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PROBLEMS Unattended Operation 14.1

A CNC machining center has a programmed cycle time = 25.0 min for a certain part. The time to unload the finished part and load a starting work unit = 5.0 min. (a) If loading and unloading are done directly onto the machine tool table and no automatic storage capacity exists at the machine, what are the cycle time and hourly production rate? (b) If the machine tool has an automatic pallet changer so that unloading and loading can be accomplished while the machine is cutting another part, and the repositioning time = 30 sec, what are the total cycle time and hourly production rate? (c) If the machine tool has an automatic pallet changer that interfaces with a parts storage unit whose capacity is 12 parts, and the repositioning time = 30 sec, what are the total cycle time and hourly production rate? Also, how long does it take to perform the loading and unloading of the 12 parts by the human worker, and what is the time the machine can operate unattended between parts changes? Solution: (a) Tc = 25.0 + 5.0 = 30.0 min/pc Rc = 60/30 = 2.0 pc/hr (b) Tc = Max(25.0, 5.0) + 0.5 = 25.5 min/pc Rc = 60/25.5 = 2.35 pc/hr (c) Tc = Max(25.0, 5.0) + 0.5 = 25.5 min/pc Rc = 60/25.5 = 2.35 pc/hr Time to load/unload = 12(5.0) = 60 min UT = 12(25.5) - 60 = 306 - 60 = 246.0 min = 4.1 hr

Determining Workstation Requirements 14.2

A total of 7000 stampings must be produced in the press department during the next three days. Manually operated presses will be used to complete the job and the cycle time is 27 sec. Each press must be set up before production starts. Setup time for this job is 2.0 hr. How many presses and operators must be devoted to this production during the three days, if there are 7.5 hours of available time per day? Solution: The workload consists of 7000 stampings at 27 sec per piece WL = 7000(27/60 min) + 2(60)n = 3150 + 120 n (min) = 52.5 + 2 n (hr) Time available per press during the three days AT = 3(7.5) = 22.5 hr 52.5  2n n= 22.5 22.5n = 52.5 + 2n 20.5n = 52.5 n = 52.5/20.5 = 2.56 rounded up to 3 presses and operators

14.3

A stamping plant must be designed to supply an automotive engine plant with sheet metal stampings. The plant will operate one 8-hour shift for 250 days per year and must produce 15,000,000 good quality stampings annually. Batch size = 10,000 good stampings produced per batch. Scrap rate = 5%. On average it takes 3.0 sec to produce each stamping when the presses are running. Before each batch, the press must be set up, and it takes 4 hr to accomplish each setup. Presses are 90% reliable during production and 100% reliable during setup. How many stamping presses will be required to accomplish the specified production? Solution: Production: WL =

15,000,000( 3 / 3600) = 13,157.9 hr/yr 1  0.05

AT = 250(8)(0.90) = 1800 hr/yr per press 15,000 ,000 Setup: number batches/yr = = 1500 batches = 1500 setups 10,000 WL = 1500(4) = 6000 hr/yr AT = 250(8) = 2000 hr/yr per press. n= 14.4

13,157.9 6000  = 7.31 + 3.0 = 10.31  11 presses 1800 2000

A new forging plant must supply parts to the automotive industry. Because forging is a hot operation, the plant will operate 24 hr per day, five days per week, 50 weeks per year. Total output from the plant must be 10,000,000 forgings per year in batches of 1250 parts per batch. Anticipated scrap rate = 3%. Each forging cell will consist of a furnace to heat the parts, a forging press, and a trim press. Parts are placed in the furnace an 94

hour prior to forging; they are then removed and forged and trimmed one at a time. On average the forging and trimming cycle takes 0.6 min to complete one part. Each time a new batch is started, the forging cell must be changed over, which consists of changing the forging and trimming dies for the next part style. It takes 2.0 hr on average to complete a changeover between batches. Each cell is considered to be 96% reliable during operation and 100% reliable during changeover. Determine the number of forging cells that will be required in the new plant. Solution: Production: WL =

10,000,000( 0.6 / 60) = 103,092.8 hr/yr 1  0.03

AT = 50(5)(3)(8)(0.96) = 5760 hr/yr. Setup: number batches/yr =

10,000,000 = 8000 batches/yr = 8000 setups/yr 1250

WL = 8000(2) = 16,000 hr/yr per cell AT = 50(5)(3)(8) = 6000 hr/yr. n= 14.5

103,092.8 16,000  = 17.90 + 2.67 = 20.57  21 forging cells 5760 6000

A plastic injection molding plant will be built to produce 6 million molded parts per year. The plant will run three 8-hour shifts per day, five days per week, 50 weeks per year. For planning purposes, the average batch size = 6000 moldings, average changeover time between batches = 6 hrs, and average molding cycle time per part = 30 sec. Assume scrap rate = 2 percent, and average uptime proportion (reliability) per molding machine = 97 %, which applies to both run time and changeover time. How many molding machines are required in the new plant? Solution: Production: WL =

6,000,000( 30 / 3600) = 51,020.4 hr/yr 1  0.02

Setup: number batches/yr =

6, 000, 000 = 1000 batches = 1000 setups 6000

WL = 1000(6) = 6000 hr/yr AT = 3(5)(50)(8)(0.97) = 5820 hr/yr per machine 51, 020.4  6000 n= = 8.766 + 1.031 = 9.797  10 molding machines 5820 14.6

A plastic extrusion plant will be built to produce 30 million meters of plastic extrusions per year. The plant will run three 8-hour shifts per day, 360 days per year. For planning purposes, the average run length = 3000 meters of extruded plastic. The average changeover time between runs = 2.5 hr, and average extrusion speed = 15 m/min. Assume scrap rate = 1%, and average uptime proportion per extrusion machine = 95% during run time. Uptime proportion during changeover is assumed to be 100%. If each extrusion machine requires 500 sq. ft of floor space, and there is an allowance of 40% for aisles and office space, what is the total area of the extrusion plant?

30,000,000m / yr = 33,670.0 hr/yr 15( 60) m / hr (1  0.01) AT = 360(3)(8)(0.95) = 8208 hr/yr. 30,000,000m / yr Changeover: number runs/yr = = 10,000 runs/yr = 10,000 changeovers/yr 3000m / run WL = 10,000(2.5) = 25,000 hr/yr AT = 360(3)(8) = 8640 hr/yr per machine 33,670 25,000  n= = 4.102 + 2.894 = 6.995  7 machines. 8208 8640 A = 7(500)(1 + 40%) = 4900 ft2 Solution: Production: WL =

14.7

Future production requirements in a machine shop call for several automatic bar machines to be acquired to produce three new parts (A, B, and C) that have been added to the shop’s product line. Annual quantities and cycle times for the three parts are given in the table below. The machine shop operates one 8-hour shift for 250 days per year. The machines are expected to be 95% reliable, and the scrap rate is 3%. How many automatic bar machines will be required to meet the specified annual demand for the three new parts?

95

Part A B C

Annual demand 25,000 40,000 50,000

Machining cycle time 5.0 min 7.0 min 10.0 min

Solution: AT = 250(8)(0.95) = 1900 hr/yr per machine WL =

25000(5 / 60) 40000( 7 / 60) 50000(10 / 60)   = 2147.7 + 4811.0 + 8591.1 1  0.03 1  0.03 1  0.03

WL = 15,549.8 hr/yr n = 15,549.8/1900 = 8.184  9 machines 14.8

A certain type of machine will be used to produce three products: A, B, and C. Sales forecasts for these products are: 52,000, 65,000, and 70,000 units per year, respectively. Production rates for the three products are, respectively, 12, 15, and 10 pc/hr; and scrap rates are, respectively, 5%, 7%, and 9%. The plant will operate 50 weeks per year, 10 shifts per week, and 8 hr per shift. It is anticipated that production machines of this type will be down for repairs on average 10 percent of the time. How many machines will be required to meet demand? Solution: AT = 50(10)(8)(1 - 0.10) = 3600 hr/yr per machine

52,000 65,000 70,000   = 4561.4 + 4659.5 + 7692.3 = 16,913.2 hr/yr 12(1  0.05) 15(1  0.07) 10(1  0.09) n = 16,913.2/3600 = 4 67  5 machines WL =

14.9

An emergency situation has occurred in the milling department, because the ship carrying a certain quantity of a required part from an overseas supplier sank on Friday evening. A certain number of machines in the department must therefore be dedicated to the production of this part during the next week. A total of 1000 of these parts must be produced, and the production cycle time per part = 16.0 min. Each milling machine used for this emergency production job must first be set up, which takes 5.0 hr. A scrap rate of 2% can be expected. (a) If the production week consists of 10 shifts at 8.0 hr per shift, how many machines will be required? (b) It so happens that only two milling machines can be spared for this emergency job, due to other priority jobs in the department. To cope with the emergency situation, plant management has authorized a three-shift operation for six days next week. Can the 1000 replacement parts be completed within these constraints? Solution: (a) WL =

1000(16 / 60) = 272.1 hr/wk (1  0.02)

AT = 10(8) - 5 = 80 - 5 = 75 hr/wk per machine n = 272.1/75 = 3.63  4 milling machines (b) AT = 6(3)(8) - 5 = 139 hr/wk per machine n = 272.1/139 = 1.96  2 milling machines. Yes, the job can be completed. 14.10 A machine shop has dedicated one CNC machining center to the production of two parts (A and B) used in the final assembly of the company’s main product. The machining center is equipped with an automatic pallet changer and a parts carousel that holds ten parts. One thousand units of the product are produced per year, and one of each part is used in the product. Part A has a machining cycle time of 50 min. Part B has a machining cycle time of 80 min. These cycle times include the operation of the automatic pallet changer. No other changeover time is lost between parts. The anticipated scrap rate is zero. The machining center is 95% reliable. The machine shop operates 250 days per year. How many hours must the CNC machining center be operated each day on average to supply parts for the product? Solution: Annual WL = 1000(50 + 80)/60 = 2166.67 hr/yr At 250 days per year, daily workload WL = 2166.67/250 = 8.67 hr/day.

Machine Clusters 14.11 The CNC grinding section has a large number of machines devoted to grinding shafts for the automotive industry. The grinding machine cycle takes 3.6 min. At the end of this cycle an operator must be present to unload and load parts, which takes 40 sec. (a) Determine how many grinding machines the worker can service if it takes 20 sec to walk between the machines and no machine idle time is allowed. (b) How many seconds during the work cycle is the worker idle? (c) What is the hourly production rate of this machine cluster?

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Solution: (a) n =

3.6( 60)  40 = 256/60 = 4.27 40  20

Use n1 = 4 grinding machines

(b) Worker idle time IT = 256 - 4(60) = 256 - 240 = 16 sec

 60   = 56.25 pc/hr  4.267 

(c) Tc = 256 sec = 4.267 min Rc = 4 

14.12 A worker is currently responsible for tending two machines in a machine cluster. The service time per machine is 0.35 min and the time to walk between machines is 0.15 min. The machine automatic cycle time is 1.90 min. If the worker's hourly rate = $12/hr and the hourly rate for each machine = $18/hr, determine (a) the current hourly rate for the cluster, and (b) the current cost per unit of product, given that two units are produced by each machine during each machine cycle. (c) What is the % idle time of the worker? (d) What is the optimum number of machines that should be used in the machine cluster, if minimum cost per unit of product is the decision criterion? Solution: (a) Co = $12 + 2($18) = $48.00/hr (b) Tc = Tm + Ts = 1.90 + 0.35 = 2.25 min/cycle

 60   = 106.67 pc/hr  2.25 

Rc = 2(2) 

Cpc =

$48 / hr = $0.45/pc 106.67pc / hr

(c) Worker engagement time/cycle = 2(Ts + Tr) = 2(0.35 + 0.15) = 1.0 min Idle time IT = (d) n =

2.25  1.0 = 0.555 = 55.5% 2.25

1.90  0.35 = 2.25/0.5 = 4.5 machines 0.35  0.15

n1 = 4 machines: Cpc(4) = 0.5(12/4 + 18(2.25/60) = $0.394/pc n2 = 5 machines: Cpc(5) = 0.5(12 + 18x5)(0.50/60) = $0.425/pc Use n1 = 4 machines 14.13 In a machine cluster, the appropriate number of production machines to assign to the worker is to be determined. Let n = the number of machines. Each production machine is identical and has an automatic processing time Tm = 4.0 min. The servicing time Ts = 12 sec for each machine. The full cycle time for each machine in the cell is Tc = Ts + Tm. The repositioning time for the worker is given by Tr = 5 + 3n, where Tr is in sec. Tr increases with n because the distance between machines increases with more machines. (a) Determine the maximum number of machines in the cell if no machine idle time is allowed. For your answer, compute (b) the cycle time and (c) the worker idle time expressed as a percent of the cycle time? Solution: (a) n =

4.0(60)  12 252  12  5  3n 17  3n

n(17+3n) = 252 3n2 + 17 n -252 =0 Use quadratic equation to find n. 172  4( 3)( 252) = -2.83  9.59 = 6.76 or -12.43 2( 3) Use n = 6.76 (ignore n = -12.43)  n1 = 6 machines

n = 17 

(b) Tc = 252 sec (c) Worker portion of cycle = 6(17 + 3 x 6) = 210 sec Worker idle time IT =

252  210 = 0.167 = 16.7% 252

14.14 An industrial robot will service n production machines in a machine cluster. Each production machine is identical and has an automatic processing time Tm = 130 sec. The robot servicing and repositioning time for each machine is given by the equation (Ts + Tr) = 15 + 4n, where Ts is the servicing time (sec), Tr is the repositioning time (sec), and n = number of machines that the robot services. (Ts + Tr) increases with n because more time is needed to reposition the robot arm as n increases. The full cycle time for each machine in the cell 97

is Tc = Ts + Tm. (a) Determine the maximum number of machines in the cell such that machines are not kept waiting. For your answer, (b) what is the machine cycle time, and (c) what is the robot idle time expressed as a percent of the cycle time Tc? Solution: (a) Given Tm = 130 sec, Tc = Tm + Ts = 130 + Ts 130  Ts n= 15  4n We can deduce that if there were only one machine (n = 1), then repositioning time would be zero (Tr = 0). Thus, for n = 1, (Ts + Tr) = (Ts + 0) = Ts = 15 + 4(1) = 19 sec Substituting this value into the equation for n, we have 130  19 149 n= = 15  4n 15  4n 15n + 4n2 = 149 4n2 + 15n - 149 = 0 Use quadratic equation to find n. n=

-15  152  4  4   149  2  4

= -1.875  6.38 = 4.51 or -8.26

Use n = 4.51 (ignore n = -8.26)  For no machine waiting, n1 = 4 machines (b) Tc = 130 + 19 = 149 sec (c) Robot work time = n(Ts + Tr) = 4(15 + 4 x 4) = 4(31) = 124 sec Robot idle time % = (149 – 124)/149 = 0.168 = 16.8% 14.15 A factory production department consists of a large number of work cells. Each cell consists of one human worker performing electronics assembly tasks. The cells are organized into sections within the department, and one foreman supervises each section. It is desired to know many work cells should be assigned to each foreman. The foreman’s job consists of two tasks: (1) provide each cell with a sufficient supply of parts that it can work for 4.0 hr before it needs to be resupplied and (2) preparing production reports for each workcell. Task (1) takes 18.0 min on average per workcell and must be done twice per day. The foreman must schedule the resupply of parts to every cell so that no idle time occurs in any cell. Task (2) takes 9.0 min per workcell and must be done once per day. The plant operates one shift which is 8.0 working hr, and neither the workers nor the foreman are allowed to work more than 8.0 hr per day. Each day, the cells continue production from where they stopped the day before. (a) What is the maximum number of work cells that should be assigned to a foreman, with the proviso that the work cells are never idle? (b) With the number of work cells from part (a), how many idle min does the foreman have each day? Solution: Since the foreman must resupply each cell twice in an 8.0 hour day and also prepare a report on each cell once each day, the foreman’s day can be modeled as follows: 2 n(18) + 1 n (9) = 8(60) 36 n + 9 n = 480 45 n = 480 n = 10.67. Use n1 = 10 cells. (b) Foreman’s idle time each day = 480 - 45 n = 480 - 450 = 30 min

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