Two-Stage Paced LinesActive Learning – Module 2

Dr. Cesar Malave

Texas A & M University

Background Material Any Manufacturing systems book has a

chapter that covers the introduction about the transfer lines and general serial systems.

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

Readiness Assessment Test (RAT)

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 (s1 ,s2 ,z) si 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(s1) = 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: probability = (1 - α1)(1 - α2) WW WR: probability = (1 - α1)α2

WW RW: probability = α1(1 - α2) Z, does not change

Station 1 is being repaired while station 2 is working

RW WW: if Z = 0, then probability = b1 if Z = x > 0, then probability = b1(1 - α2), Z = x - 1

RW RW: if Z = 0, then probability = (1 - b1) if Z = x > 0, then probability = (1 - b1)(1 - α2), Z = x - 1

RW RR: if Z = x > 0, probability = (1 - b1)α2, Z = x – 1

Example: P(WW0) = (1 - α1 - α2) P(WW0) + b1 P(RW0) + b1 P(RW0)

Steady-state equation obtained by the Chapman-Kolmogorov Result

Station 1 is working, while station 2 is repaired

WR WW: if Z = 0, then probability = b2

if Z > x = 0, then probability = (1 – α1) b2, Z = x + 1 WR WR:

if Z = 0, then probability = 1 – b2

if Z > x = 0, then probability = (1 – α1)(1 - b2 ), Z = x + 1 WR RR:

if Z > x = 0, probability = α1 (1 - b2), Z = x + 1

Both stations are being repaired

RR WR: if Z = x = 0, probability = b1 (1 - b2), Z = x + 1

RR RW: if Z = x = 0, probability = (1 – b1) b2, Z = x + 1

RR RR: if Z = x = 0, probability = (1 – b1) (1 - b2), Z = x + 1

System Effectiveness for a buffer of maximum size z can be calculated as

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

maximum size z :

where

and xi = αi / bi (ratio of average repair time to uptime), s = x2 / x1, r = a2 / a1

Z

x

Z

xz RWxPWWxPE

10

)()(

1

)1()1(1)[21(

)1()1(1

1)1(1

1

222

21

21

sxZbxbrx

xZbxbr

ssCxx

sC

EZ

Z

Z

)())((

)())((

2121122121

2121212121

bbbbb

bbbbbC

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.

SolutionGiven: Cycle time C = 3 min (paced), m = 8, αi = 0.01, b = 0.25

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

Now, α1 = 0.4, Z =10 and α2 = 0.4

Effectiveness can be found from Buzacott’s closed form

expression,

where x = 0.04/0.25 = 0.16 and s = 1, r = 1

Hence, the number of parts/hr = E10 * Cycles/hr = 16.40

82.0)1()]1(1)[21(

)1()1(12

22

2110

xZbxbrx

xZbxbrE

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

and bj = b, provided all stations must stop if

any individual station stops

Single Line

α1+α2 = α3+α4+α5

Feeder Line (unbuffered)

m

i

ij '

α1 α2 α5α4α3

α2 α3α1 α5α4

α 1+ α 2+ α 3+ α 6α 4+ α 5

α6

Combination Rule(Contd..)

Feeder Line (buffered)

=

α2+α3+α4

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

α4α3α2 α1

α1

1

11min

bi

Mi

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

Reversibility If the direction of flow is reversed in a serial line, production

rate remains the same.

M

rii

r

ii

M

rii

r

ii

2

1

111

,max,max

M

rii

r

ii

M

rii

r

ii ,max,max

1

111

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?

SolutionGiven: α1 = 0.01, α2 = 0.005, α3 = 0.01, α4 = 0.02, b = 0.125 for all i

(a) Thus, E0 = [1 + b-1Σ αi ]-1 = [1 + 8(0.045)]-1 = 0.735

(b) According to the median location rule, the buffer should be

located after station 3.

Now, α1 = 0.025 (0.01+0.005+0.01), Z = 5 and α2 = 0.02

Effectiveness can be found from Buzacott’s closed form

expression,

where s = 0.8, x1 = 0.2, x2 = 0.16 and C = 0.9153

Now, E5 – E0 = 0.029 < 0.04, and do not include the buffer

764.0)1(1

1

21

5

Z

Z

sCxx

sCE

Homework A 10 stage transfer line is being considered with failure rates

i = 0.004 ( i =1,2,..,10) and bj = 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.