Question

An IE performed a time study on a hydraulic valve installation & test process. Each observed installation was performed by a different production worker, so the activity factor (performance rating) varied in each observation. A factor > 1.00 meant the worker was performing better than normal in their observed time (OT) (i.e, faster), while a factor < 1.00 meant the worker was performing less normal in their observed time (OT) (ie, slower). The allowance for personal delay and fatigue as was 13% for all data points. Using the data in the table, determine:

Observation	Observed Time (OT) (MINUTES)	Activity Factor/Performance Rating	Normal Time (NT)	Allowance for Delay/Fatigue and Enviromental factors	Standard Time
1	40	1.03		13%
2	55	0.90		13%
3	50	0.95		13%
4	58	1.00		13%
5	72	0.85		13%
6	55	0.90		13%
7	60	0.91		13%
8	55	0.97		13%
9	50	0.97		13%
10	57	0.92		13%
				Total STD TIME
				# of OBSERVATIONS
				AVG STD TIME/VALVE

Questions  

A. What is the average standard time per valve?

B. From the time study problem above: if a worker was able to install 7 valves during an 8 hour period, what would the efficiency be?

C. From the OEE problem above, suppose an industrial engineer determined the lighting (once adequate for accurately and quickly reading print) had become too low due to the installation of new equipment that cast shadows on the ground level. In addition, the drawing formats and orientation were not standardized across the classes of ships being produced due to outsourcing efforts. By improving the lighting, standardizing the engineering drawings and re-training all workers on how to read the new drawings, the total produced parts during a similar production period was increased from 800 to 1120, while still only producing 40 defective parts. What would the new OEE value be with these improvements?

Answer 1

Observation	OT (mins)	Rating	Normal Time	Allowance	Standard Time
1	40	1.03	41.2	13%	46.4
2	55	0.90	49.5	13%	56.65
3	50	0.95	47.5	13%	54
4	58	1.00	58	13%	65.54
5	72	0.85	61.2	13%	70.56
6	55	0.90	49.5	13%	56.65
7	60	0.91	54.6	13%	62.4
8	55	0.97	53.35	13%	60.5
9	50	0.97	48.5	13%	55
10	57	0.92	52.44	13%	59.85
				Total std time	587.55
				No of observations	10
				Avg std time/valve	58.755

using the formaula

Standard time = Normal time + allowance

Normal Time = Average Time X Rating Factor

A) Average Std Time per valve = 58.755 mins= 60/58.755 = 1.02 valves per hour

B) Efficiency = output/input = 7/8= 87.5%

C) Increase in production/ performance = 1120/800= 1.4

Good count= Total production- defective parts produced= 1120-40= 1080

Quality = 1080/1120= 0.964

OEE = Performance x Quality = 1.4 x 0.964 = 1.3496