VII- Flow Manufacturing Systems نظام التصنيع المتدفق

VII- Flow Manufacturing Systemsالمتدفق التصنيع نظام

2- Flow System Performance Analysis

المتدفق النظام خطوط أداء تحليل

IE 469 Manufacturing Systemsصنع نظم التصنيع 469

1a- Introduction مقدمة

The manners of performance analysis تحليل أنماطيتعلق A manner related to production technology -1 األداء نمط

اإلنتاج بتقنية

This is concerned with the design of production line and include: Manufacturing process & equipment التصنيع اسلوب

ومعداتها Manufacturing material التصنيع مواد

There are many problems related to this concern such as:

المثال سبيل وعلى النمط لهذا المسائل من العديد وهناك:التالي

Manufacturability للتصنيع قابلية Machinability - قابليةالتجميع Assemble-ability – التشغيل قابلية

Manufacturing economics التصنيع اقتصاديات Tooling and Fixture design والمثبتات األدوات تصميم Time study and process design األساليب وتصميم الوقت دراسة Operation and process automation واساليبها العمليات أتمتة Equipment high production rate ومعدالت المعدات تقنيات

انتاجها material handling design وتقنياتها المناولة طرق تصميم

1b- Introduction مقدمة

The manners of performance analysis تحليل أنماطالخط A manner related to line capability -2 األداء بمقدرة يتعلق نمط

This is concerned with the design of production line and include: Line design الخط تصميم Line operation الخط تشغيل

There are many problems related to this concern such as:المثال سبيل وعلى النمط لهذا المسائل من العديد وهناك

التالي:• line balancing الخط اتزان• Material flow المواد تدفق• Resource requirement (No, of Machines,…) المصادر متطلبات• Layout design المخططات تصميم• Line reliability الخط موثوقية• Buffer design (no. and capacity) المؤقت التخزين تصميم• Loading, scheduling, and sequencing of machines وجدولة تحميل

الخط لمعدات والتسلسل

2a- General Terminology عامة تعاريفi) Ideal Cycle Time TC = TO + TH [1]

where; TC = Cycle Time TO = Longest Operation Time TH = Transfer Time

TO TH

TCt

ii) Machine Cycle Time TC = TO + TH + Ti [2] where; Ti = Idle Time

TO TH

TCt

Ti

iii) Actual Production Time TP = TC + Td [3] where; Td = Average Down Time

TO TH

TC tTi Td

TP

This time occurs due to reasons such as: من العديد نتيجة الزمن هذا يحدثالتالي مثل :األسباب

Tool failure المستخدمة األداة أو Tool Adjustment --- إخفاق األداة تعديلالجدولة Scheduled tool change---تغيرها خطة مقيد Part jam --- تغير --- جزءFeed mechanisms failure تغذية آلية أعمال Maintenance work --- إخفاقمكون Malfunction of a electrical or mechanical component --- صيانة قصور

ميكانيكي أو كهربي

2b- General Terminology عامة تعاريف

vi) Line Efficiency

E = TC / TP = TC / (TC + TD) [7] vii) Line Deficiency

D = TD / TP = TC / (TC + TD) [8]

v)Average Down Time

TD = F x Td [6]• F = Downtime Frequency, Line stops/cycle. • Td = Down Time/line stop,min.

viii)Cost of piece

Cpc = Cm + Cl x TP + Ct [9] Cm = Material cost Cl = Operation cost

Ct = Tooling and nonproductive cost

iv) Production rate

a) Ideal Production Rate RC = 1/TC [4]

b) Actual Production Rate RP = 1/TP [5]

3a- Transfer Line Performance االنتقال خط أداء تحليل للتصنيعBreak-Down Occurrence

Work-piece removal

from station

Not Removed

Upper bound

It estimates the upper limit on the frequency of line stops per cycle.

i.e. ; Feed mechanism failure, Tool changing, Modification of Fixture

Lower bound

Removed

It estimates the lower limit on the expected number of line stops per cycle.

i.e. ; Part damage failure, Tool breakdown, …etc.

3b- Transfer Line Performance االنتقال خط أداء تحليلللتصنيع

A) The performance Analysis with upper bound

i) The downtime frequency of line stops per cycle, F:

stations ofNumber ------

station particular aat jam y willProbabilit where;

10

21

1

n

i

nF

ni

i

n

ii

ii) The production Rate, RP:

11 1 PP TR

3c- Transfer Line Performance االنتقال خط أداء تحليلللتصنيع

B) The performance Analysis with lower bound

ii) The production Rate, RP:

13 1 PP TFR

i) The downtime frequency of line stops per cycle, F:

12 1111

,....,2,1 , 1

111......11.....11

1

1

211

F

ni

ni

n

i

i

n

i

iinn

Probability that the part will pass through from station1 to station n

3d- Example (1) Transfer Line Performance

Station 1 2 3 4 5 6 7 8 9 10 11 12

Process time, min 0.6 1.1 1.3 0.8 1.1 0.8 0.8 0.5 1.1 0.7 0.5 0.9

Downtime Occurrence 0 20 23 27 37 14 28 52 35 26 36 0

Transfer time, min. 0.1

12 station transfer line performs the machining operations according the data given in the table below. For a period of which 3000 parts are produced, the average repair time is 7.0 minutes.Using upper bound and lower bound approaches, determine the following:-

a. Average production rate. b. Line efficiency. c. How many hours were required to produce the 3000 parts?

3e- Example (1) Transfer Line Performance

SolutionCycle Time (TC) =1.3+0.1=1.4 min.Since the line downtime occurrence is given for period of producing 3000 parts, then The downtime frequency of failure of station (pi) = Downtime occurrence/Number of Parts

Solution 1A (Upper bound)

Total frequency of failure 09933.03000

29812

1 i ipF

Production Time(TP)= TC + F x TD =1.4 + 0.09933x7 =2.095332.09533 min.

A)Production Rate (RP ) = 1/ TP =1/2.09533 = 0.47730.4773 part/min.

B)Efficiency (E0) = TC / TP = 1.4/2.09533 = 0.66820.6682

C)Total Production Time=NP x TP =3000x2.009533/60 = 104.767104.767 hr.

3f- Example (1) Transfer Line Performance

Solution 1A (Lower bound) ,Total downtime frequency of failure/line stop i

ipF

11

12

1Station 1 2 3 4 5 6 7 8 9 10 11 12

Process time,min 0.6 1.1 1.3 0.8 1.1 0.8 0.8 0.5 1.1 0.7 0.5 0.9

Downtime Occurrence 0 20 23 27 37 14 28 52 35 26 36 0

Frequency failure, (pi) 10-3

0 6.7 7.7 9 12.3 4.7 9.3 17.3 11.7 8.7 12 0

0951.09049.1

988.09913.09883.09827.09907.09953.09877.0991.09923.09933.01

F

Production Time (TP) = TC + F x TD =1.4 + 0.0951x7 =2.0662.066 min.

A) Production Rate (RP ) = (1-F)/ TP =1/2.066 = 0.4380.438 part/min.

B) Efficiency (E0) = TC * RP = 1.4/2.2831 = 0.61420.6142

C) Total Production Time = NP x TP = 3000x2.2831/60 = 114.16114.16 hr.

3g- Transfer Line Performance خط أداء تحليلللتصنيع االنتقال

A) Relationship between line efficiency( E) and number of work-stations (n) for various work-station break-down rate

10 20 30 40 50n

0.2

0.4

0.6

0.8

1.0E

001.0

005.0

01.0

From figure the efficiency reduces as number of station increases and/ or probability of failure increases

وزيادة المحطات عدد بزيادة كفاءته تقل االنتقال خط أن الشكل من يتضح , يؤثر حيث بعضها على المحطات العتماد نظرا وذلك اعطاله معدالت

. العمل عن الخط توقف على المحطات إحدى فشل

4a- Transfer Line Break-down concept خط في التعطل االنتقال

A) Reasons of line failure during operation

العمل عن ا الخط التشغيل ويتعطل فترة لألسباب خاللالتالية:-

• Machine failure من مجموعة أو أحد في المعدة فشلالعمل محطات

• transfer mechanism failure بين االنتقال آلية فشلالمحطات

• Control devices failure المحطة في التحكم أنظمة محطة B) The causes of failure of a station (i) فشل فشل حالة تسببالتالي :في

• StarvingStarving التعطش التعطش حالة Stations (i+1) to n stop because no part حالةis arrived as station (i) fails

• BlockingBlocking االنسداد االنسداد حالة Stations (1) to (i-1) stops because no حالةpart can be passed as station (i) fails

4b- Transfer Line Break-down concept خط في التعطل االنتقال

C) Line efficiency:

The line efficiency depends on the time to failure and time of repair (i.e. Failure and repair Rates). Hence the efficiency can be defined as follow:

اإلصالح وزمن للفشل التشغيل زمن على الخط كفاءة تعتمد : , ( كالتالي( تعرف وعليه واالصالح االعطال معدالت

Timedown Time UpValue Expected

Time UpValue Expectedlim

tQ

tqE

t

ttQ

ttq

t

over timecapacity lTheoretica over timeoutput Actual

e)cycles(tim total toe)cycles(tim productive combaringby timeEstimated

4c- Transfer Line Break-down concept في التعطلاالنتقال خط

D) Method of improving efficiency تحسين كيفية دراسة يتمالتالي -:الكفاءة

• Study Failure and repair Rates to reduce stopping timeزمن لتقليل للمعدات واالصالح االعطال معدالت دراسة

العطل• Design the line to operate efficiently such as design of Buffer

storage (Number and Size)المحطات وتعطل تشغيل استقالل وطرق الخط تصميم

بينها المؤقتة المخازن وحجم عدد أو :E) Machine /equipment conditionبتصميم للمعدة حالتين هناكهما -:الماكينة

• Operation State Operation State التشغيل التشغيل حالة It is the Up time or Working حالةcondition and it is considered as (1). It represents the production of machine state

• Failure State Failure State الفشل الفشل حالة It is Down time or Stopping حالة=condition and it is considered as (0). It represents the break-down state.

5a- Unreliable Machine Analysis المعدة تحليل تتعطل

.a احتماالت تحليل يتمواالصالح بإيجاد العطل

أتناء االحتمالي التوزيعالتشغيل ألزمنة فترات

التوقف أو.b نظام تستخدم

الديناميكية االحتماالت)( العشوائية العملية

تستخدم ثم ومنإليجاد ماركوف سلسة

الحالة معادالتالفرق بمعادالت

التفاضلية والمعادالتالسلوك يعتمد حيث

على المستقبلىدون فقط الحاضر

الماضي اعتبارفي النظام وصف أنظرفقرة النمذجة موضوع

) ب4(

A) Analyze Repair and Failure by finding the probability distribution of each during operation time and downtime

B) Use of Probabilistic dynamic system analysis (Stochastic Process) and use of (Markov chain/process) to develop the state equations (difference equations or differential equations) based that future behavior id dependent on the present without considering the past state.

See system description In modelling sec (4b)

5b- Unreliable Machine Analysis معدة تحليلتتعطل

tP ,

ijP

tPi

a 1 j

jiji tPPtP

b allfor , 1 ttPi

1 OR 0

Probability of being in state at time tزمن عند الحالة كون tاحتمال

Probability of the state equal to observed i at time tعند مالحظة قيمة تساوي الحالة كون احتمال

tزمن

Transition Probability (Conditional probability) of state (i) at time (t+1) given probability of state (j) at time t

لحالة االنتقالى زمن )i(االحتمال معطى )t+1(عندالحالة زمن )j(احتال tعند

Assume the following probabilities االحتماالت وبفرض:التالية

5c- Unreliable Machine Analysis تتعطل معدة تحليل

A) Geometrical Distribution

ptPrtPtP ,11,01,0

Then the set of difference equations with constant coefficients are:-

0 1

p

r

1 -p1 -rWhere:P =The probability that a failure occurs during an operation (while the machine is

up) أثناء الفشل حدوث احتمالالمعدة عمل

r =The probability that a repair is done during a time unit (while the machine is

down) أثناء االصالح اتمام احتمالالمعد توقف

P(t)

1 2 t t+1

In time (t) or (t+1), one of the two state stop (0) OR operation (1) can be stated

ptPrtPtP 1,1,01,1

5d- Unreliable Machine Analysis معدة تحليلتتعطل

p

tP

tPr

tP

tP

,0

,11

,0

1,0

; ; ThenaXtPAssume t

pa

arX

0

11 pra

aX 1

1

0

&&

pr

tP

tP

tP

tP

1

,1

,0

,1

1,1

&&

Solve for X and a0 ,a1 and the solutionthe solution will be:

tt rppr

prpPtP

1110,0,0

tt rppr

rrpPtP

1110,1,1

rp

pP

0

rp

rP

1&&

At steady state operation for long time the solution will be: t

tPtPtPtPei ,11,1 & ,01,0 ..

1111 & 1100 pPrPPpPrPP

= E (efficiency)

5e- Unreliable Machine Analysis تتعطل معدة تحليل

B) Exponential Distribution

Then the set of difference equations are:

0 1p

r

ttptPtrtPdttP 0,11,0,0

In time (t) or (dt), the state either stop (0) or operation (1)

ttptPtrtPdttP 01,1,0,1

=The probability that a failure occurs during an interval (while the machine is up) المعدة عمل أثناء الفشل حدوث احتمال

tp tt

=The probability that a repair is done during a time unit (while the machine is down) المعد توقف أثناء االصالح اتمام احتمال

tt tr

=The probability that an operation is completed during an interval of (while the machine is up) عمل أثناء الفشل حدوث احتمال المعدة tt

tu

5f- Unreliable Machine Analysis معدة تحليلتتعطل

ptPrtPdt

tdP ,1,0

,0&& ptPrtP

dt

tP ,1,0

,1

The solutionThe solution will be:

tprepr

pP

pr

ptP

0,0,0

tPtP ,01,1

rp

pP

0

rp

rP

1&&

tPtPtPtPei ,11,1 & ,01,0 .. 1111 & 1100 pPrPPpPrPP

The average production rate is:- rp

ruuP

1

= E (efficiency)

At steady state operation for long time the solution will be: t

5g- Unreliable Machine Analysis تتعطل معدة تحليلSimplified Explanation Of The Analysis,

Assume the that the machine operate for a time period of (T) and during this period several operating cycle (t) occurs.

Each cycle has operating and stopping condition as given in the figure:

التحليل , زمنية ولتبسيط لفنرة تشغيل دورة خالل تعمل المعدة أن أفرض)T( بزمن تشغيلية دورات عدة حدثت الفترة هذه لها )t( واثناء دورة كل ؛

: الشكل في مبين كما تشغيل حالتي

Operating period T المعدة تشغيل فترة

Operating cycle tتشغيل دورة زمن

Considering one cycle, the up time (u) and downtime (d) are as shown in the figure

التشغيل زمن واحدة العطل )u(ولدورة الشكل )d(وزمن في مبين كما

Down

Up

Uptime u تشغيل زمن Downtime d العطل زمن

5h- Unreliable Machine Analysis تتعطل معدة تحليل

A) Determination of number of pieces of number of operation cycle a) In case of no failure زمن أثناء المعدة تعطل عدم حالة في

المعدة تشغيلNumber of parts produced during time (t) N = t/TC ; TC = Cycle Time النموذجية المعدة عمل دورة زمنThis represents the Number of operational cycles of the machine عدد تمثل

للمعدة التشغيل دورات

b) In case of failure تشغيل زمن أثناء المعدة تعطل حالة فيالمعدة

Number of parts produced during uptime (u) / Number of operational cycles

of the machine N = u/TC

Number of failure cycles during Down time (d) الفشل دورات d/TC عدد

Down

UpOperating cycle t دورة زمن تشغيلية

Uptime u تشغيل زمن Downtime d زمن العطل

5i- Unreliable Machine Analysis تتعطل معدة تحليل

B) Finding Times & Rates ومعدالتها األزمنة إيجاد

a) Operating cycle time (t) عمل دورة زمن

operating cycle time = uptime + downtime t = u + d

b) Operating Time (T) العمل فترة زمن

Total uptime (U) = Σ (ui); i = 1,……, m (m = number of breakdown)

Total downtime (D) = Σ (di); i = 1,……, m (m = number of breakdown)

Operating time (T) = U + D ……………………………….. (1)

Down

UpOperating cycle t دورة زمن تشغيلية

Uptime u تشغيل زمن Downtime d زمن العطل

5j- Unreliable Machine Analysis تتعطل معدة تحليل

c) Average times األزمنة متوسطBy defining the times as average values; the times become Mean Time Between Failure; MTBF= (T/m) Mean Time To Failure; MTTF= [Cycle Uptime (u)] = (U/m) Mean Time To Repair; MTTR= [Cycle downtime (d)] = (D/m) Operating cycle time (t) = u + d = MTBF = MTTF + MTTR Total up time (U) = m * MTTF Total downtime (D) = m * MTTR Total operating Time (T) = U+D = U*[1+(D/U)

= U*[1+ (MTTR/MTTF)]

5j- Unreliable Machine Analysis تتعطل معدة تحليل

d) Failure rate & repair rate اإلصالح ومعدل العطل معدلFailure rate (p) = (1/MTTF) ……………………………. (3a)

Repair rate (r) = (1/MTTR) ……………………………... (3b)

e) Finding times as a function of failure and repair rates

Cycle Uptime (u) = (1/p)

Cycle downtime (d) = (1/r)

Operating cycle time (t) = u + d = u * [ 1 + (d/u)] = u * [ 1 + (p/r)] Total up time (U) = m * (1/p)

Total downtime (D) = m * (1/r)

Total operating Time (T) = U + D = U * [1+(D/U)] = U * [ 1 + (p/r)]

f) Machine efficiency, E

E = U/T = U/(U+D) = r/[1 + p] OR

E = {MTBF –MTTR}/MTBF

5a- Unreliable system Analysis معدات نظام تحليلتتعطل

يتكون المستمر خط ألن ونظرامن متتالية مجموعة من

كفاءة , )n(الماكينات حساب يتمحالتين بأحد الخط اتاحة أو

-: وهما للخط تحدد

For flow line composing of serial number of machines the efficiency or availability can be obtained as follow:

: الفشل اعتماد األولى الحالةالتشغيل على

فشل الحالة هذه في يفترضويتم فقط التشغيل أثناء المعدةاإلنتاج زمن دورة نهاية فيليس أنه ذلك ويعني ؛ للمعدة ) األعلى الحد معيوب منتج هناك

لإلنتاج.) مهم فرض فشل وهذا حيث

أخرى معدة على يعتمد ال المعدةالمنتجة المشغوالت عدد وأن

التشغيل زمن دورات خالل تتمكاملة.

يمكن الفرض هذا على وبناءالكلي العطل زمن حساب

: كالتالي والكفاءة

Case one: Operation Dependent Failure (ODF)In this case, assume machines fail only during operation and at the end of a production cycle –{(i.e. no part inside the machine [no part defect due to failure]}. This case is upper bound case.This assumption emphasize that failure of machines is independent for each machine and number produced during cycle period has no scrap defect

Efficiency is calculated as follow:

Total down time =

Efficiency =

5b- Unreliable system Analysis معدات نظام تحليلتتعطل

الثانية : الحالة الفشل اعتمادالزمن على

فشل الحالة هذه في يفترض , ال أم تعمل كانت سواء المعدةمستقل الفشل أن ذلك ويعني

المعدة تشغيل كيفية عنمما التشغيل زمن على ومعتمدا

واالصالح الفشل أن يعنيمستقلة صورة يسلك للمعدةاألخرى المعدات عن ذاته بحد

متوقفة أو عاملة سواءالمعدة تفشل الحالة هذه فييسبب مما داخلها ومشغولة

عدد وتقليل فيها عيب ) الحد المنتجة المشغوالت

األدنى), يمكن الفرض هذا على وبناءحاصل بأنها الكفاءة حساب : كالتالي المعدات كفاءة ضرب

Second case: Time Dependent Failure (TDF) Assume failure can happen during machine work or not. This means that failure is independent upon machine operation and depend on operating time. Each machine failure and repair is independent of other machines.In this case the machine can fail and part still in the machine causing defect and reducing number of part produced (Lower bound). The efficiency is as follows:

Efficiency =

5c- Example (1)

20 machines-Flow line work with efficiency of 0.98 each. Find the line efficiency in cases of operational dependent failure and time dependent failure!

Case: Operation Dependent Failure (ODF)

Case: Time Dependent Failure (TDF)

5d- Example (2)A line consists of two stations. The first fail every 10 cycles, the second fails every 15 cycles and repair takes 2 cycles. Find the efficiency in cases of operational dependent failure and time dependent failure!

10 12

10 1 fail

17 19

15 2 fail

24 26

20 1 fail

36 40

301&2 fail

Cycles

Output

Case: Operation Dependent Failure (ODF)

Case: Time Dependent Failure (TDF)

6a- Line Analysis With Buffers مؤقت بتخزين خط تحليل

وسعة عدد تصميم يعتمدالتحسن مقدار على المخازن

تكون, حيث كفاءة الممكنالخط

مخازن • وجود عدم حالة فياألدنى حد

مخازن • حالة بين )n-1( فينهائية )n( المحطات ال بسعة

األقصى حد

OE

E

بين المؤقت التخزين تأثير المحطات:

؛ مراحل إلى الخط بتقسيمالخط كفاءة تحسين يتمالمحطات عدد تقليل نتيجة

. مرحلة كل في

الخط بتقسيم يبدأ وعادةمخزن بينهما مرحلتين إلىعدد إلى تزداد ثم ومن واحدعدد تساوي المراحل من

من )n(المحطات بعدد )n-1(المخازن

ال بسعة المخازن وتكونمحدودة بسعة أو نهائية

The effect of buffers between workstations

By dividing the line to stages, the line efficiency will improve as result of reducing the number of workstations in each stage.

Usually, for line with (n) workstations can be divided into two stages with one buffer and then increase number of stages until it reach (n) stages with (n-1) buffers between two workstations.

The buffer capacity can be Infinite Capacity or Finite Capacity

The design of number and capacity of buffers depends on the possible improvement can be achieved. Thus the line efficiencyin case of no buffers, the Lower Limit isin case of (n-1) buffers with infinite capacity between (n) workstation, Upper Limit is

OE

E

6b- Line Analysis With Buffers بتخزين خط تحليلمؤقت

1 2 3 4 5 6 7

1 2 3 4 5 6 7b

Stage 1 Stage 2

1 2 3 4 5 6 7b b b b b b

1 2 bb 3 4 5 6 7

Stage 1 Stage 2 Stage 3

Example:

6c- Line Analysis With Buffers مؤقت بتخزين خط تحليلFrom the previous example, when buffers are available the line efficiency ranges between lower and upper limit according the number of buffers.

الحدين بين الكفاءة تتراوح مؤقتة مخازن اتاحة عند انه المثال من يتبنالمخازن لعدد وفقا واألقصى األدنى

مالحظات:الكفاءة • قيمتي تقارب عند

, فائدة تكون واألقصى األدنىمما قليل المؤقتة المخازن زيادةوجود في يتم التقسيم أن يعني

. معنوي فارقتكون • أن يجب الخط تقسيم عند

جدا متقاربة المراحل كفاءة ( يتوازن( لكي تقريبا متساوية

بعنق يعرف ما يسبب وال اإلنتاجالزجاجة.

كفاءة • أعلي تحقيق امكانيةبالتالي:

األعطال • معدالت تكونالمختلفة للمراحل واالصالحذلك يؤدي حيث متساوية

. المراحل كفاءة لتساويإذا • بحيث المخازن عدد تصميم

في الفارق يكون العدد زاد. معنوي غير الخط كفاءة

إذا • بحيث المخازن سعة تصميمفي الفارق يكون السعة زادت ) قليل معنوي غير الخط كفاءة

جدا).

Remarks: When lower and upper limit efficiency

values are close, the benefits from increasing buffer is limited. This mean more stages are done if there is significant difference between upper and lower limits.

When dividing to stages, the efficiency of stages should be close to balance production and not causing bottleneck.

High efficiency can be reached by:• Equal the stages failure and repair rates.

This result of equal efficiency of all stages.• suitable number of buffer is obtained

when no significant increase of efficiency can be achieved.

• Also, suitable capacity of buffer is obtained when no significant increase of efficiency can be achieved.

6d- Line Analysis With Buffers بتخزين خط تحليلمؤقت

Example: Line stages مراحل إلى الخط لتقسيم مثال 16 station-flow line operate with cycle time of TC=10 sec. when it. fails the repair time is Td=2 min. and the probability of stations failure frequency are as given in the table. Find the efficiency of line when it is divided to 2, 3, 4 stages.

1 0.01

2 0.02

3 0.01

4 0.03

5 0.02

6 0.04

7 0.01

8 0.01

9 0.03

10 0.01

11 0.02

12 0.02

13 0.02

14 0.01

15 0.03

16 0.01

Station pi Station pi

6e- Line Analysis With Buffers مؤقت بتخزين خط تحليل

Solution: B) Two Stages

Solution: C) Three Stages

0.5814

0.5556

0.5319

Ej

1

2

3

1-5

6-10

11-16

0.09

0.1

0.11

Stage Stations Fj

0.6410

0.6098

0.6098

0.7410

Ej

1

2

3

4

1-4

5-8

9-12

13-16

0.07

0.08

0.08

0.07

Stage Stations Fj

Solution: D) Three StagesSolution: A) Single Stage

1 2 3 4

0.15

0.3

0.45

0.6

0.75

Stage

Efficiency

7a- Two Stage Line Analysis With Finite Capacity Bufferمحددة تخزين وبسعة بمرحلتين خط تحليل

The following figure show a two stages flow line with a buffer has a capacity of (b). مؤقت مخزن المرفق الشكل مرحلتين (A) يبين بين

مقدارها وبسعة االنتقال لخط1 2 3 4 5 6 7

b

Stage 1 Stage 2

Operation states of the line

Stages operate until a unit in one of two stages the fail The state can be written as follow: تفشل أن إلى المرحلتين تعمل

التشغيلية والحالة فيهما الوحدات احدى :كالتالي

o In case Stage 2 fails: stage 1 operate until buffer is full at capacity Z and stage 1 is blocked if the stage 2 is not repaired.

o In case Stage 1 fails: stage 2 operate until buffer is empty with capacity 0 and stage 2 stops as it is starved for parts if the stage 1 is not repaired

للخط التشغيلية الحالة

تفشل أن إلى المرحلتين تعملوالحالة فيهما الوحدات احدى

: كالتالي التشغيليةo وحدات احدى حالةفشل في

الثانية , المرحلة المرحلة تستمريمتلئ أن إلى العمل في األولى

القصوى سعته عند ، Zالمخزناألولى المرحلة تتوقف وعندهالم إذا لالنسداد نظرا العمل عنالمرحلة في الوحدة إصالح يتم

الثانية.o وحدات احدى فشل حالة في

األولى المرحلة , المرحلة تستمريفرغ أن إلى العمل في الثانية

تتوقف وعندها ، المخزننظرا العمل عن الثانية المرحلة

إصالح يتم لم إذا للتعطش. األولى المرحلة في الوحدة

7b- Two Stage Line Analysis With Finite Capacity Bufferمحددة تخزين وبسعة بمرحلتين خط تحليل

For long operation time at steady state, the efficiency is

Where:

o Eo = overall efficiency for the line as one stage

o E = overall efficiency for a two-stage line with a buffer with capacity (b).

o h(b) = the proportion of the down time D’1 ( when stage 1 is down) that stage 2 could be up and operating within the buffer capacity (b)

o D’1 = is the proportion of total time that stage 1 is down,

o Eo & D’1 are defined as follow:

&&

7c- Two Stage Line Analysis With Finite Capacity Bufferمحددة تخزين وبسعة بمرحلتين خط تحليل

h(b) are computed using Markov Chain Analysis سلسلة تحليل. ماركوف

It is expressed by set of equations covering several downtime distributions based on the assumption that both stages are never down at the same time.ومبنية, العطل لزمن توزيعات عدة تغطي بمعادالت عنها عبر وقد. الوقت نفس في للمرحلتين عطل حدوث احتمال عدم فرض على[Four of these equation are represented in table 18.2 Groover book]

These equations are:

Where:B is the largest integer satisfying the relation,

L is representing the leftover units, The number by which b exceeds

7d- Two Stage Line Analysis With Finite Capacity Bufferمحددة تخزين وبسعة بمرحلتين خط تحليل

For constant Repair Distribution: each down occurrence is assumed to require a constant repair times (Td)

For Geometric Repair Distribution: assumes that the probability that repair are completed during any cycle is independent of repairs began. time since repair began

7e- Example10-station transfer line is divided to two stages of 5-station each. The cycle time of each stage is TC=1 min. All stations have the same probability of stopping, p= 0.02. the downtime is constant, Td=5 min. compute the line efficiency for several buffer capacities.

Solution: B) Two Stages with Infinite capacity buffer

Solution: A) Single Stage without buffer

7e- Example

Solution: C) Two Stages with finite capacity buffer b= 1

Solution: E) Two Stages with finite capacity buffer b= 100

B = 20 , L = 0 , h(100) = 0.952 , E = 0.6587

Solution: D) Two Stages with finite capacity buffer b= 10

B = 2 , L = 0 , h(10) = 0.6667 , E = 0.6111

7e- Example

40 80 120 160

0.5

0.55

0.6

0.65

0.7

Buffer Size

Efficiency

200

7f- Two Stage Line Analysis With Finite Capacity Buffer using Transition states of the system وبسعة بمرحلتين خط تحليل

محددة تخزين

1 2 3 4 5 6 7b

Stage 1 Stage 2

Usually three case can be visualize for production rate as follow:وهي :- مرحلة لكل اإلنتاج لمعدل حاالت ثالث وهناك

1. The two stages have Equal production rate (Q1=Q2)

2. Stage 1 production rate > Stage 2 production rate (Q1>Q2)

3. Stage 1 production rate < Stage 2 production rate Q1<Q2

The line is analyzed by developing a Stochastic model. The line is described by set of steady state equations describing the operating states of the line. These equation are used to find the production throughput and cost.

بمعادالت الخط بوصف عشوائي نموذج تطوير يتم الخط ولتحليلالتشغيلية الخط الحاالت من الخارجة اإلنتاج معدالت اليجاد

التصنيع نظام وتكلفة

7f1- Analyze two stage line with equal production rate Q1=Q2

1 2 3 4 5 6 7b

State description: الخط حاالت assume Stage 1 as S1 & Stage 2 as S2) وصف

4. S1 Fails & S2 operates. The buffer level

9. S1&S2 operates. The buffer level

6. S1 fails & S2 operate. The buffer level

7. S1 operate & S2 fails. The buffer level

8. S1&S2 operates. The buffer level

1. S1&S2 fail and buffer level . Repair is carried on S1

2. S1&S2 fail and buffer level . Repair is carried on S2

3. S1&S2 operate and buffer level .

5. S2 Fails & S1 operates. The buffer level


Graphical representation of the states to develop the equations

3

6

7

5

8

4

9

12

Notations


The following conditions holds for instant of time:-• Flow in –flow out =0• Sum of state probabilities =1

Notations


State equations معادالتالخط حاالت

With initial solution conditions caused by the flows from state 8 to 5 and from state 9 to 4

36

75

8

4

912


1- The availability can be defined as:-

2- The Production Rate can be defined as:- Qs = A q

3- The Expected Storage Level can be defined as:-

4- The Total System Cost can be defined as:-

8a- Assembly Line Performance التجميع خط أداء تحليل

Assume;qi = probability of the component is defectivemi = probability that a defective component cause jam

When a component is fed, three events may occur at a particular work-station (i):

1. The component is defective and cause a station jam2. The component is defective and does not cause a station jam3. The component is not defective

A) Multi stations analysis

1 42 3 5 6BasePart

FinishProd.

Components

8b- Assembly Line Performance التجميع خط أداء تحليل

1- For the first event; The probability that the defective part cause a jam

2- For the second event; The probability that the defective part does not cause a jam iii qm 1

3- For the third event; The probability that the part is not defective ii q1

111 y,probabilit equalFor

111

stations; For -5

1

n

iiiii

n

i

qqmmq

qqmqm

n

8c- Assembly Line Performance التجميع خط أداء تحليل

nap

iii

n

iap

mqqP

qmqP

n


1

line theoff comingproduct acceptable The stations; For -6

1

nqp

iii

n

iqp

mqqP

qmqP


11

defect oneleast at contain that proportionassembly The -7

1

nmqF

qmFn

iii

n

ii

y,probabilit equalFor

cycleper occurance downtime offrequency The -8

11

8d- Assembly Line Performance التجميع خط أداء تحليل

dCP

n

idiiCP

nmqTTT

TqmTT


assembly per timeproduction Average The -9

1

P

n

ap

P

qp

P

iii

n

iap

PP

T

mqqR

T

P

T

qmqR

TR


1

1 rate, production avarage The -10

1

PD

PCTT DTTE

,Deficiency The12 ,Efficiency The-11

ap

tPLmPC

P

CTCCC

Cost, The-13

8e- Assembly Line Performance خط أداء تحليلالتجميع

1BasePart

FinishProd.

B) Single station analysis

dCP

n

idiiCP

nmqTTT

TqmTT


assembly per timeproduction Average The -15

1

meElement ti timeHandling

where;

assembly per timecycle The -14

1

ei

h

n

ieihC

TT

TTT

8f- Example (1) التجميع خط لتحليل مثال

10-station in-line assembly machine has an ideal cycle time =6 sec. The base part is automatically loaded prior to first station, and components are added at each stations. For jamming rates (m=0,0.5,1.0) and defect rates (q=0,.01,.02), find:a) Production Rate of line---b) Yield of good assemblies---c) Production Rate of good assemblies---d) Uptime Efficiency---e) Cost per unit

0.5

0.3

0.1 0.0710060016001.00

0.2133.320012001.00.01

0.352012011201.00.02

0.3

0.2

0.1

0.02 0.5 200 0.9044 180.9 33.3 0.2322

0.01 0.5 300 0.9511 285.3 50 0.1472

0 0.5 600 1 600 100 0.07

0.1

0.1

0.1

0.02 0 600 0.8171 490.2 100 0.0857

0.01 0 600 0.9044 542.6 100 0.0774

0 0 600 1 600 100 0.07

Tp= (TC+ nmqTd) , Min

q mRp=1/ Tp

assbly/hrYield Pap = (1-q+mq)n

Rap= Rp* Pap

E=TC /Tp %

Cost, Cpc = CO * Tp / Pap ,$/ass

8g- Example (2) التجميع خط لتحليل مثالA single station assembly machine performs five work elements to assemble four components to a base part. The elements are listed in the table below, together with fraction defect rate (q), and probability of a station jam (m) for each components added.

Time to load the base part is 3 sec, and time to unload the completed assembly is 4 sec, given a total load/unload time of Th = 7 sec. when jam occurs, it takes an average of 1.5 min to clear the jam and restart the machine. Determine:a) Production Rate of all product,---b) Yield of good assemblies---c) Production Rate of good assemblies---d) Uptime Efficiency of the assembly machine.

Element operation Time, sec q m p

1 Add gear 4 0.02 1.0

2 Add spacer 3 0.01 0.6

3 Add gear 4 0.015 0.8

4 Add gear and mesh 7 0.02 1.0

5 Fasten 5 0 NA 0.012

8g- Example (2) التجميع خط لتحليل مثال

Solution

a) Production rateThe ideal cycle time of assembly machine,

TC = 7 + (4+3+4+7+5) = 30 sec = 0.5 min

Frequency of downtime occurrences

F = 0.02*1.0 + 0.01*0.6 + 0.015*0.8 + 0.02*1.0 + 0.012 = 0.07

Adding the average downtime due to jams

Tp = 0.5 + 0.07*1.5 = 0.605

Production rate

Rp = 60/0.605 = 99.2 total assemblies/hr

b) Yield of good productPap = 1.0 * 0.996 * 0.997 * 1.0 = 0.993

c) Production rate of only good assemblies Rap = 99.2 * 0.993 = 98.5 good assemblies/hr

d) Uptime efficiency

E = 0.5/0.605 = 0.8264 = 82.64 %

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