Earliness and Tardiness Penalties

Chapter 5

Elements of Sequencing and Schedulingby Kenneth R. Baker

Byung-Hyun Ha

R1

2

Outline Introduction Minimizing deviations from a common due date

Four basic results Due date as decisions

The restricted version Different earliness and tardiness penalties Quadratic penalties Job dependent penalties Distinct due dates Summary

3

Introduction Until now

Basic single-machine model with regular measures of performance, which are nondecreasing in job completion times

Among regular measures, total tardiness criterion has been a standard way of measuring conformance to due dates

• The measure does not penalize jobs completed early

Just-In-Time (JIT) production “Inventory is evil” Earliness, as well as tardiness, should be discouraged

E/T criterion in basic single-machine model Earliness and tardiness

• Ej = max{0, dj – Cj} = (dj – Cj)+

• Tj = max{0, Cj – dj} = (Cj – dj)+

Linear penalty function with unit earliness (tardiness) penalty j (j)• f(S) = j=1

n (j(dj – Cj)+ + j(Cj – dj)+) = j=1n (jEj + jTj)

Nonregular measure

4

Introduction Variations in E/T criterion

Decision variables• Job sequence with due dates given• Due dates and job sequence

Setting due dates internally, as targets to guide the progress of shop floor activities

Due dates• Common due dates (dj = d)

Several items constitute a single customer’s order Assembly environment where components should all be ready at the same

time• Distinct due dates

Penalties• Common penalties (j = , j = )• Distinct penalties Role of penalty functions

• Guiding solutions toward the target of meeting all due date exactly• Measuring suboptimal performance of nonideal schedules

5

Minimizing Deviations from a Common Due Date Basic E/T problem

Minimizing sum of absolute deviations of job completion times from common due date (dj = d, j = j = 1)

f(S) = j=1n |Cj – dj| = j=1

n (Ej + Tj) Due date can be in the middle of jobs?

Tightness of due date d Restricted version vs. unrestricted version

d

d

6

Basic E/T Problem, Unrestricted Theorem 1

In the basic E/T model, schedules without inserted idle time constitute a dominant set.

Theorem 2 In the basic E/T model, jobs that complete on or before the due date can

be sequenced in LPT order, while jobs that start late can be sequenced in SPT order.

V-shaped schedule

Exercise Prove Theorem 1 using proof by contradiction. Prove Theorem 2 using proof by contradiction.

7

Basic E/T Problem, Unrestricted Theorem 3

In the basic E/T model, there is an optimal schedule in which some job completes exactly at the due date.

Proof sketch of Theorem 3 (proof by contradiction) Suppose S is an optimal schedule where Ci – pi d Ci . Let b (a) denote the number of early (tardy) jobs in sequence. Case 1 (a b)

• Consider S' where S is shifted earlier by t = Ci – d.• Increase in earliness (decrease in lateness) penalty is bt (at).• Hence, f(S) f(S'), because at bt.

Case 2 (a b)• Consider S' where S is shifted later by t = d – (Ci – pi).• Decrease in earliness (increase in lateness) penalty is bt (at).• Hence, f(S) f(S'), because at bt.

Therefore, in either case a schedule with the property of the theorem is at least as good as S.

8

Basic E/T Problem, Unrestricted Properties of optimal schedule by Theorem 1, 2, 3

Optimum is describable by a sequence of jobs and a start time of 1st job V-shaped schedule 2n candidates instead of n! candidates

Analysis on optimal schedule Notations

• A (B) -- set of jobs completing after (on or before) the due date• a = |A|, b = |B|• Ai (Bi) -- ith job in A (B)

Earliness penalty for job Bi -- EBi = pB(i+1) + pB(i+2) + ... + pBb

Total penalty for the jobs in B• CB = i=1

b EBi = i=1b (pB(i+1) + pB(i+2) + ... + pBb)

= 0pB1 + 1pB2 + ... + (b – 2)pB(b–1) + (b – 1)pBb . Total penalty for the jobs in A

• CA = apA1 + (a – 1)pA2 + ... + 2pA(a–1) + 1pAa .

f(S) = CA + CB minimized by assigning jobs regarding processing times

9

Basic E/T Problem, Unrestricted Algorithm 1: Solving the Basic E/T Problem

1. Assign the longest job to set B.2. Find the next two longest jobs. Assign one to B and one to A.3. Repeat Step 2 until there are no jobs left, or until there is one job left, in

which case assign this job to either A or B. Finally, order the jobs in B by LPT and the jobs in A by SPT.

Exercise: solve basic E/T problem with jobs below and d = 24.

Job j 1 2 3 4 5 6

pj 1 3 4 6 7 9

10

Basic E/T Problem, Unrestricted Algorithm 1*

Considering secondary measure: minimum total completion time Same as Algorithm 1 except that, in Step 2, shorter job is assigned to B

and, in Step 3, if n is even, assign the shortest job in A

Theorem 4 In the basic E/T model, there is an optimal schedule in which the bth job

in sequence completes at time d, where b is the smallest integer greater than or equal to n/2.

Due date for unrestricted version Supposing jobs are indexed SPT order The problem is unrestricted for d , where

= pn + pn–2 + pn–4 + ... For unrestricted problem, Algorithm 1* will produce optimal schedule Exercise: When d = 18, is it unrestricted? When d = 17?

Job j 1 2 3 4 5 6

pj 1 3 4 6 7 9

11

Basic E/T Problem, Unrestricted Due dates as decision

One way of finding an optimal solution• Set d = and utilize algorithm 1*

optimaltotal

penaltyf(S)

common due date d

12

Restricted Version Basic E/T problem, restricted (d )

Optimal solution may contain a straddling job Theorem 1 and 2 hold, but Theorem 3 does not

• V-shaped schedules still constitute a dominant set

Should optimal schedule start at time zero always? Three jobs with p1 = 1, p2 = 1, p3 = 10, and d = 5 Optimal schedule, in which either

• the schedule starts at time zero, or• some job completes exactly at the due date

NP-hardness A dynamic programming technique (Hall et al., 1991)

• Solving problems with several hundreds of jobs

13

Restricted Version An effective heuristic: S-A heuristic (Sundararaghavan and Ah

med, 1984) Assuming p1 p2 ... pn .

1. Let L = d and R = i=1n pi – d. Let k = 1.

2. If L R, assign job k to the first available position in sequence and decrease L by pk.

Otherwise, assign job k to the last available position in sequence and decrease R by pk.

3. If k n, increase k by 1 and go to Step 2. Otherwise, stop.

Exercise Find good sequence for the jobs below with d = 90.Job j 1 2 3 4 5 6

pj 1 10 11 48 50 53

14

Restricted Version Adjustment of start time

Delay of start time leads to reduction in total penalty, when e n/2• where e is number of jobs that finish before due date

Schedule 6-3-2-1-4-5 of jobs below with d = 90

Job j 1 2 3 4 5 6

pj 1 10 11 48 50 53

15

Different Earliness and Tardiness Penalties A generalization of basic model

Minimize f(S) = j=1n (Ej + Tj) where

-- holding cost (endogenous), -- tardiness penalty (exogenous)

Properties of optimal solution Theorem 1, 2, and 3 hold

Components of objective function CB = 0pB1 + 1pB2 + ... + (b – 2)pB(b–1) + (b – 1)pBb . CA = apA1 + (a – 1)pA2 + ... + 2pA(a–1) + 1pAa .

Algorithm 2: E/T with different earliness and tardiness penalties1. Initially, sets B and A are empty, and jobs are in LPT order.2. If |B| (1 + |A|), then assign the next job to B; otherwise, assign the n

ext job to A.3. Repeat Step 2 until all jobs have been scheduled.

Exercise: consider jobs below with = 5, = 2, and d = 24.Job j 1 2 3 4 5 6

pj 1 3 4 6 7 9

16

Different Earliness and Tardiness Penalties Generalization of Theorem 4

In the basic E/T model with earliness penalty and tardiness penalty , there is an optimal schedule in which the bth job in the sequence completes at time d, where b is the smallest integer greater than or equal to n/( + ).

Criterion for unrestricted version = pB1 + pB2 + ... + pB(b–1) + pBb

Condition for delaying start of schedule e n/( + )

Effectiveness of modified S-A heuristic Tested by randomly generated problems

=

Problem Size Average Error No. of Optima Average Error No. of Optima

n = 8n = 10n = 12n = 15

0.40%0.24%0.26%0.32%

10944

1.52%0.84%0.66%0.07%

557

10

17

Quadratic Penalties Avoiding large deviations from due date

Minimize f(S) = j=1n (Cj – d)2 = j=1

n (Ej2 + Tj

2)

Due date d as decision variable d = = j=1

n Cj /n

Quadratic E/T problem, unrestricted f(S) = j=1

n (Cj – )2

Problem of minimizing variance of completion times, but not easily solvable

A heuristic solution (Vani and Raghavachari, 1987)• Neighborhood search using pairwise interchanges

18

Job Dependent Penalties Permitting each job to have its own penalties

f(S) = j=1n (jEj + jTj)

NP-hardness A dynamic programming technique (Hall and Posner, 1991)

• Solving problems with hundreds of jobs in modest run times

Generalization of Theorem 1–41. There is no inserted idle time.2. Jobs that complete on or before the due date can be sequenced in non-i

ncreasing order of the ratio pj /j, and jobs that start late can be sequenced in non-decreasing order of the ratio pj /j .

3. One job completes at time d.4. In an optimal schedule the bth job in sequence completes at time d, whe

re b is the smallest integer satisfying the inequalityiB (j + j) j=1

n j

19

Distinct Due Dates Different due dates in job set

f(S) = j=1n (j(dj – Cj)+ + j(Cj – dj)+) = j=1

n (jEj + jTj) NP-hardness

• T-problem reduces to this problem

A solution technique Decomposing into two subproblems

• Finding a good job sequence• Scheduling inserted idle time

• Solvable in polynomial time Refer to p. 74 of Pinedo, 2009

A neighborhood search (Armstrong and Blackstone, 1987) A branch-and-bound procedure (Darby-Dowman and Armstrong, 1986)

20

Summary Earliness/tardiness problem

From JIT concepts Nonregular performance measure

Properties Optimum is describable by a sequence of jobs and a start time of 1st job V-shaped schedule 2n candidates instead of n! candidates

Restricted vs. unrestricted versions Difficulties in finding good schedules with tight due date

Extended models Job-dependent penalty and due dates ...