Unidad Vi Programacion 2011-1

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Unidad VI

Programacion y control de actividades en procesos intermitentes

Unidad VI

Programacion y control de actividades en procesos intermitentes

© 2006 Prentice Hall, Inc.

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Importancia Estrategica de la Programacion a Corto Plazo

La programacion eficiente y efectiva puede proporcionar a la empresa una ventaja competitiva.

El movimiento rapido de bienes a traves de la instalacion de produccion significa mejor uso de los activos y menos costos.

Capacidad adicional resulta de la entrega rapida de productos a los clientes.

Buenos programas resultan en entregas mas confiables

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Criterios de la programacion de la produccion

1. Minimizar el tiempo para completar un trabajo

2. Maximizar la utilizacion de las instalaciones

3. Minimizar el inventario de trabajo en proceso (WIP)

4. Minimizar el tiempo de espera de los clientes.

Optimizar el uso de los recursos Optimizar el uso de los recursos de tal manera que se cumplan los de tal manera que se cumplan los

objetivos de produccion. objetivos de produccion.

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Programacion o Secuenciacion de trabajos

Especificar el orden en el cual los trabajos deberan ser ejecutados en los centros de trabajo.

Reglas de prioridad seran usadas para despachar o secuenciar los trabajos.

FCFS: Primeros en llegar primeros en ser atendidos

SPT: Tiempo de procesamiento mas corto

EDD: Fecha de vencimiento mas temprana

LPT: Tiempo de procesamiento mas largo

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Ejemplo de Secuenciacion

Trabajo Tiempo de procesamiento (Dias)Fecha de entega

(Dias)

A 6 8

B 2 6

C 8 18

D 3 15

E 9 23

Aplique las cuatro reglas de secuenciacion anteriores a estos cinco trabajos

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Secuencia de trabajos

Tiempo de procesamiento

Tiempo de flujo

Fecha de entrega Tardanza

A 6 6 8 0

B 2 8 6 2

C 8 16 18 0

D 3 19 15 4

E 9 28 23 5

28 77 11

FCFS: Secuencia A-B-C-D-EFCFS: Secuencia A-B-C-D-E

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Job SequenceJob Work (Processing)

Time Flow Time Job Due Date Job Lateness

A 6 6 8 0

B 2 8 6 2

C 8 16 18 0

D 3 19 15 4

E 9 28 23 5

28 77 11

FCFS: Secuencia A-B-C-D-EFCFS: Secuencia A-B-C-D-E

Average completion time = = 77/5 = 15.4 daysTotal flow timeNumber of jobs

Utilization = = 28/77 = 36.4%Total job work time

Total flow time

Average number of jobs in the system = = 77/28 = 2.75 jobs

Total flow timeTotal job work time

Average job lateness = = 11/5 = 2.2 daysTotal late daysNumber of jobs

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B 2 2 6 0

D 3 5 15 0

A 6 11 8 3

C 8 19 18 1

E 9 28 23 5

28 65 9

SPT: Sequence B-D-A-C-ESPT: Sequence B-D-A-C-E

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B 2 2 6 0

D 3 5 15 0

A 6 11 8 3

C 8 19 18 1

E 9 28 23 5

28 65 9

SPT: Sequence B-D-A-C-ESPT: Sequence B-D-A-C-E

Average completion time = = 65/5 = 13 daysTotal flow timeNumber of jobs


Total flow time




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B 2 2 6 0

A 6 8 8 0

D 3 11 15 0

C 8 19 18 1

E 9 28 23 5

28 68 6

EDD: Sequence B-A-D-C-EEDD: Sequence B-A-D-C-E

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B 2 2 6 0

A 6 8 8 0

D 3 11 15 0

C 8 19 18 1

E 9 28 23 5

28 68 6

EDD: Sequence B-A-D-C-EEDD: Sequence B-A-D-C-E



Total flow time




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E 9 9 23 0

C 8 17 18 0

A 6 23 8 15

D 3 26 15 11

B 2 28 6 22

28 103 48

LPT: Sequence E-C-A-D-BLPT: Sequence E-C-A-D-B

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E 9 9 23 0

C 8 17 18 0

A 6 23 8 15

D 3 26 15 11

B 2 28 6 22

28 103 48

LPT: Sequence E-C-A-D-BLPT: Sequence E-C-A-D-B



Total flow time




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Rule

Average Completion Time

(Days) Utilization (%)Average Number of Jobs

in System

Average Lateness

(Days)

FCFS 15.4 36.4 2.75 2.2

SPT 13.0 43.1 2.32 1.8

EDD 13.6 41.2 2.43 1.2

LPT 20.6 27.2 3.68 9.6

Summary of RulesSummary of Rules

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Comparison of Sequencing Rules

No one sequencing rule excels on all No one sequencing rule excels on all criteriacriteria

SPT does well on minimizing flow time and SPT does well on minimizing flow time and number of jobs in the systemnumber of jobs in the system

But SPT moves long jobs to the end which But SPT moves long jobs to the end which may result in dissatisfied customersmay result in dissatisfied customers

FCFS does not do especially well (or FCFS does not do especially well (or poorly) on any criteria but is perceived as poorly) on any criteria but is perceived as fair by customersfair by customers

EDD minimizes latenessEDD minimizes lateness

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Critical Ratio (CR)

An index number found by dividing the An index number found by dividing the time remaining until the due date by the time remaining until the due date by the work time remaining on the jobwork time remaining on the job

Jobs with low critical ratios are Jobs with low critical ratios are scheduled ahead of jobs with higher scheduled ahead of jobs with higher critical ratioscritical ratios

Performs well on average job lateness Performs well on average job lateness criteriacriteria

CR = =CR = =Due date - Today’s dateDue date - Today’s date

Work (lead) time remainingWork (lead) time remaining

Time remainingTime remaining

Workdays remainingWorkdays remaining

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Critical Ratio Example

Job Due DateWorkdays Remaining Critical Ratio Priority Order

A 30 4 (30 - 25)/4 = 1.25 3

B 28 5 (28 - 25)/5 = .60 1

C 27 2 (27 - 25)/2 = 1.00 2

Currently Day Currently Day 2525

With CR < 1, Job B is late. Job C is just on schedule With CR < 1, Job B is late. Job C is just on schedule and Job A has some slack time.and Job A has some slack time.

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Critical Ratio Technique

1.1. Helps determine the status of specific Helps determine the status of specific jobsjobs

2.2. Establishes relative priorities among Establishes relative priorities among jobs on a common basisjobs on a common basis

3.3. Adjusts priorities automatically for Adjusts priorities automatically for changes in both demand and job changes in both demand and job progressprogress

4.4. Dynamically tracks job progressDynamically tracks job progress

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Sequencing N Jobs on Two Machines: Johnson’s Rule

Works with two or more jobs that Works with two or more jobs that pass through the same two pass through the same two machines or work centersmachines or work centers

Minimizes total production time and Minimizes total production time and idle timeidle time

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Johnson’s Rule

1.1. List all jobs and times for each work List all jobs and times for each work centercenter

2.2. Choose the job with the shortest activity Choose the job with the shortest activity time. If that time is in the first work center, time. If that time is in the first work center, schedule the job first. If it is in the second schedule the job first. If it is in the second work center, schedule the job last.work center, schedule the job last.

3.3. Once a job is scheduled, it is eliminated Once a job is scheduled, it is eliminated from the list from the list

4.4. Repeat steps 2 and 3 working toward the Repeat steps 2 and 3 working toward the center of the sequencecenter of the sequence

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Johnson’s Rule Example

JobWork Center 1 (Drill

Press) Work Center 2 (Lathe)

A 5 2

B 3 6

C 8 4

D 10 7

E 7 12

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A 5 2

B 3 6

C 8 4

D 10 7

E 7 12

DD AABB CCEE

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A 5 2

B 3 6

C 8 4

D 10 7

E 7 12

BB AACCDDEE

TimeTime 00 33 1010 2020 2828 3333

B ACDEWC 1

WC 2

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A 5 2

B 3 6

C 8 4

D 10 7

E 7 12

BB AACCDDEE

TimeTime 00 33 1010 2020 2828 3333

TimeTime 0 0 11 33 55 77 99 1010 1111 1212 1313 1717 1919 21 22 2321 22 23 2525 2727 2929 3131 3333 3535

B ACDE

B ACDE

WC 1

WC 2

BB EE DD CC AA

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Limitations of Rule-Based Dispatching Systems

1.1. Scheduling is dynamic and rules Scheduling is dynamic and rules need to be revised to adjust to need to be revised to adjust to changeschanges

2.2. Rules do not look upstream or Rules do not look upstream or downstreamdownstream

3.3. Rules do not look beyond due Rules do not look beyond due datesdates

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Scheduling Service Employees With Cyclical Scheduling

Objective is to meet staffing Objective is to meet staffing requirements with the minimum requirements with the minimum number of workersnumber of workers

Schedules need to be smooth and Schedules need to be smooth and keep personnel happykeep personnel happy

Many techniques exist from simple Many techniques exist from simple algorithms to complex linear algorithms to complex linear programming solutionsprogramming solutions

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Cyclical Scheduling Example

1.1. Determine the staffing requirementsDetermine the staffing requirements

2.2. Identify two consecutive days with the Identify two consecutive days with the lowest total requirements and assign lowest total requirements and assign these as days offthese as days off

3.3. Make a new set of requirements Make a new set of requirements subtracting the days worked by the first subtracting the days worked by the first employeeemployee

4.4. Apply step 2 to the new row Apply step 2 to the new row

5.5. Repeat steps 3 and 4 until all Repeat steps 3 and 4 until all requirements have been metrequirements have been met

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Cyclical Scheduling ExampleCyclical Scheduling Example

MM TT WW TT FF SS SS

Employee 1Employee 1 55 55 66 55 44 33 33

Capacity (Employees)Capacity (Employees)

Excess CapacityExcess Capacity

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Employee 7Employee 7 11



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Employee 7Employee 7 11

Capacity (Employees)Capacity (Employees) 55 55 66 55 44 33 33

Excess CapacityExcess Capacity 00 00 00 00 00 11 00

Unidad Vi Programacion 2011-1

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