Haeryip Sihombing BANK SOAL – BMFP 4542 (TEST -1)
1. Identify and briefly describe the five commonly traded organizational currencies.
•
Task-related; the ability to contribute to others accomplishing their work,
•
Position-related; the ability to enhance others' positions within their organization,
•
Inspiration-related; the ability to enhance people's desire to make a difference and
add meaning to their lives,
•
Relationship-related; the ability to form relationships that transcend normal
professional boundaries and extend into friendship,
•
Personal-related; the extent to which one can help others feel a sense of importance
and personal worth.
2. Identify and compare the old-fashioned view of managing projects and the new
perspective of managing projects.
The old-fashioned view of managing projects emphasized directing and controlling
subordinates; the new perspective emphasizes managing project stakeholders and
anticipating change as the most important jobs.
3. Under what conditions would the top-down approach to estimating project times
and costs be the best choice?
In the early stages of a project to help develop the initial plan, in making strategic decisions,
in projects of high uncertainty, in small internal projects, or in projects with an unstable
scope.
4. Under what conditions would the bottom-up approach to estimating project times
and costs be the best choice?
When low cost, efficient estimates are needed, when time and cost are important, when
working on a fixed price contract, or when the customer wants details.
5. In the network computation process what is a forward pass and what three things
does it determine?
Starting with the first activity, each path is traced forward through the network, adding times
until the end of the project. This determines 1. how soon each activity can start, 2. how soon
each activity can finish, and 3. how soon the entire project can be completed.
6. What is a risk profile and what benefits does it provide to risk management?
A risk profile is a list of questions that address traditional areas of uncertainty on a project.
The questions have been developed and refined from previous, similar projects. These
profiles are generated and maintained by the project office and are updated and refined
during the life of the project. This historical file assists in identifying risks for future projects.
1
Haeryip Sihombing 7. What is Change Control Management and what function does it perform?
Change Control Management is the formal process for making and tracking changes once a
project has started. Any changes must be detailed and accepted by the project team. Risks
associated with making changes are thus assessed and documented.
8. Why would a project manager use heuristics rather than a mathematical solution
to level resources?
Mathematical models work on small networks and a small number of resources. Larger
projects have massive data requirements. Heuristics tend to yield a good solution without the
heavy burden imposed by math models.
9. Identify and briefly describe five reasons for attempting to reduce the duration of a
project.
1. imposed durations made by top management, 2. market demands created by competition
and rapid technology advances, 3. incentive contracts that pay for early project completion,
4. recovery of unforeseen project delays, 5. to reduce project costs by reducing charges
created by high overhead costs, 6. to reassign key resources to other projects.
10. Why is the project duration with the lowest direct costs seldom the optimum
duration for a project?
Indirect costs, those accumulating each day, will at some point outweigh the reduction in
direct costs leading to an increase in total costs. The Project Cost-Duration graph is a model
that demonstrates this.
2
Haeryip Sihombing Problem Statement 1
Prem Binny Appliances wants to establish an assembly line to manufacture its new
product, the Mini-Me Microwave Oven. The goal is to produce five Mini-Me
Microwave Ovens per hour. The tasks, task-times, and immediate predecessors for
producing one Mini-Me Microwave Oven are as follows:
Task
A
B
C
D
E
F
Time (min)
10
12
8
6
6
6
Immediate Predecessors
A
A, B
B, C
C
D, E
a. What is the theoretical minimum for the smallest number of workstations
that Binny can achieve in this assembly line?
Output rate:
Prem Binny Appliances need to produce 5 Mini-Me Microwave Ovens per hour
The amount of time each workstation is allowed to complete its tasks:
Limited by the bottleneck task (the longest task in a process):
Theoretical Minimum (TM) = number of station needed to achieve 100%
efficiency
Therefore, the theoretical minimum for the smallest number of workstations that
Binny can achieve in this assembly line is 4 stations.
3
Haeryip Sihombing b. Graph the assembly line and assign workers to workstations. Can you
assign them with theoretical minimum?
Workstation
1
2
3
4
5
Eligible Task
A,B
A
C
D
E
F
Task Selected
B
A
C
D
E
F
Task Time
12
10
8
6
6
6
Idle Time
0
2
4
6
0
6
As conclusion, the workers are failed to assign with theoretical minimum.
c. What is the efficiency of your assignment?
Hence, there is left a balance delay of:
100% - 80% = 20%
Problem Statement 2
Paul Silver, owner of Sculptures International, just initiated a new art project. The
following data are available for the project.
Activity
Activity Time (days)
Immediate Predecessor(s)
A
B
4
1
4
Haeryip Sihombing C
D
E
3
2
3
A
B
C, D
a. Draw the network diagram for the project.
b. Determine the project’s critical path and duration.
The critical path is the longest path through the project, defining the minimum
completion time for the overall project. The critical path in this project is A-C-E,
determining that the project can be completed in 10 days.
Critical path duration, A-C-E
= 4 days + 3 days + 3 days
= 10 days
c. What is the total slack for each activity?
Calculation for slack time of each activity
Activity
Early Start
Early Finish
Late Start
Late Finish
Slack
A
B
C
D
E
Day 0
Day 0
Day 4
Day 1
Day 7
Day 4
Day 1
Day 7
Day 3
Day 10
Day 0
Day 4
Day 4
Day 5
Day 7
Day 4
Day 5
Day 7
Day 7
Day 10
0
4
0
4
0
Slack for the individual activities is calculated by taking the difference between the
late-start and early-start times (or, alternatively, between the late-finish and earlyfinish times) for each activity. If the difference is zero, then there is no slack; the
activity is totally defined as to its time-position in the project and must therefore be
5
Haeryip Sihombing a critical path activity. For other activities, the slack defines the flexibility in start
times, but only assuming that no other activity on the path is delayed.
Problem Statement 3
Given the activities whose sequence is described by the following table, draw the
appropriate activity-on-arrow (AOA) network diagram.
Activity
A
B
C
D
E
F
G
H
I
Immediate Predecessor(s)
A
A
B
B
C
E, F
D
G, H
Time (days)
5
2
4
5
5
5
2
3
5
a. Which activities are on the critical path?
The critical path is A-C-F-G-I
Activity
Duration
ES
EF
LS LF
Slack
Critical Path
A
5
0
5
0
5
0
*
B
2
5
7
6
8
1
C
4
5
9
5
9
0
D
5
7
12
8
13
1
E
5
7
12
9
14
2
F
5
9
14
9
14
0
*
G
2
14
16
14
16
0
*
H
3
12
15
13
16
1
I
5
16
21
16
21
0
6
*
*
Haeryip Sihombing b. What is the length of the critical path?
Length of critical path, A-C-F-G-I
= [5 + 4 + 5 + 2 + 5] days = 21 days
Problem Statement 4
Dr. Lori Baker, operations manager at Nesa Electronics, prides herself on excellent
assembly-line balancing. She has been told that the firm needs to complete 96
instruments per 24-hour day. The assembly-line activities are:
Task
Time (min)
Predecessors
A
3
-
B
6
-
C
7
A
D
5
A, B
E
2
B
F
4
C
G
5
F
H
7
D, E
I
1
H
J
6
E
K
4
G, I, J
Total
50
a. Draw the precedence diagram.
A
7
4
5
C
F
G
5
B
D
7
1
2
H
I
E
6
J
7
K
Haeryip Sihombing b. If the daily (24-hour) production rate is 96 units, what is the highest
allowable cycle time?
The highest allowable cycle time is
Therefore, the highest allowable cycle time is 15 min/unit.
c. If the cycle time after allowances is given as 10 minutes, what is the daily
(24-hour) production rate?
Therefore, the daily production rate is 57 units per day.
d. With a 10-minute cycle time, what is the theoretical minimum number of
stations with which the line can be balanced?
Cycle time = 10 minutes
Theoretical Minimum (TM) = number of station needed to achieve 100%
efficiency
Therefore, the theoretical minimum number of workstations which the line can
be balanced is 5 stations.
e. With a 10-minute cycle time and six workstations, what is the efficiency?
Number of workstation = 6 workstations
f. What is the total idle time per cycle with a 10-minute cycle time and six
workstation?
Cycle time = 10 minutes
Number of workstation = 6 workstations
8
Haeryip Sihombing Therefore, total idle time per cycle with a cycle time of 10 minutes and 6
workstations is 16.67%.
g. What is the best work station assignment you can make without exceeding a
10-minute cycle time and what is the efficiency?
Workstation
Eligible Task
Task Selected
Task Time
Idle Time
1
A, B, C, D, E
2
B, D, E, F
3
D, E, G
4
E, H, I, J
5
J, K
A
C
B
F
D
G
E
H
I
J
K
3
7
6
4
5
5
2
7
1
6
4
7
0
4
0
5
0
8
1
0
4
0
Therefore, the best workstations assignment is with the theoretical minimum number
of workstation which is 5 stations without exceeding the 10 minutes cycle time and
reached an efficiency of 100%. The assignment is shown in table above.
Problem Statement 5
Registration at Delhi University has always been a time of emotion, commotion, and
lines. Students must move among four stations to complete the trying semiannual
process. Last semester’s registration, held in the north campus, is described in
Figure below. You can see, for example, that 450 students moved from the
paperwork station (A) to advising (B), and 550 went directly from A to picking up
their class cards (C). Graduate students, who for the most part had preregistered,
proceeded directly from A to the station where registration is verified and payment
collected (D). The layout used last semester is also shown in Figure below. The
registrar is preparing to set up this semester’s stations and is anticipating similar
numbers.
9
Haeryip Sihombing Advising
station
(B)
Pick up
paperwork
and forms
(A)
--350
0
0
Paperwork/forms (A)
Advising (B)
Class cards (C)
Verification/payment
(D)
Pick up
class
cards
(C)
550
200
--0
450
--0
0
Verification of
status and
payment (D)
50
0
750
---
Existing Layout
A
B
C
30’
30’
D
30’
Figure. Registration flow of students
a)
What is the “load x distance,” or “movement cost,” of the layout shown?
From (A) to (B) = 450 x 30’ = 13 500
From (A) to (C) = 550 x 60’ = 33 000
From (A) to (D) = 50 x 90’ = 4 500
From (B) to (A) = 350 x 30’ = 10 500
From (B) to (C) = 200 x 30’ = 6 000
From (C) to (D) = 750 x 30’ = 22 500
b)
Provide an improved layout and compute its movement cost.
B
A
30’
C
30’
D
30’
From (A) to (B) = 450 x 30’ = 13 500
From (A) to (C) = 550 x 30’ = 16 500
From (A) to (D) = 50 x 60’ = 3 000
From (B) to (A) = 350 x 30’ = 10 500
From (B) to (C) = 200 x 60’ = 12 000
From (C) to (D) = 750 x 30’ = 22 500
10
Haeryip Sihombing Problem Statement 6
Recently, you were assigned to manage a project for your company. You have
constructed a network diagram depicting the various activities in the project (Figure
3.15). In addition, you have asked your team to estimate the amount of time that
they would expect each of the activities to take. Their responses are shown in the
following table.
Activity
A
Time estimates (days)
Optimistic
Most likely
5
8
Pessimistic
11
B
4
8
11
C
5
6
7
D
2
4
6
E
4
7
10
C
A
Start
D
Start
E
B
Figure. Network diagram for your project
Answer
teA
=
(a + 4m + b) / 6 = (5 + 4(8) + 11) / 6 = 8
teB
=
(4 + 4(8) + 11) / 6 = 7.8
teC
=
(5 + 4(6) + 7) / 6 = 6
teD
=
(2 + 4(4) + 6) / 6 = 4
teE
=
(4 + 4(7) + 10) / 6 = 7
σteA2
=
[(b – a) / 6]2 = [(11 – 5) / 6]2 = 1
σteB2
=
[(11 – 4) / 6]2 = 1.36
σteC2
=
[(7 – 5) / 6]2 = 0.11
σteD2
=
[(6 – 2) / 6]2 = 0.44
σteE2
=
[(10 – 4) / 6]2 = 1
11
Haeryip Sihombing Activity Predecessor Optimistic
A
B
C
D
E
A
A
B, D
0
5
4
5
2
4
8
0
A
0
8
3.8
Most
likely
8
8
6
4
7
8
0
0
Expected
time
8
7.8
6
4
7
11
11
7
6
10
Variance
1
1.36
0.11
0.44
1
14
0.8
C
8.8
6
8
0
Pessimistic
14.8
12
0
D
3.8
4
7.8
7.8
B
7.8 7.8
7.8
14.8
0
E
7.8
7
14.8
a) What is the expected completion time of the project?
Critical path = A – D – E
Duration
= 8 + 4 + 7 = 19 days
b) What is the probability of completing the project in 21 days?
Z=
= 1.280
p = 0.89973
The probability is 89.973%
c) What is the probability of completing the project in 17 days?
Z=
= -1.280
p = 0.10027
(by using Calculator fx 570, is: MODE – MODE – 1 – SHIFT -3 -1, -1.280 ENTER)
The probability is 10.027%
12
Haeryip Sihombing Problem Statement 7
Consider a project with the following information. Construct the project activity
network using AOA methodology and label each node and arrow appropriate.
Identify all dummy activities required to complete the network.
Activity
Duration
Predecessors
A
B
C
D
E
F
G
H
3
5
7
3
5
4
2
5
A
A
B, C
B
D
C
E, F, G
Activity
Duration
ES
EF
LS
LF
Slack
A
B
C
D
E
F
G
H
3
5
7
3
5
4
2
5
0
3
3
10
8
13
10
17
3
8
10
13
13
17
12
22
0
5
3
10
12
13
15
17
3
10
10
13
17
17
17
22
2
4
5
-
a. What is the critical path of the project?
A–C–D–F–H
b. Draw the graphic of the project
13
Haeryip Sihombing Problem Statement 8
Subsidiary Manufacturing produces custom-built pollution control devices for
medium-size steel mills. The most recent project undertaken by Subsidiary
Manufacturing requires 14 different activities.
a. The managers would like to determine the total project completion time (in
days) and those activities that lie along the critical path. The appropriate data
are shown in the following table.
b. What is the probability of being done in 53 days?
c. What date results in a 99% probability of completion?
Activity
Immediate
Predecessor(s)
Optimistic
Time
Most
Likely
Time
Pessimistic
Time
A
B
C
D
E
F
G
H
I
J
K
L
M
N
A
A
B, C
D
D
E, F
G, H
I
I
J
K
L, M
4
1
6
5
1
2
1
4
1
2
8
2
1
6
6
2
6
8
9
3
7
4
6
5
9
4
2
8
7
3
6
11
18
6
8
6
8
7
11
6
3
10
14
Haeryip Sihombing Answer:
a. The total project completion time (in days) and those activities that lie along the
critical path.
TE = (O + 4M + P) ÷ 6
Variance = [(P – O)/6]2
Where
O - Optimistic time ; M - Most likely time ; P - Pessimistic time ; TE Expected time
`Activity
Immediate
Optimistic Most Likely
Predecessor(s)
Time
Time
A
A
B, C
D
D
E, F
G, H
I
I
J
K
L, M
A
B
C
D
E
F
G
H
I
J
K
L
M
N
4
1
6
5
1
2
1
4
1
2
8
2
1
6
Pessimistic Expected
Variance
Time
Time
6
2
6
8
9
3
7
4
6
5
9
4
2
8
7
3
6
11
18
6
8
6
8
7
11
6
3
10
Critical path = A – C – E – H – I – K – M – N (55.92 days)
b. What is the probability of being done in 53 days?
Where
= critical path duration
= scheduled project duration
Z = probability (of meeting scheduled duration)
15
8.75
3
9
8
9.17
3.33
6.17
4.33
5.5
4.83
9.17
4
2
8
0.25
0.11
0
1
8.03
0.44
1.36
0.11
1.36
0.69
0.25
0.44
0.11
0.44
Haeryip Sihombing = 10.55
= 3.248
c. What date results in a 99% probability of completion?
When P = 0.9901, Z = 2.33
Problem Statement 9
Tailwind, Inc., produces high quality but expensive training shoes for runners. The
Tailwind shoes, which sells for $210, contains both gas and liquid-filled
compartments to provide more stability and better protection against knee, foot, and
back injuries. Manufacturing the shoes requires 10 separate tasks. There are 400
minutes available for manufacturing the shoes in the plant each day. Daily demand
is 60. The information for the task is as follows:
Task
Performance
Time(min)
Task Must Follow Task Listed
Below
A
1
-
B
3
A
C
2
B
D
4
B
E
1
C, D
F
3
A
G
2
F
H
5
G
I
1
E,H
J
3
I
16
Haeryip Sihombing a) Draw a precedence diagram.
C
120
A
B
60
180
E
D
60
240
F
J
60
180
H
G
180
I
300
120
b) Assign tasks to the minimum feasible number of workstations according to the
“ranked positioned weight” decision rule.
Cycle time =
= 400 sec per unit
Task
A
Position weight
1080
Rank
1
Performance time
60
B
840
2
180
C
420
5
120
D
540
4
240
E
300
6
60
F
840
2
180
G
660
3
120
H
540
4
300
I
240
7
60
J
180
8
180
Total standard time
Workstation =
1500
≈ 4 workstations
Cycle time = 400 second per unit
Stations
1
2
3
4
Time remaining
400, 340, 160,40
400, 220 160, 0
400, 280, 40
400, 100, 40
17
Tasks
A, B, C
F, I, J
G, D
H, E
Haeryip Sihombing c) What is the efficiency of the process?
Efficiency of the process = Ratio of productive time to total time
= [1500/4(400)] x 100%
= 93.7%
d) What is the idle time per cycle?
Stations
1
Time remaining
400, 340, 160,40
Tasks
A, B, C
Idle time per cycle
40 seconds
2
400, 220 160, 0
F, I, J
0 second
3
400, 280, 40
G, D
40 seconds
4
400, 100, 40
H, E
40 seconds
Problem Statement 10
The project manager of Good Public Relations gathered the data shown in table
shown for a new advertising campaign.
Time Estimates (days)
Optimistic
Most likely
Pessimistic
Activity
A
8
10
12
Immediate
Predecessor(s)
-
B
5
8
17
-
C
7
8
9
-
D
1
2
3
B
E
8
10
12
A, C
F
5
6
7
D, E
G
1
3
5
D, E
H
2
5
8
F,G
I
2
4
6
G
J
4
5
8
H
K
2
2
2
H
a. How long is the project likely to take?
Expected time,
Activity
=
2
, Variance,
Optimistic
A
Immediate
predecessor(s)
-
B
-
2
=
Pessimistic
8
Most
Likely
10
5
8
18
Variance
12
Expected
time(Days)
10
17
10
4
0.44
Haeryip Sihombing C
-
7
8
9
8
0.11
D
B
1
2
3
2
0.11
E
A, C
8
10
12
10
0.44
F
D, E
5
6
7
6
0.11
G
D, E
1
3
5
3
0.44
H
F, G
2
5
8
5
1
I
G
2
4
6
4
0.44
J
H
4
5
8
5
0.44
K
H
2
2
2
2
0
36
2.43
Total
Critical path is A-E-F-H-J
The project is has to take 36 days to be done.
b. What is the probability that project will take more than 38 days?
Z=
-
P( X > 38 )
/
= 1 - P( X < 38 )
= 1= 1- 0.8997 = 0.1003
The probability of the project will more than 38 days is 10.03%
c. Consider the path A-E-G-H-J what is the probability that this path will
exceed the expected project duration?
19
Haeryip Sihombing Expected time for A-E-G-H-J = 33days
Variance for A-E-G-H-J = 2.76
P( X > 33 ) = 1 - P( X < 33 )
= 1= 1- 0.0351 = 0.9649
The probability A-E-G-H-J path will exceed the expected project duration is
96.5%
Problem Statement 11
The optical disk project team has started gathering the information necessary to
develop the project network-predecessor activities and activity times in weeks. The
results of their meeting are found in the following table:
Activity
Description
Duration
Predecessor
1
Define scope
6
None
2
Define customer problems
3
1
3
Define data records and relationship
5
1
4
Mass storage requirements
5
2,3
5
Consultant needs analysis
10
2,3
6
Prepare installation network
3
4,5
7
Estimate costs and budget
2
4,5
8
Design section “point” system
1
4,5
9
Write request proposal
5
4,5
10
Compile vendor list
3
4,5
11
Prepare management control system
5
6,7
12
Prepare comparison report
5
9,10
13
Compare system “philosophies”
3
8,12
14
Compare total installation
2
8,12
15
Compare cost of support
3
8,12
16
Compare customer satisfaction level
10
8,12
17
Assign philosophies points
1
13
18
Assign installation cost
1
14
19
Assign support cost
1
15
20
Assign customer satisfaction points
1
16
21
Select best system
1
11,17,18,19,20
22
Order system
1
21
20
Haeryip Sihombing The project team has requested that you create a network for the project, and
determine if the project can be completed in 45 weeks.
The project can complete within 45 weeks. This is because the project can complete
in 44 weeks according to the network below
Problem Statement 12
Boiling Electronics manufactures DVD players for commercial use. W. Blaker
Boiling, president of Boiling Electronics, is contemplating producing DVD players
for home use. The activity necessary to built an experimental model and related data
given in the following table:
Activity Normal Time
(weeks)
A
3
Crash Time
(weeks)
2
Normal Cost
($)
1,000
Crash Cost
($)
1,600
Immediate
Predecessor (s)
-
B
2
1
2,000
2,700
-
C
1
1
300
300
-
D
7
3
1,300
1,600
A
E
6
3
850
1,000
B
F
2
1
4,000
5,000
C
G
4
2
1,500
2,000
D,E
21
Haeryip Sihombing a) What is the project completion date?
The project completion date is in 14 weeks.
b) Crash this project to 10 weeks at the least cost.
Crash
Time
(weeks)
2
Crash –
Normal
(weeks)
1
Normal
Cost ($)
Crash
Cost ($)
Crash – Normal($)
Time (weeks)
A
Normal
Time
(weeks)
3
1,000
1,600
600
B
2
1
1
2,000
2,700
700
C
1
1
0
300
300
0
D
7
3
4
1,300
1,600
75
E
6
3
3
850
1,000
50
F
2
1
1
4,000
5,000
1000
G
4
2
2
1,500
2,000
250
Activity
10,950
So, the project completion date is 10
weeks. The cost as below:
4 X D = 4 X 75 = $300
2 X E = 2 X 50 = $100
TOTAL = $300 + $100 + $10950 =
$11350
22
Haeryip Sihombing c) Crash this project to 7 weeks (which is the maximum it can be crash) at
the least cost.
So, the project completion date is 7
weeks. The cost as below:
7 X D = 7 X 75 = $525
5 X E = 5 X 50 = $250
TOTAL = $525 + $250 + $10950 =
$11725
Problem Statement 13
A firm with four departments has the following closeness matrix and the current
block plan shown in Figure below
1
2
3
A
B
4
5
6
C
D
Conway Consulting’s Block
Plan
Department
A
B
C
D
Current Block Plan
Closeness Matrix
Trips Between Departments
A
B
C
12
10
20
-
D
8
6
0
-
a) What is the weighted-distance score for the current layout (assuming
rectilinear distance)?
The weighted-distance score for the current layout is
84.
Department Pair
Number of trips
Distance
A, B
A, C
A, D
12
10
8
1
1
2
23
Current weighteddistance score
12
10
16
Haeryip Sihombing B, C
B, D
C, D
20
6
0
2
1
1
Total:
40
6
0
84
b) Develop a better layout. What is its weighted-distance score? After proposed
a better layout, the total weighted-distance score is 72.
Department
Pair
A, B
A, C
A, D
B, C
B, D
C, D
B
C
A
D
Number of
trips
12
10
8
20
6
0
Distance
1
2
1
1
2
1
Total:
proposed weighted-distance
score
12
20
8
20
12
0
72
Problem Statement 14
A small renovation of a Hard Rock Café gift shop has six activities (in hours). For
the following estimates of a, m, and b, calculates the expected time and the standard
deviation for each activity:
Activity
A
B
C
D
E
F
a
11
27
18
8
17
16
m
15
31
18
13
18
19
b
19
41
18
19
20
22
Weighted average activity time, te =
Where
a = optimistic activity time (1 chance in 100 of completing the activity earlier under
normal conditions)
b = pessimistic activity time (1 chance in 100 of completing the activity later under
normal conditions)
m = most likely activity time
24
Haeryip Sihombing Standard deviation for the activity,
Activity
A
B
C
D
E
F
a
11
27
18
8
17
16
m
15
31
18
13
18
19
b
19
41
18
19
20
22
te
15
32
18
13.17
18.17
19
1.33
2.33
0
1.83
0.50
1.00
Problem Statement 15
Rezin Enterprises, a machine shop, is planning to move to a new, larger location.
The new building will be 60 feet long by 40 feet wide. Rezin envisions the building
as having six distinct production areas, roughly equal in size. He feels strongly about
safety and intends to have marked pathways throughout the building to facilitate the
movement of people and materials. See the following building schematic.
1
2
3
4
5
6
His foreman has completed a month-long study of the number of loads of material
that have moved from one process to another in the current building. This
information is contained in the following matrix.
To
Materials
(M)
Welding
(W)
Drills
(D)
Lathes
(L)
Materials (M)
0
100
50
0
0
50
Welding (W)
25
0
0
50
0
0
Drills (D)
25
0
0
0
50
0
Lathes (L)
0
25
0
0
20
0
Grinders (G)
50
0
100
0
0
0
Benders (B)
10
0
20
0
0
0
From
Grinders Benders
(G)
(B)
Finally, Rezin has developed the following matrix to indicate distances between the
work areas shown in the building schematic.
25
Haeryip Sihombing 1
2
3
4
5
6
1
-
Distance between Work Areas
2
3
4
5
20
40
20
40
20
40
20
60
40
20
6
60
40
20
40
20
What is the appropriate layout of the new building?
Current building schematic (Current Layout):
M
W
D
L
G
B
Proposed Layout
wd Score
wd Score
Distance (2)
Distance (3)
(1) x (2)
(1) x (3)
20
2000
20
2000
40
2000
20
1000
60
3000
40
2000
20
500
20
500
40
2000
40
2000
40
1000
20
500
40
2000
20
1000
40
1000
40
1000
20
400
20
400
40
2000
40
2000
40
4000
20
2000
60
600
40
400
20
400
20
400
wd = 20900
wd
Current Layout
Department
Pair
M,W
M,D
M,B
W,M
W,L
D,M
D,G
L,W
L,G
G,M
G,D
B,M
B,D
Number of
Trips (1)
100
50
50
25
50
25
50
25
20
50
100
10
20
= 15200
New building schematic (Proposed Layout):
M
D
B
W
G
L
26
Haeryip Sihombing Problem Statement 16
Barbara Gordon, the project manager for Web Ventures, Inc., compiled a table
showing time estimates for each of the company’s manufacturing activities of a
project, including optimistic, most likely, and pessimistic.
a. Calculate the expected time, te, for each activity.
b. Calculate the variance, σ², for each activity.
Activity
A
B
C
D
E
Optimistic
(a)
3
12
2
4
1
Most
Likely (m)
8
15
6
9
4
Pessimistic
(b)
19
18
16
20
7
9
15
7
10
4
7.11
1
5.44
7.11
1
Problem Statement 17
You are the manager of a project to improve a billing process at your firm. Table
below contains the data you will need to conduct a cost analysis of the project.
Indirect costs are $1,600 per week, and penalty costs are $1,200 per week after week
12,
a. What is the minimum-cost schedule for this project?
b. What is the difference in total project costs between the earliest completion
time of the project using “normal” times and the minimum-cost schedule
you derived in part (a)?
Immediate
Normal Crash Normal
Crash
Activity predecessor(s)
time
time
cost
cost
A
4
1
5,000
8,000
B
-
5
3
8,000
10,000
C
A
1
1
4,000
4,000
D
B
6
3
6,000
12,000
E
B, C
7
6
4,000
7,000
F
D
7
6
4,000
7,000
a.
B
A
D
C
F
E
B + D + F = (8,000+10,000) + (6,000+12,000) + (4,000+7,000) = 47,000
A + C + E = (5,000+8,000) + (4,000+4,000) + (4,000+7,000) = 32,000
27
Haeryip Sihombing B + E = (8,000+10,000) + (4,000+7,000) = 29,000
So, from the calculation above, the minimum-cost schedule for this project is for B +
E that is only 29,000.
b. The total project costs between the earliest completion time of the project using
“normal” times is the total cost of Normal cost only without adding the Crash cost.
So, the total cost of Normal Cost for this project is 31,000. While for the minimumcost schedule for this project is the Normal Cost adding with the Crash Cost. But
after we have calculated all, we have to only take the minimum value only. The total
minimum-cost schedule for this project is 29,000.
Problem Statement 18
After an extensive product analysis using group technology, Bob Buerlein has
identified a product he believes should be pulled out of his process facility and
handled in a work cell. Bob has identified the following operations as necessary for
the work cell. The customer expects delivery of 250 units per day, and the work day
is 420 minutes.
a. What is the Takt Time?
b. How many employees should be cross-trained for the cell?
c. Which operations may warrant special consideration?
Operation
Shear
Bend
Weld
Clean
Paint
Standard Time (min)
1.1
1.1
1.7
3.1
1.0
a. 250 units per day
420 minutes
= 0.5952 unit per minute
The takt time is 0.5952 unit per minute
b. 2 employees should be cross-trained for the cell that is for Shear and Bend
operation. This is because the operations are quite same and the standard time
also same for both of the operations that is 1.1 min.
c. The operation that may warrant special consideration is Clean operation. This
due to the standard time that is larger than the others operation. The standard
time is 3.1 min while others are only 1.0, 1.1, and 1.7. So, it means that this
operation are harder than the others operation.
28
Haeryip Sihombing Problem Statement 19
Development of a new deluxe version of a particular software product is being
considered by Ravi Bedara’s software house. The activities necessary for the
completion of this project are listed in the following table:
Maximum
Crash
Direct cost
Normal
Time
(weeks)
Cost
4
$2,000
Crash
Time
(weeks)
3
Cost
$2,600
2200
1
2800
3
500
3
500
4
8
2300
4
2600
100
3
6
900
3
1200
F
1200
1
3
3000
2
4200
G
300
2
4
1400
2
2000
Activity
A
Slope
600
Time
1
B
600
1
2
C
0
0
D
75
E
a) What is the project completion date?
=
Duration
of
A+D+G
=4+8+
4 weeks
= 16
weeks
In normal cost
b) What is the total cost required for completing this project on normal time?
= Cost A + B + C + D + E + F + G
= $ 2,000 + 2,200 + 500 + 2,300 + 900 + 3,000 + 1,400 = $ 12,300
c) If you wish to reduce the time required completing this project by 1 week,
which activity should be crashed, and how much will this increase the total
cost?
29
Haeryip Sihombing Activity C is the lowest cost, activity c is being chosen.
Increase the total cost: $12,300 - $75 = $12,225
Activity C should be crashed and $12,225 will increase the total cost.
d) What is the maximum time that can be crashed? How much would costs
increase?
In crash cost
Maximum time can be crashed:
= project completion date – ( Crash time of A + D + G )
= 16 weeks – ( 3 + 4 + 2 weeks) = 7 weeks
Costs increase:
= (slope * maximum crash time) of D + A + G
= (4*75) + ( 1*600) + (2*300) = $1500
The maximum time can be crashed in 7 weeks by increase $1,500.
Problem Statement 20
The associate administrator at Getwell Hospital wants to evaluate the layout of the
outpatient clinic. Table below shows the interdepartmental flows (patients/day)
between departments; Figure below shows the current layout.
Table: Closeness Matrix
Trips between Departments
Department
1. Reception
2. Business Office
3. Examining room
1
2
3
4
5
6
-
25
35
5
10
15
-
5
10
15
-
20
30
30
7
8
20
15
20
10
Haeryip Sihombing 4. X-ray
-
5. Laboratory
25
15
25
-
20
25
6. Surgery
-
40
7. Postsurgery
15
8. Doctor’s office
-
4
6
5
7
2
8
3
1
Figure Current Layout
a) Determine the effectiveness of the current layout, as measured by the total
wd score, using rectilinear distances. = 790
Department
pair
1,2
1,3
1,4
1,5
1,6
1,8
2,3
2,4
2,5
2,8
3,4
3,5
3,6
3,8
4,5
4,6
4,7
5,6
5,8
6,7
7,8
Number of
tips (1)
25
35
5
10
15
20
5
10
15
15
20
30
20
10
25
15
25
20
25
40
15
Current layout
Proposed layout
Distance Wd score
(2)
(1) * (2)
3
75
1
35
4
20
2
20
3
45
2
40
2
10
1
10
3
45
1
15
3
60
1
30
2
40
1
10
2
50
1
15
3
45
1
20
2
50
2
80
3
45
Current layout, wd =
790
Distance
Wd score
(3)
(1) * (3)
3
75
1
35
4
20
3
30
2
30
2
40
2
10
1
10
2
30
1
15
3
60
2
60
1
20
1
10
1
25
2
30
3
75
1
20
1
25
1
40
3
45
Proposed layout, wd =
705
4
5
6
7
2
8
3
1
Proposed Layout
Current layout, total wd score is 790.
31
Haeryip Sihombing b) Try to find the best possible layout based on the same effectiveness measure.
4
5
6
7
2
8
3
1
Proposed Layout
Proposed layout, total wd score is 705.
c) What is the impact on your new solution if it must be revised to keep
department 1 at its present location?
There is no impact on the new solution for the proposed layout, the total wd
still remain same which is 705.
d) How should the layout developed in part (c) be revised if the
interdepartmental flow between the examining room and the X-ray
department is increased by 50 percent? Decreased by 50 percent?
The proposed layout did not change the location of examining room and Xray department, so will not affect the new proposed layout. For the increased
by 50% of the interdepartmental flow between the examining room and the
X-ray department, the total wd is 735 which is increase 30
from the
previous proposed layout. For the decrease 50%, the total wd is 675 which is
decrease 30 from the previous proposed layout.
Problem Statement 21
From the following information, draw the project network. Compute the early, late,
and slack times for each activity. Identify the critical path. (Hint: Draw the finish-tostart relationship first.)
5
10
15
5
Finish-to-start
Predecessor
NONE
A
A
B
Finish-to-Start
Lag
0
0
0
5
20
15
10
20
B
D
C
F
0
0
10
0
ID
Duration
A
B
C
D
E
F
G
H
32
Additional lag
Lag
Relationship
None
0
None
0
Start-finish C to D
20
Start-start D to E
5
Finish-finish D to E
Finish-finish E to F 25
None
0
Finish-finish G to F 0
None
10
Haeryip Sihombing Problem Statement 22
The Maser is a new custom-designed sports car. An analysis of the task of building
the Maser reveals the following list of relevant activities, their immediate
predecessors, and their duration:
Job
Description
Letter
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T
U
Immediate
predecessor(s)
Start
design
order special accessories
Build frame
Build doors
Attach axles, wheels gas tank
Build body shell
Build transmission and drive train
Fit door to body shell
Build engine
Bench-test engine
Assemble chassis
Road-test Chassis
Paint body
Install wiring
Install interior
Accept delivery of special accessories
Mount body and accessories on chassis
Road test car
Attach exterior trim
Finish
Normal
time (days)
A
B
B
B
D
B
B
G,E
B
J
F,H,K
L
I
N
N
C
M,O,P,Q
R
S
T
a) Draw a network diagram for the project.
b) Mark the critical path and state its length.
c) If the Maser had to be completed 2 days earlier, would it help to:
i) Buy preassembled transmissions and drive train?
ii)Install robots to halve engine-building time?
iii)Speed delivery of special accessories by 3 days
d) How might resources be borrowed from activities on the non-critical
path to speed activities on the critical pa
33
0
8
0.1
1
1
1
2
3
1
4
2
1
0.5
2
1
1.5
5
1
0.5
1
0
Lag exercise 18 (QUESTION 39):
Lag 5
Lag 20 Lag 5 Lag 25 Lag 10 Lag 10 The critical paths of the network are shown by dashed-lines in the diagram, or
as below:
i.
ii.
iii.
A, C, G, F, and H.
A, C, D, E, F, and H
A, B, D, E, F, and H
b) Critical path is marked as black arrow along the flow of the process, as shown at
the network diagram above. Total length of critical path (A, B, J, K, L, M, R, S,
T, and U) is 18 weeks without slack time.
c) I) No. Since the process H (build transmission and drive train) is not a process
within the critical path, hence it wouldn’t help to complete the project 2 days
earlier with preassembled transmission and drive train.
II) No. It is not helpful to install robots to halved engine-building time. This is
because of other process path would prevent the project to be finished 2 weeks
earlier. The installation of robot would only make the complete time of project
earlier by 1 week.
III) No. Process Q is not within the critical path. Even the special accessories
are accepted 3 days earlier, the project complete time is still restricted by the
process time of critical path as stated above.
d) In order to speed up the activities on the critical path, some recourses from the
non-critical path, especially those activities with many slack time, can be
borrowed to facilitate the activities on the critical path. The activities with many
slack time here might be process d, f, and h, can be holden for awhile and the
resources used such as the assembly material, tools and manpower can be
borrowed to perform the activities on the critical path. As long as the time of the
resources stated are not borrowed to critical path more than the slack time as
indicated in the box of network diagram above, the activities on the critical path,
and hence the whole system, will be speeded up.
Haeryip Sihombing Problem Statement 23
Summit Arora is developing a program in leadership training for middle-level managers.
Sumit has listed a number of activities that must be completed before a training program of
this nature could be conducted. The activities, immediate predecessors and times appear in
the accompanying table:
Activity
A
B
C
D
E
F
G
Immediate
Predecessor(s)
B
A,D
C
E,F
Time (days)
2
5
1
10
3
6
8
a) AON network:
E
A
(3)
(2)
START
0
B
D
G
(5)
(10)
(8)
C
F
(1)
(6)
2
13
A
13
2
0
0
B
0
5
0
11
C
11
1
15
18
0
E
15
15
3
5
5
15
0
D
5
5
6
1
1
12
18
15
7
11
F
12
6
18
b) Critical Path: B,D,E,G
37
18
26
0
G
18
8
26
END
Haeryip Sihombing c) Total project completion time= 5+10+3+8 = 26 days
d) Slack time for activity A = 15-2 = 13 days
B,D,E,G = No Slack Time = 0
C&F
= 18-7 = 11 days
Problem Statement 24
Suppose production requirements in solved Problem 9.2 increase and require a reduction in
cycle time from 8 minutes to 7 minutes. Balance the line once again, using the new cycle
time. Note that it is not possible to combine task times so as to group task into the minimum
number of workstation. This condition occurs in the actual balancing problems fairly often.
Task
A
B
C
D
E
F
G
H
Performance
Time (mins)
5
3
4
3
6
1
4
2
28
Task Must Follow This
Task
A
B
B
C
C
D,E,F
G
New line balancing
4
6
C
E
4
2
G
H
5
3
1
A
B
F
WS5
3
WS1
WS2
D
WS4
WS3
The theoretical minimum number of workstation (WS) is:
=
= 4 workstation.
By theory minimum number of 4 workstations can be group together but after trial and error
for the new line balancing to group into necessary workstation for 7 minutes cycle time, it is
proven that 4 workstation can’t be achieve as theory indicated but instead it only can be
group into 5 workstations.
38
Haeryip Sihombing Problem Statement 25
Chunmum Furniture Inc., produces all types of office furniture. The “Executive Secretary” is
a chair that has been designed using ergonomics to provide comfort during long work hours.
The chair sells for $130. There are 480 minutes available during the day and the average
daily demand has been 50 chairs. There are 8 tasks:Task
Performance
Time (minutes)
Task Must Follow Task
Listed Below
A
B
C
D
E
F
G
H
4
7
6
5
6
7
8
6
A, B
C
D
E
E
F, G
a. Draw a precedence diagram of this operation
A
C
D
E
F
H
B
G
b. What is the cycle time for this operation?
Cycle time,
The desired output rate is 50 Chairs 480 minutes per day.
C = 1/r
C= 480/50
= 9.6 minutes/units
c. What is the theoretical minimum number of work station?
Theoretical minimum ( TM )
TM = ∑ t / C
TM = 49 / 9.6
= 5.1 @ 5 stations.
d. Assign tasks to work stations
39
Haeryip Sihombing How much total idle time is present each day?
Idle time = nc - ∑t
= 8(9.6) – 49 = 27.8 seconds
Percentage of idle time = idle time per cycle X 100
Nactual X Cycle Time
= 27.8 X 100
8X9.6
= 36.2 %
f. What is the overall efficiency of the assembly line?
Efficiency
= 100 % - 36.2 % = 63.8 %
Problem Statement 26
The diagram in Figure below was developed for a project that you are managing. Suppose
that you are interested in finding ways to speed up the project at minimal additional cost.
Determine the schedule for the completing the project in 25 days at minimum cost. Penalty
and project overhead costs are negligible. Time and cost data for each activity are shown in
Table below.
Activity
A
B
C
D
E
F
G
H
I
Normal
Time (days)
12
13
18
9
12
8
8
2
4
Crash
Cost ($)
1, 300
1, 050
3, 000
2, 000
650
700
1, 550
600
2, 200
Time (days)
11
9
16
5
10
7
6
1
2
40
Cost ($)
1, 900
1, 500
4, 500
3, 000
1, 100
1, 050
1, 950
800
4, 000
Haeryip Sihombing D
Start
A
E
B
F
G
C
Activity
A
B
C
D
E
F
G
H
I
Slope
600
112.5
750
250
225
350
200
200
450
I
Maximum
Crash
Time
1
4
2
4
2
1
2
1
2
Day 25
Initial total direct cost $13050
A
12
Start
Finish
H
Direct Cost
Normal
Time
Cost
12
1,300
13
1,050
18
3,000
9
2,000
12
650
8
700
8
1,550
2
600
4
2,200
Crash
Time
11
9
16
5
10
7
6
1
2
Cost
1,900
1,500
4,500
3,000
1,100
1,050
1,950
800
4,000
D
AC
9
DU
E
G
12
8
B
F
H
13
8
2
C
I
18
4
41
Finish
Haeryip Sihombing Day 24
Initial total direct cost $13800
Activity Change
C
D
750
9
A
12
Start
E
G
12
8
B
F
H
13
8
2
C
I
17
4
Finish
Day 23
Initial total direct cost $12450
Activity Change
D
I
9
450
A
12
Start
Project duration
25
24
23
E
G
12
8
B
F
H
13
8
2
C
I
17
4
Direct Cost +
13050
13800
14250
42
Indirect Cost
10050
7050
4050
=
Total Cost
23,100
20,850
18,300
Finis
Haeryip Sihombing The project will actually materialize as planned any movement away from this tie option
duration will increase projects cost. The optimum cost-time duration is 24 time units and
$20,850. The movement from 25 time units occurs, in this range the absolute slopes of the
indirect costs are greater than the direct cost slopes.
Problem Statement 27
The following is a table of activities associated with a project at Bill Figg Enterprise, their
durations and what activities each must precede:
Activity
A (start)
B
C
E
F (end)
Duration (weeks)
1
1
4
2
2
Precedes
B, C
E
F
F
-
a) Draw an AON diagram of the project, including activity durations
1
0
0
A
0
1
2
2
1
B
1
2
1
1
1
0
C
1
4
4
1
E
3
3
2
5
5
5
5
7
0
F
5
2
7
b) Define the critical path, listing all critical activities in chronological order
Critical path are A, C, F
c) What is the project duration (in weeks)?
Project duration (in week) is 7 weeks
d) What is the slack (in weeks) associated with any and all noncritical paths through
the project?
Activity
Early start
A
B
C
E
F
0
1
1
2
5
Early
finish
1
2
5
4
7
Late start
Late finish
Slack
0
2
1
3
5
1
3
5
5
7
0
1
0
1
0
The slack (in weeks) happen for activity B and E.
43
Haeryip Sihombing Problem Statement 28
Use trial and error to balance the assembly line described in the following table and Figure
8.19 so that it will produce 40 units per hour.
a.
b.
c.
d.
What is the cycle time?
What is the theoretical minimum number of workstations?
Which work elements are assigned to each workstation?
What are the resulting efficiency and balance delay percentages?
Work
element
A
B
C
D
E
F
G
H
I
J
K
Time (sec)
40
80
30
25
20
15
60
45
10
75
15
Total
Immediate Predecessor
(s)
None
A
A
B
C
B
B
D
E,G
F
H, I, J
415
D
25
B
F
80
15
H
45
A
G
40
J
60
75
E
I
20
10
C
30
a. Cycle time = available
output rate
= 60 x 60 = 90
40
b. Theoretical minimum number of workstations
= 415 = 4.611 @ 5 = 5 stations
90
44
K
15
Haeryip Sihombing D
25
B
F
45
H
15
80
A
G
40
J
60
K
75
15
C
30
E
I
20
10
c. Workstation 1: A, C, E
Workstation 2: B
Workstation 3: F, J
Workstation 4: D, H
Workstation 5: G, I, K
d. Efficiency = ratio of productive time to total time
= 415 / 5 (90) x 100
= 92.2 %
Balance Delay = amount by which efficiency falls short of 100%
= 100% – 92.2 % = 7.8 %
Problem Statement 29
McGee Carpet and Trim installs carpet in commercial offices. Andrea McGee has been very
concerned with the amount of time it took to complete several recent jobs. Some of her
workers are very unreliable. A list of activities and their optimistic completion time, the most
likely completion time, and their pessimistic completion time (all in days) for a new contract
are given in the following table:
Activity
A
B
C
D
Time (days)
a
3
2
1
6
m
6
4
2
7
45
b
8
4
3
8
Immediate
predecessor (s)
C
Haeryip Sihombing E
F
G
H
I
J
K
a)
b)
c)
d)
2
6
1
3
10
14
2
4
10
2
6
11
16
8
6
14
4
9
12
20
10
B,D
A,E
A,E
F
G
C
H,I
Determine the expected completion time and variance for each activity
Determine the total project completion time and the critical path for the project
Determine ES, EF, LS, LF and slack for each activity
What is the probability that McGee carpet and Trim will finish the project in 40
days or less?
a)
Activity
A
B
C
D
E
F
G
H
I
J
K
Time (days)
a
m
b
Immediate
predecessor (s)
3
2
1
6
2
6
1
3
10
14
2
6
4
2
7
4
10
2
6
11
16
8
8
4
3
8
6
14
4
9
12
20
10
C
B,D
A,E
A,E
F
G
C
H,I
te =
a + 4m + b
6
5.833
3.667
2.000
7.000
4.000
10.000
2.167
6.000
11.000
16.333
7.333
variance
{(b − a ) / 6}2
b) Total completion time
= 5.883 + 3.667 + 2 + 7 + 4 + 10 + 2.167 + 6 + 11 + 16.333 + 7.333 = 75.333
= 75 day
c) Critical path of the project
= C, D, E, F, H, K = 2 + 7 + 4 + 10+6+7.33 = 36.33 ≈ 37 day
∑σ Te = √ (0.111 + 0.111 + 0.444 + 1.778 + 1.000 + 1.778) = √ 5.222 = 2.28517
Activity
A
B
C
D
E
F
G
H
I
ES
0
0
0
2
9
13
13
23
15
EF
6
4
2
9
13
23
15
29
36
LS
-9
-5
-10
-8
-1
3
6
13
8
46
LF
3
4
-8
-1
3
13
2
19
19
Slack
9
5
10
10
10
10
7
10
7
0.694
0.111
0.111
0.111
0.444
1.778
0.250
1.000
0.111
1.000
1.778
Haeryip Sihombing J
K
0
6
9
A
-9
6
0
5
B
-5
4
9
18
10
2
29
13
G
3
6
2
4
13
10
F
3
10
-1
2
19
15
7
13
18
37
2
15
I
8
8
11
23
23
13
10
D
d)
-1 Probability
4 3
-8
7
-1
0
10
26
7
9
E
18
37
19
29
10
H
13
6
0
10
C
-10
2
29
37
10
K
19
19
8
2
2
-8
37
18
0
J
2
16
18
Ts = 40 day
Te = 37 day
Z = (40 – 37)/ 2.28517 = 1.312813
0.90538
[by using CASIO fx 570 : mode-mode-1-shift-3-1-1.312813]
Problem Statement 30
The department of engineering at a university in New Jersey must assign six faculty members
to their new offices. The closeness matrix shown indicates the expected number of contact
per day between professors. The available office spaces (1-6) for the six faculty members are
show in figure 8.18. Assume equal-sized offices. The distance between office 1 and 2 (and
between offices 1 and 3) is 1 unit.
Professor
A
B
C
D
E
F
Closeness matrix
Contact between professor
A
B
C
D
4
12
2
-
E
F
10
7
4
-
a) Because of their academic positions, professor A must be assigned to office 1, professor C
must be assigned to office 2 and professor D must be assigned to office 6. Which faculty
47
Haeryip Sihombing members should be assigned to office 3, 4 and 5 respectively to minimize the total weighted
distance score (assuming rectilinear distance)?
1
2
3
4
5
6
b) What is the weighted distance score of your solution?
Professor
pair
AC
BD
BF
CD
CE
DF
Current plan
Number Distance
of trip (1)
(2)
4
1
12
1
10
2
2
1
7
1
4
1
A
B
C
D
E
F
wd score
(1) X (2)
4
12
20
2
7
4
Total wd = 49
Current layout
a) To minimize the total weighted distance score, I was choosing to arrange the position of
the professor. That is professor F must be assigned to office 3, professor E must be assigned
to office 4 and professor B must be assigned to office 5.
A
C
F
E
B
D
New layout
Professor
pair
AC
BD
BF
CD
Solution plan
Number of
Distance
trip (1)
(2)
4
1
12
1
10
1
2
2
48
wd score
(1) X (2)
4
12
10
4
Haeryip Sihombing CE
DF
7
4
1
2
7
8
Total wd = 45
b) The solution of the layout shows that the weighted distance score are 45. The old layouts
of the weighted distance score are 49. The differences of the weighted distance score are 4.
That is show that the new layouts are more effective and less weighted distance score.
Problem Statement 31
A. Mach 10 is a none-person sailboat designed to be used in the ocean. Manufactured by
Creative Leisure, Mach 10 can handle 40-mph winds and seas over 10 feet. The final
assembly plant is in Cupertino, California. At this time, 200 minutes are available each day to
manufacture Mach 10. The daily demand is 60 boats. Given the following information:i) Draw precedence diagram and assign task to the fewest workstations possible
ii) What is the efficiency of the assembly line?
iii) What is the theoretical minimum number of workstation?
iv) What is the idle time?
Task
A
B
C
D
E
F
G
H
I
Performance
time (min)
1
1
2
1
3
1
1
2
1
Task must follow
task listed below
A
A
C
C
C
D, E, F
B
G, H
i)
A
B
H
C
D
E
F
iii)
Desired output rate, r = 60 units /day
49
I
G
Haeryip Sihombing Time available = 200 minutes/day
r = 60 units/200 minutes
= 0.3 units/minute
Cycle time, C = 1/r = 1/0.3 = 3.33 minutes/unit
TM
iv)
ii)
= ∑t / C
= 13 / 3.33
= 3.9 ≈ 4 stations
Idle time
= nc - ∑t
= 4 (3.33) – 13
= 0.32 minutes
Percentage idle time = ((idle time / (TM x C)) x100
= ((0.32 / (4x3.33)) x100
= 2.40 %
Efficiency
= 100% - percentage idle time
= 100% - 2.40 %
= 97.6%
Problem Statement 32
A. Because of the expected high demand for Mach 10, Creative Leisure has decided to
increase manufacturing time available to produce the Mach 10:i) If demand remained the same and 300 minutes were available each day, how
many work stations would be needed?
ii) What would be the efficiency of the new system?
iii) What would be the impact or the system if 400 minutes were available?
i) Desired output rate, r = 60 units /day
Time available = 300 minutes/day
r = 60 units/300 minutes
= 0.2 units/minute
Cycle time, C = 1/r
= 1/0.2 = 5 minutes/unit
TM
= ∑t / C
= 13 / 5 = 2.6 ≈ 3 stations
ii)
Idle time
= nc - ∑t
= 3 (5) – 13 = 2 minutes
Percentage idle time = ((idle time / (TM x C)) x100
= ((2 / (3x5)) x100 = 13.33 %
Efficiency
= 100% - percentage idle time
= 100% - 13.33 % = 86.67%
iii) Desired output rate, r = 60 units /day
Time available = 400 minutes/day
r = 60 units/400 minutes
= 0.15 units/minute
50
Haeryip Sihombing Cycle time, C = 1/r
= 1/0.15 = 6.67 minutes/unit
TM
Idle time
= ∑t / C
= 13 / 6.67 = 1.95 ≈ 2 stations
= nc - ∑t
= 2 (6.67) – 13 = 0.34 minutes
Percentage idle time = ((idle time / (TM x C)) x100
= ((0.34 / (2x6.67)) x100 = 2.55 %
Efficiency
= 100% - percentage idle time
= 100% - 2.55 % = 97.45%
B. Draw the activity on node (AON) project network associated with the following activities
for Dave Carhart’s Consulting Company project. How long should it take Dave and his team
to complete this project? What are the critical path activities?
Activity
A
B
C
D
E
F
G
H
0
3
0
A
0
3
3
3
Immediate
Predecessor
A
A
B
B
C
D
E, F
7
0
B
3
4
3
0
C
3
6
Time
(days)
3
4
6
6
4
4
6
8
7
13
0
D
7
7
6
9
7
9
11
2
E
9
4
9
Critical Path = A-B-E-H
21 Days.
51
13
13
13
0
F
9
4
13
13
19
0
G
13
6
13
19
21
0
H
13
8
21
Haeryip Sihombing Problem Statement 33
Xinshang Electronics wants to establish an assembly line for producing a new product, the
Personal Little Assistant (PLA). The tasks, task times, and immediate predecessors for the
tasks are as follows:
Task
Time
(sec)
12
15
8
5
20
A
B
C
D
E
Immediate
Predecessors
A
A
B, C
D
Xinshang’s goal is to produce 180 PLAs per hour.
a) What is the cycle time
Given r = desired output rate = 180 / hour
1 hour = 180 unit
c = 1 / r = (1 / 180)*3600sec =20 sec/unit
b) What is the theoretical minimum for the number of workstations that Xinshang can
achieve in this assembly line?
TM = ∑ of all work element
Cycle time
= 12+15+8+5+20
60 sec
=
20 sec/unit
=
= 3 workstation
20 sec/unit
15
B
12
5
20
A
D
E
C
8
c) Can the theoretical minimum actually be reached when workstations are assigned?
Yes, the theoretical minimum actually can be reached when workstations are assigned.
Problem Statement 34
Jason Ritz, district manager for Gumfull Food Inc., is in charge of opening a new fast-food
outlet in the college town of Clarity. His major concern is the hiring of a manager and a cadre
52
Haeryip Sihombing of hamburger cooks, assemblers, and dispensers. He also has to coordinate the renovation of
a building that was previously owned by a pet supplies retailer. He has gathered the data
shown in Table 3.8.
Activity
Description
A
B
C
Interview at college for new manager
Renovate building
Place ad for employees and interview
applicants
Have new manager prospects visit
Purchase equipment for new outlet and
install
Check employee applicant references and
make final selection
Check reference for new manager and
make final selection
Hold orientation meetings and do payroll
paperwork
D
E
F
G
H
Immediate
Predecessor(s)
-
Time (wk)
a
m
b
2
5
7
4
8
9
6
11
17
A
B
1
2
2
4
3
12
C
4
4
4
D
1
1
1
E,F,G
2
2
2
Top management told Ritz that the new outlet is to be opened as soon as possible. Every
week that the project can be shortened will save the firm $1,200 in lease costs. Ritz thought
about how to save during the project and came up with two possibilities. One was to employ
Artic, Inc., a local employment agency to locate some good prospects for the manager’s job.
This approach would save three weeks in activity A and cost Gumfull foods $2,500. The
other was to add a few workers to shorten the time for activity B by two weeks at an
additional cost of $2,700. Help Jason Ritz by answering the following questions.
a) How long is the project expected to take?
Activity times and variances table
Activity
A
B
C
D
E
F
G
H
a
2
5
7
1
2
4
1
2
m
4
8
9
2
4
4
1
2
b
6
11
17
3
12
4
1
2
53
te
4
8
10
2
5
4
1
2
[{b-a}/6]2
0.44
1
2.78
0.11
2.78
0
0
0
Haeryip Sihombing 0
4
4
7
A
7
4
0
1
B
1
8
0
0
C
0
10
6
7
D
11
11
2
8
8
13
1
E
9
9
4
10
10
10
13
14
6
7
0
G
13
1
14
14
16
0
H
14
2
16
14
0
F
10
4
14
Critical path is= C + F + H = 10 + 4 + 2 = 16week
b) Suppose that Ritz has a personal goal of completing the project in 14 weeks. What is
the probability will happen this quickly?
a) Probability of completing the Project
TS = 14
Standard deviation for the project, √ ∑ σte2 = √ 2.78+0+0= √ 2.78
Z = (TS – TE)/ √ ∑ σte2
= -1.20
= ( 14
– 16) / √ 2.78
0.13
c) What additional expenditures should be made to reduce the projects duration? Use
the expected time for each activity as though it were certain.
i.
Additional cost
Activity
Description
A
Interview at college for new manager
B
Renovate building
Total
Cost
$2500
$2700
$5200
The new project duration of activity A is 1 week and activity B is 6 week after
reduction of project duration. This reduction duration involving additional cost that
show in table above.
ii. Lease cost saved
Lease Cost saved = 3 (1200) + 2 (1200) = $6000
iii. Cost saving = $6000 - $5200 = $800
54
Haeryip Sihombing From the calculation, there is no need additional expenditures should be made to reduce the
project duration. The cost rearrangement proved the reduction of project duration can save
company expenditures worth $800.
Problem Statement 35
Given the following information, draw the project network. Compute the early, late and slack
times for the project network. Which activities on the critical path have only the start or finish
of the activity on the critical path?
ID
Duration
Finish-to-start
Predecessor
Finish to
Start lag
A
B
C
D
E
2
4
6
8
18
F
G
H
I
J
2
5
5
14
15
None
A
A
A
B
C
D
F
None
E
G, H
0
0
0
0
0
10
0
0
0
0
0
Additional lag
relationships
Lag
None
None
Finish-finish C to F
None
Finish-finish E to G
0
0
7
0
9
None
Start-start G to H
None
Finish-finish I to J
None
10
0
5
0
a.) Which activities on the critical path have only the start or finish of the activity on the
critical path?
Based on graph of project network, critical part is from A, D, F, G,J.
2
6
2
B
4
4
18
36
10
E
8
18
8
26
Lag 10
0
2
0
A
2
2
4
2
C
13
6
2
19
10
22
D
24
8
32
50
0
I
36
14
50
Lag 5
Lag 10
8
11
36
35
Lag 9
Lag 7
10
12
0
F
32
2
35
55
40
0
H
35
5
12
40
17
0
G
35
5
40
40
55
0
J
40
15
55
Haeryip Sihombing Problem Statement 36
J.Wold, project manager of Print Software, Inc, wants you to prepare a project network,
compute the early, late, and slack activity times; determine the planned project duration; and
identify the critical path. His assistant has collected the following information for the Color
Printer Drivers Software Project:
ID
A
B
C
D
E
F
G
H
I
J
K
L
Description
External specifications
Review design features
Document new features
Write software
Program and test
Edit and publish notes
Review manual
Alpha site
Print manual
Beta site
Manufacture
Release and ship
8
10
0
B
8
2
10
Predecessor
None
A
A
A
B
C
D
E,F
G
H,I
J
K
10
Time
8
2
3
60
60
2
2
20
10
10
12
3
70
0
E
10
60
70
70
0
90
H
70 20
0
8
0
A
0
8
8
8
11
57
C
65
3
8
10
68
68
D
18 60
78
11
90
90
13
57
F
68
2
0
70
10
G
78
2
70
10
80
J
90
70
68
100
80
100
0
I
80 10
10 100
90
112
K
100 12
112
112
115
0
L
112 3
i) Project duration = 8 + 2 + 60 + 20 + 10 + 12 + 3 = 115
ii) Critical Path = A, B, E, H, J, K, L
56
115
Haeryip Sihombing Problem Statement 37
You are considering the decision of whether or not to crash your project. After asking your
operations manager to conduct an analysis, you have determined the precrash and “post
crash” activity duration and costs, hown in table below. (Assume all activities are on the
critical path)
Activity
A
B
C
D
E
F
G
Normal
Durations
Cost
6 days
$ 1000
5 days
$ 2500
3 days
$ 800
7 days
$ 3500
2 days
$ 500
5 days
$ 2000
10 days
$ 5000
Crashed
Duration
Cost
4 days
$ 2000
5 days
$ 2500
2 days
$ 1200
3 days
$ 7000
1 days
$ 5000
4 days
$ 3000
6 days
$ 6300
a.) Calculate the per day cost for crashing each activity.
Formula:
Crash Cost per Day: Normal Duration – Crash Duration
Crash Cost – Normal Cost
Activity
Activity A
Activity B
Activity C
Day
2 days
0 days
1 days
Cost
$ 1000
$ 0
$ 400
Computation cost for crashing per day
$ 1000/2 days = $ 500 per day
$ 0/0 days = $ 0 per day
$ 400/1 days = $ 400 per day
Activity D
4 days
$ 3500
$ 3500/4 days = $ 875 per day
Activity E
Activity F
1 days
1 days
$ 4500
$ 1000
$ 4500/1 days = $ 4500 per day
$ 1000/1 days = $ 1000 per day
Activity G
2 days
$ 1300
$ 1300/2 days = $ 650 per day
(b) Which are the most attractive candidates for crashing? Why
Acitivity C the most attractive candidates for crashing because the project can be crashed by
one day, and the project cost will be increasing by RM400.
Problem Statement 38
Dream Team Productions a firm hired to coordinate the release of the movie Paycheck
(starring Uma Thurman and Ben Affleck) identified 16 activities to be completed before the
release of the film.
Activity
A
B
Immediate
Predecessors
-
Optimistic Most Likely
Time
Time
1
2
3
3.5
57
Pessimistic Expected Time
Time
(O + 4M+P) /6
4
2
4
4
Haeryip Sihombing C
D
E
F
G
H
I
J
K
L
M
N
O
P
A
B
C
C
C
D
E
F,G,H
J,K,L
I,M
N
10
4
2
6
2
5
9.9
2
2
2
5
1
5
5
13
7
5
8
5.5
9
12
5
6
6
6.5
2
8
9
12
5
4
7
4
7.7
10
4
4
4
6
1.1
7
7
12
5
4
7
4
8
10
4
4
4
6
1
7
7
a.) How many weeks in advance of the film release should Dream Team have started its
marketing campaign? What is the critical path? The tasks (in time units of weeks) are
as above.
The Marketing campaign should be started as early as 24 weeks before the ‘Paycheck’ film
been release by Dream Team. The Critical Path for this project is C-J-N-P; with project
completion time is 24 weeks
Network Diagram
A
F
B
G
M
H
C
O
I
J
D
K
E
L
N
P
(b) If activities I and J were not necessary, what impact would this have on the critical
path and the number of weeks needed to complete the marketing campaign.
Critical path of the project is defined as the sequence of tasks from beginning to end that
takes the longest time. If the activities I and J were removed from the project, it would have a
greater impact on the critical path. The new critical path after the changes is D-K- N-P
and
the time taken for the project completion is 17 weeks compared to previous project
completion 24 weeks.
58
Haeryip Sihombing Problem Statement 39
Assume that the activity in problem 3.15 have the following cost A= RM 300/week, B= RM
100/week, C= RM 200/week, E= RM100/week and F= RM400/week. Assume also that you
can crash an activity down to 0 week in duration and every week you can shorten the project
in worth RM250 to you. The following is a table of activities associated with a project at Bill
Figg Enterprise, their durations and what activities each must precede:
Activity
A (start)
B
C
E
F (end)
Duration (weeks)
1
1
4
2
2
Precedes
B, C
E
F
F
-
What activity is total crashing cost.
The activity would be crash is activity A,C and F because that is critical path
Activity
Duration
A
B
C
E
F
1
1
4
2
2
10
Duration
reduce to
0
3
1
7
59
Crashing
cost
300
100
200
100
400
Total crashing
cost
300
200
400
900
Haeryip Sihombing A= 1 week
B= 1 week
C= 1 week
TOTAL
Project shorten worth
= RM 250
= RM 250
= RM 250
= RM 750
Total Crashing Cost = Project shorten worth – crashing cost
= RM 750 – RM 900 = RM – 250
Bill fig enterprise should not crashing the project because it will more costly compare to the
current project duration.
1
B
2
2
1
0
A
1
1
1
E
4
2
5
1
2
1
3
3
1
C
5
5
0
0
0
F
7
2
7
0
1
4
5
5
Problem Statement 40
Johnson cogs want to set up a line to serve 60 customer per hour. The worker element and
their precedence relationship are show in the following table
Work Element
Time (sec)
Immediate Precedes
A
B
C
D
E
F
G
H
I
J
40
30
50
40
6
25
15
20
18
30
A
A
B
B
C
C
D,E
F,G
H,I
Total = 274sec
a.) What is the theoretical minimum numbers of station?
60 customer/ hour
=1 customer/ minutes = 1 customer/ 60 sec
Cycle time = 60 sec/customer
N minimum = Total time / cycle time
= 274/60 = 4.56 ~ 5 work station
60
Haeryip Sihombing b.) How many station are required using trial and error to find a solution
Figure 1 show 6 work station require using trail and error.
c.) Suppose that a solution requiring five station is obtain. What is its efficiency
Suppose have a 5 from the theoretical minimum numbers of station but in actual
condition 5 work station cannot be archive because don’t have a solution to crate 5
work station with archive 60 seconds cycle time.
Total idle time for 6 work station is 20+30+10+20+2+4 = 86
Efficiency = Total task time/ no work station x cycle time = 76.1%
Problem Statement 41
Mach 10 is a one-person sailboat designed to be used in the ocean. Manufactured by Creative
Leisure, March 10 can handle 40 mph winds and seas over 10 feet. The final assembly plant
is in Cupertino, California. At tis time, 200 minutes are available each day to manufacture
Mach 10. The daily demands is 60 boats. Given the following information,
61
Haeryip Sihombing Task
A
B
C
D
E
F
G
H
I
Performance Time
1
1
2
1
3
1
1
2
1
Task Must follow task listed below
A
A
C
C
C
D,E,F
B
G,H
a) Draw the precedence diagram and assign task to the fewest workstation possible.
b) What is the efficiency of the assembly line?
% of idle time = [idle time per cycle time/ (Workstation X Cycle time)] X 100
= [3/ (4X4)] X 100 = 18.75%
Efficiency
= 1 - 0.19 = 0.81
c) What is the theoretical minimum number of workstations?
Cycle time= operating time/demand = 200/60 = 3.33 @4minute/boat
Work station = total of all work element/cycle time = 13/4 = 3.25 @ 4 station
d) What is the idle time?
Workstation
ABC
DHG
EF
I
IDLE TIME
0
0
0
3
62
Haeryip Sihombing Problem Statement 42
The information in Table 3.11 is available about a large project.
Activity
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Activity
time (days)
3
4
2
5
3
4
6
2
1
4
3
3
2
4
1
5
Activity
(Cost)
100
150
125
175
150
200
75
50
100
75
150
150
100
175
200
150
Immediate
Predecessor(s)
A
B
B
C,D
C
C,D,E
E
D,E
F,G
G,H,I
I,J
K,M
H,M
N,L,O
a) Determine the critical path and the expected completion of the project.
Critical path = BDFKNP
63
Haeryip Sihombing Expected completion time = 25 days
Task
Performance Time
(sec)
Predecessors
b) Plot the total project cost, starting from day 1 to the expected completion date of
the project, assuming the earlier start time for each activity. Compare that result
to similar plot for the latest start times. What implication does the time different
have for cash flow and project scheduling.
Problem Statement 41
The Toys r There Company has decided to manufacture a new toy tractor, the production of
which is broken into six steps. The demand for the tractor is 4,800 units per 40-hour
workweek:
64
Haeryip Sihombing A
B
C
D
E
F
20
30
15
15
10
30
None
A
A
A
B,C
D,E
a) Draw a precedence diagram of this operation.
B
E
30
A
C
20
15
10
D
F
15
30
b) Given the demand, what is the cycle time for this operation?
r = 4,800 hour/week
r = 4,800 = 120 units/hour
40
Cycle time = 1min/60sec = 1min/unit = 60 sec/unit
c) What is the theoretical minimum number of workstations?
= 2 stations
TM = 120 sec
60 sec
d) Assign tasks to workstations.
Station
1
2
3
Work elements assigned
A, B, E
C, D
F
A
20
B
C
10
15
D
15
65
E
30
F
30
Haeryip Sihombing e) How much total idle time is present each cycle?
Station
1
2
3
Work elements
assigned
A, B, E
C, D
F
Cumulative
Time
60
30
30
Idle time
c = 60 sec
0
30
30
Total idle time = 30 + 30 = 60 sec
f) What is the overall efficiency of the assembly line with five stations; and with six
stations?
•
5 stations: Efficiency = [ 120 sec/ 5(60) ] X 100
= 0.4 X 100 = 40%
•
6 stations: Efficiency = [ 120 sec/ 6(60) ] X 100
= 0.333 X 100 = 33.33 %
Problem Statement 42
Reliable Garage is completing production of J2000 kit car. The following data are available
for the project.
Activity
A
B
C
D
E
F
G
H
Activity Time
(days)
2
6
4
5
7
5
5
3
Immediate Predecessor(s)
a) Draw the network diagram for the project.
66
A
B
C
C
C
F
D,E,G
Haeryip Sihombing b) Determine the project’s critical path and duration.
Critical path = A, B, C, E, H
The duration was 22 days
c) What is the total slack for each activity?
The total slack for each activity = 2 + 2 + 2 + 2 + 0 + 3 + 0 + 3 = 14 days
Problem Statement 43
The head of the information systems group at Conway Consulting must assign six new
analysts to offices. The following closeness matrix shows the expected frequency of the
contact between analysts. The block plan in Figure 1 shown the available office locations (16) for six analysts (A-F). Assume equal-sized offices and rectilinear distance. Owing to their
tasks, analyst A must be assigned to location 4 and analysts D to location 3.
What are the best locations for the other four analysts?
What is the wd score for your layout?
Closeness Matrix
Contacts Between Analysts
Analyst
A
B
C
D
E
Analyst A
6
Analyst B
12
Analyst C
2
7
Analyst D
Analyst E
Analyst F
F
4
-
1(A)
2(B)
3(C)
C
F
D
4(D)
5(E)
6(F)
A
E
B
Figure 2:
New Conway Consulting’s Block Plan
Figure 1:
Conway Consulting’s Block Plan
Analyst/Activity
Number of
trips
6
12
2
7
4
Total Wd Score
Distance
Wd Score
A,C
B,D
C,D
C,E
D,F
2
2
3
2
2
12
24
6
14
8
64
67
New
Distance
1
1
2
2
1
New Wd
Score
6
12
4
14
4
40
Haeryip Sihombing The best locations for the other four analysts are analyst B be assigned to location 6, analysts
C to location 1, analyst E to location 5 and analysts F to location 2. See the Figure 2. The
total wd score for my new layout is 40, reduced total 24 point if compares with the original
lay out.
Problem Statement 43
The activities described by the following table are given for the Standard Task Corporation:
Activity
A
B
C
D
E
F
G
H
I
Immediate Predecessor (s)
A
A
B
B
C
E,F
D
G,H
Total Time =
Time
9
7
3
6
9
4
6
5
3
52
1. Draw the appropriate AON PERT diagram for Standard Corporation’s management
team
D
H
B
E
A
I
G
C
F
2. Find the critical path
i) A,B,D,H,and I
ii) A,B,E,G,and I
iii) A,C,F,G,and I
3. What is the project completion time?
i.) A,B,D,H,and I, total completion time are 9+7+6+5+3= 30sec
ii.) A,B,E,G,and I, total completion time are 9+7+9+6+3= 34sec
iii.) A,C,F,G,and I, total completion time are 9+3+4+6+3= 25sec
68