network scheduling with limited_resources- revised1

Network Scheduling with Limited Resources

Introduction Early start schedule (EST ) or Late start schedule (LST)generated by two sets of data

1.Activity precedence relation2.Activity Duration (based on PERT or CPM )

•Assumption - Resources available in unlimited supply• In practical situations fixed manpower , certain number of machine , and limited budget

Schedule Graph , Unlimited Resources

18 10 12 6 6 1

Labor availableLabor available

Project completion

time

Let assume following constraint 10 labor available on any one day

10 10 10 10 6 6 1

Schedule Graph , Resource Limit Of 10 Men

Cont..

Complexity Of Project Scheduling With Limited Resource

•Scheduling project with limited resource is combinatorial problem (to find the best possible discrete solution)

•Large combination of activity start time , each combination representing a different schedule

• Integer programming can be applied to solve the problem

Heuristic Programs Heuristic – standard procedure to solve problem

but not rule

Resource Leveling Program – • It reduces peak resource requirement• smooth out period assignment within a

constraint on project duration

Resource Allocation Program – • It allocates resources to project activities in an

attempt to find the shortest project schedule consistent with fixed resource limit

• Project duration may change

Heuristic method for Resource Leveling of Project Schedule

1. Utilize resources at more constant rate

2. Scheduler may use activity slack as means of smoothing peak resource requirement

3. Draw EST and LST schedule

4. Concentrate on Activity having maximum slack

Early start schedule graph

Manpower loading chart

Maximum slack 7 days so postpone

activity 9

Maximum slack 7 days so postpone

activity 9

Peak requirement 24Peak requirement 24

Number over activity indicates the

activity number as

well as resource.

Schedule Graph

Manpower loading chart .

Peak requirement reduce to 15

Cont..

Schedule Graph

Manpower Loading Chart

Activity 8 is delay by full slack of 2 days

Peak requirement occur at days 1, 2

Peak requirement occur at days 1, 2

Cont..

Schedule Graph

Manpower Loading Chart.

Best result obtain by delaying

activity 3 by 2 days

Smoothen histogram

Cont..

Smoothing Peak Resource Requirement

Heuristic Method for Resource Allocation in Project Scheduling

1. Allocate resources serially in time2. Start on the first day and schedule all

jobs possible , repeat for second day, and so on

3. When Several jobs compete for the same resource , give preference to the job having least slack

4. Reschedule non-critical jobs ,if possible , to free the resource for scheduling critical or non-slack jobs

Day 1 - only one job 7 is available to start in this period and there are sufficient men to schedule it.Schedule job 7 , 3 men remain available

Only 10 men are available for the

project on any day.

Resource

Day

Day 2- Job 1,3,4,5 can be started on day 2 , but there are not enough men to schedule themThen second heuristic is applied for scheduling the jobs with least slack first

Schedule job 3( slack=0); 7 men remainSchedule job 4( slack=0); 3 men remainSchedule job 1( slack=3); 2 men remain

Job 5 must be delayed , as there are just [2=10-(4+3+1)] men unassignedPostponed the job 5 ( slack =4)

Cont..

Day 3- continue job 3(slack =0); 7 men remaincontinue job 4(slack =0); 3men remaincontinue job 1(slack =3); 2men remainPostpone job 5 (slack =3)

Cont..

Day 4Continue job 4 (slack =0); 6 men remain Continue job 1 (slack =3); 5 men remain Job 6 has no slack and hence is critical , but only 5 men are still unassigned

Then 3rd heuristic is applied – job 1 could be postponed without delaying the project Reschedule job 1 ; 6 men remain Schedule job 6 (slack =0); 0 men remain Postpone Job 5 (slack =2)

Cont..

Cont..

Day

Manpower/resource

requirement

Day 5Continue job 6 (slack=0); 4 men remain Schedule Job 2 (slack =0); 2 men remainSchedule Job 1 ( slack = 0); 1 men remain Postpone job 5 (slack =1)

Day 6 Continue Job 6 (slack =0); 4 Men remain .Continue Job 2 (slack =0); 2 Men remain .Continue Job 1 (slack =0); 1 Men remain .Postpone job 5 (slack =0)Now job 5 is critical but no active job can be postponed without delaying project

Cont..

Cont..

Day

Manpower

Day 7 Continue Job 1 (slack =1); 9 Men remain Schedule Job 5 ( slack =0); 4 Men remain Postpone job 9 ( slack =1)

Day 8Continue Job 5 (slack =0); 5 men remain Postpone job 9 (slack =0); No noncritical job can be rescheduledPostpone Job 8 (Slack =2)

DayManpower

Day 9– Schedule job 9 ( slack =0); 1 man remain . Postpone job 8 (slack =2)

Day 10 – Schedule job 10 ( slack =0); no man remain.Postpone Job 8 (slack =1)

Day 11 –Continue job 10 (slack =0); no men remain .Postpone Job 8 (slack =0)

Day 12— Continue job 10 (slack =1); no men remain

Postpone Job 8( slack =0)

Day 13— Schedule job 8 ( slack =0); 2 men remain

Day 14 – Continue job 8 ( slack = 0); 2 men remain

Cont..

1 2 3 4 5 6 7 8 9 10 11 12 13 14

7 7 7 10 9 9 6 6 9 10 10 10 8 8Manpower

Day

Final schedule limited resource

Other Heuristic Approaches• First schedule the critical activity

•If it fails to do so then schedule it in next period

•Rule for crew size selection critical activity schedule first with maximum crew size ( i.e. crashing ) if it fails then

•Schedule it at minimum crew size

•If activity can not schedule at minimum crew size then its Early start date is delayed and schedule it on next day

•For non-critical jobs also delayed if sufficient resources not available for scheduling

•Borrow form active job- resources borrowed from the activity only when the stretching of the job will not delay the entire project

Cont..

• Add on Unused Resource

1.list activities which required resources and which have crew size less than maximum

2.Job arrange in ascending order of their slack

3.Proceeding down the list and allocate resources

Cont..

Conceptual Problem of Critical Path Analysis When Resource Are Limited

Early start schedule

Late Start schedule

Cont..

1 2 3 4 5

18 10 10 8 7

1 2 3 4 5

4 4 5 18 22

DayResource

Day

Resource

If resources are limited up to 10 men then optimal early start schedule and Late start schedule is as follows

Early start schedule

Day

Resource

1 2 3 4 5 6

10 10 10 9 7 7

Late Start schedule

1 2 3 4 5 6

10 10 8 10 7 8

Day

Resource

ObservationsSlack concept is a measure of flexibility in a project schedule

Slack depends on precedence ordering and resource availability

Resource limitation reduce the amount of slack in a schedule

Slack is conditional upon the scheduling rules for creating early start and late start schedules

Conclusion

1. Ordinary PERT and CPM• No explicit consideration for resources

2. Scheduling program • For each resource , the program calculates a

total resource requirement profile by summing period by period , requirements for that resource of all activities as they occur in an early start schedule

3. Resource leveling program •Using resource requirement profile of EST schedule , the program attempts to reduce peak requirements by shifting slack jobs to nonpeak periods •Resource limits are not specified , but peak requirements are leveled as much as possible without delaying the specified due date.

Cont..

4. Resource allocation Program • Fixed amount of resource allocated to available

jobs according to certain scheduling heuristic that determine which job will be postponed if total requirements for a given period exceed resources available

• The completion date of the project may be pushed ahead to keep within specified resource limits

• No heuristic program can guarantee an optimum schedule

Cont..

Integer Programming Formulation – Project scheduling with limited

Resource

• S- Resource s= 1,2,3,……………,m• d -day (time period) d=1,2,……………….,z• j -Job (activity) j= 1,2,………………,n• P- immediate predecessor of j ; p ε P j

= [ all immediate predecessors of j]

• Objective function Minimize

( 2)1 1 1

4 ......n n n

jk j k s jsj j j

X X R X

where k is some number such that

1T<k<z and R 4Rs s

Variables : = activity of job j on day d ; Constrained = 1 ( if job j is active ) or

0 (if job j is inactive)

Constraints

1.0 ≤ X jd ≤ 1 (x is constrained to equal either 0 or 1); all jd

a Men available in shop s on day d.

C Crew size ,men of shop s required on job j.

t time length of job j, in days.

T=Minimum project completion time, given unlimited resources.

sd

sj

j

jdX

2. Job will be performed :

3. Capacity of shop will not be exceeded

4.No job will be started before its predecessors are completed

d = 1,……,z AND j= 1,….,n except

beginning job

1

t j= 1,....,n. s

jd jd

X

1

d=1,....,n

s=1,.....,m

s

sj jd sdj

C X a

1

1

all p d

p jd jd j jj

t X X t P

5. No jobs will be split :

( 1)2

t t ts

j jd j j d ji ji d

X X X

1,...,n AND d= 1,...,zj