Question

A time study engineer recorded the cycle time (in minute) for machining of a component. The recorded time study data is given below. The performance rating of the worker is 110%. The standard time (in minute) by assuming 10% allowance is \(\underline{\hspace{2cm}}\). [round off to nearest integer] \[ \begin{array}{|c|c|} \hline \text{Cycle time (min)} & \text{Frequency} \\ \hline 42 & 1\\ 43 & 2\\ 44 & 3\\ 45 & 2\\ 46 & 1\\ \hline \end{array} \]

Hint:

Standard time = Average observed time × performance rating × (1 + allowance).

Answer 1

Correct Answer: 53 - 54

Step 1: Compute the average cycle time \[ \bar{t} = \frac{42(1) + 43(2) + 44(3) + 45(2) + 46(1)}{1 + 2 + 3 + 2 + 1} \] \[ \bar{t} = \frac{42 + 86 + 132 + 90 + 46}{9} = \frac{396}{9} = 44.0\ \text{min} \]

Step 2: Apply performance rating Rating = 110% → multiplier = 1.10 \[ t_{\text{normal}} = 44.0 \times 1.10 = 48.4\ \text{min} \]

Step 3: Include 10% allowance \[ t_{\text{standard}} = 48.4 \times 1.10 = 53.24\ \text{min} \] Rounded result is between 53 and 54 minutes.