18.3 Predetermined Time Systems, Work Sampling, and Learning Curves

Work measurement extends beyond direct stopwatch timing. When operations are not yet physically installed on the shop floor, or when timing long non-repetitive cycles via stopwatch is economically infeasible, industrial engineers utilize Predetermined Motion Time Systems (PMTS), Work Sampling, and Learning Curve Models. These three complementary methodologies enable establishing standard times in advance of manufacturing, auditing system-wide machine/operator utilization, and forecasting productivity gains across high-volume production programs. On the FE exam, these topics are tested through precise unit conversions, sample size calculations, sequence model evaluations, and log-linear regressions.

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1. Predetermined Motion Time Systems (PMTS) Foundations

A Predetermined Motion Time System (PMTS) establishes labor standards by analyzing basic human motions (reaching, moving, grasping, positioning) and retrieving established normal times from empirical research tables. Synthesized from millions of frames of high-speed industrial motion pictures, PMTS relies on the premise that standard manual motions require consistent durations under identical physical conditions.

                      Direct Time Study vs. PMTS

 Direct Stopwatch Study:                    PMTS (MTM / MOST):
 [ Physical Work on Floor ]                 [ Engineering Blueprint / CAD ]
             │                                             │
             ▼                                             ▼
 [ Stopwatch Observation ]                  [ Decompose into Micro-motions ]
             │                                             │
             ▼                                             ▼
 [ Subjective Pace Rating ]                 [ Lookup Standard Table Times ]
             │                                             │
             ▼                                             ▼
 [ Apply Allowances = Standard Time ]       [ Sum Tabular TMUs = Standard Time ]

Advantages and Limitations of PMTS

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2. Methods-Time Measurement (MTM) and the TMU

Developed in 1948 by Maynard, Stegemerten, and Schwab, Methods-Time Measurement (MTM) is the most internationally recognized PMTS. Its primary variants include MTM-1 (high-precision micro-motion level), MTM-2 (aggregated motion blocks for cycle times $> 1\text{ minute}$), and MTM-UAS (universal analyzing system for batch production).

The Time Measurement Unit (TMU)

To avoid awkward fractions of hours, minutes, or seconds, MTM introduced the Time Measurement Unit (TMU), defined as one hundred-thousandth of an hour:

$$1\text{ TMU} = 0.00001\text{ hour} = 10^{-5}\text{ hour}$$

Mandatory Conversion Equivalents

On the FE exam, rapid and error-free conversion between TMUs, hours, minutes, and seconds is essential:

$$1\text{ TMU} = 0.0006\text{ minutes} = 6 \times 10^{-4}\text{ minutes}$$ $$1\text{ TMU} = 0.036\text{ seconds}$$ $$1\text{ hour} = 100,000\text{ TMU}$$ $$1\text{ minute} = \frac{1}{0.0006} = 1,666.67\text{ TMU}$$ $$1\text{ second} = \frac{1}{0.036} = 27.78\text{ TMU}$$

                      TMU Conversion Quick Reference

                 ┌───────────────────────────────────────┐
                 │              1.00000 Hour             │
                 ├───────────────────────────────────────┤
                 │             100,000 TMU               │
                 ├───────────────────┬───────────────────┤
                 │    60 Minutes     │   3,600 Seconds   │
                 ├───────────────────┼───────────────────┤
                 │ 1 min = 1,666.7 TMU│ 1 sec = 27.78 TMU │
                 │ 1 TMU = 0.0006 min│ 1 TMU = 0.036 sec │
                 └───────────────────┴───────────────────┘

MTM-1 Motion Taxonomy

MTM-1 deconstructs tasks into basic motion codes. For example, motion code $R12C$ represents:

• $R$ = Reach motion
• $12$ = Distance of 12 inches
• $C$ = Case C condition (reaching to an object jumbled in a container with other objects; requires search and select, demanding 14.2 TMU compared to Case A at 9.6 TMU).

Core MTM-1 motion categories include: Reach ($R$), Move ($M$), Turn ($T$), Apply Pressure ($AP$), Grasp ($G$), Position ($P$), Release ($RL$), Disengage ($D$), Eye Travel ($ET$), and Body Motions (Walk $W$, Bend $B$, Kneel $K$).

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3. Maynard Operation Sequence Technique (MOST)

Developed by Kjell Zandin in 1980, MOST is an advanced, high-level PMTS roughly 40 times faster to apply than detailed MTM-1. MOST observes that object displacements follow structured, repetitive activity patterns.

Index Numbers and TMU Calculation

In MOST, motion variables are assigned standardized Index Numbers from a fixed geometric progression:

$$\text{Index Scale: } 0, \, 1, \, 3, \, 6, \, 10, \, 16, \, 24, \, 32, \, 42, \, 54$$

The Multiplier Rule: To determine total Time Measurement Units (TMU) from any MOST sequence model, sum all index numbers in the sequence and multiply by 10:

$\text{Total TMU} = 10 \times \sum (\text{Index Numbers})$

The Three BasicMOST Sequence Models

                         BasicMOST Sequence Models

 1. General Move (Unconstrained Spatial Displacement):
    [  A    B    G  ]   [  A    B    P  ]   [  A  ]
       Get Object          Put Object       Return

 2. Controlled Move (Constrained Physical Path - Slide/Rotate):
    [  A    B    G  ]   [  M    X    I  ]   [  A  ]
       Get Object         Move Process       Return

 3. Tool Use (Hand Tools / Gauges / Fasteners):
    [  A    B    G  ]   [  A    B    P  ]   [ Tool Action ]   [  A    B    P  ]   [  A  ]
       Get Tool            Put Tool          Work with Tool      Put Away Tool     Return

1. General Move Sequence ($A\ B\ G\ A\ B\ P\ A$): Used when an object moves freely through the air under unconstrained manual control:
   • $A$ = Action Distance (reach or move: $A_0 \le 2\text{ in.}$; $A_1 = \text{within reach}$; $A_3 = 1-2\text{ steps}$; $A_6 = 3-4\text{ steps}$)
   • $B$ = Body Motion ($B_0 = \text{none}$; $B_3 = \text{bend and arise } 50\%$; $B_6 = \text{bend and arise fully}$)
   • $G$ = Gain Control ($G_1 = \text{light/simple grasp}$; $G_3 = \text{interlocked/heavy/blind grasp}$)
   • $P$ = Place ($P_0 = \text{toss/drop}$; $P_1 = \text{lay aside/loose fit}$; $P_3 = \text{adjust/tight fit}$; $P_6 = \text{careful alignment}$)
   • Subphases: $A\ B\ G$ (Retrieve object) $\to$ $A\ B\ P$ (Place object) $\to$ $A$ (Return to workstation).
2. Controlled Move Sequence ($A\ B\ G\ M\ X\ I\ A$): Used when an object is constrained along a fixed physical track (sliding, cranking, pushing a lever, pivoting a toggle):
   • $M$ = Move Controlled (push/pull distance: $M_1 \le 12\text{ in.}$; $M_3 > 12\text{ in.}$)
   • $X$ = Process Time (machine-controlled time duration)
   • $I$ = Alignment (aligning with visual index marks: $I_1 = \text{single point}$; $I_3 = \text{two points}$)
3. Tool Use Sequence ($A\ B\ G\ A\ B\ P\ [\text{Tool Action}]\ A\ B\ P\ A$): Used when handling hand tools. Tool Action parameters include:
   • $F$ = Fasten (threading bolts)
   • $L$ = Loosen (unthreading nuts)
   • $C$ = Cut (snips, knife)
   • $S$ = Surface Adjust (sandpaper, deburring file)
   • $M$ = Measure (calipers, micrometer)
   • $R$ = Record (writing on clipboard)
   • $T$ = Think (visual inspection, decision pause)

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4. Work Sampling (Tippett's Ratio-Delay Study)

Invented by British statistician L.H.C. Tippett in 1934, Work Sampling is a statistical measurement technique that estimates the proportion of time operators or machines spend in various defined activity categories (e.g., active cutting, setting up, idle, waiting for material) through a large number of instantaneous, randomly scheduled observations.

Statistical Foundation

Work sampling relies on the Binomial Distribution. If an event occurs with true probability $p$, the probability that $x$ observations out of $N$ random observations capture the event follows the binomial distribution. When sample size $N$ is sufficiently large ($N p \ge 5$ and $N(1 - p) \ge 5$), the sampling distribution of the sample proportion $\hat{p} = x / N$ is normally distributed with mean $\mu_{\hat{p}} = p$ and standard error $\sigma_{\hat{p}}$:

$$\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{N}}$$

Sample Size Determination ($N$)

To determine the total number of random observations $N$ required to estimate true proportion $p$ within an absolute error tolerance $e$ at confidence level $(1 - \alpha)$:

$$\ne = z_{\alpha/2} \cdot \sigma_{\hat{p}} = z_{\alpha/2} \sqrt{\frac{p(1 - p)}{N}} \quad \implies \quad N = \frac{z_{\alpha/2}^2 \cdot p(1 - p)}{e^2}$$

Where:

• $N$ = required total random observations
• $z_{\alpha/2}$ = critical value of standard normal distribution ($1.96$ for $95\%$ confidence)
• $p$ = estimated true proportion of the activity (if completely unknown, use conservative $p = 0.50$, which maximizes the variance $p(1 - p) = 0.25$ and yields the maximum sample size)
• $e$ = allowable absolute margin of error (e.g., $\pm 2\% = 0.02$; $\pm 3\% = 0.03$)

Relative Error Formulation: If the allowable error is specified as a relative percentage of $p$ ($e = k \cdot p$, e.g., $\pm 5\%$ of $p$): $N = \frac{z_{\alpha/2}^2 \cdot (1 - p)}{k^2 \cdot p}$

Random Observation Scheduling and Bias Prevention

1. Random Timing: Observations must occur at truly random calendar instants generated via pseudo-random number algorithms to prevent cyclical alignment with periodic operator breaks, shift changes, or machine cycles.
2. Instantaneous Observation: The analyst must record the exact state observed at the very instant the workstation enters visual sight. Looking away and re-checking introduces severe observational bias.
3. Control Limits (p-Charts): To ensure work patterns remain statistically stable across the study window, daily observation proportions ($p_i$) are plotted on a $p$-control chart with limits based on daily sample size $n_{\text{daily}}$: $$UCL / LCL = \bar{p} \pm 3 \sqrt{\frac{\bar{p}(1 - \bar{p})}{n_{\text{daily}}}}$$ Points falling outside control limits indicate non-random systemic anomalies (e.g., severe power outage, major material stockout) and must be investigated.

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5. Learning Curve Models: Wright's vs. Crawford's

First quantified by T.P. Wright in 1936 during aircraft production, the Learning Curve Phenomenon (or progress curve) establishes that as cumulative production quantity doubles, direct labor hours required per unit decrease at a constant, predictable percentage rate.

                      The Log-Linear Learning Curve

 Direct Labor Hours (Y_x)                   Linear Log-Log Transformation
  ▲                                          ▲ ln(Y_x)
  │                                          │
K │─┐                                      ln(K)─┐
  │  \                                       │    \  Slope = b = ln(s) / ln(2)
  │   \                                      │     \   (b is always negative)
  │    \_                                    │      \
  │      \___                                │       \
  │          \_______                        │        \
  └──────────────────────► Cumulative Units  └─────────► ln(x)
  0   1   2   4   8    16      (x)           0

The Learning Rate Percentage ($s$)

The parameter $s$ ($0 < s < 1$) represents the learning rate percentage. A learning rate of $s = 0.80$ (an $80\%$ curve) means that every time cumulative production volume doubles ($1 \to 2$, $2 \to 4$, $4 \to 8$, $8 \to 16$), direct labor hours required drop to $80\%$ of their previous level (a $20\%$ reduction in unit time).

Wright's Log-Linear Model (Cumulative Unit Time Model)

Wright's formulation models the direct labor hours required to produce unit number $x$ ($Y_x$):

$$Y_x = K \cdot x^b$$

Taking natural logarithms yields a linear equation:

$$\ln(Y_x) = \ln(K) + b \cdot \ln(x)$$

Where:

• $x$ = cumulative unit number ($x = 1, 2, 3, \dots$)
• $Y_x$ = time (or cost) required to fabricate unit $x$
• $K$ = time (or cost) required to fabricate the first unit ($x = 1$)
• $b$ = learning curve slope exponent: $$b = \frac{\ln(s)}{\ln(2)} = \frac{\log_{10}(s)}{\log_{10}(2)}$$ (Note: Because $0 < s < 1$, $\ln(s)$ is negative, meaning $b$ is strictly negative).

Standard Industrial Learning Exponents ($b$)

Cumulative Total Time ($T_n$) and Cumulative Average Time ($\bar{Y}_n$)

• Exact Total Labor Time for First $n$ Units: $$T_n = \sum_{x=1}^n Y_x = \sum_{x=1}^n K \cdot x^b$$
• Continuous Calculus Approximation (for large $n$): $$T_n \approx \int_0^n K \cdot x^b \, dx = \left[ \frac{K}{b + 1} x^{b+1} \right]_0^n = \frac{K}{b + 1} n^{b+1}$$
• Cumulative Average Time per Unit ($\bar{Y}_n$): $$\bar{Y}_n = \frac{T_n}{n} \approx \frac{K}{b + 1} n^b$$

Crawford's Cumulative Average Model

In Crawford's model (popularized in defense procurement), the log-linear equation models the cumulative average time directly: $\bar{Y}_x = K \cdot x^b$. Total time is $T_x = x \cdot \bar{Y}_x = K \cdot x^{b+1}$. Individual unit time is computed by successive differences: $Y_x = T_x - T_{x-1}$. On the FE exam, Wright's unit model ($Y_x = K x^b$) is the standard formulation unless Crawford is explicitly named.

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6. Step-by-Step Worked Engineering Calculations

Worked Example 18.3.1: Work Sampling Sample Size & Control Limits

Problem: An industrial engineer conducts a work sampling study in a sheet-metal fabrication shop to determine the percentage of time a robotic plasma cutter is down due to torch nozzle slag fouling. Preliminary maintenance records suggest nozzle fouling accounts for approximately $12\%$ of operating time ($p = 0.12$).

1. Calculate the required total number of random observations ($N$) to estimate true downtime proportion within an absolute precision limit of $\pm 2.5\%$ ($e = 0.025$) at a $95\%$ confidence level ($z = 1.96$).
2. Over a 10-day study, exactly $n_{\text{daily}} = 70\text{ observations}$ are recorded each day. Calculate the upper and lower control limits ($UCL$ and $LCL$) for daily monitoring.
3. On Day 4, the observer records 16 nozzle fouling events ($p_4 = 16 / 70 = 0.2286$). Determine whether Day 4 was in statistical control.

Solution:

Step 1: Calculate Required Total Sample Size ($N$)

$$N = \frac{z_{\alpha/2}^2 \cdot p(1 - p)}{e^2}$$

• Calculate numerator: $$z^2 = (1.96)^2 = 3.8416$$ $$p(1 - p) = 0.12 \times (1 - 0.12) = 0.12 \times 0.88 = 0.1056$$ $$\text{Numerator} = 3.8416 \times 0.1056 = 0.405673$$
• Calculate denominator: $$e^2 = (0.025)^2 = 0.000625$$
• Compute $N$: $$N = \frac{0.405673}{0.000625} = 649.08 \approx 650\text{ observations}$$

Step 2: Calculate Daily $p$-Chart Control Limits With $\bar{p} = 0.12$ and $n_{\text{daily}} = 70$:

$$\sigma_{p} = \sqrt{\frac{\bar{p}(1 - \bar{p})}{n_{\text{daily}}}} = \sqrt{\frac{0.12 \times 0.88}{70}} = \sqrt{\frac{0.1056}{70}} = \sqrt{0.0015086} \approx 0.03884$$ $$UCL = \bar{p} + 3 \sigma_p = 0.12 + 3(0.03884) = 0.12 + 0.1165 = 0.2365\text{ (or } 23.65\%\text{)}$$ $$LCL = \bar{p} - 3 \sigma_p = 0.12 - 3(0.03884) = 0.12 - 0.1165 = 0.0035\text{ (or } 0.35\%\text{)}$$

Step 3: Evaluate Day 4 Statistical Control On Day 4, sample proportion was $p_4 = 16 / 70 = 0.2286$ ($22.86\%$).

• Comparing against control limits: $$LCL (0.0035) \le p_4 (0.2286) \le UCL (0.2365)$$
• Engineering Conclusion: Day 4 falls just beneath the Upper Control Limit ($0.2286 < 0.2365$). The operation was technically in statistical control, reflecting extreme random variation rather than an assignable structural failure.

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Worked Example 18.3.2: Wright's Learning Curve Cost Modeling

Problem: An aerospace defense contractor is preparing a competitive bid to manufacture 16 unmanned aerial vehicle (UAV) composite wing assemblies. Production data from fabricating prototype Unit 1 indicates a direct labor requirement of $K = 400\text{ hours}$. Based on historical composite fabrication data, an $80\%$ learning curve ($s = 0.80$) applies.

1. Determine the learning curve slope parameter $b$.
2. Calculate the projected direct labor hours required to manufacture Unit 2, Unit 4, Unit 8, and Unit 16 using the doubling property.
3. Calculate the projected labor hours required to fabricate Unit 5.
4. Estimate the total cumulative direct labor hours ($T_{16}$) required to manufacture all 16 units using the integration approximation.
5. If direct labor is billed at $\text{\textdollar}75.00/\text{hour}$, calculate the estimated total labor cost for the 16-unit contract.

Solution:

Step 1: Calculate Learning Curve Slope Exponent ($b$)

$$b = \frac{\ln(s)}{\ln(2)} = \frac{\ln(0.80)}{\ln(2)} = \frac{-0.22314355}{0.69314718} \approx -0.321928$$

Step 2: Projected Labor Hours via Doubling Property Because volume doubles, unit time drops by $20\%$ ($s = 0.80$):

• Unit 1: $Y_1 = K = 400.0\text{ hours}$
• Unit 2: $Y_2 = 400.0 \times 0.80 = 320.0\text{ hours}$
• Unit 4: $Y_4 = 320.0 \times 0.80 = 256.0\text{ hours}$
• Unit 8: $Y_8 = 256.0 \times 0.80 = 204.8\text{ hours}$
• Unit 16: $Y_{16} = 204.8 \times 0.80 = 163.84\text{ hours}$

Step 3: Projected Labor Hours for Unit 5 Using Wright's equation $Y_x = K \cdot x^b$:

$$Y_5 = 400.0 \times (5)^{-0.321928}$$ $$\ln(Y_5) = \ln(400) + (-0.321928) \times \ln(5) = 5.99146 - 0.321928 \times 1.609438 = 5.99146 - 0.518123 = 5.47334$$ $$Y_5 = e^{5.47334} \approx 238.25\text{ hours}$$

(Sanity check: $Y_5 = 238.25\text{ hr}$ is logically bounded between $Y_4 = 256.0\text{ hr}$ and $Y_8 = 204.8\text{ hr}$).

Step 4: Cumulative Total Hours ($T_{16}$) via Integration Approximation

$$T_n \approx \frac{K}{b + 1} n^{b+1}$$

• Calculate exponent: $b + 1 = -0.321928 + 1.0 = 0.678072$
• Calculate coefficient: $\frac{K}{b + 1} = \frac{400.0}{0.678072} \approx 589.9079$
• Evaluate $n^{b+1}$ at $n = 16$: $$16^{0.678072} = e^{0.678072 \times \ln(16)} = e^{0.678072 \times 2.772589} = e^{1.880015} \approx 6.5536$$
• Compute $T_{16}$: $$T_{16} \approx 589.9079 \times 6.5536 \approx 3,866.0\text{ total labor hours}$$ (Summing the 16 discrete unit hours exactly gives $\sum_{x=1}^{16} 400 x^{-0.321928} = 3,568.1\text{ hr}$. The integral overestimates by about $8.4\%$ because it integrates from $x = 0$, where the unit-time curve $K x^{b}$ is unbounded. For contract bidding, prefer the exact discrete sum; use the integral only as a fast sanity bound.)

Step 5: Estimated Contract Labor Cost

$$\text{Total Labor Cost} = 3,866.0\text{ hours} \times \$75.00/\text{hour} = \$289,950$$

(Based on the exact discrete summation, the defensible bid figure is $3,568.1 \times \text{\textdollar}75.00 = \text{\textdollar}267,608$).

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7. NCEES Reference Handbook Tips & Realistic Exam Traps

• TMU Unit Conversions: Forgetting the difference between minutes and seconds when converting TMUs is an ubiquitous trap:
  • $1\text{ TMU} = 0.036\text{ SECONDS}$
  • $1\text{ TMU} = 0.0006\text{ MINUTES}$
  • To convert TMU to minutes: $\text{min} = \text{TMU} \times 0.0006$ (or $\text{TMU} / 1666.7$).
  • To convert TMU to seconds: $\text{sec} = \text{TMU} \times 0.036$ (or $\text{TMU} / 27.78$).
• BasicMOST 10x Multiplier: When evaluating MOST sequences, students frequently sum the index numbers ($1 + 0 + 1 + 1 + 0 + 3 + 0 = 6$) and mistakenly report 6 TMU. Always multiply by 10 $\to 60\text{ TMU}$!
• Learning Curve Exponent Sign: In $Y_x = K \cdot x^b$, the exponent $b = \ln(s)/\ln(2)$ is always negative. If you forget the negative sign and calculate $b = +0.3219$, unit times will explode upward ($Y_4 = 625\text{ hr}$ instead of $256\text{ hr}$).
• Work Sampling Absolute vs. Relative Precision: Read the problem carefully:
  • If absolute error is given ("estimate proportion within $\pm 2\%$"): $e = 0.02$, use $N = z^2 p(1-p)/e^2$.
  • If relative error is given ("estimate proportion within $\pm 5\%$ of the true value"): $e = 0.05 \times p$, use $N = z^2 (1-p) / (k^2 p)$.