2.810 Manufacturing Processes and Systems Quiz #2. November 15, minutes

Transcription

1 2.810 Manufacturing Processes and Systems Quiz #2 November 15, minutes Open book, open notes, calculators, computers with internet off. Please present your work clearly and state all assumptions. Print Name: Solutions (TA) Problems: 1. TPS Cell Rate Increase 2. Waterjet Scheduling 3. Additive Mfg Vs Machining 4. Additive Mfg Systems 5. Readings / 30 points / 20 points / 15 points / 25 points / 10 points / 100 points 1

2 1) (30 pts) Toyota Cell Production Rate Increase Consider machining the following aluminum part in a TPS cell with a process plan, times and configuration as shown below. The part is machined from bar stock 2.5in X 2.0in Process Operation Step #1 Cold Saw to 4.5in (no facing required) #2 Mill: Large hog out #3 Drill & Mill: Shallow drill and radius #4 Drill & Mill: Deep drill and radius #5 Mill: Small hog out #6 Inspection and deburr Approx. vol. removed Tool changes (no tool change needed) 10 in3 1 end mill (no tool change) 0.6 in3 3 (center drill, form drill, end mill) 2.4 in3 3 (center drill, form drill, end mill) 3.6 in3 1 end mill (no tool change) Process time Manual time (minutes) (minutes) 2 2 (includes taking material from raw stock) (includes depositing finished part into adjacent bin) 2

3 The drilling and milling steps are depicted in the figures below. 3

4 The TPS Cell that uses the process plan above is shown below. The Manual Time/Machine Time for each process are given below in minutes. The walking time between each station is 1/12 minute. Raw Materials #1 #2 #3 #6 #5 #4 Finished Parts a) Currently this cell is operated with one operator traveling from #1 around the loop to #6 and then back to #1. Please determine the production rate for the cell (as parts per hour) as well as the inventory in the cell and the time it takes for one part to move through the cell (hours). Step Walking Time Manual Time Machine Time / / / / / /12 TOTAL

5 (+3 pts) The first thing we need to do is check whether the manual + walking time or the manual + machine time dictates the cycle time. max(manual Time + Machine Time) = = 10 minutes (from 2nd machine) Walking Time + Manual Time = = 12.5 minutes Therefore, Walking Time + Manual Time > max(manual Time + Machine Time) and the walking/manual time determines the cycle time. (+3 pts) Calculate the production rate of the system λ = parts time = 1 part 60 min parts = min 1 hr hr (+3 pts) Find the average number of parts in the system L = = parts (+3 pts) Use Little s Law to calculate average time in the system L = λw w = w = L λ 6.99 parts = hours 4.8 parts/hr b) We would like to increase the production rate for this cell by using two operators. Please show the best arrangement for the highest production rate by only adding a second operators. Add decouplers (additional inventory) if necessary. Please draw on the diagram below to show how you have arranged the operators. What is the production rate now? What limits the production rate? Please make three suggestions on how to improve on this production rate. Rank order them in terms on the costs that might be incurred and give the estimated new improved production rate with this improvement. State all assumptions clearly. 5

6 (+3 pts) Choosing an operator configuration that reduces the production rate. Actually most choices will work as long as the choice leaves you with machine #2 as the limiting factor. The best choice for easiest explanation and calculation is for Operator #1 to be responsible for machines 1 -> 2 -> 6 (but stop at 5 to pick up the part from the decoupler) and Operator #2 to be responsible for machines 3 -> 4 -> 5 (but stop at 2 to pick up the part from the decoupler) (+3 pts) Depending on your operator configuration, decouplers should be added after the last machine that each operator uses. There needs to be a decoupler to account for each handoff or intersection of the operator s paths. And the definition of a decoupler is that it only holds one unique part. In this configuration, a decoupler is placed right after Machine #2 (for Operator #1 to pick up) and Machine #5 (for Operator #2 to pick up). Adding extra decouplers for no reason was a common mistake here. More decouplers would increase the inventory and number of parts in the system and not be optimal. There was also an issue of parts not being in the correct location for the other worker to pickup or only being available at certain times during the loop. 6

7 Operator 1: Step Walking Time Manual Time Machine Time / /12 6 1/ /12 TOTAL Operator 2: Step Walking Time Manual Time Machine Time 3 1/ / / /12 TOTAL (+3 pts) By inspection of the new tables, it is clear that Operator #1 has the longer times, and so we should investigate further. As before, let s compare the Walking Time + Manual Time to the max(manual Time + Machine Time). max(manual Time + Machine Time) = = 10 minutes (from 2nd machine) Walking Time + Manual Time = = 6.33 minutes Therefore, Walking Time + Manual Time < max(manual Time + Machine Time) and Machine #2 limits the cycle time because it is the slowest operation inside Operator #1 s loop. (+3 pts) Calculate the production rate using the Machine #2 limitation. 7

8 λ = parts time = 1 part 60 min parts = 6 10 min 1 hr hr Various suggestions that could be ranked (+2 pts for each of the two suggestions and +2 for ranking them). Extrude or bandsaw the part rather than milling in Machine #2. Notice that we are milling 10 in^3 and that is a lot of material, which could be removed much faster in cross-section. This would substantially reduce the time of Machine #2 (probably to less than a minute). Offload some of the milling volume in Machine #2 to a different machine so that the machine times are more balanced. Make sure this makes sense with the process plan steps and order of operations. Increase the MRR (or feed rate or other factors that would positively influence the MRR) and then account for a more often tool change Better train the workers at Machine #2 or have a pre-set fixture that reduces manual time to below 0.33 minutes (for those that calculated the walking time of 8.33 minutes) Add a second machine (very expensive but would drop the cycle time down to walking time + manual time again since now the machining step for #2 would take only 4 minutes (and this would give a max time of 6 including the manual time which is less than the walking + manual time of 6.33) Because we are already limited by the Machine #2, adding another worker will not help us no matter where they are placed in the system (unless we already duplicate Machine #2 or make one of these other changes above). All answers must be somehow related to either the manual or machine time associated with Step #2. Better answers definitely took into account the process plan on the first page. 8

9 2) (20 pts) Waterjet scheduling a) We all know that the water jet is a popular machine in This problem is concerned with the efficient scheduling of this machine. Assuming that the teams are scheduled to arrive every two hours at the waterjet. To ensure a timely progress of each project, each team is asked to ensure that their waterjet cutting can be accomplished in 1.5 hours. However, we know that some projects are ambitious and they will take longer than 1.5 hours. Model this problem as a queuing problem assuming the average processing rate is one project every 1.5 hours but there is significant variation in the processing times and arrival times (i.e. exponentially distributed). How long do you think it would take to process eight projects by this method? (+2 pts) (+2 pts) λ = arrival rate = 1 team = 0.5 teams/hr 2 hours μ = service rate = 1 team = 0.67 teams/hr 1.5 hours (+2 pts) MM1 Queue (exponentially distributed) (+2 pts) (+2 pts) L = λ μ λ = 0.5 = teams w = L λ teams = = 6.25 hours 0.5 teams/hr Total process time = w 8 = 6.25 hours 8 = 50 hours Full credit was given for small changes due to rounding. Most people lost points for misinterpreting the service rate and arrival rates. Credit was passed through if those errors propagated through the problem. Other students multiplied the number of projects incorrectly due to a bad misinterpretation of the wait time and process time. 9

10 b) One of the students in suggests that we can redesign our waterjet projects by rescheduling the teams to arrive every 2.5 hours and pick an average processing rate that ensures that the average time for each project is two hours. What is the average processing rate that ensures this outcome? (+2 pts) λ = arrival rate = 1 team = 0.4 teams/hr 2.5 hours (+2 pts) μ = service rate = 1 team x =? (+2 pts) MM1 Queue (exponentially distributed) (+2 pts) (+2 pts) w = 2 hours w = 2 = 1 μ 0.4 μ = service rate = = 0. 9 teams/hr μ = service rate = 1 team, x = 1.1 hours x 10

11 3) (15 pts) Additive Manufacturing Vs Machining a) Figure 1b in the paper on additive manufacturing provided by Prof. John Hart shows the cost to make a family of planar stainless steel parts of increasing complexity by both CNC milling and selective laser melting (SLM). The figure suggests that as complexity increases there is a cross over and the additive manufactured parts appear to be less expensive than the machined parts. This suggests that for moderately complex parts, additive manufacturing (AM) may displace machining in production manufacturing. Please give your thoughts on this matter. For example, discuss other parameters besides cost that might be important in promoting this cross over. In particular, please address quality and rate. More credit is given to answers that use numerical estimates to support the arguments. Use the nature of the parts shown in the figure as a basis for your answers. Be sure to make clear whether a particular aspect of the trade-off between the two processes would be in favor of AM, or against it. (There is NO NEED to concern yourself with attempting to use the complexity metric used in the paper. DON T WASTE TIME ON IT!) 11

12 (+3 pts) Nature of the part. Based on the hint from the prompt with the picture of the various part, we see that the complexity alludes to the nature of the part (what it looks like). Because SLM is an additive process and machining is a subtractive process, the nature of the part (whether it is hollow or nearly a solid part) as well as its function has an extremely large impact on all of the parameters (cost, rate, quality, etc). (+2 pts) 2 pts of credit was given for any other interesting comments given either about flexibility (SLM can be more flexible than machining in case the product changes or becomes more complex), about the graph itself (there is a wide variance of data between the two bureaus so it seems like there is some skill involved or various types of techniques within CNC and SLM), or more detail about cost (some of the simplest parts can be very difficult to machine such as a sphere and some of the simplest parts can be very difficult for SLM if they require several support structures, etc). Rate (+3 pts) The more detail the better. In general, SLM is slower than machining. But why? Rate in additive manufacturing is limited by the heat transfer needed to melt the material. In machining, you can overcome some of the heat constraints by using harder/stronger tools, changing the material, and using coolant. (+2 pts) As asked for, more credit was given to numerical answers that mentioned a print rate of kg/hr (Hart AM Slide 17) for SLM. Quality (+3 pts) The more detail the better. In general, SLM has worse resolution than machining. For the vast majority of current products, post-processing (machining) is actually needed to achieve the requisite surface finish. In addition, there is anisotropic strength properties depending on the axis of printing. This further slows down the effective rate and increases total cost. Meanwhile, machining can have very good resolution on its own by using different tools, cutting speeds, and milling directions. Since it is a subtractive process, strength can be much more isotropic. (+2 pts) As asked for, more credit was given to numerical answers that mentioned a resolution of 0.1mm (Hart AM Slide 17) for SLM. 12

13 4) (25 pts) Additive Manufacturing System Performance Consider the following SLM manufacturing system made up of 3 sequential steps. In the first step parts are printed; in the second, parts are washed (to remove the support structures); and in the third steps parts are fully cured. In the system shown below there are 3 parallel printing machines. The parts moving through the system have a volume of 50cm3. Print 1.1 Print 1.2 Wash Cure Print 1.3 Machine Capacity (parts) Rate or Time Printer (single machine) cm3/hr Wash minutes/batch Cure minutes/batch Consider the operation of the system under the following conditions: Parts are printed 120 at a time (40 in each print machine) then they move to the subsequent steps. Once the printers finish the 120 parts another 120 parts are loaded. 13

14 a) Please draw the system boundaries around the entire system and determine the average number of parts in the system, the average time for the parts in the system, and the average production rate. (+2) Calculate the production rates of each step (drawing the system boundary around each step) to determine the bottleneck and the production rate of the total system. Print volume = 40 parts * 50 cm^3/part = 2000 cm^3 Print time = (2000 cm^3) / (100 cm^3/hr) = 20 hours Print production rate = 40 parts / 20 hours = 2 parts/hr Since there are 3 printers, total print production rate = 2 parts/hr * 3 = 6 parts/hr Wash production rate = 60 parts / 0.25 hours = 240 parts/hr Cure production rate = 60 parts / 1 hour = 60 parts/hr (+3) The bottleneck in the system is the print stations. Therefore, the production rate of the system is 6 parts/hr. Many students confused how 3D printing works in batches compared to other operations and also how print volume relates to print time. This was penalized heavily as well as not utilizing Little s Law for the system. (+2 pts) Calculate the average parts in the system. 2 pts were given for a non-averaged attempt to sum the parts based solely on the batch sizes for each station ( for instance for a total of 240 parts). (+3 pts) The more exact answer uses a weighted average for the number of parts in the system. See below for the drawing of the system where we track 240 parts going through the system in steady state across 40 hours. Using this analysis, we can account for how long each batch of parts stay in the system or how many parts are in each location during that timeline. 14

15 L = = parts (+5 pts) Use Little s Law around the entire system to calculate the average time in the system. w = L λ = parts 6 parts/hr = hours Many students did not use Little s Law, which was crucial for this question, and ended up making more than one average quantity that ultimately did not abide by Little s Law. Also the bottleneck had to be analyzed first and so the print rate could not be calculated last. b) What inventory holding space is required in this system? Please draw it on the figure and give the part volume size needed. (+1 pt) There are two inventory stations needed, one because of the batch bottleneck and one because of the buildup of the differential between wash and cure times. (+2 pts) The inventory needed in front of the wash station is 60 parts (or 3000 cm^3 of space) because out of the 120 parts that are finished printing, only 60 parts can enter the wash at a time. (+2 pts) The inventory needed in front of the cure station is 60 parts (or 3000 cm^3 of space) because the wash station finishes those 60 parts and then another 60 parts from the original print batch before the cure station is ready again. 15

16 If you tried to say something else, you must have been extremely clear with your assumptions because the system explained above would create these by default. c) What is the shortest time to produce one part in the system? How does this production rate compare with the average production rate for the system operating with 120 parts at a time? (+3 pts) If only analyzing one part, then we need to redo the print calculations. Print volume = 1 part * 50 cm^3/part = 50 cm^3 Print time = (50 cm^3) / (100 cm^3/hr) = 0.5 hours Wash time = 0.25 hours Cure time = 1 hour Shortest time to produce one part = 0.5 hr hr + 1 hr = 1.75 hours (105 minutes) (+2 pts) Compare this to the 120-part system. This is a much shorter time to wait for a single part. Therefore, this system is good for a small number of parts if the lead time is constrained. Otherwise, the 120-part system produces parts much faster but as a much longer lead time. 16

17 5) (10 pts) Readings Interchangeable parts played an outsized role in the development of modern manufacturing. However, its historical development appears to have been largely misunderstood. In your literature readings, the current understanding of this development was clearly articulated. Please answer the following questions and identify in your reading where you learned these facts; (+2 pts) Citations from Hounsell Introduction (pgs. 3 and 4) a) Who conclusively demonstrated that Eli Whitney did not produce products with interchangeable parts? (+2 pts) Woodbury convincingly argued that it wasn t Eli Whitney. Edwin Battinson solidly confirmed this point. Students that wrote Merritt Roe Smith were awarded 1 pt. b) Who then was identified as the prime mover of the development of interchangeable parts? (+2 pts) The United State Ordinance Department was the prime mover in the development of interchangeable parts. Students that wrote John Hall were awarded 1 pt. c) What was the product that demonstrated interchangeability? (+2 pts) Small arms or the breech-loading rifle was the product that demonstrated interchangeability. Students that only wrote guns or something similar were given 1 pt if they also provided some greater detail. d) Did interchangeability reduce the cost to make this product? (+2 pts) No, the unit cost of Springfield small arms was significantly higher than that of arms produced by more traditional methods. It wasn t until much later that interchangeability reduced costs in manufacturing. Students that did not explicitly write No were given 1 pt if they mentioned that eventually it reduced cost. 17