/5 = -0.8
5
Area under the normal curve for the region Z ≤ -0.8
P (Z≤ -0.8) = 0.5 - ϕ (-0.8)
= 0.5 - 0.21186 = P= 28.81%
Assuming that the expected times are normally distributed, find the probability of meeting the
schedule time as given for the network. Scheduled project completion time is 30 days. Also find
the date on which the project manager can complete the project with a probability of 0.90
Activity Days
(i-j)
Optimistic Most
pessimistic
likely
1-2
2
5
14
1-3
9
12
15
2-4
5
14
17
3-4
2
5
12
4-5
6
6
12
3-5
8
17
20
Solution:
Activity
t0
tm
te = (t0 + 4tm +tp_)/6
σ2 = (tp-t0/6)2
1-2
1-3
2-4
3-4
3-5
4-5
σ2 = 1+4= 5
2
9
5
2
6
8
5
12
14
5
6
17
6
12
13
5
16
7
4
1
4
1
4
1
σ=2.236
The probability of completing the
project in 30 days
Expected project duration = 28 days
Critical Path = 1-3-5
Project length variance , σ2 =5
σ = 2.236
D = Ts – Te / σ = 30 – 28 / 2.236 =
0.8944
area under the normal curve for the region Z≤0.8944
P(Z≤0.8944) = 0.5 + ϕ (0.89) = 0.5 + 0.3133 =0.8133 = 81.33%
probability of completing the project is 0.9 the corresponding value of Z=1.29 Ts= 30.88weeks
4
Module 5: Queuing Theory
1
A box office ticket window being manned by a single server. Customer arrives to purchase ticket
according to Poisson input process with a mean rate of 30/hr. the time required to serve a
customer has an ED with a mean of 90 seconds
determine:
a) Mean queue length.
b) Mean waiting time in the system.
c) The probability of the customer waiting in the queue for more than 10min.
d) The fraction of the time for which the server is busy.
Solution:
Mean arrival rate =λ= 30/ hour
Mean service rate = μ = 1/90 x60 x 60 = 40 / hour.
Mean queue length.
𝜆2
302
Lq = μ(μ −𝜆) = 40(40 −30) = 2.25 customers
(b) Mean waiting time in the system.
1 𝐿𝑞
1
2.25
1
W =Wq + μ = 𝜆 + μ = 30 + 40 = 0.1/hr
(c) The probability of the customer waiting in the queue for more than 10min.
𝜆
30
1
Probability = μ 𝑒 (𝜆 −μ )𝑤 = 40 𝑒 (30 −40 ) (6) = 𝟎. 𝟏𝟒𝟏𝟔
(d) The fraction of the time for which the server is busy.
𝜆
30
ρ = μ = 40 = 0.75hr
2
A company distributes its products by trucks loaded at its only loading station. Both company’s
trucks and contractor’s trucks are used for this purpose. It was found out that on an average
every five minutes, one truck arrived and the average loading time was three minutes. 50% of the
trucks belong to the contractor. Find out
a) The probability that a truck has to wait,
b) The waiting time of truck that waits, and
c) The expected waiting time of contractor’s trucks per day, assuming a 24 hours shift.
Solution:
It is given that λ= 60/5 = 12 tucks/ hour
Average service rate = μ = 60 / 3 = 20 trucks/ hour
i.
i. Probability that a truck has to wait is given by the probability of a busy system
Ρ= λ/ μ = 12/20 = 0.6
ii.
ii. The waiting time of a truck that waits is given by
W = 1 / (μ- λ) = 1 / (20-12) = 1/8 = 7.5 minutes
iii.
iii. Given that 50% of the total trucks belong to the contractor. Hence the expected waiting
time of contractor's trucks per day (assuming 24 hours shift)
= (no of trucks per day) x (contractor's percentage) x (Expected waiting time of a truck).
= 12 x 24 x (50/100) x (12 / 20(20-12) )
= 288 x 0.5 x (12 / (20 x 8))
= 10.8 hours per day
5
3
Jobs arrive at an inspection station according to Poisson process at a mean rate of 2/hr and are
inspect one at a time on a FIFO basis. The quality control engineer both inspects and makes minor
adjustments. The total service time for the job appears to be ED with a mean of 25mins. Jobs that
arrive but cannot be inspected immediately by the engineer must be stored until the engineer is
free to take them. Each job requires 1 sq mts space determine
a) The waiting line length
b) The waiting time
c) % of idle time of the engineer
d) The floor space to be provided in the quality control room.
Solution:
Mean arrival rate =λ= 2/ hour
Mean service rate = μ = 1/25 x 60 =2.4 / hour.
a) The waiting line length
𝜆2
𝐿𝑞 = μ(μ −𝜆) = 4.16
b) The waiting time
𝐿𝑞
Wq = 𝜆 = 2.08
c) % of idle time of the engineer
𝜆
po =1- μ x100 = 16.66%
d) The floor space to be provided in the quality control room.
𝜆
2
L =Lq + μ = 4.16 + 2.4 = 4.993
L = 5 x 1m2
6