Lecture 2

Chapter 8 - Project Management 1

Lecture 2

Today’s lecture covers the followings:

1. To study “project crashing” concept2. LP formulation for project

management problem3. The use of QM (Try it yourself!)

Tutorial: Chapter 8: 8th Ed:Q26 and Q30 9tj Ed: Q21 and Q23

(to p2)

(to p26)


Project Crashing Basic Concept

In last lecture, we studied on how to use CPM to

determine solution for a project problem

There, we determine its critical path and completion

time.

Question: Can we cut short its project completion time?

If so, how! (to p3)


Project Crashing Solution!

Yes, the project duration can be reduced by assigning more resources to project activities

But, doing this would somehow increase our project cost!

How do we strike a balance? (to p4)


Trade-off concept

Here, we adopt the “Trade-off” concept• ie, we attempt to “crash” some “critical”

events by allocating more sources to them, and also to maintain a balance that the shortening time is not less than the normal activities

• How to do that:• Question: What criteria should it be based

on when deciding to crashing critical times?(to p11)

(to p5)


Example – crashing (1)

The critical path is 1-2-3, the completion time =11How? Path: 1-2-3 = 5+6=11 weeks

Path: 1-3 = 5 weeksNow, how many days can we “crash” it?

1 3

25 (1) 6(3)

5(0)

Normal weeks

Max weeks can be crashed

(to p6)



1 3

25 (1) 6(3)

5(0)

The maximum time that can be crashed for:Path 1-2-3 = 1 + 3 = 4Path 1-3 = 0

Total weeks can be crashed = 4 + 0 = 4

Are we to use up all these 4 weeks? (to p7)



1 3

25 (1) 6(3)

5(0)

If we used all 4 days, then path 1-2-3 has (5-1) + (6-3) = 7 completion weeks

Now, we need to check if the completion time for path 1-3 has lesser than 7 weeks (why?)

Now, path 1-3 has (5-0) = 5 weeksSince path 1-3 still shorter than 7 weeks, we used up all 4 crashed weeks

Question: What if path 1-2 has, say 8 week completion time?

4(0) 3(0)

(to p8)



1 3

25 (1) 6(3)

8(0)

Such as

Now, we cannot use all 4 days (Why?)Because path 1-2-3 will not be critical path anymore aspath 1-3 would now has longest hour to finish

Rule: When a path is a critical path, it will stay as a critical path

So, we can only reduce the path 1-2-3 completion time to the same timeAs path 1-2. (HOW?)

(to p9)



1 3

25 (1) 6(3)

8(0)

Solution:

We can only reduce total time for path 1-2-3 = path 1-2,

that is 8 weeks

If the cost for path 1-2 and path 2-3 is the same thenWe can random pick them to crash so that its completionTime is 8 weeks (to p10)



1 3

25 (1) 6(3)

8(0)

Solution:

1

2

3

5 (1) 6(3)

8(0)

OR

4(0) 4(1)

3(0)

Now, paths 1-2-3 and 1-3 are both critical paths(to p4)


Time-cost Trade-offIn this subject, the decision for “crashing” the project is based on the trade-off between

“time and cost”

The method is called “Time-cost Trade-off”

How it works?– We determine an average crash cost for each event

• How to do that?

– Procedural step. (to p13)

(to p12)


Project Crashing and Time-Cost Trade-OffExample Problem (1 of 3)

Table 8.5Normal Activity and Crash Data for the Network in Figure 8.16

A B C DE=(A-B)

F= (D-C)/(A-B)

Note: A,B,C,D are given Note: we will use F values to decideWe need to compute E and F which path to crash!

F

(to p11)


Time-Cost Trade-OffSteps:

1. use “normal cost” to determine the critical path2. for each event, compute their average crash cost3. for each section of critical path, crash their maximum time by retaining this section be part of the “critical” path.4. compute total crashing costs and completion time

Example (to p14)


Example: trade-off• Consider the same example as show in below• Step 1 determine it critical path• Step 2 determine all average unit crash cost • Step 3 crashing events with minimum costs• Step 4 compute crashed weeks and costs

(to p15)

(to p16)

(to p21)

(to p20)

(to p17)

More example!


Step 1

• Using CPM, the critical path is

1-2-3-4-6-7

(to p14)


Step 2

(to p14)


First, we cluster each segment of critical path into sections that can be crashed and to consider to crash them one section at a time

Step 3:

Section1 Section 2 Section 3 Section 4

(to p18)


We now add the normal and crashed time and cost to each segment

Step 3:


12(5) $400

8(3) $500

12(3) $$7000 4(1) $7000

4(1) $3000

4(3) $$200

4(3) $$200

(to p19)


We now crashed them one section at a time as follows:

Step 3:


12(5) $400

8(3) $500

12(3) $$7000 4(1) $7000

4(1) $3000

4(3) $$200

4(3) $$200

7(0)

5(0)

9(0) 3(0)

(to p14)


We now crashed them one section at a time as follows:

Step 4:

12(5) $400

8(3) $500

12(3) $$7000 4(1) $7000

4(1) $3000

4(3) $$200

4(3) $$200

7(0)

5(0)

9(0) 3(0)

Total crash cost=(5*$400)+(3*$500)+(3*$7000)+(1*$7000)= 31,000Total crashed weeks= 5+3+3+1=12Note: critical path is 1-2-3-4-6-7Completion time = 7+5+0+9+3 = 24

(to p14)


Crashed cost

1

2

3 4

10(5) 5(4)

4(2)4(1)

How to solve this problem?(to p22)


Further detail steps

1. Determine the critical path

2. Crash the critical path to the level where other non-critical paths become a critical one

3. Consider for further crashing until all possible crashing resources were consumed!

(to p23)


Critical path

1

2

3 4

10(5) 5(4)

4(2)4(1)

The critical path is 1-3-4, completion time is 10+5 = 15 (to p24)


Crash to a level to which other non-critical path is introduced

1

2

3 4

10(5) 5(4)

4(2)4(3)

The non-critical path is 1-2-4, has the processing time = 4+4 = 8So, we try to reduce the critical path to this level !

5(0) 3(2)

Both criticalPaths = 8

(to p25)


Crash all resources until no further can be reduced!

1

2

3 4

10(5) 5(3)

4(2)4(3)

Stop, since no more resources can be reduced in path 1-3-4

5(0) 3(1)

Both criticalPaths = 7

3(2)

2(0)

(to p1)


Formulating the CPM/PERT Network as a Linear Programming Model

- The objective is to determine the earliest time the project can be completed

(i.e., the critical path time).

• normal CPM• crashing model

(to p27)

(to p30)

(to p1)


LP formulation

General linear programming model is:

minimize Z = cixi

subject to

xj - xi tij for all activities i j

xi, xj 0

where xi = earliest event time of node i

xj = earliest event time of node j

tij = time of activity i j

• LP formulation for the project management(to p28)


LP for the CPM• Let consider a simple problem as outlined as

follows:

Let xi be denote as each node iAnd segment of say path 1-2 as x2-x1

Then (to p29)


Objective is Minimize Z = x1 + x2 + x3 + x4 + x5 + x6 + x7

Subject to x2 - x1 12 (for path 1-2)x3 - x2 8 (for path 2-3)x4 - x2 4 (for path 2-4)x4 - x3 0 (for path 3-4)x5 - x4 4 (for path 4-5)x6 - x4 12 (for path 4-6)x6 - x5 4 (for path 5-6)x7 - x6 4 (for path 6-7) xi, xj 0

Do you know how to read the results from the LP output?(to p26)


General concept• All formulation of CPM is used, except we

need one more variable to represent the crashed cost per unit of each path

• Example! (to p31)


Consider again the following crashed cost as an example

- Objective is to reduce the project duration from 36 to 30 weeks at the minimum possible crash cost.

We now y to represent theseOur objective is to min these

How?(to p32)


Min 400y12 + 500y23 + 3000y24 + 0y34 + 200y45 + 7000y46 + 200y56 + 7000y67

And all yij <= their total allowance crash time

A complete model is shown in next slide (to p33)


The CPM/PERT Network as a Linear Programming ModelExample Problem Project Crashing - Model Formulation

xi = earliest event time of node i

xj = earliest event time of node j

yij = amount of time by which activity i j is crashed (i.e., reduced)

minimize Z = $400y12 + 500y23 + 3000y24 + 200y45 + 7000y46 + 200y56 + 7000y67

subject to y12 5 y12 + x2 - x1 12

y23 3 y23 + x3 - x2 8

y24 1 y24 + x4 - x2 4

y34 0 y34 + x4 - x3 0

y45 3 y45 + x5 - x4 4

y46 3 y46 + x6 - x4 12

y56 3 y56 + x6 - x5 4

y67 1 x67 + x7 - x6 4

x7 30 xj, yij 0

New set of equations

Max crashing time for critical pathi.e. total allowable crashed time

CPM value

(to p26)

Lecture 2

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Transcript of Lecture 2