ENGG*3070 – Assignment No. 4 (Due Nov. 24, 2011) Flexible Manufacturing systems (FMS) 1. A FMS consists of three stations plus a load/unload station. Station 1 loads and unloads parts using two...

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ENGG*3070 – Assignment No. 4 (Due Nov. 24, 2011) Flexible Manufacturing systems (FMS) 1. A FMS consists of three stations plus a load/unload station. Station 1 loads and unloads parts using two servers (material handling workers). Station 2 performs horizontal milling operations with two servers (identical CNC horizontal milling machines). Station 3 performs vertical milling operations with three servers (identical CNC vertical milling machines). Station 4 performs drilling operations with two servers (identical drill presses). The machines are connected by a part handling system that has three work carriers and a mean transport time = 3.5 min. The FMS produces four parts, A, B, C, and D, whose part mix fractions and process routings are presented in the table below. The operation frequency fijk = 1.0 for all operations. Determine (a) maximum production rate of the FMS, (b) utilization of each machine in ths system, and (c) average utilization of the system sU . Part j Part mix pj Operation k Descriptio n Station i Process time tijk A 0.2 1 Load 1 4 min 2 H. Mill 2 15 min 3 V Mill 3 14 min 4 Drill 4 13 min 5 Unload 1 3 min B 0.2 1 Load 1 4 min 2 Drill 4 12 min 3 H. Mill 2 16 min 4 V. Mill 3 11 min 5 Drill 4 17 min 6 Unload 1 3 min C 0.25 1 Load 1 4 min 2 H. Mill 2 10 min 3 Drill 4 9 min 4 Unload 1 3 min D 0.35 1 Load 1 4 min 2 V. Mill 3 18 min 3 Drill 4 8 min 4 Unload 1 3 min 2. Use the extended bottleneck model to solve Problem No. 1 with the following number of parts in the system: (a) N = 5 parts, (b) N = 8 parts, and (c) N = 12 parts. Also determine the manufacturing lead time for the three cases of N in (a), (b), and (c). 3. A flexible manufacturing system is used to produce four parts. The FMS consists of one load/unload station and two automated processing stations (processes X and Y). The number of servers for each station type is to be determined. The FMS also includes an automated conveyor system with individual carts to transport parts between servers. The carts move the parts from one server to the next, drop them off, and proceed to the next delivery task. Average time required per transfer is 3.5 minutes. 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, which is 1 → 2 → 3 → 1. The product mix and processing times at each station are presented in the table below: Product j Product mix pj Station 1 Station 2 Station 3 Station 1 A 0.1 3 min 15 min 25 min 2 min B 0.3 3 min 40 min 20 min 2 min C 0.4 3 min 20 min 10 min 2 min D 0.2 3 min 30 min 5 min 2 min Required production is 10 parts per hour, distributed according to the product mix indicated. 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. Inventory Management 4. A manufacturing firm located in Guelph produces an item in a three-month time supply. An analyst, attempting to introduce a more logical approach to selecting run quantities, has obtained the following estimates of characteristics of the item ?? = 400 ??????????/???? ?? = $5 ?? = $4 ?????? 100 ?????????? ?? = 0.25$/$/???? Note: Assume that the production rate is much larger than D. a) What is the economic order quantity of the item? b) What is the time between consecutive replenishments of the item when the EOQ is used. c) The production manager insists that the ?? = $5 figure is only a guess. Therefore, he insists on using his simple three-month supply rule. Indicate home much you would find the range of ?? values for which the EOQ (based on ?? = $5) would be preferabel (in terms of a lower total replenishement and carrying costs) to the three-month supply. 5. The supplier of a product wants to discourage large quantity purchases. Suppose that all of the assumption of the basic EQQ apply except that a reverse quantity discount is applicable, that is, the unit variable cost is given by ?? = � ???? 0 <>< (1="" +="" )="" ≤="" where=""> 0 a) Write an expression (or expressions) for the total relevant cost per year as a function of the order quantity ??. Introduce (and define) whatever other symbols you feel are necessary. b) Using graphical sketches, indicate the possible positions of the best order quantity (as is done on Slide 18 - pages 11-12 for the regular discount case). c) What is the best order quantity for an item with the following characteristics. Demand rate = 50,000 units/yr Fixed setup cost per replenishment = $10 ?? = $1.0/unit, ?? = 0.005 Selling price = $1.44/unit Carrying charge = 0.20$/$/yr ???? = 1500 units Note: The variable cost ?? applies to all-units (not incrementally) Flexible Manufacturing systems (FMS) Inventory Management
Answered Same DayNov 21, 2021

Answer To: ENGG*3070 – Assignment No. 4 (Due Nov. 24, 2011) Flexible Manufacturing systems (FMS) 1. A FMS...

Sonam answered on Nov 22 2021
124 Votes
2.
MLT1 = 7.0 + 8.7 + 10.5 + 11.25 + 12.6 = 50.05 minutes
Rp* = 0.1778 pc/min from Problem1
N*
= 0.1778(50.05) = 8.9
(a) For N = 5 < N* = 8.9,
MLT1 = 50.05 minutes, and Tw = 0 Rp = 5/50.05 = 0.0999 pc/min = 5.99 pc/hour
U1 = (7.0/2)*(0.0999) = 0.350 = 35.0%
U2 = (8.7/2)*(0.0999) = 0.435 = 43.5%
U3 = (10.5/3)*(0.0999) = 0.350 = 35.0%
U4 = (11.25/2)*(0.0999) = 0.562 = 56.2%
U5 = (12.6/3)*(0.0999) = 0.420 = 42.0%
Us = [(2* 0.35) + (2* 0.435) + (3*0.35) + (2*0.562)]/9 = 0.416 = 41.6%
(b) N = 8 < N* = 8.9,
Rp = 8/50.05 = 0.1598 pc/min = 9.59 pc/hour
MLT1 =50.05 minutes, and Tw = 0.
U1 = (7.0/2)(0.1598) = 0.559 = 55.9%
U2 = (8.7/2)(0.1598) = 0.695 = 69.5%
U3 = (10.5/3)(0.1598) = 0.559 = 55.9%
U4 = (11.25/2)(0.1598) = 0.899 = 89.9%
U5 = (12.6/3)(0.1598) = 0.671 = 67.1%
Us = [(2* 0.559) + (2* 0.695) + (3* 0.559) +...
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