1. Approximate Three-Stage Model: Active Learning – Module 3 Dr. Cesar Malave Texas A & M University

2. 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.

3. 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

4. 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

5. 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 2 M (Z+1) M-1 states

6. 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 where x i = α i / b i

7. Three-Stage Model (Contd..) Deeper analysis into the model:  Consider a line without buffers  For every unit produced, station i is down for x i cycles  x i 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 α 0 and β 0,we have

8. 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

9. Model Analysis (contd..):  Let us define h ij (Z 1,Z 2 ) as the proportion of time station j operates when i is under repair for the specified buffer limits E Z 1 Z 2 = E 00 + P 1 h 12 (Z 1,Z 2 ) + P 3 h 32 (Z 1,Z 2 ) - 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 rateα 2 when station 3 is down and buffer 2 is full - with a failure rate {α 3 [1 – h 32 (Z 1,Z 2 )]}

10. Model Analysis (contd..):  Stations 2 and 3 along with the connecting buffer are replaced by a pseudo station 2’ with a failure rate α 2 ’ = α 2 +α 3 [1 – h 32 (Z 1 Z 2 )] - Eq 2 ~ α 2 +α 3 [1 – h 32 (Z 2 )] as h 32 () will not depend on Z 1  Hence for a two-stage line, effectiveness can be written as E Z = E 0 + P 1 h 12 (Z) - Eq 3 = E 0 + P 2 h 21 (Z) where P i is the probability that station i is down as referred before h 12 (Z) is nothing but P 1 h 12 (Z 1 Z 2 )

11. Model Analysis (contd..):  Had we known h 32 (Z 1 Z 2 ), 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 h 32 (Z 1 Z 2 ) ? 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 α 2 when station 1 is down and buffer 1 is empty - with a failure rate {α 1 [1 – h 12 (Z 1,Z 2 )]}

12. Model Analysis (contd..):  Stations 1, 2 and the connecting buffer can be replaced by a pseudo workstation 1’with a failure rate α 1 ’ = α 2 +α 1 [1 – h 12 (Z 1 Z 2 )] - Eq 4 ~ α 2 +α 1 [1 – h 12 (Z 1 )] as h 12 () will not depend on Z 2  Station 3 will have a failure rate α 3  The two-stage pseudo line can be solved by estimating h 32 (Z 1 Z 2 ) from h 21 (Z) of Eq 3  Solving for Case 1, i.e. estimating h ij () 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.

13. Solution Procedure: 1. Initialize h 12 (Z 1,Z 2 ) at, say, 0.5. Denote stages 1and 2 in any pseudo two-stage approximation as 1’ and 2’, respectively. Calculate E 00, the effectiveness for the unbuffered line. 2. Solve the two-stage line with α 1’ given by - Eq 4. Estimate h 32 (Z 1,Z 2 ) = h 2’1’ (Z) from Eq 3. α 2’ = α 3 3. Solve the two-stage line with α 2’ given by - Eq 2. Estimate h 12 (Z 1,Z 2 ) = h 1’2’ (Z) from Eq 3. α 1’ = α 1 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 E Z 1 Z 2 = E 00 + P 1 h 12 (Z 1,Z 2 ) + P 3 h 32 (Z 1,Z 2 )

14. 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.

15. Solution Step 1: Set h 12 (15,15) = 0.5 Step 2: Combine stations 1 and 2 α 1 ’ = α 2 +α 1 [1 – h 12 (15,15)] = 0.01 + 0.005[.5] = 0.0125 α 2 ’ = α 3 = 0.005. Hence, we find that x 1 ’ = 0.125, x 2 ’ = 0.05, s = x 2 ’ / x 1 ’ = 0.4. Using Buzacott ’ s expression with s ≠ 1,we find C = 0.951898 and E 15 = 0.8751. Now, using Eq 3 with P 2 = x 2 ’ / (1+x 1 ’ +x 2 ’ ), we find h 32 (15,15) ≈ (E 15 – E 0 )/P 2 = 0.564

16. Solution (contd..) Step 3: Combine stations 2 and 3 α 2 ’ = α 2 +α 3 [1 – h 32 (15,15)] = 0.01 + 0.005[.436] = 0.01218 α 1 ’ = α 1 = 0.005. Hence, we find that x 1 ’ = 0.05, x 2 ’ = 0.1218, s = x 2 ’ / x 1 ’ = 2.436. Using Buzacott ’ s expression with s ≠ 1,we find C = 1.04916 and E 15 = 0.87740. Now, using the result E 15 = E 0 + P 1 h 12, estimate h 12 (15,15). Now P 1 = x 1 ’ / (1+x 1 ’ +x 2 ’ ) = 0.04267, we find h 12 (15,15) ≈ (E 15 – E 0 )/P 1 = 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 h 12 (15,15) = h 12 (15,15) = 0.563 Step 4: Estimate 3-stage effectiveness E 15 15 = E 00 + P 1 h 12 (15,15) + P 3 h 32 (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

17. Homework  Consider a three-stage transfer line with buffers between each pair of stages. Stage I has a failure rate α i and repair rate b i. The maximum buffer sizes are Z 1 and Z 2, respectively. Assume geometric failure and repair rates and ample repair workers. How many states are there for the system? Consider state (RWWz 1 0) where 0<z 1 < Z 1. Write the balance equation for this state.

18. 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.