Approximate Three-Stage Model:Active Learning – Module 3

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 3.4) of Modeling and Analysis of

Manufacturing Systems, by Ronald G.Askin and Charles R.Stanridge, John Wiley & Sons, 1993.

Chapter 3 of Manufacturing Systems Engineering, by Stanley B.Gershwin, Prentice Hall, 1994.

Lecture Objectives At the end of the lecture, each student should

be able to Evaluate the effectiveness (availability) of a

three-stage transfer line given the Buffer capacities Failure rates for the work stations Repair rates for the work stations

Time Management Introduction - 5 minutes Readiness Assessment Test (RAT) - 5 minutes Lecture on Three Stage Model - 15 minutes Team Exercise - 15 minutes Homework Discussion - 5 minutes Conclusion - 5 minutes Total Lecture Time - 50 minutes

Approximate Three-Stage Model Introduction

Markov chains can be used to model transfer lines with any number of stages

The number of states to be considered increases with the number of stages, say M stages with intermediate buffers of capacity Z require 2M(Z+1)M-1 states

Readiness Assessment Test (RAT) Consider a three-stage line with two buffers

Assume that a maximum of one station is down at a time.

Determine the probability for station i to be down

M

m

m

i

i

x

xP

1

1where xi = αi / bi

Three-Stage Model (Contd..) Deeper analysis into the model: Consider a line without buffers

For every unit produced, station i is down for xi cycles xi is the ratio of average repair time to uptime From stations i = 1,…,M, all the other stations are

operational except station i Considering the pseudo workstation 0 with cycle failure

and repair rates α0 and β0 ,we have

11

0

M

i

iPE

Model Analysis:

Let us consider the station 2 and the three types of states it can produce Production is there when all stations are up Production is there when station 1 is down, but station 2

operates because of storage utilization from the buffer 1 Production is there when station 3 is down, but station 2

operates because of storage utilization from the buffer 2

1 1 2 2

# Workstation # # Buffer #

3

Model Analysis (contd..): Let us define hij(Z1,Z2) as the proportion of time station j operates

when i is under repair for the specified buffer limits

EZ1Z2 = E00 + P1h12(Z1,Z2) + P3h32(Z1,Z2) - Eq 1

Effectiveness of the line can be calculated by converting the three

stage model into a two-stage model with the help of a pseudo work station

Case 1: From buffer 1, there are two possibilities: Line is down when station 2 is down with failure rateα2

when station 3 is down and buffer 2 is full - with a failure rate {α3[1 – h32(Z1,Z2)]}

Model Analysis (contd..): Stations 2 and 3 along with the connecting buffer are

replaced by a pseudo station 2’ with a failure rate

α2’ = α2 +α3[1 – h32(Z1Z2)] - Eq 2 ~ α2 +α3[1 – h32(Z2)] as h32() will not depend on Z1

Hence for a two-stage line, effectiveness can be written as EZ = E0 + P1h12(Z) - Eq 3 = E0 + P2h21(Z) where

Pi is the probability that station i is down as referred before

h12(Z) is nothing but P1h12(Z1Z2)

Model Analysis (contd..): Had we known h32(Z1Z2), we could have solved the

two pseudo station line using the equations defined for estimating the effectiveness of two-staged lines with buffers and calculated the effectiveness of the three-stage line by substituting the values obtained in Eq 1.

The question is do we know the value of h32(Z1Z2) ? The answer is no !

Case 2: From buffer 2, there are two possibilities – Line is down when station 2 is down with a failure rate α2

when station 1 is down and buffer 1 is empty - with a failure rate {α1[1 – h12(Z1,Z2)]}

Model Analysis (contd..):

Stations 1, 2 and the connecting buffer can be replaced by a pseudo workstation 1’with a failure rate

α1’ = α2 +α1[1 – h12(Z1Z2)] - Eq 4 ~ α2 +α1[1 – h12(Z1)] as h12() will not depend on Z2

Station 3 will have a failure rate α3 The two-stage pseudo line can be solved by estimating

h32(Z1Z2) from h21(Z) of Eq 3 Solving for Case 1, i.e. estimating hij() factor is involved as

an input for Case 2 and vice versa. Thus by utilizing these two cases, the effectiveness of the three-stage model can be found.

Solution Procedure:

1. Initialize h12(Z1,Z2) at, say, 0.5. Denote stages 1and 2 in any pseudo two-stage approximation as 1’ and 2’, respectively. Calculate E00 , the effectiveness for the unbuffered line.

2. Solve the two-stage line with α1’ given by - Eq 4. Estimate h32(Z1,Z2) = h2’1’(Z) from Eq 3. α2’ = α3

3. Solve the two-stage line with α2’ given by - Eq 2. Estimate h12(Z1,Z2) = h1’2’(Z) from Eq 3. α1’ = α1

If suitable convergence criteria is satisfied, go to step 4, otherwise go to step 2.

4. Finally, effectiveness for a three-stage line is estimated by

EZ1Z2 = E00 + P1h12(Z1,Z2) + P3h32(Z1,Z2)

Team Exercise

A 20-stage transfer line with two buffers is being considered. Tentative plans place buffers of size 15 after workstations 10 and 15. The first 10 workstations have a cumulative failure rate of α = 0.005. Workstations 11 through 15 have a cumulative failure rate of α = 0.01 and workstations 16 through 20 together yield an α = 0.005. Repair of any station would average 10 cycles in length.

Estimate the effectiveness of this line design.

SolutionStep 1:

Set h12(15,15) = 0.5

Step 2: Combine stations 1 and 2

α1’ = α2 +α1[1 – h12(15,15)] = 0.01 + 0.005[.5] = 0.0125

α2’ = α3 = 0.005. Hence, we find that x1’ = 0.125, x2’ = 0.05,

s = x2’/ x1’ = 0.4. Using Buzacott’s expression with s ≠ 1,we find

C = 0.951898 and E15 = 0.8751. Now, using Eq 3 with

P2 = x2’/ (1+x1’+x2’), we find h32(15,15) ≈ (E15 – E0)/P2 = 0.564

83333.0)005.001.0005.0(101

1

1

13

1

1

00

i

ibE

Solution (contd..)Step 3: Combine stations 2 and 3

α2’ = α2 +α3[1 – h32(15,15)] = 0.01 + 0.005[.436] = 0.01218

α1’ = α1 = 0.005. Hence, we find that x1’ = 0.05, x2’ = 0.1218,

s = x2’/ x1’ = 2.436. Using Buzacott’s expression with s ≠ 1,we find

C = 1.04916 and E15 = 0.87740. Now, using the result E15 = E0 + P1h12 ,

estimate h12(15,15). Now P1 = x1’/ (1+x1’+x2’) = 0.04267,

we find h12(15,15) ≈ (E15 – E0)/P1 = 0.563 As our new estimate of 0.563 differs from our initial guess of 0.5, we return to step 2. As we continue the process, we find that

h12(15,15) = h12(15,15) = 0.563

Step 4: Estimate 3-stage effectiveness

E15 15

= E00 + P1h12(15,15) + P3h32(15,15)

≈ 0.08333 + [0.05/(1+0.05+0.1+0.05)]*0.563 + [0.05/(1+0.05+0.1+0.05)]*0.563

= 0.88

Homework Consider a three-stage transfer line with

buffers between each pair of stages. Stage I has a failure rate αi and repair rate bi. The maximum buffer sizes are Z1 and Z2 , respectively. Assume geometric failure and repair rates and ample repair workers. How many states are there for the system? Consider state (RWWz10) where 0<z1< Z1. Write

the balance equation for this state.

Conclusion Markov chain models can be used to

determine the increase in output for a single buffer.

Accurate output determination for a general line with many buffers is a difficult problem.