Economic Lot Scheduling
1. Summary Machine Scheduling
2. ELSP (one item, multiple items)
3. Arbitrary Schedules
4. More general ELSP
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Operations Scheduling
Topic 1
Summary Machine Scheduling Problems
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Machine Scheduling

Single machine
1|| Cmax

1||  C j
1|| Lmax
1|| U j
1 ||  w j C j
1| prec | hmax
1||  w jU j
1| rj |  C j
1| rj | Lmax
1||  T j
1| rj , prmp |  C j
1| rj , prmp | Lmax
1||  w jT j
Parallel machine
Pm || Cmax
Pm ||  C j
Pm | prec | Cmax
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3-partition
partition
P 2 || Cmax P | prec | Cmax
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Why easy?

Show polynomial algorithm gives optimal solution

Often by contradiction

–
See 1 ||
–
See
w C
j
j
1| prec | hmax
week 1
week 1
By induction
–
E.g. 1||
U
j
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Why hard?
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Why hard?
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Why hard?
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Why hard?


Some problem are known to be hard problems.
–
Satisfiability problem
–
Partition problem
–
3-Partition problem
–
Hamiltonian Circuit problem
–
Clique problem
Show that one of these problems is a special case of your
problem
–
Often difficult proof
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Partition problem

Given positive integer a1, …, at and b with ½ j aj = b
do there exist two disjoint subsets S1 and S2 such that
jS aj = b ?


Iterative solution:
–
Start with set {a1} and calculate value for all subsets
–
Continue with set {a1, a2} and calculate value for all subsets; Etc.
–
Last step: is b among the calculated values?
Example: a = (7, 8, 2, 4, 1) and b = 11
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Some scheduling problems


Knapsack problem
1| d j  d |  wjU j
Translation:
n = t; pj = aj; wj = aj; d = b; z = b;
Question:
exists schedule such that j wjUj  z ?
Two-machine problem
P 2 || Cmax
Translation:
n = t; pj = aj; z = b;
Question:
exists a schedule such that Cmax  z ?
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3-Partition problem

Given positive integer a1, …, a3t and b with ¼b < aj < ½b
and j aj = t b do there exist pairwise disjoint three element
subsets Si such that
jS aj = b ?
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Topic 2a
Economic Lot Scheduling Problem:
One machine, one item
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Lot Sizing



Domain:
–
large number of identical jobs
–
setup time / cost significant
–
setup may be sequence dependent
Terminology
–
jobs = items
–
sequence of identical jobs = run
Applications
–
Continuous manufacturing: chemical, paper, pharmaceutical, etc.
–
Service industry: retail procurement
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Objective

Minimize total cost
–
setup cost
–
inventory holding cost

Trade-off

Cyclic schedules but acyclic sometimes better
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Scheduling Decisions

Determine the length of runs
–

Determine the order of the runs
–

gives lot sizes
sequence to minimize setup cost
Economic Lot Scheduling Problem (ELSP)
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Overview

One type of item / one machine
–


without setup time
Several types of items / one machine
–
rotation schedules
–
arbitrary schedules
–
with / without sequence dependent setup times / cost
Generalizations to multiple machines
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Minimize Cost

Let x denote the cycle time

Demand over a cycle
= Dx

Length of production run needed
= Dx / Q

Maximum inventory level
= (Q - D) Dx / Q
=x
= (Q - D)  x
= (1 - ) D x
Inventory
(Q  D) Dx
Q
Time
x
idle time
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Optimizing Cost
book



Solve
min
c 1 
D2 x 
 h  Dx 

x 2 
Q 
shorter
min
c 1
 h 1    Dx
x 2
Optimal Cycle Time
2Qc
x
hD(Q  D)
x
Optimal Lot Size
2cDQ
gx 
h(Q  D)
2cD
gx 
h(1   )
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2c
h(1   ) D
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With Setup Time

Setup time s

If s  x(1-) above optimal

Otherwise cycle length
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g
   utilizatio n
q
s
x
1 
is optimal
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Topic 2b
Economic Lot Scheduling Problem:
Lot Sizing with Multiple Items
With Rotation Schedule
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Multiple Items

Now assume n different items

Demand rate for item j is Dj

Production rate of item j is Qj

Setup independent of the sequence

Rotation schedule: single run of each item
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Scheduling Decision

Cycle length determines the run length for each item

Only need to determine the cycle length x

Expression for total cost / time unit
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Optimal Cycle Length
 cj


  aj x
j 1  x

n

Average total cost
with

Solve as before
aj 
x
j
x
cj

 a  x
j
j
1
hj (1   j )D j
2
n
c
j 1
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

j
 n

aj 
 j 1 
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Example
1
2
3
4
_qj
400
400
500
400
_gj
50
50
60
60
_hj
20
20
30
70
_cj
2000
2500
800
0
Items
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Data of example 7.3.1
c(j)
h(j)
D(j)
Q(j)
rho(j)
a(j)
t(j)
1
2000
20
50
400
0.125
437.5
2.14
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2
2500
20
50
400
0.125
437.5
2.39
3
800
30
60
500
0.12
792
1.01
4
0
70
60
400
0.15
1785
0.00

j
c j  5300

j
a j  3452
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Solution


x   c j   ½ h j (1   j ) g j 
j 1
 j 1

n
n
1
1
7
44
34 

  2 10. .50  15. .60  35. .60  5300
8
50
40 


 3452 
1
5300  1.5353  1.24 months
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


j
cj
j
aj
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Solution

The total average cost per time unit is
2155  2559 1627  2213  $8554
a
(alternative 1:
(alternative 2:

c
j
j
 3452
 5300
2  5300
AC 
1.24
AC
2  5300  3452 )
or
2  3452 1.24 )
How can we do better than this?
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With Setup Times

With sequence independent setup costs and no setup times the
sequence within each lot does not matter


Only a lot sizing problem
Even with setup times, if they are not job dependent then still
only lot sizing
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Job Independent Setup Times

If sum of setup times < idle time then our optimal cycle length
remains optimal

Otherwise we take it as small as possible
n
x
s
j 1
n
j
1   j
j 1
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Job Dependent Setup Times

Now there is a sequencing problem

Objective: minimize sum of setup times

Equivalent to the Traveling Salesman Problem (TSP)
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Long setup

If sum of setups > idle time, then the optimal schedule has the
property:
–

Each machine is either producing or being setup for production
An extremely difficult problem with arbitrary setup times
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Topic 3
Arbitrary Schedules
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Arbitrary Schedules

Sometimes a rotation schedule does not make sense
(remember problem with no setup cost)

For the example, we might want to allow a cycle 1,4,2,4,3,4 if
item 4 has no setup cost

No efficient algorithm exists
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Problem Formulation

Assume sequence-independent setup

Formulate as a nonlinear program
min
min COST
 sequences  lot sizes
s.t.
demand met over the cycle
demand is met between production runs
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Notation

Setup cost and setup times

All possible sequences
c jk  ck , s jk  sk .
S  ( j1 , j2 ,..., jh ) : h  n
q l  q jl  qk

Item k produces in l-th position

Setup time sl, run time (production) tl, and idle time ul
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Inventory Cost

Let x be the cycle time

Let v be the time between production of k
q l t l qk t l
v l 
g
gk

Total inventory cost for k is
l

 l
1 l l
q
l
h (q  g ) l  t
2
g 

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2
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Mathematical Program
l

1  (S ) 1 l l
Q
l
min min
h
(
Q

D
) l


l
l
S
x , ,u
x  l 1 2
D
Subject to
 l 2  (S ) l 
 ( )   c 
l 1


Qk   j  Dk x,
k  1,..., n
jI k
l

Q
j
j
j
(


s

u
) l

jLl
D
 l
 ,

l  1,..., ( S )
 (S )
j
j
j
(


s

u
)x

j 1
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Two Problems

Master problem
–

Subproblem
–

finds the best sequence
finds the best production times, idle times, and cycle length
Key idea: think of them separately
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Subproblem (lot sizing)
l

1  1 l l
Q
l
min
h
(
Q

D
) l


l l
x , ,u
x  l 1 2
D
Subject to
 l 2  l
 ( )   c 
l 1


l

Q
j
j
j
(


s

u
) l

jLl
D
 l
 ,

l  1,...,

j
j
j
(


s

u
)x

j 1
But first: determine a sequence
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Master Problem

Sequencing complicated

Heuristic approach

Frequency Fixing and Sequencing (FFS)

Focus on how often to produce each item
–
Computing relative frequencies
–
Adjusting relative frequencies
–
Sequencing
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Step 1: Computing Relative Frequencies

Let yk denote the number of times item k is produced in a cycle

We will
–
simplify the objective function by substituting
1
ak  hk (qk  g k )  k
2
–
drop the second constraint  sequence no longer important
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Rewriting objective function
l
1  1 l l
l Q
min
h (Q  D )  l


l l
x , ,u
x  l 1 2
D
 l 2  l
 ( )   c 
l 1


item k: yk times
 Qk
1 n
1
   yk hk (Qk  Dk ) 
x  k 1 2
 Dk
substitute: ak , ρk

 k2 n
1 n
   ak yk 2   ck yk 
x  k 1
ρ k k 1

substitute:  k

 2 n
 k   ck yk 
k 1


n
n
yk
x
  ak
  ck
yk k 1 x
k 1
Assumption: for each item
production runs of equal length and evenly spaced
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Mathematical Program
n
min
yk , x
ak x n ck yk


x
k 1 yk
k 1
n
Subject to
sk yk
 1 ρ

x
k 1
Remember: for each item
production runs of equal length and evenly spaced
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43
Solution
ak
yk  x
ck  sk

Using Lagrange multiplier:

Adjust cycle length for frequencies

Idle times then  = 0

No idle times, must satisfy

ak
 sk

ck   sk
k 1 
n
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
  1  

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Step 2: Adjusting the Frequencies

Adjust the frequencies such that they are
–
integer
–
powers of 2
–
cost within 6% of optimal cost
–
e.g. such that smallest yk =1
yk' , and

 k'
New frequencies and run times
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Step 3: Sequencing

Variation of LPT

Calculate


'
ymax
 max y1' ,..., yn'

'
'
y
y
Consider the problem with max machines in parallel and k jobs
'
of length  k

'
'
List pairs ( yk ,  k ) in decreasing order

Schedule one at a time considering spacing

Only if for all machines assigned processing time <
x
y
'
max
then
equal lot sizes possible
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46
Topic 4
Lot Sizing on Multiple Machines
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Multiple Machines

So far, all models single machine models

Extensions to multiple machines
–
parallel machines
–
flow shop
–
flexible flow shop
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48
Parallel Machines

Have m identical machines in parallel

Setup cost only

Item process on only one machine

Assume
–
rotation schedule
–
equal cycle for all machines
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Decision Variables

Same as previous multi-item problem

Addition: assignment of items to machines

Objective: balance the load

Heuristic: LPT with
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k 
gk
qk
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Different Cycle Lengths

Allow different cycle lengths for machines

Intuition: should be able to reduce cost

Objective: assign items to machines to balance the load

Complication: should not assign items that favor short cycle to
the same machine as items that favor long cycle
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Heuristic Balancing

Compute cycle length for each item

Rank in decreasing order

Allocation jobs sequentially to the machines until capacity of
each machine is reached

Adjust balance
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Further Generalizations

Sequence dependent setup

Must consider
–
preferred cycle time
–
machine balance
–
setup times

Unsolved

General schedules  even harder!

Research needed :-)
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53
Flow Shop

Machines configured in series

Assume no setup time

Assume production rate of each item is identical for every
machine
 Can be synchronized
m

Reduces to single machine problem
ck   cik
i 1
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54
Variable Production Rates

Production rate for each item not equal for every machine

Difficult problem

Little research

Flexible flow shop: need even more stringent conditions
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