= = 4.102 + 2.894 = 6.995 7 machines.
6. 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 8hour 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?
WL = = 2147.7 + 4811.0 + 8591.1
WL = 15,549.8 hr/yr
n = 15,549.8/1900 = 8.184 9 machines
7. 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
WL = = 4561.4 + 4659.5 + 7692.3 = 16,913.2 hr/yr
n = 16,913.2/3600 = 4 67 5 machines
8. 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 = = 272.1 hr/wk
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.
9. 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.
10. 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?
Solution: (a)
n = = 256/60 = 4.27 Use
n1 =
4 grinding machines
(b) Worker idle time IT = 256 - 4(60) = 256 - 240 = 16 sec
(c)
Tc = 256 sec = 4.267 min
Rc = 4 =
56.25 pc/hr
11. 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
Rc = 2(2) = 106.67 pc/hr
Cpc = =
$0.45/pc
(c) Worker engagement time/cycle = 2(Ts + Tr) = 2(0.35 + 0.15) = 1.0 min
Idle time
IT = = 0.555 =
55.5%
(d)
n = = 2.25/0.5 =
4.5 machines
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
12. 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 =
n(17+3n) = 252
3n2 + 17 n -252 =0
Use quadratic equation to find n.
n = = -2.83 9.59 = 6.76 or -12.43
Use n = 6.76 (ignore n = -12.43) n1 = 6 machines
(b) Tc = 252 sec
(c) Worker portion of cycle = 6(17 + 3 x 6) = 210 sec
Worker idle time
IT = = 0.167 =
16.7%
13. 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 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
n =
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
n = =
15n + 4n2 = 149
4n2 + 15n - 149 = 0
Use quadratic equation to find n.
n = = -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%
CHAPTER 15:
14. A product whose work content time = 47.5 min is to be assembled on a manual production line. The required production rate is 30 units per hour. From previous experience, it is estimated that the manning level will be 1.25, proportion uptime = 0.95, and repositioning time = 6 sec. Determine (a) cycle time, and (b) ideal minimum number of workers required on the line. (c) If the ideal number in part (b) could be achieved, how many workstations would be needed?
Solution: (a)
Tc = = 1.9 min
(b)
w = Minimum Integer =
25 workers
(c) n = 25/1.25 = 20 workstations
15. A manual assembly line has 17 workstations with one operator per station. Work content time to assemble the product = 28.0 min. Production rate of the line = 30 units per hour. The proportion uptime = 0.94, and repositioning time = 6 sec. Determine the balance delay.
Solution:
Tc = = 1.88 min,
Ts = 1.88 - 0.1 = 1.78 min
w = n = 17 workers and 17 stations
Eb = = 0.9253,
d = 1 - 0.9253 = 0.0747 =
7.47%
16. A manual assembly line must be designed for a product with annual demand = 100,000 units. The line will operate 50 wks/year, 5 shifts/wk, and 7.5 hr/shift. Work units will be attached to a continuously moving conveyor. Work content time = 42.0 min. Assume line efficiency = 0.97, balancing efficiency = 0.92, and repositioning time = 6 sec. Determine (a) hourly production rate to meet demand, and (b) number of workers required.
Solution: (a)
Rp = =
53.33 pc/hr
(b)
Tc = = 1.08 min,
w* = Minimum Integer = 38.89
39 workers
(c) Ts = 1.08 - 0.1 = 0.98 min
w = Minimum Integer = 46.58
47 workers
17. A single model assembly line is being planned to produce a consumer appliance at the rate of 200,000 units per year. The line will be operated 8 hours per shift, two shifts per day, five days per week, 50 weeks per year. Work content time = 35.0 min. For planning purposes, it is anticipated that the proportion uptime on the line will be 95%. Determine (a) average hourly production rate, (b) cycle time, and (c) theoretical minimum number of workers required on the line. (d) If the balance efficiency is 0.93 and the repositioning time = 6 sec, how many workers will actually be required?
Solution: (a)
Rp = =
50 pc/hr
(b)
Tc = =
1.14 min
(c)
w = Minimum Integer = 30.7
31 workers.
(d) Ts = 1.14 - 0.10 = 1.04 min
Er = 1.04/1.14 = 0.9123
w = Minimum Integer = 36.2
37 workers.
18. The required production rate = 50 units per hour for a certain product whose assembly work content time = 1.2 hours. It is to be produced on a production line that includes four workstations that are automated. Because the automated stations are not completely reliable, the line will have an expected uptime efficiency = 90%. The remaining manual stations will each have one worker. It is anticipated that 8% of the cycle time will be lost due to repositioning at the bottleneck station. If the balance delay is expected to be 0.07, determine (a) the cycle time, (b) number of workers, (c) number of workstations needed for the line, (d) average manning level on the line, including the automated stations, and (e) labor efficiency on the line.
Solution: (a)
Tc = =
1.08 min
(b) Tr = 0.08Tc, therefore, Ts = 0.92Tc, therefore, Er = 0.92
Eb = 1 - d = 1 - 0.07 = 0.93
w = Minimum Integer = 77.9
78 workers.
(c) n = 78 + 4 = 82 workstations.
(d) M = 78/82 = 0.951
(e) Labor efficiency = EEbEr = 0.90(0.93)(0.92) = 0.77 = 77%
19. A final assembly plant for a certain automobile model is to have a capacity of 225,000 units annually. The plant will operate 50 weeks/yr, two shifts/day, 5 days/week, and 7.5 hours/shift. It will be divided into three departments: (1) body shop, (2) paint shop, (3) general assembly department. The body shop welds the car bodies using robots, and the paint shop coats the bodies. Both of these departments are highly automated. General assembly has no automation. There are 15.0 hours of work content time on each car in this third department, where cars are moved by a continuous conveyor. Determine (a) hourly production rate of the plant, (b) number of workers and workstations required in trimchassisfinal if no automated stations are used, the average manning level is 2.5, balancing efficiency = 90%, proportion uptime = 95%, and a repositioning time of 0.15 min is allowed for each worker.
Solution: (a)
Rp = = 60 cars/hr.
(b)
Tc = = 0.95 min,
Ts = 0.95 - 0.15 = 0.80 min
Twc = 15(60) = 900 min of direct labor
w = Minimum Integer =
1250 workers, and
n = =
500 stations
20. The work content for a product assembled on a manual production line is 48 min. The work is transported using a continuous overhead conveyor that operates at a speed of 5 ft/min. There are 24 workstations on the line, onethird of which have two workers; the remaining stations each have one worker. Repositioning time per worker is 9 sec, and uptime efficiency of the line is 95%. (a) What is the maximum possible hourly production rate if the line is assumed to be perfectly balanced? (b) If the actual production rate is only 92% of the maximum possible rate determined in part (a), what is the balance delay on the line?
Solution: (a) w = 2(24/3) + 1(24 x 2/3) = 2(8) + 1(16) = 32 workers
w = , therefore,
Ts =
If the line is perfectly balanced, the
Eb = 1.0.
Ts = = 1.5 min,
Tc = 1.5 + 0.15 = 1.65 min
Rp = =
34.55 pc/hr (at
E = 0.95 and
Eb = 1.0)
(b) Actual Rp = 0.92(34.55) = 31.78 pc/hr
Tc = = 1.7935,
Ts = 1.7935 - 0.15 = 1.6435 min
d = = 0.087 =
8.7%
21. Work content time for a product assembled on a manual production line is 45.0 min. Production rate of the line must be 40 units/hr. Work units are attached to a moving conveyor whose speed = 8 ft/min. Repositioning time per worker is 8 sec, line efficiency is 93%, and manning level = 1.25. Owing to imperfect line balancing, it is expected that the number of workers needed on the line will be about 10% more workers than the number required for perfect balance. If the workstations are arranged in a line, and the length of each station is 12 ft, (a) how long is the entire production line, and (b) what is the elapsed time a work unit spends on the line?
Solution: (a)
Tc = = 1.395 min
Ts = 1.395 - 0.133 = 1.262 min
For perfect balance,
Eb = 1.0.
w = Minimum Integer = 35.7
With 10% more workers,
w = 1.1(35.7) = 39.2 40 workers.
n = = 32 stations
L = (32 stations)(12 ft/station) = 384 ft.
(b) Given
L = 384 ft and
vc = 8 ft/min,
ET = =
48.0 min
22. Show that the two statements of the objective function in single model line balancing in Eq. (15.20) are equivalent.
Solution:
nTs in the first expression = in the second expression, since
Ts is a constant
(Ts = Max{Tsi} in Eq. (15.11))
And
Twc in the first expression = in the second expression, according to Eq. (15.15).
Therefore
nTs -
Twc = Q.E.D.
23. The table below defines the precedence relationships and element times for a new model toy. (a) Construct the precedence diagram for this job. (b) If the ideal cycle time = 1.1 min. repositioning time = 0.1 min, and uptime proportion is assumed to be 1.0, what is the theoretical minimum number of workstations required to minimize the balance delay under the assumption that there will be one worker per station? (c) Use the largest candidate rule to assign work elements to stations. (d) Compute the balance delay for your solution.
|
|
Work element
|
Te (min)
|
Immediate predecessors
|
|
|
1
|
0.5
|
-
|
|
|
2
|
0.3
|
1
|
|
|
3
|
0.8
|
1
|
|
|
4
|
0.2
|
2
|
|
|
5
|
0.1
|
2
|
|
|
6
|
0.6
|
3
|
|
|
7
|
0.4
|
4,5
|
|
|
8
|
0.5
|
3,5
|
|
|
9
|
0.3
|
7,8
|
|
|
10
|
0.6
|
6,9
|
Solution: (a) Precedence diagram:
![]()
(b) Ts = Tc - Tr = 1.1 - 0.1 = 1.0 min
With
M = 1.0,
n =
w = Minimum Integer = 4.3, Use
n = 5 stations
(c) Line balancing solution using the largest candidate rule.
|
|
List of elements by Te value
|
|
Allocation of elements to stations
|
|
|
Element
|
Te (min)
|
Predecessors
|
|
Station
|
Element
|
Te
|
Te
|
|
|
3
|
0.8
|
1
|
|
1
|
1
|
0.5 min
|
|
|
|
6
|
0.6
|
3
|
|
|
2
|
0.3 min
|
|
|
|
10
|
0.6
|
6, 9
|
|
|
4
|
0.2 min
|
1.0 min
|
|
|
1
|
0.5
|
-
|
|
2
|
3
|
0.8 min
|
|
|
|
8
|
0.5
|
3, 5
|
|
|
5
|
0.1 min
|
0.9 min
|
|
|
7
|
0.4
|
4, 5
|
|
3
|
6
|
0.6 min
|
|
|
|
2
|
0.3
|
1
|
|
|
7
|
0.4 min
|
1.0 min
|
|
|
9
|
0.3
|
7, 8
|
|
4
|
8
|
0.5 min
|
|
|
|
4
|
0.2
|
2
|
|
|
9
|
0.3 min
|
0.8 min
|
|
|
5
|
0.1
|
2
|
|
5
|
10
|
0.6 min
|
0.6 min
|
|
|
|
|
|
|
|
|
|
4.3 min total
|
(d) Balance delay
d = = 0.14 = 14%
24. Solve the previous problem using the Kilbridge and Wester method in part (c).
Solution: (a) Precedence diagram same as in Problem 15.11.
(b) Same as in Problem 15.11: n = 5 stations.
(c) Line balancing solution using the Kilbridge & Wester method:
|
|
List of elements by precedence columns
|
|
Allocation of elements to stations
|
|
|
Element
|
Te (min)
|
Column
|
|
Station
|
Element
|
Te
|
Te
|
|
|
1
|
0.5
|
I
|
|
1
|
1
|
0.5 min
|
|
|
|
2
|
0.3
|
II
|
|
|
2
|
0.3 min
|
|
|
|
3
|
0.8
|
II
|
|
|
4
|
0.2 min
|
1.0 min
|
|
|
4
|
0.2
|
III
|
|
2
|
3
|
0.8 min
|
|
|
|
5
|
0.1
|
III
|
|
|
5
|
0.1 min
|
0.9 min
|
|
|
6
|
0.6
|
III
|
|
3
|
6
|
0.6 min
|
|
|
|
7
|
0.4
|
IV
|
|
|
7
|
0.4 min
|
1.0 min
|
|
|
8
|
0.5
|
IV
|
|
4
|
8
|
0.5 min
|
|
|
|
9
|
0.3
|
V
|
|
|
9
|
0.3 min
|
0.8 min
|
|
|
10
|
0.6
|
VI
|
|
5
|
10
|
0.6 min
|
0.6 min
|
|
|
|
|
|
|
|
|
|
4.3 min total
|
(d) Same as in Problem 15.11: d = 0.14 = 14%
25. Solve the previous problem using the ranked positional weights method in part (c).
Solution: (a) Precedence diagram same as in Problem 15.11.
(b) Same as in Problem 15.11: n = 5 stations.
(c) Line balancing solution using the Kilbridge & Wester method:
|
|
Elements by ranked positional weights
|
|
Allocation of elements to stations
|
|
|
Element
|
Te (min)
|
RPW
|
|
Station
|
Element
|
Te
|
Te
|
|
|
1
|
0.5
|
4.3
|
|
1
|
1
|
0.5 min
|
|
|
|
3
|
0.8
|
2.8
|
|
|
2
|
0.3 min
|
|
|
|
2
|
0.3
|
2.4
|
|
|
5
|
0.1 min
|
0.9 min
|
|
|
5
|
0.1
|
1.9
|
|
2
|
3
|
0.8 min
|
|
|
|
4
|
0.2
|
1.5
|
|
|
4
|
0.2 min
|
1.0 min
|
|
|
8
|
0.5
|
1.4
|
|
3
|
8
|
0.5 min
|
|
|
|
7
|
0.4
|
1.3
|
|
|
7
|
0.4 min
|
0.9 min
|
|
|
6
|
0.6
|
1.2
|
|
4
|
6
|
0.6 min
|
|
|
|
9
|
0.3
|
0.9
|
|
|
9
|
0.3 min
|
0.9 min
|
|
|
10
|
0.6
|
0.6
|
|
5
|
10
|
0.6 min
|
0.6 min
|
|
|
|
|
|
|
|
|
|
4.3 min total
|
(d) Same as in Problem 15.11: d = 0.14 = 14%
26. A manual assembly line is to be designed to make a small consumer product. The work elements, their times, and precedence constraints are given in the table below. The workers will operate the line for 400 min per day and must produce 300 products per day. A mechanized belt, moving at a speed of 1.25 m/min, will transport the products between stations. Because of the variability in the time required to perform the assembly operations, it has been determined that the tolerance time should be 1.5 times the cycle time of the line. (a) Determine the ideal minimum number of workers on the line. (b) Use the Kilbridge and Wester method to balance the line. (c) Compute the balance delay for your solution in part (b).
|
|
Element
|
Time Te
|
Preceded by:
|
Element
|
Time Te
|
Preceded by:
|
|
|
1
|
0.4 min
|
-
|
6
|
0.2 min
|
3
|
|
|
2
|
0.7 min
|
1
|
7
|
0.3 min
|
4
|
|
|
3
|
0.5 min
|
1
|
8
|
0.9 min
|
4, 9
|
|
|
4
|
0.8 min
|
2
|
9
|
0.3 min
|
5, 6
|
|
|
5
|
1.0 min
|
2, 3
|
10
|
0.5 min
|
7, 8
|
Solution: (a)
Tc = = 1.333 min/asby. No
Tr value given. Assume
Tr = 0.
Twc = = 0.4 + 0.7 + . . . + 0.5 = 5.6 min
w = Minimum Integer = 4.2
w =
5 workers
(b) Line balancing solution using Kilbridge & Wester method. Assume M = 1.0.
![]()
|
|
List of elements by precedence columns
|
|
Allocation of elements to stations
|
|
|
Element
|
Te (min)
|
Column
|
|
Station
|
Element
|
Te
|
Te
|
|
|
1
|
0.4
|
I
|
|
1
|
1
|
0.4 min
|
|
|
|
2
|
0.7
|
II
|
|
|
2
|
0.7 min
|
1.1 min
|
|
|
3
|
0.5
|
II
|
|
2
|
3
|
0.5 min
|
|
|
|
4
|
0.8
|
III
|
|
|
4
|
0.8 min
|
1.3 min
|
|
|
5
|
1.0
|
III
|
|
3
|
5
|
1.0 min
|
|
|
|
6
|
0.2
|
III
|
|
|
6
|
0.2 min
|
1.2 min
|
|
|
7
|
0.3
|
IV
|
|
4
|
7
|
0.3 min
|
|
|
|
9
|
0.3
|
IV
|
|
|
9
|
0.3 min
|
0.6 min
|
|
|
8
|
0.9
|
V
|
|
5
|
8
|
0.9 min
|
0.9 min
|
|
|
10
|
0.5
|
VI
|
|
6
|
10
|
0.5 min
|
0.5 min
|
|
|
|
|
|
|
|
|
|
5.6 min total
|
Note: In the above solution, the Kilbridge & Wester method was followed very precisely, so that the solution consisted of a total of six stations. A better solution, but one that violates the K&W algorithm, would be to combine elements 9 and 8 at station 4 and elements 7 and 10 at station 5, for a total of only 5 stations.
(c) Use cycle time of 1.3 min (station 2) rather than 1.333 min. Using the six stations in the K&W solution,
balance delay
d = = 0.282 =
28.2%
27. Solve the previous problem using the ranked positional weights method in part (b).
Solution: (a) Same solution as in Problem 15.14: (Tc = 1.333 min, Tr = 0, Twc = 5.6 min) w = 5 workers.
(b) Line balancing solution using ranked positional weights method. Assume M = 1.0.
|
|
Elements by ranked positional weights
|
|
Allocation of elements to stations
|
|
|
Element
|
Te (min)
|
RPW
|
|
Station
|
Element
|
Te
|
Te
|
|
|
1
|
0.4
|
5.6
|
|
1
|
1
|
0.4 min
|
|
|
|
2
|
0.7
|
4.5
|
|
|
2
|
0.7 min
|
1.1 min
|
|
|
3
|
0.5
|
3.4
|
|
2
|
3
|
0.5 min
|
|
|
|
5
|
1.0
|
2.7
|
|
|
4
|
0.8 min
|
1.3 min
|
|
|
4
|
0.8
|
2.5
|
|
3
|
5
|
1.0 min
|
|
|
|
6
|
0.2
|
1.9
|
|
|
6
|
0.2 min
|
1.2 min
|
|
|
9
|
0.3
|
1.7
|
|
4
|
9
|
0.3 min
|
|
|
|
8
|
0.9
|
1.4
|
|
|
8
|
0.9 min
|
1.2 min
|
|
|
7
|
0.3
|
0.8
|
|
5
|
7
|
0.3 min
|
|
|
|
10
|
0.5
|
0.5
|
|
|
10
|
0.5 min
|
0.8 min
|
|
|
|
|
|
|
|
|
|
5.6 min total
|
(c) Use cycle time of 1.3 min (station 2) rather than 1.333 min. Using the five stations in the RPW solution,
balance delay
d = = 0.138 =
13.8%
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REVIEWQUESTIONS-2.docx
0
