Foundations — Reliability — MTTF, MTBF, exponential failure model

Before you touch MTTF, MTBF, or the exponential model on the parent note, you need every symbol it silently assumes. This page builds each one from nothing — a plain-words meaning, the picture it stands for, and why the topic can't do without it. Read top to bottom; each block leans on the one above it.

1. Time, and the survivor's clock — $t$

Picture a stopwatch that begins the instant the satellite switches on. Every question in this topic is of the form "at this reading on the stopwatch, what's true?" The stopwatch never runs backwards, so $t$ only increases: $t\ge 0$.

Why the topic needs it: every other quantity — reliability, failure rate, average lifetime — is measured against this clock. Without an agreed zero-point, "5-year survival" is meaningless.

2. Probability — a number between 0 and 1

Look at the bar above: it's a full unit of "certainty" that we slice up. If a coin lands heads with probability $0.5$, half the bar is shaded. A satellite surviving with probability $0.11$ shades just a sliver.

See Probability and Statistics Fundamentals if this feels shaky.

3. Reliability — the shrinking survival number $R(t)$

Combine the two ideas above: $R$ is a probability (Section 2), and it depends on the clock reading $t$ (Section 1). So $R(t)$ is not one number — it's a whole curve, one probability for each moment.

Read the figure left to right:

• At $t=0$ the curve starts at height $1$ — the part definitely works when you switch it on. So $R(0)=1$.
• As $t$ grows, the curve only ever goes down or stays flat, never up. A dead part can't come back to life (until we allow repair, much later). This "never rises" property is called monotonically decreasing.
• The curve stays inside the band $0\le R(t)\le 1$, because it's a probability.

4. Rate of change — the slope $\frac{dR}{dt}$

The parent note writes $\frac{dR}{dt}$. That symbol frightens people, so let's earn it.

In the figure, the dashed straight line just kisses the curve at one point — that's the tangent line, and its steepness is $\frac{dR}{dt}$ there. Since the curve heads downward, this slope is negative.

Why the topic needs a derivative and not just subtraction. We want "failures per hour right now," not "over the whole mission." A slope answers a this-instant question — exactly what an instantaneous rate is. That's why calculus enters: it's the only tool that measures change at a single point.

5. Failure rate — $\lambda$ (lambda)

The parent's definition: $\lambda (t)=-\frac{1}{R(t)}\frac{dR}{dt}$

Decode it piece by piece using what we built:

• $\frac{dR}{dt}$ (Section 4) is how fast reliability drops — but that counts drops as a share of the whole original population.
• Dividing by $R(t)$ rescales it to a share of the survivors so far (only they can still fail).
• The minus sign flips the negative slope into a positive rate.

Constant $\lambda$ — a flat failure rate that never changes — is the special assumption that makes the whole exponential model work. Physically it's the middle of the Bathtub Curve: past infant defects, before wear-out. When $\lambda$ isn't constant, you graduate to the Weibull Distribution.

6. The exponential $e^{-\lambda t}$

The whole model lands on $R(t)=e^{-\lambda t}$. Two symbols to unpack: $e$, and raising it to a power.

The figure shows $e^{-\lambda t}$: starts at $1$, swoops down, and flattens toward (but never reaches) $0$. Compare two curves — a bigger $\lambda$ (steeper, less reliable) vs a smaller $\lambda$ (gentler, more reliable).

The product $\lambda t$ must be a pure number. Units: $\left(\frac{\text{failures}}{\text{hour}}\right)\times (\text{hours})=\text{failures}$, a bare count. You can never raise $e$ to something with units, so this is a sanity check on every calculation.

7. The logarithm $\ln$ — the undo button

Why the topic needs it. Example 2 in the parent asks "what $\lambda$ gives 95% reliability?" The unknown $\lambda$ is trapped inside the exponent of $e$. The only key that opens that lock is $\ln$: $0.95=e^{-\lambda t}\stackrel{\ln }{\to }\ln (0.95)=-\lambda t\Rightarrow \lambda =-\frac{\ln (0.95)}{t}$

Taking $\ln$ of both sides pulls $\lambda$ out into the open where we can solve for it.

8. Expected value $E[T]$ and the integral $\int$

Why both are needed for MTTF. Failure can happen at any real instant, not just whole hours. To average over a continuous smear of possible failure times, plain addition won't do — you need the integral, the continuous version of "sum up." That's why MTTF is written ${\int }_0^{\infty }tf(t)dt$: each possible failure time $t$, weighted by how likely it is, all summed. The parent grinds this out to the clean result $\text{MTTF}=1/\lambda$.

Here $f(t)$ is the probability density — how the failure likelihood is spread across time; it's the slope of the "already-failed" curve, $f(t)=-\frac{dR}{dt}$.

How it all feeds the topic

Each foundation box points into the ones that depend on it, and everything funnels into the reliability topic. From here you're ready for Series vs. Parallel System Reliability, Redundancy Design, and the Poisson Process view of the same maths.

Equipment checklist

Test yourself — cover the right side and answer each before revealing.