operational research & managementoperations scheduling economic lot scheduling 1.summary machine...

Operational Research & Management Operations Scheduling

Economic Lot Scheduling

1. Summary Machine Scheduling

2. ELSP (one item, multiple items)

3. Arbitrary Schedules

4. More general ELSP


Topic 1

Summary Machine Scheduling Problems

Operational Research & Management Operations Scheduling 3

Machine Scheduling

Single machine

Parallel machine

jjCw||1

max1|| C

1| , |j jr prmp C

1| |j jr Cmax1| |prec h

max1| , |jr prmp L

max1| |jr L

1|| jU1|| j jw U1|| jT1|| j jw T

max||Pm C

max| |P prec Cmax| |Pm prec C

|| jPm C3-partition

max1|| L1|| jC

max2 ||P C

partition


Why easy?

Show polynomial algorithm gives optimal solution

Often by contradiction

– See week 1

– See week 1

By induction

– E.g.

jjCw||1

max1| |prec h

1|| jU


Why hard?


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


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

jS 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


Some scheduling problems

Knapsack problem

Translation: n = t; pj = aj; wj = aj; d = b; z = b;

Question: exists schedule such that j wjUj z ?

Two-machine problem

Translation: n = t; pj = aj; z = b;

Question: exists a schedule such that Cmax z ?

1| |j j jd d w U

max2 ||P C


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

jS aj = b ?


Topic 2a

Economic Lot Scheduling Problem:

One machine, one item


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


Objective

Minimize total cost

– setup cost

– inventory holding cost

Trade-off

Cyclic schedules but acyclic sometimes better


Scheduling Decisions

Determine the length of runs

– gives lot sizes

Determine the order of the runs

– sequence to minimize setup cost

Economic Lot Scheduling Problem (ELSP)


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


Minimize Cost

Let x denote the cycle time

Demand over a cycle = Dx

Length of production run needed = Dx / Q = x

Maximum inventory level = (Q - D) Dx / Q = (Q - D) x

= (1 - ) D x Inventory

Time

x

( )Q D Dx

Q

idle time


Optimizing Cost

book shorter

Solve

Optimal Cycle Time

Optimal Lot Size

21min

2

c D xh Dx

x Q

2

( )

Qcx

hD Q D

2

( )

cDQgx

h Q D

1min 1

2

ch Dx

x

2

(1 )

cx

h D

2

(1 )

cDgx

h


With Setup Time

Setup time s

If s x(1-) above optimal

Otherwise cycle length is optimal1

sx

nutilizatioq

g


Topic 2b

Economic Lot Scheduling Problem:

Lot Sizing with Multiple Items

With Rotation Schedule


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


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


Optimal Cycle Length

Average total cost

with

Solve as before

1

nj

jj

ca x

x

1 1

n n

j jj j

x c a

1(1 )

2j j j ja h D

jjjj

ca x

x


Example

Items 1 2 3 4

_qj 400 400 500 400

_gj 50 50 60 60

_hj 20 20 30 70

_cj 2000 2500 800 0


Data of example 7.3.1

1 2 3 4c(j) 2000 2500 800 0h(j) 20 20 30 70D(j) 50 50 60 60Q(j) 400 400 500 400

rho(j) 0.125 0.125 0.12 0.15a(j) 437.5 437.5 792 1785t(j) 2.14 2.39 1.01 0.00

5300jjc

3452jja


Solution

1

1 1

1

1

½ (1 )

7 44 342 10. .50 15. .60 35. .60 5300

8 50 40

3452 5300 1.5353 1.24 months

n n

j j j jj j

x c h g

jj

jj

c

a


Solution

The total average cost per time unit is

(alternative 1: )

(alternative 2: )

How can we do better than this?

8554$2213162725592155

34522 5300 3452

5300

j

j

aAC

c

2 5300

or 2 3452 1.241.24

AC


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


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

jj

n

jjs

x

1

1

1


Job Dependent Setup Times

Now there is a sequencing problem

Objective: minimize sum of setup times

Equivalent to the Traveling Salesman Problem (TSP)


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


Topic 3

Arbitrary Schedules


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


Problem Formulation

Assume sequence-independent setup

Formulate as a nonlinear program

sequences lot sizesmin min COST

s.t.

demand met over the cycle

demand is met between production runs


Notation

Setup cost and setup times

All possible sequences

Item k produces in l-th position

Setup time sl, run time (production) tl, and idle time ul

. , kjkkjk sscc

nhjjjS h :),...,,( 21

kjl qqq

l


Inventory Cost

Let x be the cycle time

Let v be the time between production of k

Total inventory cost for k is

k

lk

l

ll

g

tq

g

tqv

2)(2

1 ll

llll t

g

qgqh


Mathematical Program

( ) ( )2

, , 1 1

1 1min ( )n

2mi ) (

l l

lS Sl l l l l

ll l

S x u

Qh Q D c

x D

)

1

(

( )

, 1,...,

( ) , 1,...,

( )

k

l

jk k

j I

lj j j l

lj L

j j jS

j

Q D x k n

Qs u l

D

u x

S

s

Subject to


Two Problems

Master problem

– finds the best sequence

Subproblem

– finds the best production times, idle times, and cycle length

Key idea: think of them separately


Subproblem (lot sizing)

Subject to

2

, , 1 1

1 1min ( ) ( )

2l l

ll l l l l

lx u l l

Qh Q D c

x D

1

( ) , 1,...,

( )

l

lj j j l

lj L

j j j

j

Qs u l

D

s u x

But first: determine a sequence


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


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

– drop the second constraint sequence no longer important

1( )

2k k k k ka h q g


Rewriting objective function

2

1 1

2

21 1

1 1

ρ

1 1( )

2

1

n nk

k k k k k k kk kk

n nk

k k k kk k

n nk

k kk

k

k k

Qy h Q D c y

x D

a y c yx

yxa c

y x

item : timeskk y

substitute: , ρk ka

substitute: k

2

, , 1 1

1 1min ( ) ( )

2l l

ll l l l l

lx u l l

Qh Q D c

x D

Assumption: for each item production runs of equal length and evenly spaced


Mathematical Program

,1 1

mink

n nk k k

y xk kk

a x c y

y x

1

1 ρn

k k

k

s y

x

Subject to

Remember: for each item production runs of equal length and evenly spaced


Solution

Using Lagrange multiplier:

Adjust cycle length for frequencies

Idle times then = 0

No idle times, must satisfy

kk

kk sc

axy

1

1n

kk

k k k

as

c s


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

New frequencies and run times

' ', and k ky


Step 3: Sequencing

Variation of LPT

Calculate

Consider the problem with machines in parallel and jobs of length

List pairs in decreasing order

Schedule one at a time considering spacing

Only if for all machines assigned processing time < then

equal lot sizes possible

''1

'max ,...,max nyyy

'maxy

'k

'ky

' '( , )k ky

'max

x

y


Topic 4

Lot Sizing on Multiple Machines


Multiple Machines

So far, all models single machine models

Extensions to multiple machines

– parallel machines

– flow shop

– flexible flow shop


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


Decision Variables

Same as previous multi-item problem

Addition: assignment of items to machines

Objective: balance the load

Heuristic: LPT with k

kk q

g


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


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


Further Generalizations

Sequence dependent setup

Must consider

– preferred cycle time

– machine balance

– setup times

Unsolved

General schedules even harder!

Research needed :-)


Flow Shop

Machines configured in series

Assume no setup time

Assume production rate of each item is identical for every machine

Can be synchronized

Reduces to single machine problem

m

iikk cc

1


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

operational research & managementoperations scheduling economic lot scheduling 1.summary machine...

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