18.2 Direct Stopwatch Time Study: Observed Time, Performance Rating, and Allowances

Direct stopwatch time study, formulated by Frederick Winslow Taylor in 1881, remains the foundational work measurement technique in modern industrial operations. It establishes the Standard Time required for a qualified, thoroughly trained operator, working at a normal pace under standard conditions, to complete a specified task. Setting rigorous standard times is vital for production scheduling, assembly line balancing, manufacturing capacity planning, wage incentive systems, and standard product costing. On the FE exam, mastery of stopwatch mechanics, statistical sample size sizing, pace rating dynamics, and allowance mathematical formulations is mandatory.

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1. Direct Stopwatch Time Study Methodology

A stopwatch study follows a standardized, eight-step engineering procedure:

                    Stopwatch Time Study Procedure
                    
 1. Standardize Method ──► Ensure standard tooling, feeds, and workplace layout
          │
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 2. Select Operator ──► Qualified, representative, working at average pace
          │
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 3. Break Down Elements ──► 0.04 to 0.50 min, distinct visual/auditory break points
          │
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 4. Record Observed Times ──► Continuous or snapback timing across n cycles
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 5. Pace Rate Operator ──► Assign Performance Rating (PR) to each manual element
          │
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 6. Compute Normal Time ──► NT = OT * (PR / 100) * frequency
          │
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 7. Apply PF&D Allowances ──► Personal, Fatigue, and Delay adjustments
          │
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 8. Determine Standard Time ──► ST = NT * (1 + A_NT) or ST = NT / (1 - A_shift)

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2. Timing Methods: Continuous vs. Snapback

Two distinct recording methods are utilized during direct stopwatch observation:

                       Continuous vs. Snapback Timing

 Continuous Timing:
 Watch runs:   0.00 ──────► 0.32 ──────────► 1.15 ────────────► 1.58 (min)
 Breakpoints:               Point 1          Point 2            Point 3
 Recorded:                  R1 = 0.32        R2 = 1.15          R3 = 1.58
 Subtracted:                t1 = 0.32        t2 = 0.83          t3 = 0.43

 Snapback Timing:
 Watch runs:   0.00 ──► 0.32 | 0.00 ──► 0.83 | 0.00 ──► 0.43 (min)
 Reset:                Snap!            Snap!            Snap!
 Recorded:             t1 = 0.32        t2 = 0.83        t3 = 0.43

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3. Work Element Breakdown Principles

Dividing a job into discrete work elements is essential because:

• Operators do not perform at a uniform pace across an entire cycle (an operator may move quickly during light manual transfers but slowly during delicate alignments).
• It enables segregating operator-controlled manual elements from machine-controlled elements.
• It facilitates building Standard Data systems for future synthetic time setting.

Rules for Element Definition

1. Homogeneity: Group related motions into homogeneous tasks. Never combine a manual handling element with a machine-controlled cut.
2. Element Duration: Elements must not be so short that human reaction time introduces observational error ($> 0.04\text{ minutes}$ or $2.4\text{ seconds}$), nor so long that performance rating varies significantly across the element ($< 0.50\text{ minutes}$ or $30\text{ seconds}$). A target elemental length is typically $0.10\text{ to }0.30\text{ minutes}$.
3. Distinct Break Points: The transition between elements must feature an unmistakable visual or auditory break point (e.g., "part clinks in tote," "press clutch clicks," "operator releases drill handle").
4. Cyclical vs. Irregular (Periodic) Elements:
   • Cyclical Elements: Occur once every production cycle (occurrence frequency $f = 1/1 = 1.0$).
   • Irregular Elements: Occur periodically once every $N$ cycles (e.g., retrieving a tote of blanks every 25 parts, $f = 1/25 = 0.04$; sharpening a cutting tool every 50 parts, $f = 1/50 = 0.02$).
5. Foreign Elements: Unscheduled occurrences (e.g., dropped part, tangled wire, supervisor interruption, brief conveyor stall). The analyst must record the foreign event, note its duration, and either discard it as an non-representative outlier or apportion it into the unavoidable delay category.

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4. Statistical Sample Size Determination

Because time study is an empirical sampling process, the observed mean $(\bar{x})$ is an estimate of the true population mean $(\mu)$. To ensure the calculated standard time meets legal, industrial, and contractual standards, the analyst must compute the required number of cycles ($n$).

The NCEES Sample Size Formula

Assuming observed elemental times follow a normal distribution, the required number of observations $n$ to achieve an estimate within a specified relative precision limit $k$ at a confidence level of $(1 - \alpha)$ is:

$n = \left( \frac{z_{\alpha/2} \cdot s}{k \cdot \bar{x}} \right)^2$

Where:

• $n$ = required total number of timed cycles (always rounded up to the nearest integer)
• $z_{\alpha/2}$ = two-tailed critical value of the standard normal distribution:
  • For $90%$ confidence: $z_{0.05} = 1.645$
  • For $95%$ confidence: $z_{0.025} = 1.960$
  • For $99%$ confidence: $z_{0.005} = 2.576$
• $s$ = sample standard deviation of the preliminary observed times from a pilot sample of size $n'$: $s = \sqrt{\frac{\sum_{i=1}^{n'} (x_i - \bar{x})^2}{n' - 1}}$
• $\bar{x}$ = sample arithmetic mean of the preliminary observed times
• $k$ = allowable relative precision limit / error tolerance, expressed as a decimal (e.g., $\pm 5% = 0.05$; $\pm 10% = 0.10$)

Small Pilot Sample Formulation ($n' < 30$): If the preliminary pilot study is small ($n' < 30$), the Student's $t$-distribution replaces the standard normal critical value: $n = \left( \frac{t_{\alpha/2, \, n'-1} \cdot s}{k \cdot \bar{x}} \right)^2$ If the resulting $n > n'$, additional cycles must be recorded until the cumulative sample size satisfies the criterion.

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5. Performance Rating (Pace Rating) & Normal Time

Performance Rating (or pace rating) is the process whereby an experienced time study analyst evaluates an operator's actual working pace relative to the concept of Standard Performance (100% normal pace).

Standard Performance Benchmarks

Standard normal performance ($100%$) represents the pace achievable by a qualified, thoroughly motivated operator working sustainably without cumulative physiological fatigue over an 8-hour shift. Classical industrial engineering benchmarks include:

• Walking unburdened on smooth, level ground at 3.0 miles per hour ($4.4\text{ ft/s}$ or $88\text{ ft/min}$).
• Dealing a standard 52-card deck into four separate piles (representing bridge hands placed in the corners of a 1-foot square) in 0.50 minutes ($30\text{ seconds}$).
• Assembling 30 small cylindrical pins into a standard pegboard in 0.435 minutes.

Rating Systems

1. Speed Rating: The observer compares the operator's speed of motion directly against the mental 100% benchmark and records a single rating percentage (e.g., $85%$, $100%$, $115%$, $125%$).
2. Westinghouse System of Rating: Evaluates four independent performance attributes using standardized empirical score tables:
   • Skill: Proficiency in following method (+0.15 for Superskill to -0.22 for Poor).
   • Effort: Will to work (+0.13 for Excessive to -0.17 for Poor).
   • Conditions: Environmental heat, lighting, noise (+0.06 for Ideal to -0.07 for Poor).
   • Consistency: Variation between cycle times (+0.04 for Perfect to -0.04 for Poor). The algebraic sum of the four adjustments is added to $1.00$ to compute the net rating factor.

Normal Time ($NT$) Calculation

Normal Time is the time an average operator working at $100%$ pace would require to complete the element. For any manual element $j$:

$NT_j = \overline{OT}_j \times \left( \frac{PR_j}{100} \right) \times f_j$

Where:

• $\overline{OT}_j$ = mean observed elapsed time for element $j$
• $PR_j$ = performance rating percentage assigned to element $j$
• $f_j$ = occurrence frequency per finished cycle (e.g., $f = 1.0$ for cyclical elements; $f = 0.05$ for an element occurring once every 20 units)

The total Normal Time for the entire job is the sum across all $m$ elements:

$NT_{\text{job}} = \sum_{j=1}^m NT_j$

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6. Personal, Fatigue, and Delay (PF&D) Allowances

A human operator cannot maintain normal working pace for an entire 480-minute shift without biological and operational interruptions. To establish a realistic, humane standard, industrial engineers add PF&D Allowances:

                      The PF&D Allowance Structure

                      Total Shift (480 minutes)
 ┌───────────────────────────────────────────────┬──────────────┐
 │        Productive Working Time (NT)           │  Allowances  │
 └───────────────────────────────────────────────┴──────────────┘
                                                        │
          ┌─────────────────────────────────────────────┴────────┐
          ▼                                             ▼        ▼
   Personal Needs (P)                             Fatigue (F)  Unavoidable Delays (D)
   Restroom, water, comfort                       Basic: 4-5%  Minor tool jams, machine
   Typically 5% (~24 min/day)                     Variable: 0-25% maintenance, instructions
                                                  (Heat, posture, load) Typically 2-5%

1. Personal Needs ($P$): Time necessary to maintain personal hygiene, visit the restroom, and drink water. Standard industrial practice universally assigns $5%$ (approximately $24\text{ minutes}$ in an 8-hour shift).
2. Basic Fatigue ($F$): Energy expenditure recovery. Basic fatigue for seated light assembly in an air-conditioned room is $4%\text{ to }5%$. Variable fatigue increments are added based on physical and environmental severity:
   • Awkward Posture: Bending, crouching, overhead reaching ($+2%\text{ to }+7%$).
   • Muscular Load: Heavy lifting or repetitive pushing ($+1%\text{ to }+15%$).
   • Atmospheric Conditions: High thermal heat index or humidity ($+2%\text{ to }+15%$).
   • Visual & Mental Strain: Fine microscopic alignment ($+2%\text{ to }+5%$).
3. Unavoidable Delays ($D$): Minor, unpredictable operational interruptions outside the operator's control (e.g., brief supervisor instructions, material handler delivery delays, power fluctuations). Typically $2%\text{ to }5%$.

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7. Standard Time Formulations: The Two Allowance Bases

CRITICAL EXAM ALERT: The single most common error on the FE Industrial exam is applying the incorrect allowance formula. The NCEES Reference Handbook recognizes two distinct mathematical bases for calculating Standard Time ($ST$) from Normal Time ($NT$):

                    Allowance Formulations Comparison

 Basis 1: Allowance on Normal Time (A_NT)      Basis 2: Allowance on Shift Time (A_shift)

 ┌───────────────┬───────────────┐              ┌───────────────────────────────┬──────────────┐
 │  Normal Time  │ Allowance (A) │              │      Normal Time (Productive) │Allowance (A) │
 └───────────────┴───────────────┘              └───────────────────────────────┴──────────────┘
 ◄──────── Standard Time ────────►              ◄──────────────── Total Shift ─────────────────►

        ST = NT * (1 + A_NT)                                ST = NT / (1 - A_shift)

Method 1: Allowance Based on Normal Time ($A_{NT}$)

Here, the allowance percentage is expressed as a direct markup added onto the productive Normal Time. Standard Time is computed as:

$ST = NT \times (1 + A_{NT})$

Example: If $NT = 5.0\text{ minutes}$ and allowance is $15%$ of normal time ($A_{NT} = 0.15$): $ST = 5.0 \times (1 + 0.15) = 5.75\text{ minutes}$

Method 2: Allowance Based on Total Shift / Working Day ($A_{\text{shift}}$)

Here, allowances are allocated as a fixed percentage of the total working day (e.g., $60\text{ minutes}$ of total allowance in a $480\text{ minute}$ shift $= 12.5%$). In this framework, the working shift is partitioned into productive normal time and non-productive allowance time:

$\text{Shift Time} = \text{Productive Time} + \text{Allowance Time}$ $ST \cdot (1 - A_{\text{shift}}) = NT \quad \implies \quad ST = \frac{NT}{1 - A_{\text{shift}}}$

Example: If $NT = 5.0\text{ minutes}$ and allowance is $15%$ of the working shift ($A_{\text{shift}} = 0.15$): $ST = \frac{5.0}{1 - 0.15} = \frac{5.0}{0.85} = 5.882\text{ minutes}$

Mathematical Equivalence and Conversion

Notice that for the same nominal percentage ($15%$), Method 2 yields a larger Standard Time than Method 1 ($5.88\text{ min}$ vs. $5.75\text{ min}$). To convert between the two allowance bases:

$A_{NT} = \frac{A_{\text{shift}}}{1 - A_{\text{shift}}} \quad \iff \quad A_{\text{shift}} = \frac{A_{NT}}{1 + A_{NT}}$

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8. Machine-Controlled vs. Operator-Controlled Elements

When a production operation includes automated equipment (e.g., CNC milling, robotic welding, plastic injection cure), elements must be categorized based on control:

• Operator-Controlled (Manual) Elements: The elapsed duration depends entirely on the operator's speed, effort, and dexterity. Performance rating MUST be applied ($PR \ne 100%$).
• Machine-Controlled Elements: The process duration is governed by machine kinematics, feeds, speeds, or thermodynamic cycles. The operator cannot accelerate the cut. Therefore, machine-controlled elements are NEVER pace-rated ($PR = 100%$ or pace rating is omitted).
• Applying Allowances to Machine Time: While manual fatigue allowances do not apply to machine automatic run time, personal needs ($5%$) and unavoidable machine delay allowances ($2-3%$) apply to the entire cycle duration.

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9. Step-by-Step Worked Engineering Calculations

Worked Example 18.2.1: Multi-Element Time Study and Allowance Base Sensitivity

Problem: An industrial engineer conducts a direct continuous stopwatch study on a semi-automated machining operation. An 8-hour shift contains $480\text{ minutes}$. The observed elemental times, performance ratings, and occurrence frequencies across 20 cycles are summarized in the table below:

Plant policy mandates a total PF&D allowance of $16%$ ($P = 5%$, $F = 6%$, $D = 5%$).

1. Calculate the Normal Time ($NT$) for each of the five work elements.
2. Compute the total Job Normal Time ($NT_{\text{job}}$).
3. Calculate the Standard Time ($ST$) using Basis 1: Allowance based on Normal Time ($A_{NT} = 0.16$).
4. Calculate the Standard Time ($ST$) using Basis 2: Allowance based on Total Shift ($A_{\text{shift}} = 0.16$).
5. Determine the expected daily production output (parts per 8-hour shift) under both standard time methods.

Solution:

Step 1: Calculate Elemental Normal Times ($NT_i = \overline{OT}_i \times (PR_i / 100) \times f_i$)

• Element 1 (Manual load): $NT_1 = 0.45 \times 1.15 \times 1.0 = 0.5175\text{ minutes}$
• Element 2 (CNC Machining - Machine Controlled, $PR = 100%$): $NT_2 = 2.20 \times 1.00 \times 1.0 = 2.2000\text{ minutes}$
• Element 3 (Manual unload and deburr): $NT_3 = 0.65 \times 1.10 \times 1.0 = 0.7150\text{ minutes}$
• Element 4 (Manual gauge - periodic $f = 0.20$): $NT_4 = 0.50 \times 1.05 \times 0.20 = 0.1050\text{ minutes}$
• Element 5 (Manual chip clean - periodic $f = 0.05$): $NT_5 = 0.80 \times 1.00 \times 0.05 = 0.0400\text{ minutes}$

Step 2: Calculate Total Job Normal Time ($NT_{\text{job}}$) $NT_{\text{job}} = NT_1 + NT_2 + NT_3 + NT_4 + NT_5$ $NT_{\text{job}} = 0.5175 + 2.2000 + 0.7150 + 0.1050 + 0.0400 = 3.5775\text{ minutes} \approx 3.58\text{ minutes}$

Step 3: Calculate Standard Time under Basis 1 ($A_{NT} = 0.16$) $ST_1 = NT_{\text{job}} \times (1 + A_{NT}) = 3.5775 \times (1 + 0.16) = 3.5775 \times 1.16 = 4.1499\text{ minutes} \approx 4.15\text{ minutes}$

Step 4: Calculate Standard Time under Basis 2 ($A_{\text{shift}} = 0.16$) $ST_2 = \frac{NT_{\text{job}}}{1 - A_{\text{shift}}} = \frac{3.5775}{1 - 0.16} = \frac{3.5775}{0.84} = 4.2589\text{ minutes} \approx 4.26\text{ minutes}$

Step 5: Compute Daily Shift Production Quotas Total available time in 8-hour shift = $480\text{ minutes}$.

• Under Basis 1 ($ST_1 = 4.1499\text{ min}$): $\text{Daily Output}_1 = \frac{480\text{ minutes/shift}}{4.1499\text{ minutes/part}} = 115.66 \approx 115\text{ parts/shift}$
• Under Basis 2 ($ST_2 = 4.2589\text{ min}$): $\text{Daily Output}_2 = \frac{480\text{ minutes/shift}}{4.2589\text{ minutes/part}} = 112.70 \approx 112\text{ parts/shift}$
• Engineering Impact: Confusing Basis 1 with Basis 2 results in a production quota discrepancy of 3 parts per operator per shift, illustrating why specifying the allowance denominator is vital in labor planning.

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10. NCEES Reference Handbook Tips & Realistic Exam Traps

• Pace Rating Machine Elements: If an exam problem states an operator was pace-rated at $125%$ and lists a machine cutting cycle of $3.00\text{ minutes}$, DO NOT multiply $3.00$ by $1.25$! Machine cycle times are fixed by spindle drives and motors; multiplying machine time by pace rating yields an immediate distractor answer.
• Periodic Element Frequency ($f$): Always check whether an element occurs every cycle ($f = 1.0$) or periodically. If an element occurs once every 10 parts, you must multiply its normal time by $f = 0.10$. Omitting $f$ inflates the standard time.
• The Allowance Formula Trap: Always inspect the wording of the exam question:
  • If it says: "allowances are 15% of normal time" $\to$ use $ST = NT \times (1 + A)$.
  • If it says: "allowances account for 15% of the total workday/shift" $\to$ use $ST = NT / (1 - A)$.
  • If unspecified, look for clues or check both formulas against multiple-choice options, noting that $NT / (1 - A)$ is always the larger number.
• Sample Size ($n$) Unit Consistency: In $n = ((z \cdot s) / (k \cdot \bar{x}))^2$, ensure that $s$ and $\bar{x}$ have identical units (minutes or seconds), and that $k$ is a pure decimal ($5% \to 0.05$). Do not enter $5$ for $k$!