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Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday New homework Review

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Page 1: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

Session 13University of Southern California

ISE514 October 6, 2015

Geza P. Bottlik Page 1

Outline

• Questions?

• Quiz Results

• Exam in this classroom on Thursday

• New homework

• Review

Page 2: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

Session 13University of Southern California

ISE514 October 6, 2015

Geza P. Bottlik Page 2

Quiz Results

Page 3: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

Session 13University of Southern California

ISE514 October 6, 2015

Geza P. Bottlik Page 3

Quiz Results

Page 4: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Production System Schematic

Operation or plant

Demand Supply Master Production Schedule (MPS)

MRP explosionCapacity Planning

Material Orders

Daily ScheduleStatus

OutputClosed loop check

Page 5: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Adding up the costs (example)

A plant generates 270,000 earned hours per year (established by standards for each of the products produced). 85% efficiency is assumed.

The cost of an hour of labor, including benefits, is $30

Indirect labor totals $30M per year

Materials cost $70M per year

Material overhead costs are $4M per year

What is the cost of a product containing 0.8 hours of standard labor and $20 of material?

We first calculate the total labor rate as (30,000,000+270,000*30/0.85)/270,000=$146.40/hour

Page 6: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Adding up the costs (example continued)

Material overhead = $4M/$70M= 5.7%

Direct and indirect labor =0.8* $146.40=$117.12

Material =$20.00

Material Overhead= 0.057*20 =$1.14

Total Cost =$138.26

We can separate the direct and indirect labor into:

Direct labor = 0.8*30 =$24

Indirect =$93.12

And you can see why everyone attacks overhead

If you are independent, the profit would add another 10% or so. It is very dependent on the industry and level of investment

Page 7: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Cost Distribution

Direct Labor16%

Labor OH61%

Material13%

Material OH1%

Profit9%

Cost Distrubution @ 10% Profit

Page 8: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

Session 13University of Southern California

ISE514 October 6, 2015

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Access Data Base

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ISE514 October 6, 2015

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Hayes-Wheelwright

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ISE514 October 6, 2015

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Master Production Scheduling (continued)

Creating a Master Production Schedule

– Select items to be included (levels)

– Select planning horizon

– Select method for available to promise (ATP)

• ATP = the uncommitted portion of a company’s

inventory or planned production (one method is

called “cumulative with lookahead” )

– Combine inventory, orders, forecasts etc.

Page 11: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

Session 13University of Southern California

ISE514 October 6, 2015

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Capacity Planning

Evaluated at

MPS level - Rough cut capacity planning (RCCP)

MRP level - Capacity requirements planning (CRP)

“..capacity planning is somewhat misleading. Both RCCP and CRP are information tools rather than decision making tools. They indicate which capacity constraints are violated, but they do not provide guidance for resolving the conflict… They ignore actual lead times”

from Sipper and Bulfin

Page 12: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Inventory

kC

ADQ

2

Page 13: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Inventory

Safety stock methods

Service level

% Stockout

Cost of a Stockout

Inventory Policies

(s,Q)

(s,S)

(R,S)

(R,s,S)

Page 14: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Shop Floor Control

Many MRP packages include Shop Floor Control modules

These are designed to track the progress of products through a factory

The best of them, especially when tied to real time data gathering are very good at letting managers now where things are. They are very dependent on timely input of data

The newer ones will often include scheduling software as well. It is, however, more common to have the scheduling software in a separate module

Page 15: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Benefits and shortcomingsBenefits:

Ability to evaluate the feasibility and requirements

Material plan

Identifying shortages

Shortcomings

Infinite capacity assumption

Uncertainty - deterministic system

Lead time discrepancies

Yield estimates

System nervousness due to rolling horizon

Integrity of data

Page 16: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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Introduction to SchedulingThe general job shop problem

All problems are subsets or relaxations of basic assumptions

Organized research and study of this area followed W.W.II

n jobs {J1, J2, ….Jn}

a job is a task or lot or batch that is to processed as a unit. If it is broken into several jobs, we restart the problem with each part a new job

m machines {M1, M2, ….Mm}

a machine is a processor or resource that performs a specific function required by a job

Page 17: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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Introduction to Scheduling (continued)Processing of a job on a machine is called an operation

nm operations oij is the ith job on the jth machine

Each job passes through each machine once and only once (Real jobs may repeat or skip)

Each job has a prescribed order in which it is processed by the machines - This is called its Technological Constraint (TC). Other names are processing order, routing, process plan.

General Job Shop - each job has its own processing order

Flow Shop - all jobs have the same processing order

Permutation Shop - all machines see the jobs in the same order

Page 18: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Introduction to Scheduling (continued)

A production line with all machines tied together is an example of a permutation shop

If all jobs have the same processing order, but the machines are not tied together, we have a Flow shop, because some machines may not see all jobs in the same order.

Page 19: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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Examples

J2

J1

J3

J2

J1

J3

M1

M2

M1

M2

M1

M2

Flow Permutation

Flow non-Permutation

General

J2

J1

J3

J1

J3

J2

J1

J2

J2

J1

J1

J2

J2

J1

Page 20: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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Notations and definitionsdi - due date of Job i

ri - ready time of Job i

ai - allowance = di - ri

si - slack = di - remaining operations

pij - the time required to process oij

Wik - the waiting time of Job i preceding its kth operation (not the work on Mk):

J1’s time = W11+p11+W12 +p13+W13 +p12+W14+p1 17, if the TC is M1 to M3 to M2 to M17

We designate the kth operation as oij(k)

m

kiki WW

1

Page 21: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

Session 13University of Southern California

ISE514 October 6, 2015

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Notations and definitions (continued)

Ci is the completion time of Job i

Fi is the flow time of Job i = Ci - ri

Even though the English words have an identical meaning, we distinguish between Lateness and Tardiness

Lateness Li =Ci - di, therefore maybe positive or negative depending on whether we complete a job before or after its due date

m

kkijikii pWrC

1)( )( )( )(kijikii pWrC

Page 22: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Notations and definitions (continued)

Tardiness is non- zero only if the job is completed after its due date:

Ti = max{Li, 0}

We also define Earliness as Ei = Max{-Li, 0}

The weight or importance of a job is indicated either by wi or

Some of our definitions refer to instants in time

Completion, readiness

Others refer to elapsed time

Processing, Waiting, Flow

i

Page 23: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Notations and definitions (continued)

Scheduling - the ordering of operations subject to restrictions and providing start and finishing times for each operation

Closed Shop - serves customers from inventory (make to stock)

Open shops - Jobs are made to order

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ISE514 October 6, 2015

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Measures

Optimality or goodness of schedules only makes sense if we define the measure under which we are considering optimality or goodness.

There are three broad categories of measures:

Completion time

Due dates

Inventory or utilization

We also define a general class of measures called regular measures

Page 25: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Measures (continued)

Given two sets of completions times obtained under two schedules generated for the same problem:

C and C’

if Ci <= Ci’ implies that R(C)<=R(C’) then R is regular

Page 26: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Classification notation

All problems can be classified as n/m/A/B where

n - number of jobs

m - number of machines

A - pattern

F - Flow Shop

P - Permutation

G - General Job Shop

B - Measure

Cmax, Fmax etc.

Page 27: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Some further definitions

Jobs and ready times fixed = Static

Parameters known and fixed = Deterministic

Random arrival of jobs = Dynamic

Uncertain processing times = stochastic

Page 28: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Kinds of scheduling

Taking sequences and placing them in a schedule is called time tabling (creating a Gantt chart)

Semiactive - process each job as soon as it can be (slide to the left on the chart)

Active - No operation can be started earlier without delaying some other operation

Non-delay - no machine is kept idle

Non-feasible - does not meet Technological Constraints

Number of possible schedules (including non-feasible ones) = (n!)m

Page 29: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Optimality

Since for any given problem there are a countable number of possible schedules (as long as we do not allow preemption or unnecessary delays) there must be an optimum (or optima) because we can (theoretically) compare all possible schedules and select the best one

If we look at the space that contains our schedules and attempt to locate the optimum we find that:

Page 30: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Optimality (continued)

Optimal

all possible

feasible

semi active

Active

non - delay

Page 31: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

Session 13University of Southern California

ISE514 October 6, 2015

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Kinds of scheduling

Taking sequences and placing them in a schedule is called time tabling (creating a Gantt chart)

Semiactive - process each job as soon as it can be (slide to the left on the chart)

Active - No operation can be started earlier without delaying some other operation

Non-delay - no machine is kept idle

Non-feasible - does not meet Technological Constraints

Number of possible schedules (including non-feasible ones) = (n!)m

Page 32: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Optimality

Since for any given problem there are a countable number of possible schedules (as long as we do not allow preemption or unnecessary delays) there must be an optimum (or optima) because we can (theoretically) compare all possible schedules and select the best one

If we look at the space that contains our schedules and attempt to locate the optimum we find that:

Page 33: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Optimality (continued)Non-delay is shown as containing the optimum

but this is not always true

Optimal

all possible

feasible

semi active

Active

non - delay

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ISE514 October 6, 2015

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Schedule Generation

As a start, we will define a routine that will generate an active schedule

A semiactive schedule is one that starts every job as soon as it can, while obeying the technological and scheduling sequences. Also, the set of all semiactive schedules for a problem contains the optimal schedule

Fortunately, the set of active schedules also contains the optimum and is a smaller set.

We can forget about generating semiactive schedules

Page 35: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Active scheduling

For a given problem there will be many active schedules

The routine we will use generates only one and we will have to make frequent choices. Were we to follow each of these decision paths, we would generate all the active schedules and find the optimum

However, our purpose here is to make those choices as intelligently as possible, even though it is difficult to foresee their eventual consequence

An active schedule is one in which no operation could be started earlier without delaying another operation or violating the technological constraints

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ISE514 October 6, 2015

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Definitions

First we will define some terminology useful for our routine:

Class of problems - n/m/G/B with no restrictions

Stage - step in the routine that places an operation into the schedule - there are therefore nm stages

t - counter for stages

Pt - partial schedule at stage t

Schedulable operation - an operation with all its predecessors in Pt

St - set of schedulable operations at stage t

Page 37: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

Session 13University of Southern California

ISE514 October 6, 2015

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Definitions (continued)

sigmak - the earliest time an operation ok in St could be started

phik - the earliest time that ok in St could be finished

phik = sigmak + pk

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ISE514 October 6, 2015

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Routine by Giffler and Thompson1. t = 1, S1 is the set of first operations in all jobs

2. Find min{phik in St} and designate it phi*

Designate M on which phi* occurs as M* (could be arbitrary)

3. Choose oj in St such that it satisfies these conditions:

a. It uses M*

b. sigmaj < phi*

4. a. Add oj to Pt, which now becomes Pt+1

b. Delete oj from St which now becomes St+1

c. Add the operation that follows oj in the same job to St+1

d. Increment t by 1

Page 39: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Routine by Giffler and Thompson (continued)

5. If there are operations left to schedule, go to step 2, else stop

Note well that at step 3b. sigmaj < phi*, we will often have several choices. We always have at least one, namely, phi*

These choices are an extensive topic that we will cover later

Follow the example I have taken from French

Generating these schedules is tedious work, so leave yourselves some extra time for that homework.

Page 40: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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Non-delay schedules

Non-delay schedules are a smaller set than the active schedules and therefore are a tempting set to explore

Unfortunately, they do not always contain the optimum

We will not let that deter us, because non-delay schedules have been found to be usually very good, if not optimal

A non-delay schedule is one where every operation is started as soon as it can be

Page 41: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

Session 13University of Southern California

ISE514 October 6, 2015

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Non-delay schedules (continued)

We change two steps in the procedure for active schedules to obtain a non-delay procedure:

Step 2. instead of phi, we select sigma

Find min{sigmak in St} and designate it sigma*

Designate M on which sigma* occurs as M* (could be arbitrary)

Step 3 b. sigmaj = sigma*

Page 42: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

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ISE514 October 6, 2015

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SPT

Minimizes Fbar

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Session 13University of Southern California

ISE514 October 6, 2015

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EDD

Minimizes Tmax and Lmax

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ISE514 October 6, 2015

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Moore

Minimizes number of tardy jobs

Page 45: Session 13 University of Southern California ISE514 October 6, 2015 Geza P. Bottlik Page 1 Outline Questions? Quiz Results Exam in this classroom on Thursday

Session 13University of Southern California

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Smith - modified

Minimizes Fbar subject to Tmax <=Given Constant

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ISE514 October 6, 2015

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Lawler

Minimizes the maximum of regular measures that are linearly increasing functions of Completion times