production activity control. ( 生產活動控制 ) introduction mrpcrp order ( 工單 ) release...

Production Activity Control

Production Activity Control(生產活動控制 )

•Introduction

MRP CRP

Order ( 工單 ) release planning

Order release ( 工單發放 )

Dispatching ( 派工 )

Production report ( 生產績效 )

I/O ControlPriority Control


•Objectives

(1) 確保工單能如期完成。(2) 瞭解並掌握現場所有的製造資訊 ─ 機台狀況，品質狀況，工單進度。

PAC

I/O Control

Priority Control

現場資訊收集

Orders

Priority (Dispatching)Rule

現場資訊 (機台狀況，品質狀況，工單進度 )

Routing File, Work Center File

投入 /產出報表

派工單

生產績效報表

•Goals of Priority Control

Minimal WIPShort CustomerLead Time

MaximumUtilization of

Resources

On-Time Delivery

Balance


•Meeting Due Dates - Performance Measures

•Service Level - It is the fraction of jobs whose cycle time is less than or equal to their planned lead time. (make to order)

•Fill Rate - It is defined as the fraction of demands that are met from inventory. (make to stock)

•Lateness - The difference between the order due date and the completion time of job.

•Tardiness - It is the lateness of a job is it is late and zero otherwise.


•Maximizing Utilization

Utilization

capital equipment

Return onInvestment

Cost Accounting

Revenue



Utilization

capital equipment

Inventory

Profit ???



有效產出

存貨

營運費用

生產管理目標



資源

瓶頸資源

非瓶頸資源

Note: 任何資源，只要它的產能等於或小於它的需求則稱之為瓶頸資源。



瓶頸資源非瓶頸資源

通過瓶頸的流量 = 市場需求平衡流量

瓶頸資源非瓶頸資源

瓶頸資源

非瓶頸資源組裝部門



找出系統的瓶頸

決定如何利用瓶頸

根據上述決定，調整其它一切

把系統瓶頸鬆綁

出現新瓶頸

(Maximizing Utilization)


•Maximizing Utilization - Performance Measure

Makespan - Defined as the time it takes to finish a given fixed number of jobs

Production Rate = Number of jobs / Makespan

Utilization = Production Rate / Capacity


•Reducing WIP and Cycle Time

Cycle Time (Flow Time) = WIP / Throughput

Little’s Law

Note: 1. The cycle time of a given routing is the average time from release of a job at the beginning of the routing until it reaches an inventory point at the end of the routing 2. Throughput is the average output of a production process per unit time.


•Review of Scheduling Research - Batch Flow Lines

)/(

)( ) Time,(Runout

時間數量需求率數量存貨水準耗竭時間 R

Period Product EOQ Cycle Time 每週需求量目前存貨 R

1

ABCDE

100080015001800600

1.21.01.92.01.6

250300500800300

600400

18002300600

2.41.33.6

2.8751.67

2

ABCDE

100080015001800600

1.21.01.92.01.6

250300500800300

350900

13001500300

1.43

2.61.875

1


•Review of Scheduling Research - Job Shop Scheduling

Assumptions

1. Address one, two, or possibly three machines.2. All jobs are available at the start of the problem.3. Process times are deterministic.4. Process time do not depend on the schedule (no setup).5. Machines never break down.6. There is no preemption (i.e., once a job starts processing it must finish).7. There is no cancellation of jobs.



(1) FCFS - Fist Come, First Served(2) SPT - Shortest Processing Time (Minimizing average cycle time on a single machine)

(3) EDD - Earliest Due Date (Minimizing maximum lateness on a single machine)

(Minimizing average tardiness on a single machine)

(4) CR - Critical Ratio

Priority (Dispatching) Rules - Job works in a single workstation



•Priority (Dispatching) Rules

Ex:

Job Processing Time Due DateA 2 7B 8 16C 4 4D 10 17E 5 15F 12 18

Work Center 33



(1) FCFS

Sequence ProcessingTime

Flow Time Due Date 延誤時間

A 2 2 7 0B 8 10 16 0C 4 14 4 10D 10 24 17 7E 5 29 15 14F 12 41 18 23

Total 41 120 54

平均完成時間 = 206

120

#

Jobs

總流程時間

平均延誤時間 = 96

54

#

Jobs

總延誤時間

The length of time a job is in the shop or at a particular workstationor work center



(2) SPT



A 2 2 7 0C 4 6 4 2E 5 11 15 0B 8 19 16 3D 10 29 17 12F 12 41 18 23

Total 41 108 40

平均完成時間 = 186

108

#

Jobs

總流程時間

平均延誤時間 = 67.66

40

#

Jobs

總延誤時間



(3) EDD



C 4 4 4 0A 2 6 7 0E 5 11 15 0B 8 19 16 3D 10 29 17 12F 12 41 18 23

Total 41 110 38

平均完成時間 = 33.186

110

#

Jobs

總流程時間


38

#

Jobs

總延誤時間



(4) CR

剩餘前置製造時間現在時間到期日

CR

CR(A) = (7-0)/2 = 3.5CR(B) = (16-0)/8 = 2.0CR(C) = (4-0)/4 = 1.0CR(D) = (17-0)/10 = 1.7CR(E) = (15-0)/5 = 3.0CR(F) = (18-0)/12 = 1.5

CR 愈小愈緊急



C 4 4 4 0F 12 16 18 0D 10 26 17 9B 8 34 16 18E 5 39 15 24A 2 41 7 34

Total 41 160 85

平均完成時間 = 67.266

160

#

Jobs

總流程時間


85

#

Jobs

總延誤時間



(4) CR


Sequencing jobs through two work centers

Job M1 (Machine 1) M2 (Machine 2)A 5 5B 4 3C 8 9D 2 7E 6 8F 12 15

Ex:



•Johnson’s Rule

Johnson’s rule is a technique that can be used to minimize the throughput time for a group of jobs that are to be processed ontwo machines or at two successive work centers. It alsominimizes the total idle time at the work centers.

Determination of the optimum sequence involves these steps:

1. List the jobs and their times at each work center.2. Select the job with the shortest time. If the shortest time is at the first work center, schedule that job first; if the time is at the second work center, schedule the job last. Break ties arbitrarily.3. Eliminate the job and its time from further consideration.4. Repeat steps 2 and 3, working toward the center of the sequence, until all jobs have been scheduled.


•Johnson’s Rule

Ex:

D E C F A B

C(8) F(12)

D(7) C(9)

M1

M2

Waiting for D Waiting for F

0 2 8 16 28 33 37 51

0 2 9 17 26 28 43 48 51

E(6) A(5) B(4)

E(8) F(15) A(5) B(3)

Throughput Time = 51

•Johnson Rule - Extension to Three Machines

A B C

iiii BCBA maxminor maxmin

Job A B C

1 4 5 82 9 6 103 8 2 64 6 3 75 5 4 11

Machine


Job A’ B’

1 9 132 15 16 3 10 84 9 105 9 15

Machine

Note: A’ = A + B B’ = B + C

•Johnson Rule - Extension to Three Machines



•多機狀況

(1) Analytical Approach - Branch and Bound, Dynamic Programming

(2) Heuristic Approach ( 演算法 )

(3) Simulation Approach


•Integer Programming Approach (Branch and Bound)

r

rr

rr

Jiir

iJi

Jiir

iiJi

Jiir

r

pJTime

ppJTime

pppJTime

JLBpLB

3

32

321

)(3

)(min)(2

)(min)(1

max)()(

Ex: 3 machine flow shop

Note: Jr = 已安排的 r 個 jobs



Job Machine 1 Machine 2 Machine 3

1 12 5 132 6 10 33 9 11 184 17 16 4

Ex:



0

1 2 43

31 3213 14

3412

LB(1)=57 LB(2)=62 LB(3)=58 LB(4)=71

LB(12)=64 LB(13)=61 LB(14)=70

LB(31)=58

LB(32)=65 LB(34)=62

312 314

尋找最小者作 Branch

Production Activity Control( 生產活動控制 )

Lower Bound 計算方法實例 :

64

55)418(33

594)1611(28

64)416()79(18

max)12(

57

55)4183()13512(

573)161110()512(

57)310()1796(12

max)1(

LB

LB


•Minimizing makespan in job shops

- Minimizing the time to complete n jobs with general routing through m machines

There are (n!) m possible schedules.

Branch and Bound


•Why Schedule is Hard

Slow computer: examine 1000000 sequences per secondFast computer: 1000 times faster than slow computer

Number of Jobs Computer Time56789101112131415…20

0.12 millisec0.72 millisec5.04 millisec40.32 millisec

0.36 sec3.63 sec39.92 sec7.98 min1.73 hr24.22 hr

15.14 day…

77147 years

Number of Jobs Computer Time56789101112131415…20

0.12 microsec0.72 microsec5.04 microsec40.32 microsec362.88 microsec

3.63 millisec39.92 millisec

479.00 millisec6.23 sec87.18 sec21.79 min

…77.147 years

Slow Computer Fast Computer

NP-Hard


•The Bad News

1. There are always more than two machine. - The Johnson’s rule is not directly useful.

2. Process Times are not deterministic.

3. All jobs are not ready at the beginning of the problem. - New jobs do arriving during the entire life of the plant.

4. Process times are frequently sequence dependent.

5. We cannot hope to find optimal solutions of many real-size scheduling problems.


•The Good News

The formalized scheduling problems addressed in the operations research literature are models, not reality. - The constraints assumed in these models are not necessary fixed in the real world (control the problem by controlling the environment).

Setup Time

Reduce


•Useful Insights

•Due Date - We do have some control over due dates.

•Lot splitting - large process batches and small move batches

•Feasible schedules - An optimal schedule is really only meaningful in a mathematical model. In practice what we need is a good, feasible one.

•Focus on bottlenecks - Bottleneck resources can dominate the behavior of a manufacturing system, it is typically most critical to schedule these resources well.

•Capacity - We have some control over capacity.


•Heuristic Approaches for Makespan Problem


Palmer: Give priority to jobs having the strongest tendency to progress from short times to long times in the sequence of operations

Ex: Job (j) 1 2 3 4 5

m=3 Pj1 6 4 3 9 5

Pj2 8 1 9 5 6

Pj3 2 1 5 8 6



Palmer:

Calculate a slope index, Sj, for each job

12

2,1,

)1()3(

)5()3()1(

jj

mjmjjmj

pmpm

pmpmpmS

Order job by

][]2[]1[ nSSS

Forget negative term



Palmer:

2,2,4,6,8

22

54321

13

SSSSS

ppS jjjEx:

Order jobs as follow 3-5-4-2-1

•Simulation-Based Scheduling

Simulation model of

the entire system

Various dispatching rulefor each machine

Evaluate

“Best” Schedule


•Genetic Algorithm



•I/O Control

PAC 必須平衡 work center 的工作投入量與產出量

控制投入速率

控制產出速率

工作中心

未發放工單

產出

WIP 量

( 核發工單多寡 )

( 增加減少產能 )

production activity control. ( 生產活動控制 ) introduction mrpcrp order ( 工單 ) release...

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