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Earliness and Tardiness Penalties

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### Earliness and Tardiness Penalties

Chapter 5

Elements of Sequencing and Schedulingby Kenneth R. Baker

Byung-Hyun Ha

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

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

- Earliness and tardiness

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

- Common due dates (dj = d)
- Penalties
- Common penalties (j = , j = )
- Distinct penalties
- Corresponding cost factors are substantially different

- Decision variables

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

- Guiding solutions toward the target of meeting all due date exactly

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=1n(Ej + Tj)
- Due date can be in the middle of jobs

- Tightness of due date d
- Restricted version vs. unrestricted version

d

d

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.

Basic E/T Problem, Unrestriced

- 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.

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=1bEBi = i=1b(pB(i+1) + pB(i+2) + ... + pBb)
= 0pB1 + 1pB2 + ... + (b – 2)pB(b–1) + (b – 1)pBb.

- CB = i=1bEBi = i=1b(pB(i+1) + pB(i+2) + ... + 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

- Notations

Basic E/T Problem, Unrestriced

- 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 T/T problem with jobs below and d = 24.

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?

Basic E/T Problem, Unrestriced

- Due dates as decision
- One way of finding an optimal solution
- Set d = and utilize algorithm 1*

- One way of finding an optimal solution

Optimal

Total

Penalty

due date

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

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

Restricted Version

- An effective heuristic (Sundararaghavan and Ahmed, 1984)
- Assuming p1 p2 ... pn.
1. Let L = d and R = i=1npi – 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.

- Assuming p1 p2 ... pn.
- Exercise
- Find good sequence for the jobs below with d = 90.

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

- Delay of start time leads to reduction in total penalty, when e n/2

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 penalties
1. 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.

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 the heuristic
- Tested by randomly generated problems

Quadratic Penalties

- Avoiding large deviations from due date
- Minimize f(S) = j=1n(Cj – d)2 = j=1n(Ej2 + Tj2)

- Due date d as decision variable
- d = = j=1nCj /n

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

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

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

- Generalization of Theorem 1–4
1. 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=1nj

Distinct Due Dates

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

- Decomposing into two subproblems

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