Earliness and Tardiness Penalties

Chapter 5

Elements of Sequencing and Schedulingby Kenneth R. Baker

Byung-Hyun Ha

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

3

Introduction

Until now Basic single-machine model with regular measures of performance, whic

h are nondecreasing in job completion times Among regular measures, total tardiness criterion has been a standard w

ay 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=1n (j(dj – Cj)+ + j(Cj – dj)+) = j=1

n (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 sa

me time• Distinct due dates

Penalties• Common penalties (j = , j = )

• Distinct penalties Corresponding cost factors are substantially different

5

Introduction

Primary role of penalty functions Guiding solutions toward the target of meeting all due date exactly

Ideal schedule

Different penalty functions• Suggestions for measuring suboptimal performance of nonideal schedules

6

Minimizing Deviations from a Common Due Date

Basic E/T problem Minimizing sum of absolute deviations of job completion times from com

mon 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

7

Basic E/T Problem, Unrestriced

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.

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

8

Basic E/T Problem, Unrestriced

Theorem 3 In the basic E/T model, there is an optimal schedule in which some job c

ompletes 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.

9

Basic E/T Problem, Unrestriced

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

10

Basic E/T Problem, Unrestriced

Algorithm 1: Solving the Basic E/T Problem1. 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 T/T problem with jobs below and d = 24.

Job j 1 2 3 4 5 6

pj 1 3 4 6 7 9

11

Basic E/T Problem, Unrestriced

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

12

Basic E/T Problem, Unrestriced

Due dates as decision One way of finding an optimal solution

• Set d = and utilize algorithm 1*

OptimalTotal

Penalty

due date

13

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? Two jobs with p1 = p2 = 3 and d = 2

Three jobs with p1 = 1, p2 = 1, p3 = 10, and d = 2

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

• Solving problems with several hundreds of jobs

14

Restricted Version

An effective heuristic (Sundararaghavan and Ahmed, 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

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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

16

Different Earliness and Tardiness Penalties

A generalization of basic model Minimize f(S) = j=1

n (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 next 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

17

Different Earliness and Tardiness Penalties

Generalization of Theorem 4 In the basic E/T model with earliness penalty and tardiness penalty , t

here 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 the 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

18

Quadratic Penalties

Avoiding large deviations from due date Minimize f(S) = j=1

n (Cj – d)2 = j=1n (Ej

2 + Tj2)

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

19

Job Dependent Penalties

Permitting each job to have its own penalties f(S) = j=1

n (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-increasing 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, where b is the smallest integer satisfying the inequality

iB (j + j) j=1n j

20

Distinct Due Dates

Different due dates in job set f(S) = j=1

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

NP-hardness

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, as well

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