Question 1 of 8: Continuous Time Study — Average, Normal and Standard Time

Question 1: Continuous Time Study — Average, Normal and Standard Time (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A five-element job timed by the continuous stop-watch method over ten cycles; the tabulated figures are cumulative watch readings in hundredths of a minute, not individual element times.

Find. (a) the average (select) time per element and per cycle, (b) the normal time, (c) the standard time, and (d) the number of cycles that must be timed for the study to be accurate to within the acceptable error at 95 percent confidence.

Approach. Recover the individual element times by subtracting each continuous reading from the one before it, average each element column, sum the element averages to obtain the average cycle time, apply the performance rating and then the allowance, and finally size the study from the observed element-to-element scatter.

1. Part (a) — convert the continuous readings into element times. In continuous timing the watch runs without being reset, so the reading recorded at the end of an element is the elapsed time since the study began. The time for element $j$ of cycle $i$ is therefore the difference between successive readings, $t_{ij}=R_{ij}-R_{i,j-1}$, where the predecessor of the first element of a cycle is the last reading of the previous cycle. For cycle 1: $$t_{11}=23-0=23,\quad t_{12}=36-23=13,\quad t_{13}=52-36=16,\quad t_{14}=62-52=10,\quad t_{15}=74-62=12$$ and for cycle 2 the first element is measured from the last reading of cycle 1, $t_{21}=94-74=20$ hundredths of a minute.
2. Tabulate the ten cycles of element times. Repeating the subtraction throughout gives the following observed times, all in hundredths of a minute. No reading is an obvious outlier — the largest cycle (82) exceeds the smallest (74) by only 11 percent — so every cycle is retained in the averages.

   Element times recovered by successive subtraction (hundredths of a minute)

   Cycle	1	2	3	4	5	Cycle total
   1	23	13	16	10	12	74
   2	20	14	21	10	11	76
   3	21	16	18	9	10	74
   4	19	15	17	10	13	74
   5	22	15	18	9	12	76
   6	19	13	19	12	11	74
   7	21	16	17	9	13	76
   8	23	17	18	11	12	81
   9	20	16	19	10	11	76
   10	23	17	20	9	13	82
   Mean	21.1	15.2	18.3	9.9	11.8	76.3
   Std. dev. $s_j$	1.595	1.476	1.494	0.994	1.033	2.908

3. Average the element columns and sum them. The select time for each element is the arithmetic mean of its ten observations, $\bar{t}_{j}=\frac{1}{10}\sum_{i=1}^{10}t_{ij}$, and the average cycle time is the sum of the five element averages: $$\bar{T}=\sum_{j=1}^{5}\bar{t}_{j}=21.1+15.2+18.3+9.9+11.8=76.3\ \text{hundredths of a minute}$$ The same figure is obtained by averaging the ten cycle totals directly, $763/10=76.3$, which is a useful arithmetic check on the subtraction. Converting to minutes, $$\boxed{\bar{T}=0.763\ \text{min per cycle}}$$
4. Part (b) — apply the performance rating to obtain the normal time. Normal time removes the observed operator's pace from the measurement, $NT=\bar{T}\times R$, where $R$ is the analyst's performance rating. The paper prints no rating, so the study is treated as having been made on an operator judged to be working at a normal pace, $R=1.00$ (see the check note below). Then $$NT=0.763\times1.00=\boxed{0.763\ \text{min per cycle}}$$ Had the analyst rated the operator at 90 or 110 percent, the normal time would have been 0.687 or 0.839 min respectively; the rating scales the answer linearly and nothing else in the calculation changes.
5. Part (c) — add the allowance to obtain the standard time. Standard time is the normal time inflated to cover personal needs, unavoidable delay and fatigue, $ST=NT\,(1+A)$. Taking the customary personal-delay-fatigue allowance of 15 percent of normal time, $$ST=0.763\,(1+0.15)=\boxed{0.877\ \text{min per cycle}}$$ which corresponds to a standard output of $60/0.877=68.4$ pieces per hour. If the firm instead expresses the allowance as a fraction of the working day, the equivalent form is $ST=NT/(1-A)=0.763/0.85=0.898$ min; the two conventions differ by about 2 percent and the convention in force must be stated with the standard.
6. Part (d) — size the study from the observed scatter. For a confidence level with normal deviate $z$ and an acceptable error of $a$ expressed as a fraction of the mean, the required number of observations is $$n=\left(\frac{z\,s}{a\,\bar{t}}\right)^{2}$$ with $z=1.96$ at 95 percent confidence and, in the absence of a printed tolerance, $a=0.05$ (the conventional five-percent accuracy of an industrial time study). Applying it element by element with the standard deviations tabulated in step 2 gives $n=8.8,\ 14.5,\ 10.3,\ 15.5$ and $11.8$ for elements 1 to 5.
7. Take the governing element and round up. Every element must meet the accuracy target, so the study is sized on the worst of them — element 4, whose short mean of 9.9 makes its relative scatter the largest: $$n=\left(\frac{1.96\times0.994}{0.05\times9.9}\right)^{2}=15.5\;\Rightarrow\;\boxed{n=16\ \text{cycles}}$$ Ten cycles have already been timed, so six more are required. Two useful comparisons: sized on the cycle total rather than element by element the answer falls to 3 cycles (the element variations partly cancel), and relaxing the tolerance to $a=0.10$ drops the requirement to 4 cycles. The element-wise five-percent figure is the defensible one, because element standards, not just the cycle standard, are what a time study is normally used to set.