Two-Stage Paced Lines Active Learning – Module 2

1. Two-Stage Paced LinesActive Learning – Module 2 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 3.3) 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.

3. 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.

4. 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

5. 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

6. 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.

7. 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

8. Assumptions and Conventions • Markow Chain Environment (s1 ,s2 ,z) • siis 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

9. 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.

10. 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

11. 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

12. 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

13. 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.

14. Solution Given: 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, wherex= 0.04/0.25 = 0.16 and s = 1, r = 1 Hence, the number of parts/hr = E10* Cycles/hr = 16.40

15. 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

16. α1 α2 α5 α4 α3 α2 α3 α1 α5 α4 • 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) α6 α4+ α5 α1+ α2+ α3+ α6

17. α4 α3 α2 α1 • 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 α1

18. 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.

19. 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?

20. Solution Given: 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 = 0.16and C = 0.9153 Now, E5 – E0 = 0.029 < 0.04, and do not include the buffer

21. 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

22. 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.