The normal time of the work cycle in a worker-machine system is 5.29 min. The op
ID: 1530129 • Letter: T
Question
The normal time of the work cycle in a worker-machine system is 5.29 min. The operator-controlled portion of the cycle is 0.94 min. One work unit is produced each cycle. The machine cycle time is constant. Using a PFD allowance factor of 15% and a machine allowance factor of 25%, determine the standard time for the work cycle. If a worker assigned to this task completes 95 units during an 8-hour shift, what is the worker's efficiency? If it is known that a total of 42 min was lost during the 8-hour clock time due to personal needs and delays, what was the worker's performance on the portion of the cycle he controlled? If annually you produce 100,000 units, what are the Productive FTE requirements? If non productive time is estimated at 15%, what are the Paid FTE requirements? Assume workers are replaced when not at work. The work cycle in a worker-machine system consists of external manual work elements with a total normal time of 0.42 min, a machine cycle with machine time of 1.12 min, and internal manual elements with a total normal time of 1.04 min. Determine the standard time for the cycle, using a PFD allowance factor of 15%, and a machine allowance factor of 30%. How many work units are produced daily (8-hour shift) at standard performance? How long would it take to fulfill a customer order of 2500 units assuming you only have one machine and you work 2 8-hour shifts? Solve the previous problem but assume that the machine allowance factor is 0%. A total of 1000 units of a certain product must be completed by the end of the current week. It is now late Monday afternoon, so only four days (8-hour shifts) are left. The standard time for producing each unit of the product (all manual operations) is 11.65 min. How many workers will be required to complete this production order if it is assumed that worker efficiency will be 115%?Explanation / Answer
1) Normal time of the work cycle in worker machine system = 5.29 min
Operator controlled protion of the cycle = 0.94 min
work produced in each cycle = 1 unit
machine cycle time = constant
a) PFD allowance factor = 15%
Machine allowance factor = 25%
Normal Time Tn = Average work time * rating factor
= 5.29 * 0.25 = 1.32 minutes
Standard time Tstd = Normal time / 1- allowance factor
= 1.32 / 1- 0.15 = 1.55 minutes
b) Hstd = (95 units x Tstd )/ 60
= 95 x 1.55 / 60 = 2.45 hr
Worker efficiency Ew = 2.45 / 8
= 0.30 = 30%
c) Total time lost = 42 mins
TL = (8 x 60) - 42 = 438
Worker performance = 438 x 0.94 = 411.72
Total worker performance = 411.72 / 60 = 6.86 hrs
d) Productive FTE = number of units / FTE of one person
FTE of one person = 8 hours/day x 5 days per week x 52 weeks/ year = 2080 hours/year
Productive FTE = 100,000 / 2080 = 48.07 FTE's
e) Non productive time = 15%
= 2080 x 15% = 1768 hrs
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.