HW-03-24-10s-Chs12-16

HW-03-24-10s-Chs12-16 - IE 131 Solutions for Problems due...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
IE 131 Solutions for Problems due Mar 24 (Chs 12 through 16) 12.2 A time study analyst observed a repetitive work cycle performed by a worker and then studied the task using a predetermined motion time system. During the observation, the analyst made a record of the work elements. He also made a note of the worker’s performance and rated it as 85%. Back in the office, the analyst listed the basic motion elements for the task and retrieved from his computer the corresponding normal time values. The sum of the normal times for the basic motion elements was 1.38 min. The company uses a PFD allowance factor of 15%. Determine the standard time for the task. Solution : Normal time T n = 1.38 min The performance rating of 85% is a “red herring” in this problem because the basic motion element times are already rated at 100%. Standard time T std = 1.38(1 + 0.15) = 1.587 min 12.6 Determine the personal time, fatigue, and delay (PFD) allowance to be used for computing time standards in the following situation. Second shift workers punch in at 3:30 p.m. and punch out at 12:00 midnight. They are provided one-half hour for supper at 6:00 p.m., which is not counted as part of the 8-hour shift. For purposes of determining the allowance, 30 minutes of break time (personal time and fatigue) are allowed each worker. In addition, the plant allows 35 min for lost time due to unavoidable delays. What should the PFD allowance factor be? Solution : Allowance time for 30 min of break time plus 35 min for lost time = 65 min Allowance factor A pfd = 480/(480-65) – 1 = 1.157 – 1 = 0.157 = 15.7% 13.4 The snapback timing method was used to obtain the average times and performance ratings for work elements in a manual repetitive task. See table below. All elements are worker- controlled. All elements were performance rated at 80%. Element e is an irregular element performed every five cycles. A 15% allowance for personal time, fatigue, and delays is applied to the cycle. Determine (a) the normal time and (b) the standard time for this cycle. If the worker’s performance during actual production is 120% on all manual elements for seven actual hours worked on an eight-hour shift, (c) how many units will be produced and (d) what is the worker’s efficiency? Work element a b c d e Observed time (min) 0.32 0.85 0.48 0.55 1.50 Solution : (a) Normal time T n = (0.32 + 0.85 + 0.48 + 0.55 + 1.50/5)(0.80) T n = 2.50(0.80) = 2.00 min (b) T std = 2.00(1.15) = 2.30 min (c) Given P w = 120% for 7.0 hr on an 8-hour shift T c = T n / P w = 2.00/1.20 = 1.667 min Q = 7(60)/1.667 = 252 pc (d) H std = 252(2.30)/60 = 9.66 hr and E w = 9.66/8.0 = 1.208 = 120.8% Comment : The values of worker performance P w and worker efficiency
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

HW-03-24-10s-Chs12-16 - IE 131 Solutions for Problems due...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online