What condition would cause the maximum cycle time and minimum throughput?A.tAll parts are spread evenly between stationsB.tAll parts are processed at one workstation and moved to the next station togetherC.tAll parts are processed as soon as the workstations becoming availableD.tAll parts are processed to maintain the same waiting time
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- Maximum cycle time refers to the longest time it takes for a product to pass through all the stations in a production line. - Minimum throughput refers to the lowest rate at which products are produced in a given period. Show more…
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3. Floor-On, Ltd., operates a line that produces self-adhesive tiles. This line consists of single-machine stations and is almost balanced (i.e., station rates are nearly equal). The manufacturing engineer has estimated the bottleneck rate of the line to be 2,000 cases per 16-hour day, and the raw process time to be 30 minutes. The line has averaged 1,600 cases per day, and the cycle time has averaged 4 hours. a) What would you estimate the average WIP level to be? b) How does this performance compare to the practical worst case? c) What would happen to the throughput of the line if we increased capacity at a non-bottleneck station and held WIP constant at its current level? d) What would happen to the throughput of the line if we replaced a single-machine station with four machines whose capacity equaled that of the single machine and held the WIP constant at its current level? e) What would happen to the throughput of the line if we began moving cases of tiles between stations in large batches instead of one at a time?
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Akash M.
An FMS consists of four stations. Station 1 is a load/unload station with one server. Station 2 performs milling operations with three servers (three identical CNC milling machines). Station 3 performs drilling operations with two servers (two identical CNC drill presses). Station 4 is an inspection station with one server that performs inspections on a sampling of the parts. The stations are connected by a part handling system that has two work carriers and whose mean transport time = 3.5 min. The FMS produces four parts, A, B, C, and D. The part mix fractions and process routings for the four parts are presented in the table below. Note that the operation frequency at the inspection station (f4jk) is less than 1.0 to account for the fact that only a fraction of the parts are inspected. Determine: (a) maximum production rate of the FMS, (b) corresponding production rate of each part, (c) utilization of each station in the system, and (d) the overall FMS utilization. Part j | Part Mix pj | Operation k | Description | Station i | Process Time tijk (min) | Frequency fijk A | 0.1 | 1 | Load | 1 | 4 | 1.0 | | 2 | Mill | 2 | 20 | 1.0 | | 3 | Drill | 3 | 15 | 1.0 | | 4 | Inspect | 4 | 12 | 0.5 | | 5 | Unload | 1 | 2 | 1.0 B | 0.2 | 1 | Load | 1 | 4 | 1.0 | | 2 | Drill | 3 | 16 | 1.0 | | 3 | Mill | 2 | 25 | 1.0 | | 4 | Drill | 3 | 14 | 1.0 | | 5 | Inspect | 4 | 15 | 0.2 | | 6 | Unload | 1 | 2 | 1.0 C | 0.3 | 1 | Load | 1 | 4 | 1.0 | | 2 | Drill | 3 | 23 | 1.0 | | 3 | Inspect | 4 | 8 | 0.5 | | 4 | Unload | 1 | 2 | 1.0 D | 0.4 | 1 | Load | 1 | 4 | 1.0 | | 2 | Mill | 2 | 30 | 1.0 | | 3 | Inspect | 4 | 12 | 0.333 | | 4 | Unload | 1 | 2 | 1.0
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