C723 - FORMULAS AND SAMPLES
test=A+(4*M)+P÷6
What does each letter stand for? - Answer -A is the optimistic time
MM is the most likely time
P is the pessimistic time
To determine the latest start, subtract the activity duration from the latest finish. -
Answer -LS = Duration - LF
To find the early finish, add the early start to the duration of each task. - Answer -EF =
ES + Duration of each task
Slack time (SL) is calculated using the following formula: - Answer -SL=LS−ES
How is the latest start time calculated? - Answer -Subtract the task duration from the
latest finish.
To determine the early finish, add the task duration to the early start. - Answer -EF =
Task duration + ES
________ = Cost of Items + Total Holding Cost + Total Ordering Cost - Answer -
Inventory Cost
___________ = (Crash Cost - Normal Cost) ÷ (Normal Duration - Crash Duration) -
Answer -Cost Slope
Based on Activity A as described below, what would be the cost to crash by 2 days be?
Normal days = 12
Normal cost = $18,000
Crash days = 8
Crash cost = $22,000
Maximum time reduction = 4 - Answer -(22,000 - 18,000) ÷ (8 - 12)
4,000 ÷ 4 = 1,000
1,000 x 2 = 2,000
The answer is $2,000
Calculate duration estimates for the following situations:
A construction manager received an estimate from a sub-contracted plumber for
completion of the 2.5 bathrooms. The plumber thought it could be done in as few as 20
, labor hours, although this type of design usually takes 28 hours. The plumber does
remember that the last time, it took longer than usual and the labor hours climbed to 36.
Using the beta distribution method, how many hours should the construction manager
use as an estimate for the plumbing? - Answer -The formula is:
Te = (o + 4m + p) / 6
28 hours
What would be the additional cost to crash Activity A by the maximum time reduction
period?
Normal days = 12
Normal cost = $18,000
Crash days = 8
Crash cost = $22,000
Maximum time reduction = 4 - Answer -(22,000 - 18,000) ÷ (12 - 8)
4,000 ÷ 4 = 1,000
1,000 x 4 = 4,000
The answer is $4,000
Based on the graphic below, what would be the additional cost to crash Activity C by the
maximum time reduction period?
Normal days = 20
Normal cost = $32,000
Crash days = 12
Crash cost = $38,000
Maximum time reduction = 8 - Answer -(38,000 - 32,000) ÷ (20-12)
6,000 ÷ 8 = 750
750 x 8 = 6,000
The answer is $6,000
Probability x Value = what? - Answer -The expected value of the alternative
To calculate, determine the expected value for each alternative. First, you would
multiply the value of the first condition times the probability of that condition occurring.
Next, repeat the calculation for each condition. Finally, add all of the results of the newly
calculated values together to establish the expected value of the alternative. - Answer -
= (B1 x 60%) + (C1 x 40%)
Large Plant in a favorable market of 75% would cost $650 and an Unfavorable market
of 25% it would cost $-100.
test=A+(4*M)+P÷6
What does each letter stand for? - Answer -A is the optimistic time
MM is the most likely time
P is the pessimistic time
To determine the latest start, subtract the activity duration from the latest finish. -
Answer -LS = Duration - LF
To find the early finish, add the early start to the duration of each task. - Answer -EF =
ES + Duration of each task
Slack time (SL) is calculated using the following formula: - Answer -SL=LS−ES
How is the latest start time calculated? - Answer -Subtract the task duration from the
latest finish.
To determine the early finish, add the task duration to the early start. - Answer -EF =
Task duration + ES
________ = Cost of Items + Total Holding Cost + Total Ordering Cost - Answer -
Inventory Cost
___________ = (Crash Cost - Normal Cost) ÷ (Normal Duration - Crash Duration) -
Answer -Cost Slope
Based on Activity A as described below, what would be the cost to crash by 2 days be?
Normal days = 12
Normal cost = $18,000
Crash days = 8
Crash cost = $22,000
Maximum time reduction = 4 - Answer -(22,000 - 18,000) ÷ (8 - 12)
4,000 ÷ 4 = 1,000
1,000 x 2 = 2,000
The answer is $2,000
Calculate duration estimates for the following situations:
A construction manager received an estimate from a sub-contracted plumber for
completion of the 2.5 bathrooms. The plumber thought it could be done in as few as 20
, labor hours, although this type of design usually takes 28 hours. The plumber does
remember that the last time, it took longer than usual and the labor hours climbed to 36.
Using the beta distribution method, how many hours should the construction manager
use as an estimate for the plumbing? - Answer -The formula is:
Te = (o + 4m + p) / 6
28 hours
What would be the additional cost to crash Activity A by the maximum time reduction
period?
Normal days = 12
Normal cost = $18,000
Crash days = 8
Crash cost = $22,000
Maximum time reduction = 4 - Answer -(22,000 - 18,000) ÷ (12 - 8)
4,000 ÷ 4 = 1,000
1,000 x 4 = 4,000
The answer is $4,000
Based on the graphic below, what would be the additional cost to crash Activity C by the
maximum time reduction period?
Normal days = 20
Normal cost = $32,000
Crash days = 12
Crash cost = $38,000
Maximum time reduction = 8 - Answer -(38,000 - 32,000) ÷ (20-12)
6,000 ÷ 8 = 750
750 x 8 = 6,000
The answer is $6,000
Probability x Value = what? - Answer -The expected value of the alternative
To calculate, determine the expected value for each alternative. First, you would
multiply the value of the first condition times the probability of that condition occurring.
Next, repeat the calculation for each condition. Finally, add all of the results of the newly
calculated values together to establish the expected value of the alternative. - Answer -
= (B1 x 60%) + (C1 x 40%)
Large Plant in a favorable market of 75% would cost $650 and an Unfavorable market
of 25% it would cost $-100.
