Assignment Models
Exam Duration: 45 Mins Total Questions : 25
CPM is
- (a)
time oriented technique
- (b)
activity oriented technique
- (c)
event oriented technique
- (d)
Both (b) and (c)
PERT is
- (a)
event oriented
- (b)
activity oriented
- (c)
cost oriented
- (d)
Both (a) and (b)
Critical activities are those for which
- (a)
float = + 1
- (b)
float = 0
- (c)
float < 1
- (d)
float > 1
A dummy activity
- (a)
does not require any time
- (b)
is represented by a dotted line
- (c)
is artificially introduced
- (d)
All of the above
The critical path of a network represents
- (a)
minimum time required for completion of project
- (b)
maximum time required for completion of project
- (c)
minimum cost required for completion of project
- (d)
maximum cost required or completion of project
One time estimate is
- (a)
PERT
- (b)
CPM
- (c)
Both (a) and (b)
- (d)
None of these
Which of the following is not correct?
- (a)
The critical path of a project network represents the minimum time needed to complete the project
- (b)
Critical path is the longest path in a project network
- (c)
A delay in the completion of critical activities need not cause a delay in the completion of the whole project
- (d)
The sum of the variances of the critical activity times gives the variance of the overall project completion time
If an activity has zero slack, it implies that
- (a)
it is a dummy activity
- (b)
it lies on the critical path
- (c)
there are more than one critical paths
- (d)
All of the above
An activity in the network
- (a)
represents a task which has a definite beginning and a definite end
- (b)
cannot start unless all its immediate procedures are completed
- (c)
have float zero, if it is critical, otherwise total float as positive in case of non-critical activity
- (d)
All of the above
In a 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
- (a)
1.57
- (b)
1.33
- (c)
3.2
- (d)
2.75
\(\rho =\frac { \lambda }{ \mu } =\frac { 10 }{ 15 } =\frac { 2 }{ 3 } \)
and Ls=\(\frac { \rho }{ 1-\rho } =\frac { 2/3 }{ 1-2/3 } \)=2
Lq=Ls-\(\frac { \lambda }{ \mu } =2-\frac { 2 }{ 3 } =\frac { 4 }{ 3 } \)=1.33
At a cashier counter, 8 customers arrive on an average every 5 min. While cashier can serve 10 customers in 5 min. Select correct option
Expected number of units in the system
- (a)
4
- (b)
5
- (c)
6
- (d)
8
λ=\(\frac { 80\times 60 }{ 5 } \)=96 customer per hour
μ=\(\frac { 10\times 60 }{ 5 } \)=120 customer per hour
Expected number of customers in system
\(\frac { \lambda }{ \mu -\lambda } =\frac { 96 }{ 120-96 } =\frac { 96 }{ 24 } \)
At a cashier counter, 8 customers arrive on an average every 5 min. While cashier can serve 10 customers in 5 min. Select correct option
Average number of units in the queue
- (a)
3
- (b)
3.2
- (c)
2
- (d)
4
λ=\(\frac { 80\times 60 }{ 5 } \)=96 customer per hour
μ=\(\frac { 10\times 60 }{ 5 } \)=120 customer per hour
Average number of customers in queue
=\(\frac { { \lambda }^{ 2 } }{ \mu (\mu -\lambda ) } =\frac { 96\times 96 }{ 120\times 24 } \)=3.2
At a cashier counter, 8 customers arrive on an average every 5 min. While cashier can serve 10 customers in 5 min. Select correct option
Expected time per unit in the system is
- (a)
2.5 min
- (b)
3 min
- (c)
3.2 min
- (d)
4 min
λ=\(\frac { 80\times 60 }{ 5 } \)=96 customer per hour
μ=\(\frac { 10\times 60 }{ 5 } \)=120 customer per hour
Expected time per customer in system
=\(\frac { 1 }{ \mu -\lambda } =\frac { 1 }{ (120-96) } =\frac { 1 }{ 24 } \)h
=\(\frac{60}{24}\)=2.5 min
At a cashier counter, 8 customers arrive on an average every 5 min. While cashier can serve 10 customers in 5 min. Select correct option
Expected time, a customer spend in the queue
- (a)
4 min
- (b)
3.2 min
- (c)
3 min
- (d)
2 min
λ=\(\frac { 80\times 60 }{ 5 } \)=96 customer per hour
μ=\(\frac { 10\times 60 }{ 5 } \)=120 customer per hour
Expected time, a customer spend in queue
\(\frac { \lambda }{ \mu (\mu -\lambda ) } =\frac { 96 }{ 120\times 24 } =\frac { 1 }{ 30 } \)h
=2 min
At a cashier counter, 8 customers arrive on an average every 5 min. While cashier can serve 10 customers in 5 min. Select correct option
Number of hours for which the cashier remains busy in 8 days
- (a)
6 h
- (b)
7 h
- (c)
6.4 h
- (d)
5 h
λ=\(\frac { 80\times 60 }{ 5 } \)=96 customer per hour
μ=\(\frac { 10\times 60 }{ 5 } \)=120 customer per hour
Cashier busy in 1 day shift
=\(1-\left( 1-\frac { \lambda }{ \mu } \right) =\frac { \lambda }{ \mu } \)
=\(\frac { 96 }{ 120 } =\frac { 4 }{ 5 } \)h
So, cashier busy in 8 days shift =8 x \(\frac{4}{5}\)=6.5 h
At a cashier counter, 8 customers arrive on an average every 5 min. While cashier can serve 10 customers in 5 min. Select correct option
Average length of non-empty queue is
- (a)
4
- (b)
3
- (c)
6
- (d)
5
λ=\(\frac { 80\times 60 }{ 5 } \)=96 customer per hour
μ=\(\frac { 10\times 60 }{ 5 } \)=120 customer per hour
Average length of non-empty queue
\(\frac { \mu }{ (\mu -\lambda ) } =\frac { 120 }{ (120-96) } =\frac { 120 }{ 24 } \)=5
At a cashier counter, 8 customers arrive on an average every 5 min. While cashier can serve 10 customers in 5 min. Select correct option
Average cost due to waiting on a part of cashier, if cashier is value at Rs.100/h
- (a)
Rs.15
- (b)
Rs.20
- (c)
Rs.30
- (d)
Rs.25
λ=\(\frac { 80\times 60 }{ 5 } \)=96 customer per hour
μ=\(\frac { 10\times 60 }{ 5 } \)=120 customer per hour
Average cost due to waiting on a part of cashier
= Cost per hour x Idle time
=\(100\times \left( 1-\frac { \lambda }{ \mu } \right) \)
=\(100-\left( 1-\frac { 96 }{ 120 } \right) \)
=\(\frac { 100\times 24 }{ 120 } \)=Rs.20
Consider the following network
Distance between different stations is shown on each link in km
The shortest route from 1 to 8 is
- (a)
1 ⟶ 3 ⟶ 5 ⟶ 8
- (b)
1 ⟶ 4 ⟶ 7 ⟶ 8
- (c)
1 ⟶ 2 ⟶ 5 ⟶ 7 ⟶ 8
- (d)
None of the above
A shortest route is a route in network which length is less than other route of the network.
The routes and distances are
1 ⟶ 2 ⟶ 6 ⟶ 8 = 3 + 8 + 3 = 14 km
1 ⟶ 2 ⟶ 5 ⟶ 6 ⟶ 8 = 3 +3 + 4 + 3= 13krn
1 ⟶ 2 ⟶ 5 ⟶8 = 3 + 3 + 7 =13km
1 ⟶ 2 ⟶ 5 ⟶ 7 ⟶ 8 = 3 + 3 + 2 + 3=11 km
1 ⟶ 3 ⟶ 5 ⟶ 6 ⟶8 =6 + 2 + 4 + 4 + 3 = 19km
1 ⟶ 3 ⟶ 5 ⟶8 = 6 + 4 + 7 = 17 km
1 ⟶ 4 ⟶ 7 ⟶ 8 = 5 + 7 + 3 = 15 km
Route of minimum length is
1 ⟶ 2 ⟶ 5 ⟶ 7 ⟶ 8
A network is shown in figure.
The critical path is
- (a)
1 ⟶ 2 ⟶ 5 ⟶ 6 ⟶ 7 ⟶ 8 ⟶ 9
- (b)
1 ⟶ 2 ⟶ 3 ⟶ 4 ⟶ 8 ⟶ 9
- (c)
1 ⟶ 2 ⟶3 ⟶ 5 ⟶ 6 ⟶ 7 ⟶ 8 ⟶ 9
- (d)
1 ⟶ 2 ⟶ 3 ⟶ 4 ⟶ 7 ⟶ 8 ⟶ 9
The paths and time are given below.
(I) 1⟶ 2 ⟶ 3 ⟶ 4 ⟶ 8 ⟶ 9 = 3 + 4 + 5 + 5 + 4
=21 days
(II) 1 ⟶ 2 ⟶ 3 ⟶ 4 ⟶ 7 ⟶ 8 ⟶ 9 = 3 + 4 + 5 + 6 + 4
=22 days
(111)1 ⟶ 2 ⟶ 3 ⟶ 5⟶ 6 ⟶ 7 ⟶ 8 ⟶9
=3 + 4 + 3 + 4 + 6 + 4
=24 days
(lV) 1 ⟶ 2 ⟶ 5 ⟶ 6 ⟶ 7 ⟶ 8 ⟶ 9 =3 + 2+ 3 +4+ 6+4
= 22 days
Path 1 ⟶ 2 ⟶ 3 ⟶ 5 ⟶ 6 ⟶ 7 ⟶ 8 ⟶ 9 is critical path.
Variance for critical path is as
a ⟶ b = 4
b ⟶ c = 16
c ⟶ d = 4
d ⟶ e =1
Then, standard deviation of critical path
a ⟶ e is
- (a)
6
- (b)
5
- (c)
4
- (d)
3
a ⟶ b ⟶ c ⟶ d ⟶ c
σ12=4 σ22=16 σ32=4 σ42=1
We know that
Standard deviation =\(\sqrt { { \sigma }_{ 1 }^{ 2 }+{ \sigma }_{ 2 }^{ 2 }+{ \sigma }_{ 3 }^{ 2 }+{ \sigma }_{ 4 }^{ 2 } } \)
=\(\sqrt { 4+16+4+1 } =\sqrt { 25 } \)
=5
Time estimate of a PERT activity are given below. Pessimistic time = 14 min, most likely time = 10 min, optimistic time = 8 min. Then, the expected time of activity is
- (a)
10 min
- (b)
11 min
- (c)
10.33 min
- (d)
12 min
tp = 14 min, tm = 10 min, to=8min
We know that
te=\(\frac { { t }_{ p }+4{ t }_{ m }+{ t }_{ 0 } }{ 6 } \)=\(\frac { 14+4\times 10+8 }{ 6 } \)
\(\frac { 62 }{ 6 } \)=10.33 min
Critical path of project is given below.
Activity | Precedance | Duration (in days) |
a | - | 3 |
b | - | 4 |
c | a | 5 |
d | b | 5 |
e | c,d | 7 |
f | c,d | 5 |
g | e | 2 |
h | f | 10 |
- (a)
b ⟶ d ⟶ f ⟶ h
- (b)
a ⟶ c ⟶ e ⟶ f
- (c)
b ⟶ d ⟶ g ⟶ h
- (d)
a ⟶ d ⟶ e ⟶ h
Activity | Precedance | Duration (in days) |
a | - | 3 |
b | - | 4 |
c | a | 5 |
d | b | 5 |
e | c,d | 7 |
f | c,d | 5 |
g | e | 2 |
h | f | 10 |
Now, draw the network diagram
The paths and duration are given below.
a ⟶ c ⟶ e ⟶ g = 3 + 5 + 7 + 2 = 17 days
a ⟶ c ⟶ f ⟶ h = 3 + 5 + 5 + 10 = 23 days
b ⟶ d ⟶ e ⟶ g = 4 + 5 + 7 + 2 = 18 days
b ⟶ d ⟶ f ⟶ h = 4 + 5 + 5 + 10 = 24 days
Hence, critical path is
b ⟶ d ⟶ f ⟶ h
Consider six activities as
Activity | Time(in days) | Preoperation |
a | 5 | - |
b | 6 | a |
c | 5 | b |
d | 4 | a |
e | 3 | d |
f | 4 | c,e |
The total project duration is
- (a)
15 days
- (b)
22 days
- (c)
18 days
- (d)
20 days
First to all, draw the network diagram
The paths of complete activity are given below.
a ⟶ b ⟶ c ⟶ f
a ⟶ d ⟶ e ⟶ f
The critical path tells the project completion path.
a ⟶ b ⟶ c ⟶ f = 5 + 6 + 5 + 4 = 20 days
a ⟶ d ⟶ e ⟶ f =5 + 4 + 3 + 4 = 16 days
Critical path is
a ⟶ b ⟶ c ⟶ f
Project duration = 20 days
Consider a PERT network. The optimistic pessimistic and most likely time is given in a table as
Activity | t0 | tm | tp |
1-2 | 2 | 5 | 8 |
1-4 | 3 | 3 | 3 |
2-3 | 1 | 1 | 1 |
3-5 | 0 | 6 | 18 |
4-5 | 2 | 8 | 14 |
5-6 | 7 | 7 | 7 |
Critical path duration of network is
- (a)
21 days
- (b)
20 days
- (c)
17 days
- (d)
24 days
Activity | t0 | tm | tp | te=\(\frac { { t }_{ 0 }+4{ t }_{ m }+{ t }_{ p } }{ 6 } \) |
1-2 | 2 | 5 | 8 | 5 |
1-4 | 3 | 3 | 3 | 3 |
2-3 | 1 | 1 | 1 | 1 |
3-5 | 0 | 6 | 18 | 7 |
4-5 | 2 | 8 | 14 | 8 |
5-6 | 7 | 7 | 7 | 7 |
Now, draw the network diagram
Paths are
1 ⟶ 2 ⟶ 3 ⟶ 5 ⟶ 6 = 5 + 1 + 7 + 7 = 20 days
and 1 ⟶ 4 ⟶ 5 ⟶ 6 = 3 + 8 + 7 = 18 days
Critical path duration = 20 days
Consider a PERT network. The optimistic pessimistic and most likely time is given in a table as
Activity | t0 | tm | tp |
1-2 | 2 | 5 | 8 |
1-4 | 3 | 3 | 3 |
2-3 | 1 | 1 | 1 |
3-5 | 0 | 6 | 18 |
4-5 | 2 | 8 | 14 |
5-6 | 7 | 7 | 7 |
Standard deviation of critical path is
- (a)
82
- (b)
9
- (c)
\(\sqrt{10}\)
- (d)
7
Activity on critical path
Activity | Variance \(\left( \frac { { t }_{ p }-{ t }_{ 0 } }{ 6 } \right) ^{ 2 }\) |
1-2 | 1 |
2-3 | 0 |
3-5 | 9 |
5-6 | 0 |
Standard deviation =\(\sqrt { { \sigma }_{ 1 }^{ 2 }+{ \sigma }_{ 2 }^{ 2 }+{ \sigma }_{ 3 }^{ 2 }+{ \sigma }_{ 4 }^{ 2 } } \)
=\(\sqrt { 1+0+9+0 } =\sqrt { 10 } \).