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Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 1
Outline• Questions?
• Exam results – very good
• 19 grades improved
• 6 grades decreased
• 18 stayed the same
• Class GPA increased by 0.05
• No official homework, but do exercises of your own
• Quiz on 11/24
• Project scheduling
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 2
Exam Results
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 3
Project Scheduling
Intermittent systems
Examples: Construction of a plant
Aircraft carrier
Large airplanes
Complex and large, thousands of tasks and interdependencies
Objectives: Complete on time
Minimize cost
Minimize time
Meet customers’ requirements
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 4
Project Scheduling
Analyze -- Plan -- Schedule -- represented as a network of activities
Most used:
PERT - Program Evaluation and Review Technique
CPM - Critical Path Method
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 5
Project Scheduling
PERT – Activities on arcCPM – Activities on nodes
PERT CPM
1957 - 8 1957 - 9Navy Special Projects & Booze Allan Dupont and Remington RandLockheed Overhaul and maintain chemical plantsPolaris Missile System
Network representationActivity scheduling
R&D Routine operationsTask durations as a random variable Durations established Probabilistic DeterministicOn time completion Time cost trade - offs
Arrows represent activitiesEvents described at nodes Notation written on arrows
Activities are known before project starts
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 6
My first PERT experience
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 7
Project Scheduling
Predecessor and successor relationships between activities
If there is no such relationship, the activity is independent
Durations are independent
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 8
Project Scheduling
Activity = Task = Job
Has a beginning and an end
Has a duration = elapsed time = process time
Uses resources
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 9
Project Scheduling
Planning and scheduling steps
Identify activities
Precedence constraints
Construct the network
Estimate durations
Assign starting times
Analyze resources
Once project starts, check progress against plan
Reschedule
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 10
Project Scheduling - Networks
Network = directed graph
Finite number of nodes (n) i,j, ….. = N
subset of ordered pairs (i,j) = arcs = A
To draw a network:
from each i of N, draw arrow to j, if (i,j) is in A
where
arrow = (i,j) or name of activity
i - starting event
j - ending event j endi Start
Activity
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 11
Project Scheduling - Networks (cont)
Rules:
The length of the arrow has no significance
At a node, the outgoing activity cannot start until all incoming activities are complete
A
B
C
D
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 12
Project Scheduling - Networks (cont)
Rules (continued)
Only one initial node (no predecessors) and only one terminal node (no successors)
An activity is uniquely identified by start and end events
- no duplicate node numbers
- at most one arrow between nodes
For every arrow (i,j) such that i <j
No closed loops!
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 13
Project Scheduling - Networks (cont)
Multiple paths can be avoided with dummies
Dummy
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 14
Project Scheduling - Network example
Activity Symbol Activity Description Predecessors
A Demolish old barn -
BProcure materials for brickwork and reinforcements for concrete structures -
C Sort reusable materials resulting from demolition AD Excavate for foundations AE Dig path of driveway AF Make list of necessary materials and procure them CG Pour reinforced concrete foundations B, DH Pour concrete driveway EI Erect Brick walls B, GJ Level floor with gravel and pour rough concrete floor F, GK Install wiring for electrical system F, IL Finish walls K, M, NM Erect roof F, IN Finish concrete floor JO Mount gutters and downspouts F, MP Clean up H, L, O
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 15
Project Scheduling - Example - List
List activity only if its predecessor is complete - nondecreasing i or j numbers - topological order
Activity Start Node End Node Predecessor
A 1 2 -B 1 3 -C 2 4 AD 2 5 AE 2 6 AF 4 7 CG 5 8 B, DH 6 9 EJ 8 10 F, GI 8 11 B, GN 10 13 JM 11 15 F, IK 12 14 F, IL 14 16 K, M, NO 15 17 F, MP 17 18 H, L, O
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 16
Small example - Activity on Node (AON)
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 17
Project Scheduling – Activity on Arc (AOA)
B - E - F 3 + 2 + 2 = 7
A - D - E - F 2 + 2 + 2 + 2 = 8
A - C - F 2 + 5 + 2 = 9 Critical path, similar to bottleneck idea
We’ll generate all possible schedules to get the concepts
1
2
3
3
4 5
A, 2
B, 3
D,2
C, 5F, 2
E, 2
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 18
Project Scheduling - another example (cont)
All activities start as soon as possible( plotted to scale):
Use up total slack of E
Use up free slack of B
Use up both
1
2
3 4 5
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 19
Project Scheduling - another example (cont)
Use up total slack of E and D
If we move B to the right by one unit, we will have used up all slack and everything starts at the latest start time
1
2
3 4 5
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 20
Project Scheduling - More definitions
Earliest start time ESij or ESA
delayed = start after earliest start time
Latest start time LSij or LSA
Delay without affecting start of successors = free slack = Fsij
Delay that affects start of successors - Total slack - TSij
Free slack <= Total slack
Critical activities have the least total slack, usually 0
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 21
Project Scheduling - More definitions(cont)
EF = Earliest finish
LF = Latest finish
Y = duration
Ti = earliest occurrence of node i
Forward pass to determine ES
Topological order - a task is listed only if all its predecessors have been listed
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 22
Project Scheduling - Forward Pass
Activity Topological order Duration PredecessorES = max[EF of all
immediate predecessors] EF = ES +duration
A 1 2 2 - 0 2B 1 3 3 - 0 3C 2 4 5 A max[2] = 2 7D 2 3 2 A max[2] = 2 4E 3 4 2 B,D max[3,4] = 4 6F 4 5 2 C,E max[7,6] = 7 9
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 23
Project Scheduling - More definitions(cont)
Backward Pass
Reverse topological order
Free slack = scheduling flexibility with respect to its immediate successors
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 24
Project Scheduling - Backward Pass
Activity LF = min[LS of immediate successors] LS = LF - duration
F 9 9 - 2 = 7E =LSF = 7 7 - 2 = 5
D = LSE = 5 5 - 2 = 3
C = LSF = 7 7 - 5 = 2
B = LSE = 5 5 - 3 = 2
A =min[LSc, LSD] = min[2,3] = 2 2 -2 = 0
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 25
Project Scheduling
Free slack - scheduling flexibility with respect to its immediate successors
FSij = min [ ES of all immediate successors] - EFij
FSA= min [ESD, ESC] - EFA = min[2, 2] - 2 = 0
FSB= ESE - EFB = 4 - 3 = 1
FSC= ESF - EFC = 7 - 7 = 0
FSD= ESE - EFD = 5 - 4 = 1
FSE= ESF - EFE = 7 - 6 = 1
FSF = 9 - 9 - 0
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 26
Project SchedulingTotal Slack - scheduling flexibility relative to the project
completion time
TSij = LSij - ESij = LFij - Efij
TSA = 0 - 0 = 0
TSB = 2 - 0 = 2
TSC = 2 - 2 = 0
TSD = 3 - 2 = 1
TSE = 5 - 4 = 1
TSF = 7 - 7 = 0
Note that the activities on the critical path have 0 total slack
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 27
Statistics Review
Distributions
All measurable things vary, even if we assume that they are constant. This is why we call them random variables.
A random variable can be described by its mean and its standard deviation and the shape of its distribution
Most natural phenomena are normally distributed. The normal distribution extends to plus and minus infinity, so it is not useful for variables that have definite minima and maxima
The beta distribution does have these cutoffs.
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 28
Statistics Review - Continued
We specify the beta distribution by its minimum, maximum, and two parameters, usually denoted by alpha and beta. In the equation below, we use nu1 and nu2:
Excel uses alpha and beta and allows intervals other than 0,1.
10
0,
)()(
)1()()(
21
21
1121
21
x
xxxf
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 29
Statistics Review - Continued
The mean and standard deviation of the beta distribution can be expressed in terms of its parameters:
So it is possible to find (by trial and error), the parameters from a mean and a standard deviation
)1()( 212
21
212
21
1
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 30
Beta Distribution with max and min
http://www.me.utexas.edu/~jensen/ORMM/omie/computation/unit/project/beta.html
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 31
Beta Distribution with max and min
For project analysis we may be given the mode and require values of the shape parameters, alpha and beta, to specify the Beta distribution. Formulas for two cases are below. In each case we must choose one parameter and solve for the other.
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 32
Statistics Review - Continued
One other very important statistical fact that we need is the central limit theorem:
1. The distribution of the mean of a normal population (with standard deviation s) will be distributed normally with standard deviation s/sqrt(n), where n is the sample size
2. If n is large enough this will be true even if the population is not normally distributed
This allows us to assume that the completion time of a project is normally distributed
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 33
Statistics Review - Continued
One more statistical fact:
When adding up distributions:
1. The mean of the sum is the sum of the means
2. The variance of the sum is the sum of the variances
This allows us to get a mean and a standard deviation of the critical path of a project
Note: The standard deviation is the square root of the variance.
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 34
Statistics Review - Continued
The standard normal distribution(z) is tabulated in all statistics books, but you must be careful to ascertain the exact meaning of the tables. You map back and forth from your variable (x) to the standard table with the equation:
z = (x - xbar)/s, where s is the standard deviation and xbar is the average
I have reproduced a normal table for you on the following page. Here the probability is between z =0 and z.
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 35
Area between z= 0 and z = 1.18 is 0.3810
Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.0000 0.0039 0.0078 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359
0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753
0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141
0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517
0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879
0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224
0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549
0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852
0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133
0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389
1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621
1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830
1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015
1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177
1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319
1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441
1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545
1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633
1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706
1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767
2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817
2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857
2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890
2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916
2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936
2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952
2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964
2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974
2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981
2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986
3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990
3.1 0.4990 0.4991 0.4991 0.4991 0.4992 0.4992 0.4992 0.4992 0.4993 0.4993
3.2 0.4993 0.4993 0.4994 0.4994 0.4994 0.4994 0.4994 0.4995 0.4995 0.4995
3.3 0.4995 0.4995 0.4995 0.4996 0.4996 0.4996 0.4996 0.4996 0.4996 0.4997
3.4 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4998
3.5 0.4998
4.0 0.49997
5.0 0.4999997
6.0 0.499999999
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 36
PERT
Probabilistic methods
Instead of one duration, assume a worst, most likely, and a best possible value.
You can, of course, use other ways of approximating the distribution of the duration time.
The beta distribution is the most popular for this because it can be shaped to one’s liking and has a definite minimum and maximum
The normal distribution is not a good choice because in simulations it could yield very short or very long processing times as it is not limited at either end
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 37
PERT - continued
If you assume the three values, you can then estimate the mean and the variance by (a simplification of the beta distribution):
2
6
arg
6
arg*4min
2
luesmallestvaestvalueL
estvalueLmostlikelyvalue
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 38
PERT - Example
Continuing with our previous example:
Our critical path ACF still has average length of 9, but with a standard deviation of 0.8
If we look at path ADEF, the average is 8, with a standard deviation of 1.92
Task shortest most likely Longest Average Variance
A 1 1.5 5 2 0.44B 2 3 4 3 0.11C 4 5 6 5 0.11D 0.5 1 7.5 2 1.36E 0.5 0.75 8.5 2 1.78F 1 2 3 2 0.11
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 39
PERT - Example (continued)
Calculating the probability of a completion time of 10 or less for each path:
ACF: z =(10-9)/0.8 = 1.2 P(<=10) = 0.89
ADEF: z =(10-8)/1.92 = 1.04 P(<=10) = 0.85
That is, the “shorter” path is more likely to cause a delay
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 40
PERT - Example (continued)
If we applied the Monte Carlo technique to this problem 5 times:
Criticality Indices:
Simulation Sampled duration times for activity Length of path Critical PathA B C D E F ACF ADEF BEF
1 1.41 2.46 5.16 3.92 2.93 1.72 8.29 9.98 7.11 ADEF2 3.07 2.95 5.04 1.05 0.56 2.14 10.25 6.82 5.65 ACF3 2.47 3.20 4.90 0.87 0.73 2.29 9.66 6.36 6.22 ACF4 1.18 3.58 5.52 2.10 3.75 2.24 8.94 9.27 9.57 BEF5 2.02 3.54 4.64 0.84 2.68 2.07 8.73 7.61 8.29 ACF
Activity A B C D E FIndex 4/5 1/5 3/5 1/5 2/5 5/5
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 41
Resource Allocation
Assume a single resource, that is, people for each task:
63
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2019181716151413
1211
109
87
654321
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 42
Resource Allocation
By delaying tasks D and E to their latest start time, we can level the resource usage somewhat:
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2019181716151413
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87654321
Session 25University of Southern California
ISE514 November 17, 2015
Geza P. Bottlik Page 43
Project Scheduling - References
Morton and Pentico, pages 425 to 503
Kerzner, “Project Management”, 5th Ed. ITP 1995, pages 653 - 701 (out of 1152!)
Hax and Candria “Production and Inventory Management”, Prentice Hall 1984 Pages 325 to 359