The machine allowance is zero, and the PFD allowance is 13%. Determine (a) the normalized time for the cycle and (b) the standard time for the cycle. The actual time spent working during an 8-hour shift was 7.08 hours, and worker’s he produced 400 units of output during this time. Determine (c) the worker’s performance during the operator-controlled portions of the cycle and (d) the efficiency during this shift. worker’s Element a b c d e f Observed time 0.14 0.25 0.18 0.45 0.20 0.62 (min) Solution: (a) = (0.14 + 0.25 + 0.18 + 0.20 + 0.62/5)(0.90) + 0.45 Tn = = 0.805 + 0.45 = 1.255 min Tn (b) = 0.805(1 + 0.13) + 0.45(1 + 0) = 0.91 + 0.45 = 1.36 min Tstd (c) Given = 400 pc during 7.08 hr of an 8-hour shift Q Machine time during shift = 400(0.45) = 180 min Worker time during shift = 7.08(60) 180 = 244.8 min – Cycle time for worker portion of cycle = 244.8/400 = 0.612 min Tc = 0.805/0.612 = 1.315 = 131.5% Pw (d) Worker efficiency = 400(1.36/60)/8.0 = 1.133 = 113.3% Ew
The machine allowance is zero, and the PFD allowance is 13%. Determine (a) the normalized time for the cycle and (b) the standard time for the cycle. The worker’s actual time spent working during an 8-hour shift was 7.08 hours, and he produced 400 units of output during this time. Determine (c) the worker’s performance during the operator-controlled portions of the cycle and (d) the worker’s efficiency during this shift. Element a b c d e f Observed time (min) 0.14 0.25 0.18 0.45 0.20 0.62 Solution: (a) Tn = (0.14 + 0.25 + 0.18 + 0.20 + 0.62/5)(0.90) + 0.45 Tn = = 0.805 + 0.45 = 1.255 min (b) Tstd = 0.805(1 + 0.13) + 0.45(1 + 0) = 0.91 + 0.45 = 1.36 min (c) Given Q = 400 pc during 7.08 hr of an 8-hour shift Machine time during shift = 400(0.45) = 180 min Worker time during shift = 7.08(60) – 180 = 244.8 min Cycle time for worker portion of cycle Tc = 244.8/400 = 0.612 min Pw = 0.805/0.612 = 1.315 = 131.5% (d) Worker efficiency Ew = 400(1.36/60)/8.0 = 1.133 = 113.3%
final_ws_second_term_16-7-1434.docx