Introduction to Transfer Lines Active Learning – Module 1 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(Sections 3.1 and 3.2) 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 Understand and analyze about serial production systems that are subject to machine failures and random processing times Evaluate the effectiveness (availability) of a transfer line given Failure rates for the work stations Repair rates for the work stations

Time Management Readiness Assessment Test (RAT) - 5 minutes Introduction - 5 minutes Lecture on Paced Lines without buffers - 20 minutes Spot Exercise - 5 minutes Team Exercise - 5 minutes Homework Discussion - 5 minutes Conclusion - 5 minutes Total Lecture Time - 50 minutes

Readiness Assessment Test (RAT) Solve the following problem on probability: Six red discs, numbered from 1 to 6, and 4 green discs, numbered from 7 to 10, are placed in box A. Ten blue discs, numbered from 1 to 10, are placed in box B. Two discs are drawn from Box A and two discs are drawn from Box B. The four discs are drawn at random and without replacement. Find the probability that the discs drawn are: one red disc, one green disc and two blue discs with all four discs odd numbered.

Solution Probability: The total number of occurrences of a selection/event "a", divided by the total number of occurrences (a) plus the number of failures of those occurrences "b" (i.e.. total possible outcomes) Hence, probability is given by P = a /(a+b) The problem can be solved the following way: Probability = P (red-odd)*P (green-odd)*P (blue-odd)*P (blue-odd) + P (green-odd)*P (red-odd)*P (blue-odd)*P (blue-odd) = ( 3/10 * 2/9 * 5/10 * 4/9 ) + ( 2/10 * 3/9 * 5/10 * 4/9 ) = 4/135

Introduction Transfer Line: A set of serial, automatic machine and/or inspection stations linked by a common material transfer and control system. Buffers: Means for protecting workstations from failures elsewhere in the line, thereby improving station utilization.

Introduction (Contd..) Sample transfer line with intermediate buffers 1 2 3 1 # # Workstation # Buffer # Sample transfer line with intermediate buffers Reasons for a station to be non-functional Station failure Total line failure Station blocked Station starved Wo

Introduction (Contd..) Classification of Failures Time Dependent Failures Occur with a chronological frequency Time between failures is measured in units like hours E.g.: Daily Maintenance Operation Dependent Failures Occur while the system is running Time between failures is measured in cycles E.g.: Tool Wear

Introduction (contd..) downtime) (uptime ) uptime ( + = E Effectiveness (Availability) of a line can be defined as: E (uptime): Productive cycle time ( expected value ) E (uptime + downtime): Total cycle time ( expected ) downtime) (uptime ) uptime ( + = E

Paced Lines without Buffers Operation Dependent Failures: Assumptions Mean time to failure(MTTF) follows a geometric distribution with failure rate and the density function is given by = (1 - ) Mean cycles to failure(MCTF) is 1/ The number of cycles for repair at station is geometric with mean 1/bi cycles. Assume bi = b, i = 1,2,…M

Assumptions (contd..) All uptime and downtime random variables are independent. Idle stations do not fail. Failures occur at the end of the cycle, they do not destroy the product. A maximum of one station can fail on any cycle.

Deeper insight into the assumptions: Geometric distribution(GD) application in transfer lines. Every workstation will have two states – failure and working Probability that a station has failed = and working = 1 - Probability that the station fails at the 1st cycle is Probability that the station fails at the 2nd cycle = (1 - ) Probability that the station fails at the 3rd cycle = (1 - )2 Probability that the station fails at the t th cycle = (1 - )t-1 GD assumption allows the use of discrete time and discrete state Markow chain model to solve the above. At most one station failure assumption allows us to ignore the 2nd order terms. Idle station failure distinguishes between the two types of failures.

Model Analysis: Probability that the transfer line (M stations) first fails at the end of the tth cycle : Let us define Hence the above probability equation becomes P(T = t) = b(1 - b)t-1 M station line behaves like a single station but with failure parameter b replacing ai ..

Model Analysis (contd..): By expanding the terms in b and noting that ai being small, higher order terms approach zero and hence b Effectiveness of a line can be calculated as the ratio of the expected productive cycles between failures(uptime) divide by expected total cycles between failures(uptime + downtime). where b-1 is the mean number of cycles for repair to a station.

Spot Exercise Find the line availability for a three-station line in the respective cases 1 and 2: Also analyze the variation in the results obtained. The first, second and third stations average a failure every 10, 20 and 30 cycles respectively. Average repair time is 2 cycles. The first, second and third stations average a failure every 5, 10 and 15 seconds respectively. Average repair time is 2 seconds.

Solution Case 1: and b = ½ Case 2: Analysis: Effectiveness has been reduced from 0.75 to 0.71 because stations 1 and 2 continue the aging process while idle as a result of a failure at the other station

Time Dependent Failures: Assumptions Time to failure for station i is exponential with rate parameter ai Repair times are exponential with parameter bi For an operating time of t, Expected number of failures for a station i : ait Expected repair time : ait/bi Effectiveness of a station i is given by : As stations are independent, E = E1*E2*E3*…

Team Exercise Repairmen can fix a failed workstation in 10 minutes on an average. Cycle time is 20 seconds. Stations seem to operate 250 cycles on average before failing. Estimate the daily (8-hour shift) production for a six-station line. What is the average daily workload of the operator. ( problem 3.16 – Askin & Stanridge)

Solution Given: Average Repair time = 10 min, Cycle time C = 20sec αi = 1/250, b-1 = 30 (a) Estimate the output for 8 hours = 8*60*3 = 1440 cycles E = [1 + b-1Σ αi ]-1 = [1 + 30(0.024)]-1 = 0.581 Units/shift = 0.581 * 1440 = 837 (b) Average daily workload of the repairmen The % of time repairmen are busy = (1 – E) * 100 = 41.9 % of shift = 3.35 hrs

Homework A 15 stage transfer line is being considered. Stations 1 through 10 should have the same reliability. Each station is expected to operate approximately 1000 cycles between failures. The last 5 stations are expected to break down about once every 600 cycles. Repair times will vary but should average about the equivalent of 12 cycles in duration. Estimate the availability of the line.

Conclusion Transfer Lines : Serial production systems that are very costly and subjected to random failures Failures are of two types – Time Dependent and Operation Dependent They can affect the entire line or just a single station Reliability models help in estimating system effectiveness when buffers are not used