IOE/MFG 543
Chapter 3: Single machine models
(Sections 3.3-3.5)
1
Section 3.3:
Number of tardy jobs 1||S Uj

Number of tardy jobs
– Often used as a benchmark for
managers (or % of on-time jobs)
– Some jobs may have to wait really
long
2
Characterization of an
optimal schedule for 1||S Uj

Split jobs into two sets
J=Jobs that are completed on time
Jd=Jobs that are tardy


The jobs in J are scheduled according
to an EDD rule
The order in which the jobs in Jd are
completed is immaterial (since they
are already tardy)
3
Algorithm 3.3.1 for 1||S Uj
1.
2.
Set
J=,
Jd=,
Jc ={1,…,n} (set of jobs not yet
considered for scheduling)
Determine job j* in Jc which has the EDD, i.e,
dj*=min{dj : jJc}
Add j* to J
Delete j* from Jc
Go to 3
4
Algorithm 3.3.1 for 1||S Uj
(2)
3.
4.
If due date of job j* is met, i.e., if SjJ pj ≤ dj*
go to 4,
otherwise
let k* be the job in J which has the longest
processing time, i.e., pk*=maxjJ {pj}
Delete k* from J
Add k* to Jd
If Jc= STOP
otherwise go to 2.
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Algorithm 3.3.1 for 1||S Uj
(3)

Computational complexity O(nlog(n))
– If it is implemented efficiently!

Theorem 3.3.2
– Algorithm 3.3.1 yields an optimal
schedule for 1||S Uj
– Proof: By induction
6
Example 3.3.3

Use Algorithm 3.3.1 to determine the
schedule that minimizes S Uj
– How many jobs are tardy?
job j
1
2
3
4
5
pj
7
8
4
6
6
dj
9
17
18
19
21
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Total weighted number of
tardy jobs 1||S wjUj

NP-hard
– No polynomial time algorithm exists

If all jobs have the same due dates
– i.e., d=dj for all jobs
– Knapsack problem=> pseudopolynomial

Why not use WSPT? job j
– Example 3.3.4


What happens
when WSPT is used?
What about 2-3-1?
pj
1
11
2
9
3
90
wj
dj
12
100
9
100
89
100
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Section 3.4:
The total tardiness 1||S Tj

A more practical performance measure
than the number of tardy jobs
– May schedule the tardy jobs to minimize
total tardiness

The problem is NP-hard in the ordinary
sense
– A pseudopolynomial dynamic
programming algorithm exists
9
Dynamic programming



See Appendix B
Dynamic programming is an efficient
sequential method for total
enumeration
For 1||S Tj we can use Lemmas 3.4.1.
and 3.4.3. to eliminate a number of
schedules from consideration
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Lemma 3.4.1.



If pj≤pk and dj≤dk, then there exists an
optimal sequence in which job j is scheduled
before job k
Proof: The result is fairly obvious, so we
omit the proof (Good exercise!)
Consequences:
– We can eliminate from consideration all
sequences that do not satisfy this condition
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Lemma 3.4.2.


For some job k let C'k be the latest
completion time of job in any optimal
sequence S'
Consider two sets of due dates: d1, …, dn
and d1, …,dk-1,max(dk,C'k), dk+1, dn
– S' is optimal for the first set and S'' is optimal for
the second set

Lemma: Any sequence that is optimal for
the second set of due dates is optimal for
the first set as well
– Proof: Skip
12
Enumerate the jobs by
increasing due dates


Assume d1≤d2≤…≤dn
Let job k be such that
pk=max(p1,…,pn)
– Lemma 3.4.1=> there is an optimal
sequence such that jobs 1,…,k-1 are
scheduled before job k
– The n-k jobs k+1,…,n are scheduled
either before or after job k
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Lemma 3.4.3

There exists an integer d, 0≤d≤n-k,
such that there is an optimal sequence
S in which job k is preceded by all jobs
j with j≤k+d and followed by all jobs j
with j>k+d
– Effectively reduces the number of
schedules to be considered
– Proof: Skip
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Consequences of Lemma
3.4.3.

There is an optimal sequence that
processes the jobs in this order:
1.
2.
3.

jobs 1,2,…,k-1,k+1,…,k+d in some order
job k
jobs k+d+1, k+d+2, …, k+n in some order
How do we determine this sequence?
– Algorithm 3.4.4.
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Notation for Algorithm
3.4.4.




Ck(d)=Sj≤k+dpj is the completion time of job k
J(j,l,k) is the set of jobs in {j,j+1,…,l} that
have a processing time less than or equal to
job k and excludes job k itself
V(J(j,l,k),t) is the total tardiness of the jobs in
J(j,l,k) under an optimal sequence assuming
that this set starts processing at time t
k' is the job in J(j,l,k) with the largest
processing time, i.e., pk'=max{pj : j J(j,l,k)}
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Algorithm 3.4.4. for 1||S Tj

Initial conditions
V(Ø,t)=0
V({j},t)=max(0,t+pj-dj)

Recursive relation
V(J(j,l,k),t)=mind{V(J(j,k'+d,k'),t)
+max(0,Ck'(d)-dk')+V(J(k'+d+1,l,k'),Ck'(d))}

Optimal value function:
V({1,…,n},0)
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The complexity of
Algorithm 3.4.4.
At most n3 subsets J(j,l,k)
 At most Spj time points
 =>at most n3Spj recursive equations
 Each equation requires at most O(n)
operations
 Overall complexity is O(n4Spj)
=> pseudopolynomial

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Example 3.4.5.

Given the following data, use
Algorithm 3.4.4. to solve the problem
1||S Tj
job j
pj
dj
1
121
260
2
79
266
3
147
266
4
83
336
5
130
337
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Section 3.5.
The total tardiness 1||S wjTj


Thm. 3.5.2: The problem 1||S wjTj is
strongly NP-hard
– 3 partition reduces to 1||S wjTj
A dominance result exists. Lemma 3.5.1:
– If there are two jobs j and k with dj≤dk , pj≤pk
and wj≥wk then there is an optimal sequence in
which job j appears before job k.

Branch and bound for 1||S wjTj
– Branch: Start by scheduling the last jobs
– Bound: Solve a transportation problem
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Summary of Chapter 3:
Single machine models
1||S wjCj
1||S wj(1-erCj)
1||Lmax
1|prec| hmax
1|rj|Lmax
1||S Uj
1||S Tj
WSPT rule, works for chains too
WDSPT rule, works for chains too
EDD rule
Algorithm 3.2.1.
Branch and bound
Algorithm 3.3.1.
Dynamic programming Alg. 3.4.4.
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