Approximate Three-Stage Model: Active Learning – Module 3

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 2M(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 xi = αi / bi

7. 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 • xiis 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

8. 2 2 • 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 3 1 # # Workstation # Buffer #

9. 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)]}

10. 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)

11. 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)]}

12. 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) ofEq 3 • Solvingfor 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.

13. 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 tostep 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)

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 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 3with P2 =x2’/(1+x1’+x2’), we find h32(15,15) ≈ (E15–E0)/P2 = 0.564

16. 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). NowP1 =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

17. 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.

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.