Reliability — MTTF, MTBF, exponential failure model: Exercises

This page is a self-test ladder for the parent topic. Work each problem with pen and paper first, then open the solution. The difficulty climbs from L1 Recognition (can you name the tool?) up to L5 Mastery (can you invent the approach?).

Throughout we use only tools built on the parent note:

Reliability $R (t) = e^{- λ t}$ — the probability a thing is still alive at time $t$ .
Failure rate $λ$ — failures per unit time (constant, useful-life region).
MTTF $= 1/ λ$ — average time to the first failure.
For time conversions we use $1 year = 365.25 \times 24 = 8766$ hours.

Before we begin, one picture to anchor every calculation on this page.

Figure — Reliability — MTTF, MTBF, exponential failure model

The curve above is $R (t) = e^{- λ t}$ . Notice the amber marker: at $t = MTTF$ the survival probability is only $e^{- 1} \approx 0.37$ , not one-half. Keep this shape in your head — most exercises are just "read a value off this curve" or "invert it."

Level 1 — Recognition

Goal: pick the right formula and plug in. No traps beyond units.

Exercise 1.1

A component has a constant failure rate $λ = 0.0002$ failures/hour. What is its reliability after $t = 2000$ hours?

Recall Solution 1.1

What tool? Constant failure rate $\Rightarrow$ exponential model $R (t) = e^{- λ t}$ .

Why this and not something else? "Constant failure rate" is the exact assumption that produces the exponential curve (parent note derivation). No wear-out, no infant mortality — just the flat middle of the bathtub curve.

$λ t = 0.0002 \times 2000 = 0.4$ $R = e^{- 0.4} = 0.6703$

Answer: $R \approx 0.670$ , i.e. a 67.0% chance of surviving 2000 hours.

Exercise 1.2

The same component: what is its MTTF, in hours?

Recall Solution 1.2

What tool? For the exponential model, $MTTF = 1/ λ$ .

$MTTF = \frac{1}{0.0002} = 5000 hours$

Answer: 5000 hours. Note $t = 2000$ h in Ex 1.1 is well below MTTF, which is why $R$ was comfortably above $0.37$ .

Level 2 — Application

Goal: multi-step plug-and-chug, including unit conversion and inverting the formula.

Exercise 2.1

A star tracker has $λ = 0.00003$ failures/hour. Find its reliability over a 3-year mission.

Recall Solution 2.1

Convert time: $t = 3 \times 8766 = 26, 298$ hours.

$λ t = 0.00003 \times 26, 298 = 0.78894$ $R = e^{- 0.78894} = 0.4544$

Answer: $R \approx 0.454$ , about 45.4%. Below one-half already — a single star tracker is a poor bet for 3 years.

Exercise 2.2

A battery must reach $R = 0.90$ over $t = 10, 000$ hours. What maximum $λ$ is allowed?

Recall Solution 2.2

What tool? We know $R$ and $t$ , we want $λ$ — so we must invert the exponential. The inverse of "raise $e$ to a power" is the natural log $ln$ .

Start from $R = e^{- λ t}$ , take $ln$ of both sides: $ln R = - λ t \Rightarrow λ = - \frac{l n R}{t}$

$λ = - \frac{l n 0.90}{10 , 000} = - \frac{- 0.105361}{10 , 000} = 1.0536 \times 1 0^{- 5}$

Answer: $λ \leq 1.05 \times 1 0^{- 5}$ failures/hour (equivalently $MTTF \geq 94, 912$ h).

Level 3 — Analysis

Goal: reason about what a number means, and compare scenarios.

Exercise 3.1

A gyroscope has $MTTF = 20, 000$ hours. A colleague claims "so it's basically guaranteed to reach 15,000 hours." Evaluate this claim by computing $R (15, 000)$ .

Recall Solution 3.1

First recover $λ$ : $λ = 1/ MTTF = 1/20, 000 = 5 \times 1 0^{- 5}$ /h.

$λ t = 5 \times 1 0^{- 5} \times 15, 000 = 0.75$ $R = e^{- 0.75} = 0.4724$

Interpretation: only 47.2% survive to 15,000 h. The claim is false — it's essentially a coin flip. MTTF is an average, and the exponential distribution has a long tail, so a large fraction of parts die well before the mean.

Exercise 3.2

Show numerically that for any exponential component, exactly the same fraction survives to its own MTTF, regardless of $λ$ . Compute that fraction.

Recall Solution 3.2

What we did: evaluate $R$ at $t = MTTF = 1/ λ$ .

$R (MTTF) = e^{- λ \cdot (1/ λ)} = e^{- 1} = 0.3679$

The $λ$ cancels — the answer is a universal constant. Only 36.8% of any exponential-model population survives to its MTTF; 63.2% have already failed. This is the amber marker in the figure at the top of the page.

Level 4 — Synthesis

Goal: combine reliability of several parts — series and parallel architectures. See Series vs. Parallel System Reliability and Redundancy Design.

Exercise 4.1

Three subsystems in series (all must work) have reliabilities $R_{1} = 0.95$ , $R_{2} = 0.90$ , $R_{3} = 0.98$ over the mission. Find system reliability.

Recall Solution 4.1

What tool? Series = a chain; it survives only if every link survives. For independent parts, "AND" of survivals is a product of probabilities.

$R_{sys} = R_{1} R_{2} R_{3} = 0.95 \times 0.90 \times 0.98 = 0.8379$

Answer: 83.8%. Note it is lower than the weakest single part (0.90) — chains only get less reliable as you add links.

Exercise 4.2

Two identical units in parallel (system works if at least one works), each with $λ = 0.00008$ /h, over $t = 8000$ h. Find system reliability.

Recall Solution 4.2

Single-unit reliability: $R_{1} = e^{- 0.00008 \times 8000} = e^{- 0.64} = 0.5273$

Parallel logic: the system fails only if both fail. Probability one unit fails is $1 - R_{1}$ ; both failing (independent) is $(1 - R_{1})^{2}$ . $R_{sys} = 1 - (1 - R_{1})^{2} = 1 - (0.4727)^{2} = 1 - 0.2234 = 0.7766$

Answer: 77.7% — far above the 52.7% of one unit. Redundancy buys survival.

Exercise 4.3

A subsystem needs $R \geq 0.99$ over a mission where a single unit gives only $R_{1} = 0.70$ . How many identical units in parallel are required?

Recall Solution 4.3

Set up the parallel formula for $n$ units: $R_{sys} = 1 - (1 - R_{1})^{n} \geq 0.99$ $(1 - 0.70)^{n} \leq 0.01 \Rightarrow 0.3 0^{n} \leq 0.01$

Solve for $n$ by taking logs (log is the tool that pulls $n$ down from the exponent): $n ln 0.30 \leq ln 0.01 \Rightarrow n \geq \frac{l n 0.01}{l n 0.30} = \frac{- 4.60517}{- 1.20397} = 3.825$

(The inequality flips because $ln 0.30 < 0$ .) $n$ must be an integer, so round up.

Answer: $n = 4$ units. Check: $1 - 0.3 0^{4} = 1 - 0.0081 = 0.9919 \geq 0.99$ . ✓ (Three units give only $1 - 0.3 0^{3} = 0.973$ , not enough.)

Level 5 — Mastery

Goal: no template fits directly; you must assemble the pieces.

Exercise 5.1

A spacecraft power chain is: one solar array ( $λ_{A} = 3 \times 1 0^{- 5}$ /h) in series with a parallel pair of batteries (each $λ_{B} = 1 \times 1 0^{- 4}$ /h). Find the whole chain's reliability over a 2-year mission.

Recall Solution 5.1

Step 1 — convert time. $t = 2 \times 8766 = 17, 532$ h.

Step 2 — array reliability. $R_{A} = e^{- 3 \times 1 0^{- 5} \times 17, 532} = e^{- 0.52596} = 0.5910$

Step 3 — one battery. $R_{B} = e^{- 1 \times 1 0^{- 4} \times 17, 532} = e^{- 1.7532} = 0.1732$

Step 4 — battery pair in parallel (survives if at least one battery lives): $R_{batt} = 1 - (1 - R_{B})^{2} = 1 - (0.8268)^{2} = 1 - 0.6836 = 0.3164$

Step 5 — array in series with the battery block (both blocks must work): $R_{sys} = R_{A} \times R_{batt} = 0.5910 \times 0.3164 = 0.1870$

Answer: 18.7% — poor. This is exactly the kind of number that drives an engineer to add more battery redundancy or a higher-grade array. See Mission Design Constraints.

Exercise 5.2

For a single exponential component, what mission duration $t$ (as a fraction of MTTF) gives exactly $R = 0.99$ ? Express the answer as a multiple of MTTF.

Recall Solution 5.2

Set up in terms of MTTF. Write $λ = 1/ MTTF$ , so $λ t = t / MTTF$ . Let $x = t / MTTF$ . $R = e^{- x} = 0.99 \Rightarrow x = - ln 0.99 = 0.010050$

Answer: $t \approx 0.01005 MTTF$ . To hit 99% reliability you may only use about 1% of the MTTF. This quantifies the L3 warning: high reliability demands operating far below the mean life.

Exercise 5.3

A repairable subsystem has $MTTF = 800$ h and $MTTR = 4$ h. Find (a) its MTBF and (b) its steady-state availability $A = \frac{MTTF}{MTTF + MTTR}$ . See Availability vs. Reliability.

Recall Solution 5.3

(a) MTBF. For a repairable system, $MTBF = MTTF + MTTR = 800 + 4 = 804$ h.

(b) Availability. $A = \frac{800}{800 + 4} = \frac{800}{804} = 0.99502$

Answer: MTBF $= 804$ h and availability $\approx 99.50%$ — the system is "up" about $99.5%$ of the time. Notice reliability (does it survive uninterrupted?) and availability (what fraction of time is it usable?) are different questions; availability can stay high even with frequent short repairs.

Related deep-dives: Bathtub Curve · Weibull Distribution · Poisson Process · Fault Tree Analysis · Safety-Critical Systems · Probability and Statistics Fundamentals.