2.3 First Time Yield (FTY) & Rolled Throughput Yield (RTY)

2.3 First Time Yield (FTY) & Rolled Throughput Yield (RTY)

Core Principle: Traditional quality reporting relies on Final Yield—the percentage of units that pass final inspection and ship to customers. In modern Six Sigma, Final Yield is viewed as dangerously deceptive because it conceals internal rework loops, repairs, and scrap. To uncover true process performance, Green Belts utilize First Time Yield (FTY) and Rolled Throughput Yield (RTY) to expose the "hidden factory" that drains organizational profitability.

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The Fallacy of Final Yield & The Hidden Factory

For decades, manufacturing and service organizations evaluated operational health using Traditional Yield (also called Final Yield):

$\text{Final Yield} = \frac{\text{Units Meeting Final Acceptance}}{\text{Total Units Started}}$

Consider an electronic device assembly line that starts 1,000 units on Monday morning. By Friday afternoon, 950 finished units pass final inspection and ship to customers. Management happily reports a 95% Final Yield. What this metric completely ignores is what transpired between Monday and Friday:

• 150 units failed Step 2 and were sent to an offline rework station for manual de-soldering.
• 80 units failed Step 4 testing, were disassembled, re-flashed with firmware, and re-tested twice.
• Technicians spent 120 hours of unbudgeted overtime diagnosing intermittent electrical shorts.

This unmeasured, undocumented rework loop is known as the Hidden Factory—a concept formulated by quality pioneer Dr. Armand Feigenbaum.

graph LR
    START["1,000 Units Started"] --> STEP1["Step 1: Assembly"]
    STEP1 --> STEP2["Step 2: Soldering"]
    STEP2 --> STEP3["Step 3: Programming"]
    STEP3 --> STEP4["Step 4: Final Test"]
    STEP4 --> SHIP["950 Units Shipped<br/>(95% Final Yield!)"]
    
    STEP2 -.->|"150 Defective Units"| REWORK1["HIDDEN FACTORY<br/>De-soldering & Patching"]
    REWORK1 -.-> STEP2
    
    STEP4 -.->|"80 Defective Units"| REWORK2["HIDDEN FACTORY<br/>Disassembly & Re-test"]
    REWORK2 -.-> STEP4
    
    style REWORK1 fill:#f9d5e5,stroke:#333,stroke-width:2px
    style REWORK2 fill:#f9d5e5,stroke:#333,stroke-width:2px

The Severe Costs of the Hidden Factory

1. Devoured Capacity: Machines and personnel spend 15% to 30% of their operational hours reworking products that should have been produced correctly the first time.
2. Inflated Work-in-Process (WIP): Large buffers of inventory accumulate around rework stations, increasing carrying costs and extending overall manufacturing lead time.
3. Latent Defects in the Field: Products that undergo rework and repair suffer from significantly higher field failure rates and warranty claims than products that pass through the process cleanly on the first pass. Rework often damages adjacent components, weakens solder joints, or introduces thermal stress.

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First Time Yield (FTY)

To strip away the disguise of the hidden factory, Six Sigma utilizes First Time Yield (FTY) (also known as First Pass Yield):

$\text{FTY} = \frac{\text{Units Entering Step} - \text{Scrap} - \text{Rework}}{\text{Units Entering Step}} = \frac{\text{Units Passing Initial Inspection with Zero Rework}}{\text{Total Units Entering Step}}$

Critical Rules of FTY:

• Rework is Treated as a Defect: Even if a unit is successfully repaired, calibrated, and eventually made fully functional, it cannot be counted in FTY. If it required human intervention, repair, or a second test cycle, it failed the first pass.
• Step-Specific Metric: FTY is calculated for each discrete step in a value stream.

Worked Example of Step FTY:

An insurance claims department processes 500 medical claims through Step 3 (Eligibility Verification):

• 440 claims pass verification immediately on the first attempt.
• 35 claims contain missing member IDs, are routed to an administrative queue for manual telephone lookup, and are eventually corrected.
• 25 claims are permanently rejected and scrapped due to fraudulent policy numbers.

$\text{Final Step Yield} = \frac{440 + 35}{500} = \frac{475}{500} = 95.0\%$ $\text{FTY} = \frac{500 - 25 - 35}{500} = \frac{440}{500} = 88.0\%$

While final step yield shows 95.0%, the true First Time Yield is only 88.0%. The 7.0% discrepancy represents the hidden factory of claim rework.

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Rolled Throughput Yield (RTY)

In any real-world manufacturing or transactional environment, a product or service passes through multiple sequential operations. While each individual step may exhibit a respectable FTY (e.g., 95%), the cumulative probability of a unit passing through all steps without a defect degrades exponentially.

Rolled Throughput Yield (RTY) is the probability that a single unit will pass through an entire multi-step process from start to finish without requiring rework, repair, or scrap.

The Mathematical Formula for RTY

For an $n$-step process, RTY is the mathematical product of the First Time Yields of each individual step:

$\text{RTY} = Y_1 \times Y_2 \times Y_3 \times \dots \times Y_n = \prod_{i=1}^{n} Y_i$

Where:

• $Y_i$ = First Time Yield of process step $i$ expressed as a decimal.

graph LR
    STEP1["Step 1<br/>FTY = 96.0%"] -->|"0.96"| STEP2["Step 2<br/>FTY = 94.0%"]
    STEP2 -->|"0.9024"| STEP3["Step 3<br/>FTY = 98.0%"]
    STEP3 -->|"0.8844"| STEP4["Step 4<br/>FTY = 92.0%"]
    STEP4 -->|"0.8136"| FINISH["Overall Process RTY<br/>= 81.36%"]

Comprehensive Worked Example: The 5-Step Process

Scenario: A precision medical device manufacturer produces laparoscopic surgical instruments through a five-stage value stream. The Green Belt audits each station and records initial units entering, scrapped units, and units requiring rework:

(Note: For simplified exam calculations, step yields are often given directly as round percentages. Let us evaluate a canonical exam case where each of the 5 steps has an identical FTY of 95.0%, or 0.950).

$\text{RTY} = Y_1 \times Y_2 \times Y_3 \times Y_4 \times Y_5 = 0.95 \times 0.95 \times 0.95 \times 0.95 \times 0.95 = (0.95)^5$ $\text{RTY} = 0.7738 = \mathbf{77.38\%}$

The Management Eye-Opener:

• Management sees a line where each manager reports: "My station runs at 95% yield!"
• Final inspection at the end of the line might show 98% because rejected units were reworked offline and pushed through.
• Yet only 77.38% of the units started actually navigated the five process steps without encountering an error, scrap event, or rework cycle!
• 22.62% of the entire production volume was entangled in the hidden factory.

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Poisson-Based Yield Calculation from DPU

When a product contains many defect opportunities, defects typically follow a Poisson distribution. The Poisson probability distribution gives the probability of observing exactly $x$ defects on a unit when the average defect rate is $\lambda = \text{DPU}$:

$P(x) = \frac{e^{-\text{DPU}} \times (\text{DPU})^x}{x!}$

In quality engineering, a unit is deemed "defect-free" (yield) only if it contains exactly zero defects ($x = 0$):

$P(0) = \frac{e^{-\text{DPU}} \times (\text{DPU})^0}{0!} = \frac{e^{-\text{DPU}} \times 1}{1} = e^{-\text{DPU}}$

This yields the famous Poisson Yield Equation:

$\text{Yield} = e^{-\text{DPU}}$

Where:

• $e$ = Mathematical constant (Euler's number $\approx 2.71828$).
• $\text{DPU}$ = Defects Per Unit.

Calculating Yield from DPU

If an automotive paint line has a baseline defect rate of $\text{DPU} = 0.05$ defects per car body: $\text{Yield} = e^{-0.05} \approx 0.9512 = \mathbf{95.12\%}$

If the process is a complex avionics module with $\text{DPU} = 0.40$ defects per module: $\text{Yield} = e^{-0.40} \approx 0.6703 = \mathbf{67.03\%}$

Inverting the Formula: Finding DPU from Observed Yield

Green Belts frequently know the empirical yield and need to determine the underlying DPU. Taking the natural logarithm ($\ln$) of both sides:

$\ln(\text{Yield}) = \ln(e^{-\text{DPU}}) = -\text{DPU}$ $\text{DPU} = -\ln(\text{Yield})$

Example: An audit of 5,000 software code modules reveals that 92.0% pass static security analysis without any vulnerabilities ($\text{Yield} = 0.920$). What is the underlying DPU? $\text{DPU} = -\ln(0.920) = -(-0.0834) = \mathbf{0.0834} \text{ defects per module}$

Combining Multi-Step Processes via DPU

Because exponents add when multiplying like bases ($e^a \times e^b = e^{a+b}$), Rolled Throughput Yield across $n$ steps can be calculated directly by summing step DPUs:

$\text{RTY} = e^{-\text{DPU}_1} \times e^{-\text{DPU}_2} \times \dots \times e^{-\text{DPU}_n} = e^{-\sum_{i=1}^n \text{DPU}_i} = e^{-\text{DPU}_{\text{total}}}$

This mathematical property makes DPU exceptionally powerful: while individual step yields must be multiplied, step defect rates (DPU) can simply be added!

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How Green Belts Leverage RTY in DMAIC Projects

1. Pinpointing Value Stream Bottlenecks: In the Measure and Analyze phases, calculating FTY at each station immediately highlights which process step is driving the degradation of overall RTY.
2. Building the Business Case: Executive sponsors often resist funding Six Sigma projects when final inspection yield appears high (e.g., 97%). Showing that true RTY is 68% demonstrates that 32% of operational capacity is trapped in the hidden factory, immediately justifying capital investment.
3. Validating Real Process Improvements: In the Improve phase, Green Belts track RTY to verify that an improvement at Step 2 did not inadvertently push defects downstream to Step 4.

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Common Exam Traps

• Trap 1: Adding Step Yields Instead of Multiplying. When computing RTY for a 3-step process with 90%, 90%, and 90% yields, never add them $(90 + 90 + 90) / 3 = 90%$. RTY is a compound probability: $0.90 \times 0.90 \times 0.90 = 0.729 = 72.9%$.
• Trap 2: Counting Reworked Units in First Time Yield. If 100 units enter a step, 10 are scrapped, 15 are repaired and passed, and 75 pass cleanly, the FTY is $75 / 100 = 75%$, NOT $90 / 100 = 90%$.
• Trap 3: Linear Subtraction Instead of Poisson Yield. If asked to find yield when $\text{DPU} = 0.15$, do not simply calculate $1 - 0.15 = 0.85$ (85%). The exact Poisson yield is $e^{-0.15} = 0.8607$ (86.07%). Linear subtraction ignores defect clustering.