## Background Material

### Suggested Books:

  Chapter 3(Section 3.3) of Modeling and Analysis of Manufacturing Systems, by Ronald G.Askin and Charles R.Stanridge, John Wiley & Sons, Inc, 1993. Chapter 3 of Manufacturing Systems Engineering, by Stanley B.Gershwin, Prentice Hall, 1994.

## Lecture Objectives

 At the end of the lecture, every student should be able to   Evaluate the effectiveness (availability) of a transfer line given    Buffer capacity Failure rates for the work stations Repair rates for the work stations Determine the optimal location of the buffer in any N stage transfer line.

## Time Management

        Readiness Assessment Test (RAT) - 5 minutes Lecture on Paced Lines with buffers - 10 minutes Spot Exercise - 5 minutes Paced Lines with buffers (contd..) – 10 minutes Team Exercise - 10 minutes Homework Discussion - 5 minutes Conclusion - 5 minutes Total Lecture Time - 50 minutes

### Here are the possible reasons for a station to be down. Analyze each of these reasons and the importance of buffers in every case.

    Station Failure Total Line Failure Station Blocked Station Starved

## RAT Analysis

   

Station Failure:

Fractured tool, quality out-of-control signal, missing/defective part program, or jammed mechanism. Although this station is down, other stations can operate if they are fed product by the

buffer

, and have space for sending completed product.

Total Line Failure:

All the stations are inoperative. A power outage or error in the central line controller would cause a total line failure.

Station Blocked:

On completion of a cycle, if station

i

is not able to pass the part to station

i

+1, station

i

is blocked. Failure of the handling system, failure of a downstream station prior to the next

buffer

, or failure of a downstream station with the intermediate

buffer

between these stations currently being full. Suppose station

i

+1 is down, and its input

buffer

is filled, then station

i

must remain idle while it waits for downstream space for the just completed part.

Station Starved:

Station

i

is starved if an upstream failure has halted the flow of parts into station

i

. Even if operational, the starved station will sit idle.

### Two-Stage Paced Lines with Buffers

 Two serial stages are separated by an inventory buffer.

 Buffer reduces the dependence between the stations – blocking/starving effects are reduced.

  Buffer state should be taken into account while calculating line effectiveness.

Markow Chain Model is used.

Serial Stages Buffer

### Assumptions and Conventions

         Markow Chain Environment (

s 1 ,s 2 ,z

)

s i

is the status for station

i

and

z

is the number of items in the buffer W – station in working condition (operational) R – station in repair Failures and Repairs occur at the end of a cycle When a cycle starts, if both stations are working, station 2 receives its next part from station 1.

If station 1 is down, station 2 grabs a part from the buffer If buffer is empty, station 2 is starved If station 2 is down and station 1 is up, part from station 1 is sent to the buffer If the buffer is full, station 1 becomes blocked

### Chapman – Kolmogorov result : steady-state balance equations

  S be the set of states of the system P(s) be the probability that the system is in state s 

p(u,v)

is the transition probability from state

u

(beginning) to state

v

(ending)

### P(s)*p(s,s1) The steady-state balance equations can be obtained by applying the above equation.

Transitions for Two-stage line with buffer

At time

t

, both stations are working, the transitions include     WW WW WW    WW: probability = (1 WR: probability = (1 -

α

1 )(1 -

α

1 )

α

2 RW: probability =

α

1 (1 -

α

2 )

Z

, does not change

α

2 )   Station 1 is being repaired while station 2 is working    RW  WW:   if RW 

Z

if

Z

= 0, then probability =

b

1 = RW:

x

> 0, then probability =

b

1 (1 -

α

2 ),

Z

=

x

- 1   if RW 

Z

if

Z

= 0, then probability = (1 = RR:

x b

1 ) > 0, then probability = (1  if

Z

=

x

> 0, probability = (1 -

b

1

b

1 )

α

2 ,

Z

)(1 =

x α

2 ),

Z

– 1 =

x

- 1 Example:

P

(

WW0

) = (1 -

α

1 -

α

2 )

P(WW0)

+

b

1

P(RW0)

+

b

1

P(RW0)

Steady-state equation obtained by the Chapman-Kolmogorov Result

 Station 1 is working, while station 2 is repaired    WR  WW:   if WR 

Z

if

Z

= 0, then probability =

b 2

> WR:

x

= 0, then probability = (1 –   if WR 

Z

if

Z

= 0, then probability = 1 > RR:

x

b

2 = 0, then probability = (1 –  if

Z

>

x

= 0, probability =

α

1 (1 -

b

2

α α

), 1 1

Z

)

b

2 )(1 =

x

,

Z b

2 = + 1 ),

x Z

+ 1 =

x

+ 1  Both stations are being repaired    RR  WR:  RR  if

Z

= RW:

x

= 0, probability =

b

1 (1  RR  if

Z

= RR:

x b

2 ),

Z

= 0, probability = (1 –

b

1 )

b

2 ,

Z

=

x

+ 1 =

x

+ 1  if

Z

=

x

= 0, probability = (1 –

b

1 ) (1 -

b

2 ),

Z

=

x

+ 1

  System Effectiveness for a buffer of maximum size

z

can be calculated as

E z

x Z

  0

P

(

WWx

) 

x Z

  1

P

(

RWx

) Buzacott’s closed-form expression for the effectiveness for a buffer of maximum size

z :

where

E Z

    ( 1  1 

sC Z

2 1

x

 )[

r

1  

b

1

x

1 ( 1   ( 1

x

 ) 1 

r

b

2 ( 1 

x

 2 )

sC Zb

2 ( 1

Z x

) 

Zb

 2

x

) ( 1 

x

) 2

s

 1

s

 1

C

 (  (  1 1     2 2 )(

b

1 )(

b

1 

b

2 ) 

b

2 )     1

b

2 2

b

1 (  (  1 1     2 2 

b

1 

b

1 

b

2 ) 

b

2 ) and

x i s

=

α

=

x

2

i

/

b i (ratio of average repair time to uptime)

/

x

1 ,

r

=

a

2 /

a

1 ,

## Spot Exercise

 A paced assembly line has a cycle time of 3 minutes. The line has eight workstations and a buffer of capacity 10 is placed between the fourth and fifth workstations. Each station has a 1 percent chance of breaking down in any cycle. Repairs average 12 minutes. Estimate the number of good parts made per hour.

### Solution

Given:

Cycle time C = 3 min (paced), m = 8,

α i

= 0.01,

b

= 0.25 Thus, Cycles/hr = (60 min/hr)/(3 min/cycle) = 20 cycles/hr Now,

α 1

= 0.4,

Z

=10 and

α 2

= 0.4

Effectiveness can be found from Buzacott’s closed form expression,

E

10  ( 1  2 1 

r x

)[ 1  

r b

1  ( 1

b

 (

x

) 1  

x Zb

2 )]  2 ( 1 

Zb

2 (

x

) 1 

x

) 2  0 .

82 where

x

= 0.04/0.25 = 0.16 and s = 1, r = 1 Hence, the number of parts/hr =

E

10

*

Cycles/hr = 16.40

### System Reduction

    Aggregation of a set of stations that must be jointly active or idle and which have a common repair rate, into a single station Holds effective for all serial stations as well as stations that act as feeder stations for the main line Aggregated failure rate is obtained by summing the individual failure rates System Reduction Rules:  Combination Rule   Median Buffer Location Reversibility

 Combination Rule  A set of stations without any intermittent buffers can be replaced with a single station  '

j

i m

   

i

α 1 and

b j = b,

provided all stations must stop if any individual station stops α 2 α 3 α 4 α 5 Single Line α 1 α 1 +α 2 α 2 = α 3 +α 4 +α 5 α 3 α 4 α 5 α 6 Feeder Line (unbuffered) α 1 + α 2 + α 3 + α 6 α 4 + α 5

 Combination Rule(Contd..) Feeder Line (buffered) α 2 α 3 α 4

### =

α 1 α 2 +α 3 +α 4 α 1  Median Buffer Location  If only one buffer is to be inserted, it should be placed in the middle of the line. The upper bound on availability is given by min 1 

i

M

  1 

b i

    1

 Median Buffer Location(Contd..)  A median location is the one for which, if r is the last station before buffer and the two conditions below are satisfied max max

i r

  1

i r

  1 

i

i

, ,

i M

 

r

 1 

i i M

 

r

 1 

i

 max  max

i

 1

r

  1 

i

,

i

M

 

r

2 

i i

 1

r

  1 

i

,

i M

 

r

i

 Reversibility  If the direction of flow is reversed in a serial line, production rate remains the same.

### Team Exercise

 Consider a paced assembly system with four workstations. Mean cycles to failure are estimated to be 100, 200, 100 and 50 cycles, respectively. Repair times should average 8 cycles.  Find line availability assuming no buffers.

 A buffer of size 5 would be profitable if availability increased by at least 0.04. The buffer could be located after station 2 or 3. Which location is the best? Should the buffer be included?

### Solution

Given:

α 1

= 0.01,

α 2

(a) Thus,

E 0

= 0.005,

α 3

= [1 +

b

-1 Σ

α i

] -1 = 0.01,

α 4

= 0.02, = [1 + 8(0.045)]

b

-1 = 0.125 for all = 0.735

i

(b) According to the median location rule, the buffer should be located after station 3.

Now,

α 1

= 0.025 (0.01+0.005+0.01),

Z

= 5 and

α 2

= 0.02

Effectiveness can be found from Buzacott’s closed form expression,

E

5  1 

x

1 1 

sC Z

 ( 1 

x

2 )

sC Z

 0 .

764 where

s =

0.8

, x 1

Now,

E 5

E 0 =

0.2

, x 2 =

0.16 and

C

= 0.9153

= 0.029 < 0.04, and do not include the buffer

## Homework

 A 10 stage transfer line is being considered with failure rates 

i =

0.004 (

i

=1,2,..,10) and

b j =

0.2 (

i

=1,2,..,10). Estimate the following     Effectiveness of the line without buffers Effectiveness of the line with a buffer size of 5 after the station 5 Effectiveness of the line with a buffer size of 5 after the station 3 Optimal location of the buffer

  

## Conclusion

Buffers allow for partial independence between the stages in the line, thereby improving line effectiveness against station failures The size of the buffer may have an extreme influence on a flow production system. The importance of buffers increases with the amount of variability inherent in the system.

Although buffer involvement leads to considerable investment and factory space, it is crucial to find the optimal number and distribution of the buffers within a flow production system.