Flexible Manufacturing Systems (Continue) Example 1: A semi-automated flexible manufacturing cell is used to produce three products. The products are made by two automated processing stations followed by an assembly station. There is also a load/unload station. Material handling between stations in the FMC is accomplished by mechanized carts that move tote bins containing the particular components to be processed and then assembled into a given product. The carts transfer tote bins between stations. In this way the carts are kept busy while the tote bins are queued in front of the workstations. Each tote bin remains with the product throughout processing and assembly. The details of the FMC can be summarized as follows: The product mix fractions and station processing times for the parts are presented in the table below. The same station sequence is followed by all products: 1 → 2 → 3 → 4 → 1. The average cart transfer time between stations is 4 minutes. a) What is the bottleneck station in the FMC, assuming that the material handling system is not the bottleneck? b) At full capacity, what is the overall production rate of the system and the rate for each product? c) What is the minimum number of carts in the material handling system required to keep up with the production workstations? d) Compute the overall utilization of the FMC. e) What recommendations would you make to improve the efficiency and/or reduce the cost of operating the FMC? (a) Solution WL1 = (3+2)(0.35 + 0.25 + 0.4) = 5.0 min WL2 = 9(0.35)(1.0) + 5(0.25)(1.0) + 4(0.4)(1.0) = 6.0 min WL3 = 7(0.35)(1.0) + 8(0.25)(1.0) + 6(0.4)(1.0) = 6.85 min 146 pc/min = 8.146) = 1.0) = 6.6% U3 = (6.WL4 = 5(0.07 pc/hr RpB = 0. nc = 16/6.453 = 45.0/nc = 6.35)(1.9% (e) Recommendations: 1.50 pc/hr (c) Minimum number of carts required in system: 16.0) + 8(0. .365 = 36.85/1)(0. Reduce number of servers at station 4 to 1 server.85 = 2.76) = 3.76 pc/hr RpA = 0.2 min nt = 4 for all parts.4)(1.5% U2 = (6.35(8.25)(1.0) + 5(0.2/2)(0.0/2)(0.0/3)(0.146) = 0.76) = 3. and 2.3% U5 = (16.0 = 100% U4 = (6.34 → Use 3 carts (d) U1 = (5.146) = 0.4(8.0 min (b) Bottleneck is station 3: Rp* = 1/6.85 = 0.876 = 87.76) = 2. Reduce number of servers at station 1 to 1 server.0/1)(0.25(8. WL5 = 4(4) = 16.85 Rearranging.779 = 77.146) = 0.19 pc/hr RpC = 0.146) = 0. and an automated conveyor system with an individual cart for each product. The product mix and process times for the parts are presented in the table below: Product j Part mix pj Station 1 Station 2 Station 3 Station 4 Station 1 A 0. that is. the difference being that inspections at station 4 are performed on only one part in four for each product (f4jk = 0. (a) Using the bottleneck model.3 5 min 10 min 30 min 20 min 4 min C 0. (c) With the number of carts determined in (b). an inspection station. only eight parts are allowed in the system even though the conveyor system has a sufficient number of carriers to handle more than eight.2 5 min 15 min 25 min 20 min 4 min B 0. which are 12341 or 1 2 3 1. The conveyor carts remain with the parts during their time in the system.25). two automated processing stations. and determine the overall production rate of the system.5 5 min 20 min 10 min 20 min 4 min The move time between stations is 4 min. (d) How close are your answers in (a) and (c)? Why? . and therefore the mean transport time includes not only the move time. The FMS consists of a load/unload station. (b) Determine how many carts are required to eliminate the conveyor system as the bottleneck. show that the conveyor system is the bottleneck in the present FMS configuration. use the extended bottleneck model to determine the production rate for the case when N = 8.Example 2: A flexible manufacturing system is used to produce three products. The number of servers at each station is given in the following table: Station 1 Load and unload 2 workers Station 2 Process X 3 servers Station 3 Process Y 4 servers Station 4 Inspection 1 server Transport system Conveyor 8 carriers All parts follow either of two routings. but also the average total processing time per part. 2)(1.3)(1.0) = 16.33.333 min 19.2)(1.5)(1.2)(1. .5 min 16.0 min WL2 = 15(0.0) + 10(0.625 Given N = 8.75 min 5.0) + 30(0.0) + 20(0.0/nc = 5.75 min Bottleneck Bottleneck is station 5 (transport): Rp* = 8/62 = 0.0/2 = 4. nc = 62/5. The reason is that the same bottleneck is in effect.0 min WL4 = 20(0.1875 pc/min = 11.0 min 62.0/4 = 4.0) + 20(0.0 min Station WLi/si ratio 1 (load/unload) 2 (process X) 3 (process Y) 4 (inspection) 5 (transport) 9.3 + 0.5)(1.0) + 20(0.25.625 MLT1 = 62.25 + 1 = 3.0) = 11.0) = 19. which is the number of parts allowed in the system.0/8 = 7.129 pc/min = 7.129 pc/min = 7. case 1 applies since 8 < N* = 11.2 + 0.0/1 = 5.5) = 9.0/3 = 5.0 min Rp = 8/62 = 0.0 min WL3 = 25(0.Solution (a) WL1 = (5+4)(0. With station 2 as bottleneck.333 = 11.74 pc/hr (d) Answers in (a) and (c) are identical.0) + 10(0. Rp* = 3/16 = 0.0 min nt = 1 + 1 + 0.25 pc/hr (c) N* = 0.5)(1.1875(62.63 Use nc = 12 carriers.25(4) + 9 + 16 + 19 + 5 = 62.0) = 5.3)(1.74 pc/hr (b) Next bottleneck station after transport system = station 2. whether that is determined by the available number of carriers in the FMS or by the number of parts launched into the system. WL5 = 3. 62.3)(1. Find the minimum number of carriers to eliminate the transport system as the bottleneck. 0 min WL2 = 15(0.0) + 10(0. Use the bottleneck model to determine (a) the minimum number of servers at each station and the minimum number of carts in the transport system that are required to satisfy production demand and (b) the utilization of each station for the answers above.0)/60 = 50/60 = 0.5 min WL3 = 25(0.2)(1.5) = 10.3)(1. For Rp = 10 pc/hr. Solution (a) WL1 = (3+2)(0.1)(1. the above workloads would be 10 times the above values.0) = 13.1 + 0.4 + 0.0) = 27.5 min The preceding workloads are for one unit/hr.3)(1. drop them off. which is 1 2 3 1.5 min nt = 3. Therefore.583 use 5 servers .4)(1. The time available in one hour for each server = 60 min. The carts move the parts from one server to the next.1)(1.2 3 min 30 min 5 min 2 min Required production is 10 parts per hour. WL4 = 3(3. The product mix and processing times at each station are presented in the table below: Product j A Product mix pj 0.0) = 5.833 use 1 server Station 1: n2 = 10(27.0) + 30(0. Average time required per transfer is 3.0) + 20(0.0) + 5(0. The number of servers for each station type is to be determined.3 + 0.4)(1.Example 3: A flexible manufacturing system is used to produce four parts. distributed according to the product mix indicated.4 3 min 20 min 10 min 2 min D 0.5 minutes.5)/60 = 275/60 = 4.0) + 20(0.3 3 min 40 min 20 min 2 min C 0.2)(1. The FMS consists of one load/unload station and two automated processing stations (processes X and Y).2)(1.0) + 40(0. and proceed to the next delivery task. The following table summarizes the FMS: Station 1 Load and unload Number of human servers (workers) to be determined Station 2 Process X Number of automated servers to be determined Station 3 Process Y Number of automated servers to be determined Station 4 Transport system Number of carts to be determined All parts follow the same routing. the number of servers at each station and the number of carriers in the transport system are determined as follows: Station 1: n1 = 10(5. The FMS also includes an automated conveyor system with individual carts to transport parts between servers.1 Station 1 3 min Station 2 15 min Station 3 25 min Station 1 2 min B 0. 25 use 3 servers Station 1: n4 = 10(10.Station 1: n3 = 10(13.875 = 87.5)/60 = 105/60 = 1.3% U2 = 4.75 = 75.0% U4 = 1.833 = 83.917 = 91.25/3 = 0.583/5 = 0.5)/60 = 135/60 = 2.833/1 = 0.7% U3 = 2.5% .75/2 = 0.75 use 2 carriers (b) U1 = 0.