topic 2: project management models

UCF College of BusinessFlexible MBA Program

ISM 6407.0001Decision Support Systems

Spring 2007

Supplementary Notes

Topic 2Project Management

ISM 6407.0001 Spring 2007Topic 2: Project Management Page 1/26

PROJECT SCHEDULING – THE EARLY DAYS

Project Data:

ActivityImmediatePredecessor

Time(weeks)

A - 1B A 2C A 4D B 5E C 7F D,E 1

Gantt Chart for tracking the progress of the project activities

end ofwk 0 1 2 3 4 5 6 7 8 9 10 11 12 13

wk 1 2 3 4 5 6 7 8 9 10 11 12 13

A A

B B

C C

D D

E E

F F


PROJECT SCHEDULING DEMONSTRATION PROBLEM

Data for initial view. (Deterministic time estimates; cost budgets)

ActivityImmediate

PredecessorTime

(weeks)Budgeted Cost

(x $1000)Budgeted Cost/Wk

(x $1000)A - 1 8 8B A 2 12 6C A 4 24 6D B 5 25 5E C 7 21 3F D,E 1 5 5

A-O-A diagram

A-O-N diagram

Results of start, finish, and slack time calculations

Activity ES EF LS LFTotalSlack

FreeSlack

A 0 1 0 1 0 0B 1 3 5 7 4 0C 1 5 1 5 0 0D 3 8 7 12 4 4E 5 12 5 12 0 0F 12 13 12 13 0 0

F,1

E,7

D,5

C,4

B,2

A,11 2

4

3

5 6

A,1

E,7C,4

B,2

F,1

0 10 10 0

1 51 50 0

5 125 120 0

12 1312 130 0

D,5

3 87 124 4

1 35 74 0


PROJECT MANAGEMENT:SUMMARY NOTES ON START, FINISH, AND SLACK TIME CALCULATIONS

Forward Pass

When making a forward pass to find earliest start times and earliest finish times, use caution when an activity has more than one immediate predecessor. The earliest start time for this activity will be the largest earliest finish time of all the activity’s immediate predecessors.

Backward Pass

When making a backward pass to find latest start times and latest finish times, use caution when an activity has more than one immediate successor. The latest finish time for this activity will be the smallest latest start time of all the activity’s immediate successors.

Slack

Our textbook uses the simple term “slack.” We will be referring to two types of slack, “total slack” and “free slack.” In our textbook, any references to the term “slack” will be identical to what we will call “total slack.”

Total Slack:

Total slack is a measure of how much we can delay the start of an activity and still not delay the completion of the total project. An activity’s total slack is calculated as the difference between its latest start time and its earliest start time, or

TS = LS – ES

Free Slack:

We also described another measure of slack, one that we will call “free slack.” This concept is not mentioned in our textbook, however, since it is sometimes used in the course of project management, one should be familiar with the notion that there is more than one way to measure the amount of “cushion” that exists in a project’s schedule.

Free slack is a measure of how much we can delay the completion of an activity and still not delay the start of any of that activity’s immediate successors. An activity’s free slack is calculated as the difference between its earliest finish time and the smallest of the earliest start time for its immediate successors.


GANTT CHARTS FOR PROJECT MANAGEMENT DEMONSTRATION PROBLEM

Gantt Chart for Demo Problem Using Earliest Start and Finish Times

end ofwk 0 1 2 3 4 5 6 7 8 9 10 11 12 13

wk 1 2 3 4 5 6 7 8 9 10 11 12 13

A 0 A 1

B 1 B 3

C 1 C 5

D 3 D 8

E 5 E 12

F 12 F 13

Gantt Chart for Demo Problem Using Latest Start and Finish Times

end ofwk 0 1 2 3 4 5 6 7 8 9 10 11 12 13

wk 1 2 3 4 5 6 7 8 9 10 11 12 13

A 0 A 1

B 5 B 7

C 1 C 5

D 7 D 12

E 5 E 12

F 12 F 13


Budgeted Cost Using Earliest Start TimesEnd of

week # 0 1 2 3 4 5 6 7 8 9 10 11 12 13week # 1 2 3 4 5 6 7 8 9 10 11 12 13 Total

A 8 8B 6 6 12C 6 6 6 6 24D 5 5 5 5 5 25E 3 3 3 3 3 3 3 21F 5 5

Tot/wk 8 12 12 11 11 8 8 8 3 3 3 3 5 95Cum. Tot. 8 20 32 43 54 62 70 78 81 84 87 90 95

Budgeted Cost Using Latest Start Times End of

week # 0 1 2 3 4 5 6 7 8 9 10 11 12 13week # 1 2 3 4 5 6 7 8 9 10 11 12 13 Total

A 8 8B 6 6 12C 6 6 6 6 24D 5 5 5 5 5 25E 3 3 3 3 3 3 3 21F 5 5

Tot/wk 8 6 6 6 6 9 9 8 8 8 8 8 5 95Cum. Tot. 8 14 20 26 32 41 50 58 66 74 82 90 95


MONITORING AND CONTROLLING BUDGET ASPECTS OFPROJECT MANAGEMENT DEMONSTRATION PROBLEM

Assume that we are at the end of week 6 and the following activity has transpired:Activity A is 100% complete, and we have spent $7,000 on itActivity B is 50% complete, and we have spent $8,000 on itActivity C is 100% complete, and we have spent $22,000 on itActivity D has not yet been started, and we have spent nothing on it Activity E has not yet been started, and we have spent nothing on itActivity F has not yet been started, and we have spent nothing on it

The following table would summarize our budgetary activity:

ActivityTotal

Budget%

CompletedValue of

Work DoneActualCost

ActivityDifference

Type ofDifference

A $8,000 100% $8,000 $7,000 -$1,000 Under budgetB $12,000 50% $6,000 $8,000 $2,000 Over budgetC $24,000 100% $24,000 $22,000 -$2,000 Under budgetD $25,000 0% 0 0 0 On budgetE $21,000 0% 0 0 0 On budgetF $5,000 0% 0 0 0 On budget

For the project as a whole $38,000 $37,000 -$1,000 Under budget

Note that we are over budget on activity B, under budget on activities A and C, and right on budget for the remaining activities (D, E, and F). Overall for the project, thus far we are $1,000 under budget.

That sounds good, but close inspection reveals that we are behind schedule and not likely to complete by the target time of 13. Even using latest start and finish times, note the following (remembering that we are at the end of week 6 in the project):

A is O.K. (It is finished and should be done by now.)B is O.K. (It is half finished, and that is consistent with the LS and LF times for it.)C is O.K. (It is finished and should be done by now.)D is O.K (It has not yet been started, but its LS reveals that it could be started as late as the end of week 7.)E will cause a problem (we should have been 1 week into E, but at this point in time we

haven’t even started it. This is a critical path activity, so unless we can make up some time on this, we will not complete the project by the end of week 13.)

This type of monitoring would cause some fancy footwork among project managers. They might resort to shifting company resources to E in an effort to accelerate E and get the project back on track. This whole issue of shifting resources and accelerating activities will be addressed in later class meetings devoted to project management.


DETERMINISTIC (NO UNCERTAINTY) PERT/COST--SIMPLE EXAMPLE OF ACCELERATING ACTIVITIES

Normal Crash


Time(weeks)

Cost(x $1000)

Time(weeks)

Cost(x $1000)

Acceleration Cost (per week)

A - 1 8 1 8 -B A 2 12 1 16 4C A 4 24 1 27 1D B 5 25 4 27 2E C 7 21 3 33 3F D,E 1 5 1 5 -

Path Time (using normal times)ABDF 9ACEF 13

Time(weeks) Activities Accelerated

Total Cost (x$1000)

13None (everything done on a normal basis)

95

12 C (1st week) 95+1=96

11 C (2nd week) 96+1=97

10 C (3rd week) 97+1=98

9 E (1st week) 98+3=101

8 E (2nd week) & D 101+3+2=106

7 E (3rd week) & B 106+3+4=113

6 Can’t be done!


DETERMINISTIC (NO UNCERTAINTY) PERT/COST--SOME ADDITIONAL PROBLEM SCENARIOS

Embellishments to problem on page 6, Module 2.

Normal Crash


Time(weeks)

Cost(x $1000)

Time(weeks)

Cost(x $1000)


A - 1 8 1 8 -B A 2 12 1 16 4C A 4 24 1 27 1D B 5 25 4 27 2E C 7 21 3 33 3F D,E 1 5 1 5 -


Embellishment #1. Suppose the client offers to give us a $2,000 bonus for each week we come in under 13 weeks. We can see with incremental analysis on the total cost column that it is worth spending an extra $1000 to cut the project to 12 weeks (we get a bonus of $2000). It is worth spending another $1000 to cut the project to 11 weeks (we get another bonus of $2000). The same holds true for cutting the project to 10 weeks. However, if we want to cut the project to 9 weeks we have to spend another $3000. This is not worth it, for it will only return a bonus of another $2000. We would come to the same conclusion if we placed another column on the table and tallied total cost minus bonus ($ outflow - $ inflow) and then picked the time with the lowest total.


Total Cost(x$1000)

Total Cost – Bonus(x$1000)


95 95

12 C (1st week) 95+1=96 96-2=94

11 C (2nd week) 96+1=97 97-4=93

10 C (3rd week) 97+1=98 98-6=92 (the least costly option)

9 E (1st week) 98+3=101 101-8=93

8 E (2nd week) & D 101+3+2=106 106-10=96

7 E (3rd week) & B 106+3+4=113 113-12=101

6 Can’t be done!


Embellishment #2. Ignore the previous scenario. Suppose the deal we strike with the client calls for a penalty of $2000 for each week the project is delayed beyond week 8 (apparently the client would really like delivery of the finished product by that time). We could do an incremental analysis and decide that it is wise to shorten from 13 to 12 (spend $1000, avoid $2000 penalty), wise to shorten from 12 to 11 (spend another $1000, avoid another $2000 penalty), wise to shorten from 11 to 10 (spend another $1000, avoid another $2000 penalty), but unwise to shorten from 10 to 9 (spend another $3000, avoid another $2000 penalty). We would come to the same conclusion if we placed another column on the table and tallied the total cost plus the penalty cost, giving us a slightly different look to our table from page 16 of the course pack.


Total Cost(x$1000)

Total Cost + Penalty Cost(x$1000)


95 95+10=105

12 C (1st week) 95+1=96 96+8=104

11 C (2nd week) 96+1=97 97+6=103

10 C (3rd week) 97+1=98 98+4+102 (the least costly option)

9 E (1st week) 98+3=101 101+2=103

8 E (2nd week) & D 101+3+2=106 106

7 E (3rd week) & B 106+3+4=113 113

6 Can’t be done!


Nonlinear tradeoff embellishment to the problem on page 6, Module 2.

Normal Crash


Time(weeks)

Cost(x $1000)

Time(weeks)

Max.Accel.

Acceleration Cost per week (x$1000)

A - 1 8 1 0 -B A 2 12 1 1 4C A 4 24 1 3 1, 2, 7D B 5 25 4 1 2E C 7 21 3 4 3, 3, 4, 8F D,E 1 5 1 0 -


The following table shows the full range of possibilities for project time and project cost. We could embellish this nonlinear situation with bonuses or penalties much like we did the linear case above.


Total Cost(x$1000)


95

12 C (1st week) 95+1=96

11 C (2nd week) 96+2=98

10 E (1st week) 98+3=101

9 E (2nd week) 101+3=104

8 E (3rd week) & D 104+4+2=110

7 C (3rd week) & B 110+7+4=121

6 Can’t be done!


ANOTHER EXAMPLE OF PERT/COST AND ACTIVITY ACCELERATION

Normal Crash


Time(weeks) Cost

Time(weeks)

Max.Accel.

Acceleration Cost per week

A - 4 $4,000 2 2 $100, $350B A 3 $3,000 2 1 $200C A 4 $3,000 2 2 $200, $200D B, C 6 $6,000 4 2 $1200, $1600

Path TimeABD 13ACD 14


Total Cost (x$1000)


$16,000

13 A (1st) $16,000 + $100 = $16,100

12 C (1st) $16,100 + $200 = $16,300

11 A (2nd) $16,300 + $350 = $16,650

10B (1st)C (2nd)

$16,650 + $200 + $200 = $17,050

9 D (1st) $17,050 + $1,200 = $18,250

8 D (2nd) $18,250 + $1,600 = $19,850

7 Can’t be done! -----------

A,4

C,4

B,3

D,6


PROJECT SCHEDULING WITH UNCERTAINTY IN TIME ESTIMATES

Data for initial view. (Probabilistic time estimates)


Optimistic time

a

Most likely timem

Pessimistic time

b

Expected time

te

Variance2

A - .5 1 1.5 1 1/36B A 1 2 3 2 4/36C A 3 4 5 4 4/36D B 3 5 7 5 16/36E C 6 7 14 8 64/36F D,E .5 1 1.5 1 1/36

te =(a + 4m + b) 2 = ( b-a )

2

6 6

Results of start, finish, and slack time calculations using mean time


FreeSlack

A 0 1 0 1 0 0B 1 3 6 8 5 0C 1 5 1 5 0 0D 3 8 8 13 5 5E 5 13 5 13 0 0F 13 14 13 14 0 0

Path Expected time, Te Variance, 2 Standard Deviation, ABDF 9ACEF 14 70/36 1.3944

or 1.9444

Critical path (ACEF) has an expected time of 14 and standard deviation of 1.3944


SOME ILLUSTRATIONS OF PROBABILITY CALCULATIONS WITH OURDEMONSTRATION PROJECT MANAGEMENT PROBLEM

Recall that the demonstration problem had a critical path with a mean time, t, of 14 weeks and a standard deviation, , of 1.394

1. What is the probability that the project can be completed within 16 weeks (i.e., 16 weeks or less)?

z = (16-14)/1.394 = 1.43Probability to the left of this z from the normal table = .92364


z = (18-14)/1.394 = 2.87Probability to the left of this z from the normal table = .99795


z = (13-14)/1.394 = -.72There are no negative z values in the table, but we can make use of the symmetry in the normal curve.Probability to the left of z = .72 from the normal table = .76424Probability in the tail to the right of z = .72 is 1-.76424 = .23576This tail to the right of z = .72 is exactly the same size as the tail to the left of z = -.72Therefore, the probability of completing the project with in 13 weeks is .23576

4. What is the probability that the project will be completed in somewhere between 16 weeks and 18 weeks?

This probability will be the area between a z value of 1.43 (the z for 16 weeks from above) and a z value of 2.87 (the z for 18 weeks from above).This probability is .99795-.92364 = .07431

5. What is the probability of being able to start activity F by week 14 of the project?Before you can start F you must finish all the activities on the network that lead up to F. You can treat those activities as a mini-project with two paths (ABD with a mean time of 8 and ACE with a mean time of 13). The critical path here is ACE, and it has a variance of 1.91667 (i.e., 69/36). The standard deviation would be the square root of this, or standard deviation = 1.384z = (14-13)/1.384 = .72Probability to the left of this z value from the normal table = .76424


A TWIST ON PROBABILISTIC PERT

Return to our demonstration problem, which had a mean time of 14 and a standard deviation of 1.394 for the completion of the project. Suppose that, as a project manager, you are wondering what promise completion date would have about an 80% chance of being achieved. To solve this you would have to find the z value that corresponds to an 80% probability (which means you use the z table in a reverse mode).

Search around in the z table to find a probability of .8000 (or one pretty close to that value). If you go into the table, you can find a probability of .79955 (that’s pretty close to .8000, so we’ll use that rather than interpolating between values in the table). The z value that corresponds to this probability is .84

The only unknown in the z calculation is the promise date (or due date, as the book calls it):

Z = (due date – mean)/standard deviation

or

.84 = (due date – 14)/1.394

solve this and you find that due date = 14 + (.84)(1.394) = 15.17 weeks (which is a little over 15 weeks and 1 day)


AND YET ANOTHER TWIST ON PROBABILISTIC PERT

Return to our demonstration problem, which had a mean time of 14 and a standard deviation of 1.394 for the completion of the project. Suppose that the potential customer has a strong desire to have this project completed by week 13. We have already seen (page 12 of this module) that there is only a .23576 probability (i.e., a 23.576% chance) that we will be finished by week 13. We have decided that we would like to promise that the project will be done by week 13, but we would like to improve the odds that we can accomplish this to about 98.5%. In order to increase the odds, we are going to have to shorten the project mean time by accelerating some activities.

This should be apparent by looking at the formula for the Z calculation:

Z = (due date – project mean time)/

In this formula we know the Z value (Z = 2.17, which is the Z value that would correspond to a probability of .9850), we know the due date (this is the 13 weeks that the customer wants), we know the standard deviation ( = 1.394). What we don’t know is what the project mean time should be reduced to in order to satisfy the equation. This is a simple calculation, as follows:

2.17 = (13 – project mean time)/1.394

(2.17)(1.394) = 13 – project mean time

project mean time = 13 – (2.17)(1.394) = 13 – 3.02 = 9.98 weeks (let’s call it 10 weeks)

So, the project mean time (which was 14 weeks if we did no acceleration) will have to be reduced to 10 weeks. This means we will have to accelerate the project a total of 4 weeks. To do that, we do as we did with our earlier acceleration exercises, i.e., shorten the critical path(s) down to 10 weeks.

One question that often comes up in this context is, “Won’t the process of accelerating activities change the standard deviation in addition to the mean?” The answer is NO. Accelerating activities will lower the mean time for the critical path, but it will have no impact on the standard deviation. Here is the reason: The three time estimates we made for each activity were based on the premise that we would work on these activities at a normal pace (9-5 Monday through Friday, no overtime, no extra shifts, no extra equipment, etc.). If we choose to work at a faster pace (using some of these options), we still are not sure how long each activity will take, so we will still have three time estimates; they will just be lower across the board. So, the impact of accelerating an activity will be to lower its three time estimates across the board. This will certainly lower the mean times for accelerated activities, but their variances will remain the same due to the computational formula (the difference between the pessimistic and optimistic times is the same before and after the acceleration, so the variance calculation will yield the same result).


PROJECT MANAGEMENT – REVIEW PROBLEMS USING SAME NETWORK STRUCTURE

Problem 1: Deterministic Mode (No Uncertainty in the Time Estimates)

Data gathered in preliminary steps are in bold face. Nonbold figures were calculated using these initial estimates.

ActivityImmediate

PredecessorTime

(weeks)Budgeted Cost

(x $1000)Budgeted Cost/Wk

(x $1000)A - 2 4 2B A 3 12 4C A 2 6 3D A 2 8 4E B 1 5 5F C 3 9 3G D 2 10 5H E, F, G 2 4 2

Resulting project network is drawn on next page

Results of start, finish, and slack time calculations


FreeSlack

A 0 2 0 2 0 0B 2 5 3 6 1 0C 2 4 2 4 0 0D 2 4 3 5 1 0E 5 6 6 7 1 1F 4 7 4 7 0 0G 4 6 5 7 1 1H 7 9 7 9 0 -

Budgeted Cost Using Earliest Start and Finish TimesEnd of

week # 0 1 2 3 4 5 6 7 8 9 10 11 12 13week # 1 2 3 4 5 6 7 8 9 10 11 12 13 Total

A 2 2 4B 4 4 4 12C 3 3 6D 4 4 8E 5 5F 3 3 3 9G 5 5 10H 2 2 4

Tot/wk 2 2 11 11 12 13 3 2 2 58Cum. Tot. 2 4 15 26 38 51 54 56 58


Budgeted Cost Using Latest Start and Finish TimesEnd of

week # 0 1 2 3 4 5 6 7 8 9 10 11 12 13week # 1 2 3 4 5 6 7 8 9 10 11 12 13 Total

A 2 2 4B 4 4 4 12C 3 3 6D 4 4 8E 5 5F 3 3 3 9G 5 5 10H 2 2 4

Tot/wk 2 2 3 11 11 12 13 2 2 58Cum. Tot. 2 4 7 18 29 41 54 56 58

A-O-A diagram B E

A C F H

D G

A-O-N diagram

Path TimeABEH 8ACFH 9 * Critical PathADGH 8

GD

A C F

B E

H


Problem 2: Introduce Data for Time/Cost Tradeoffs (Still Deterministic Mode)


Normal Crash


Time(weeks)

Cost(x $1000)

Time(weeks)

Cost(x $1000)


A - 2 4 2 4 -B A 3 12 2 18 6C A 2 6 1 8 2D A 2 8 2 8 -E B 1 5 1 5 -F C 3 9 2 13 4G D 2 10 1 15 5H E, F, G 2 4 1 16 12

Path TimeABEH 8ACFH 9 * Critical PathADGH 8

Find the activities to accelerate and the project cost if we were to try to get the project done in 9 weeks, 8 weeks, and 7 weeks.


Total Cost (x$1000)


58

8

Must reduce path ACFH

Choose C (1st and only week of acceleration)

58 + 2 = 60

7

Must reduce paths ABEH, ACFH, and ADGH

Choose H (1st and only week of acceleration)

60 + 12 = 72


Problem 3: No Time/Cost Tradeoffs, but We Have Probabilistic (Stochastic) Mode (There is Uncertainty in the Time Estimates)



Optimistic time

a

Most likely timem

Pessimistic time

b

Expected time

te

Variance2

A - 1 2 3 2 4/36B A 2 3 4 3 4/36C A 1 2 6 2.5 25/36D A 1 2 3 2 4/36E B .5 1 1.5 1 1/36F C 1 3 8 3.5 49/36G D 1 2 3 2 4/36H E, F, G 1 2 3 2 4/36

TimePath (using expected times)ABEH 8ACFH 10* Critical Path 2 = 2.28 = 1.51ADGH 8

Find:a. Probability of completing the project within 12 weeks.

Z = (12-10)/1.51 = 1.32 Probability = .9066

b. Probability that it will take more than 11 weeks to complete the project.

Z = (11-10)/1.51 = .66 Probability = .2546 (note that we want the area in the right tail)

c. Probability that the project will be completed in 7 weeks or less.

Z = (7-10)/1.51 = -1.99 Probability = .0233 (we must use symmetry of table to get this)


SOLUTIONS TO SOME TEXTBOOK PROJECT MANAGEMENT PROBLEMS

13-12 & 13-13 Critical path is BDEG; Time is 26.

Network diagram using activity on node convention

Network diagram using activity on arrow convention

13-14 & 13-15 Critical path is BDEG; Time is 17


A

B

C

D

E

F

G

2

1

4

3 5 6

B D

CF

EG

A

A

B

C

D

E

F

G



13-16 & 13-17 Critical paths are ACG and BEG; Time is 19



1 654

3

2A C

B D

E G

F

A

B

F

E

D

C

H

G

1

3

2

5

4

6

A

B

C

G

F

H

D

E


13-19 Critical path is ADFHJK, Mean time is 68.7, standard deviation. is 3.5; P(70)=.6443; P(80)=.9994; P(90)=.9999



13-21

Activity Total Budgeted Cost

Percent of Completion

Value of Work Completed

Actual Cost Activity Difference

A 22,000 100 22,000 20,000 -2,000B 30,000 100 30,000 36,000 +6,000C 26,000 100 26,000 26,000 0D 48,000 100 48,000 44,000 -4,000E 56,000 50 28,000 25,000 -3,000F 30,000 60 18,000 15,000 -3,000G 80,000 10 8,000 5,000 -3,000H 16,000 10 1,600 1,000 -600

After 8 weeks:Value of work completed = $181,600Actual cost = $172,000Cost under run = $9,600

A

B

C

D

E

F

I

G

L

HK

J

1

2

3

4

5

6 7

8A

C

D

E

B

G

F

H

J K

L

I


13-23

Path Normal Time Time Activities Accelerated Total CostACFH 9 15 None $308,000

ACEGH 15 13 1A+1C+1D or 1A+1E+1D or 1C+1E+1D or 2E+1D or 1A+1G or 1C+1G or 1E+1G

$311,000BDGH 14



13-24

Path Normal Time Time Activities Accelerated CostADG 14 14 None $10,950BEG 12 13 1st D $10,950+75 = $11,025CF 3 12 2nd D $11,025+75 = $11,110

11 3rd D & 1st E $11,110+75+50 = $11,22510 4th D & 2nd E $11,225+75+50 = $11,350

A C

B

F

E

D G

H

1

3

2

5

4

6 7

A

B

C

D

E

F

G

H



Network diagram using activity on arrow diagram

A

B

C

D

E

G

F

1

2

3

4 5 6

A

C

D

B E

F

G


HERE IS SOME ADDITIONAL DETAIL ON THE SOLUTIONS TO PROBLEMS 13-13 AND 13-15

13-13

Path TimeCritical Path?

AEG 13BDEG 26 * C.P.CFG 15

Activity Time ES EF LS LF TS FSCritical

Activity?A 2 0 2 13 15 13 13 noB 5 0 5 0 5 0 0 yesC 1 0 1 11 12 11 0 noD 10 5 15 5 15 0 0 yesE 3 15 18 15 18 0 0 yesF 6 1 7 12 18 11 11 noG 8 18 26 18 26 0 - yes

13-15

Path TimeCritical Path?

AFG 12ACEG 15BDEG 17 * C.P

Activity Time ES EF LS LF TS FSCritical

Activity?A 3 0 3 2 5 2 0 NoB 7 0 7 0 7 0 0 YesC 4 3 7 5 9 2 2 NoD 2 7 9 7 9 0 0 YesE 5 9 14 9 14 0 0 YesF 6 3 9 8 14 5 5 NoG 3 14 17 14 17 0 - Yes

topic 2: project management models

Business

total project

project management page

activitys total slack

course of project management

measure of slack

earliest start times

smallest latest start

types of slack