a.)
NOTE: Enter values from the problem into the yellow highlighted columns to find the solution:
A 3 None 0 3 0 3 YES
B 2 None 0 2 4.5 6.5
b.)
Paths: Length:
A-C-E-G 15.5
A-D-E-G 11
c.)
d.)
No, crashing activity C from 6 days down to 2 days will only reduce the total project to 12 days (the goal is less
The critical activities are A-C-E-G as they are the ones that make up the critical path. The entire project will
ACTIVITY
DURATION (DAYS)
ES
PREDECESSORS
EF
LS
LF
CRITICAL
PATH?
NOTE: Enter values from the problem into the yellow highlighted columns to find the solution:
A 3 None 0 3 0 3 YES
B 2 None 0 2 1 3
C
2A 3 5 3.5 5.5
ACTIVITY
DURATION (DAYS)
ES
PREDECESSORS
EF
LS
LF
CRITICAL
PATH?
a.)
NOTE: Enter values from the problem into the yellow highlighted columns to find the solution:
A 1 None 0 1 1 2
B 2 None 0 2 0 2 YES
b.)
Paths: Length:
Yes, activity F is on all paths.
A-D-F-G 9.5
A-D-F-H 7.5
B-D-F-G 10.5
c.)
d.)
No, she should not crash activity E by 1 day as it is not on the critical path; thus the total project time will not be
reduced by shortening activity E.
The critical activities are B-D-F-G as they are the activities on the critical path. The entire project will take 10.5
ACTIVITY
DURATION (DAYS)
ES
PREDECESSORS
EF
LS
LF
CRITICAL
PATH?
NOTE: Enter values from the problem into the yellow highlighted columns to find the solution:
A 1 None 0 1 1 2
B 2 None 0 2 0 2 YES
C1.5 None 0 1.5 12.5
Paths: Length:
A-D-F-G 9.5
A-D-F-H 7.5
ACTIVITY
DURATION (DAYS)
PREDECESSORS
ES
EF
LS
LF
CRITICAL
PATH?
a.)
NOTE: Enter values from the problem into the yellow highlighted columns to find the solution:
A 4 None 0 4 0 4 YES
B 3 A 4 7 14 17
C 7 A 4 11 411 YES
D 9 A 4 13 12 21
I 4 D E 13 17 21 25
J12 E12 24 13 25
b.)
Paths: Length:
A-B-F-L-N 24
c.)
d.)
e.)
NOTE: Enter values from the problem into the yellow highlighted columns and use solver (or try values in column N) to find the solution:
DURATION (DAYS)
Before
After crash
A 4 4 None 0 4 0 4 YES 0$0
B 3 3 A 4 7 11 14 0$0
C 7 4 A 4 8 4 8 YES 3 $1,000 3 $3,000
D 9 9 A 4 13 918 3 $2,500 0$0
Crash
Cost
LF
CRITICAL
PATH?
Max Days
Crashabl
Crash
Cost per
Days to
Crash
ACTIVITY
PREDECESSORS
ES
EF
LS
The critical activities are A-C-H-L-N as they make up the critical path. The entire project will take 34 days to
complete.
ACTIVITY
DURATION (DAYS)
ES
PREDECESSORS
EF
LS
LF
CRITICAL
PATH?
E, 8
D, 9
C, 7B, 3
A, 4
N, 5
M, 4
L, 5
J12 10 E12 22 12 22 YES 2 $2,000 2 $4,000
K 1 1 E 12 13 21 22 0$0
E, 8
D, 9
C, 4B, 3
a.)
NOTE: Enter values from the problem into the yellow highlighted columns to find the solution:
A 4 None 0 4 8 12
B 5 None 0 5 0 5 YES
b.)
Paths: Length:
A-D-F 13
d.)
ACTIVITY SLACK (WEEKS)
A 8
B 0
ACTIVITY
DURATION
(WEEKS)
ES
PREDECESSORS
EF
LS
LF
CRITICAL
PATH?
e.)
f.)
NOTE: Enter values from the problem into the yellow highlighted columns to find the solution:
A 4 None 0 4 1 5
B 5 None 0 5 0 5 YES
C 7 A B 5 12 512 YES
LS
LF
CRITICAL
PATH?
Path B-C-D-G-H is the only path greater than 25 weeks, so reducing one or a combination of these
ACTIVITY
DURATION
(WEEKS)
PREDECESSORS
ES
EF
the slack time available for A but did not put it on the critical path. Adding activity C as a predecessor to G had
a.)
NOTE: Enter values from the problem into the yellow highlighted columns to find the solution:
A 8 None 0 8 1 9
B 6 None 0 6 0 6 YES
Paths: Length:
A-E 14
B-E 12
b.)
NOTE: Enter values from the problem into the yellow highlighted columns and use solver (or try values in column N) to find the solution:
DURATION (WEEKS)
Before
After crash
The cheapest way to crash the project by 2 weeks is to crash activity A by 1 week and D by 2 weeks. This
The shortest time in which the project can be completed is 9 weeks. This requires maximum crashing for all
A 8 7 None 0 7 0 7 YES 3 $2,000 1 $2,000
B 6 6 None 0 6 0 6 YES 1 $4,000 $0
C 3 3 None 0 3 5 8 1 $1,000 $0
$0
$0
$0
$0
$0
$0
$0
$0
c.) Original Weeks to Complete Project: 15
Crash
Cost
LF
CRITICAL
PATH?
Weeks
Crashabl
Crash
Cost per
Weeks to
Crash
ACTIVITY
PREDECESSORS
ES
EF
LS
The critical activities are B-D-F as they are the activities on the critical path. The entire project will take 15
weeks.
Activities A, B, and C each start at least one path. Activities E and F each end at least one path. Having
ACTIVITY
DURATION
(WEEKS)
ES
PREDECESSORS
EF
LS
LF
CRITICAL
PATH?
E, 6
A, 8
a.)
NOTE: Enter values from the problem into the yellow highlighted columns and use solver (or try values in column N) to find the solution:
DURATION (WEEKS)
Before After crash
A 3.5 3.5 None 0 3.5 0 3.5 YES $0
B 2 2 None 0 2 1.5 3.5 $0
b.) Cost Per Week of Project: $3,000
Original Weeks to Complete Project: 15
Yes, you should crash the project. If you crash all three activities for a total cost of $3,550 you will save
EF
LS
LF
Weeks
Crashable
Crash Cost
per Week
Weeks to
Crash
Crash Cost
CRITICAL
PATH?
ACTIVITY
ES
PREDECESSORS
G, 2F, 3
D, 2
A, 3.5
NOTE: Enter values from the problem into the yellow highlighted columns and use solver (or try values in column N) to find the solution:
DURATION (WEEKS)
ORIGINAL NEW
A 2 2 None 0 2 1 3 $0
B 3 3 None 0 3 0 3 YES $0
F 3 3 C 6 9 10 13 $0
Crash Cost
CRITICAL
PATH?
ACTIVITY
ES
PREDECESSORS
EF
LS
LF
Weeks
Crashable
Crash Cost
per Week
Weeks to
Crash
Question 1
Activity Description
Duration
(Weeks)
A 1
B 3
C 1
Question 2
NOTE: Enter values from the problem into the yellow highlighted columns to find the solution:
A 1 None 0 1 0 1 YES
B 3 A 1 4 1 4 YES
C 1 B 4 5 6 7
Question 3
for 10 years.
Robert may want to increase his time estimates to provide extra buffer, especially as this is his first time
ACTIVITY
DURATION
(WEEKS)
ES
PREDECESSORS
EF
LS
LF
CRITICAL
PATH?
Important time milestones for this project include the earliest possible start date (September 2020), the project
deadline (March 31, 2021), registration deadline (May 15, 2021), as well as the activities listed below.
Negotiate dates and estimate costs
Develop daily schedule of trip
Make air transportation arrangements
C, 1
Make local transportation arrangements
Select accommodations
Finalize loose ends
Develop and post online information packet