Wear leveling and flash controllers: Worked examples

This page is the exercise machine for the parent topic. The parent gave you the two engines that drive every number here:

Recall The two formulas you will use on this page

The write amplification factor measures how much extra writing garbage collection forces: $WAF = \frac{data physically written to flash}{data the host asked to write} = \frac{1}{1 - u}$ where $u$ is the average fraction of still-valid pages in the blocks the controller erases. The lifetime of a drive is total endurance divided by daily physical wear: $lifetime = \frac{( capacity ) \times ( P/E rating )}{( host writes/day ) \times WAF}$

Every symbol above was earned in the parent note. If any feels unfamiliar, re-read it before continuing — this page assumes nothing new except arithmetic.

The scenario matrix

The whole topic, viewed as "what can the exam throw at me," lives inside a small grid. Each cell is a regime of the input $u$ (the valid fraction) or a different quantity being solved for. We will hit every cell.

Cell	What makes it special	Covered by
A. $u \to 0$ (mostly stale)	Best case, $WAF \to 1$	Ex 1
B. mid $u$ (typical drive)	Normal working point	Ex 2
C. $u \to 1$ (mostly valid)	Danger zone, $WAF \to \infty$	Ex 3
D. degenerate $u = 1$	Formula divides by zero — what does it mean?	Ex 4
E. solve backwards	Given WAF, find $u$	Ex 5
F. lifetime word problem	Combine WAF + endurance + daily load	Ex 6
G. over-provisioning effect	How free space lowers $u$ and buys years	Ex 7
H. dynamic-only trap	Uneven wear across hot/cold split, no formula plug	Ex 8
I. exam twist: WAF < 1	Compression breaks the "amplification only adds" instinct	Ex 9

Figure — Wear leveling and flash controllers

Cells A, B, C — the three regimes of $u$

Cell D — the degenerate input $u = 1$

Worked example Example 4 — Cell D: what happens at exactly 100% valid?

Suppose the only candidate block for GC is 100% valid ( $u = 1$ ). What does the formula say, and what does it mean physically?

Forecast: The formula has $1 - u$ in the denominator. What is $1/0$ ?

Step 1. $1 - u = 1 - 1 = 0$ . Why this step? Before blindly dividing, check the denominator — this is the "look for zero inputs" habit.

Step 2. $WAF = \frac{1}{0}$ is undefined (it diverges to $+ \infty$ ). Why this step? Division by zero has no numeric value; mathematically WAF blows up.

Step 3. Read the physics. A fully-valid block yields zero free pages after copying all $N$ of them elsewhere — so GC does work but reclaims nothing. The controller must never pick a 100%-valid block; it would be pure loss. Why this step? The undefined answer is telling us the operation is forbidden, not that the drive explodes.

Verify: Take the limit from below: $u = 0.99 \Rightarrow WAF = 100$ ; $u = 0.999 \Rightarrow WAF = 1000$ . As $u \to 1^{-}$ the value grows without bound — consistent with "undefined at exactly 1." ✓

Cell E — solving backwards (given WAF, find $u$ )

Cell F — the full lifetime word problem

Worked example Example 6 — Cell F: how long will this SSD live?

A 512 GB drive uses TLC cells rated 1500 P/E cycles. The host writes 80 GB/day. GC runs at $u = 0.60$ . Estimate the lifetime in years.

Forecast: Big drive, moderate write load — guess: a few years?

Step 1 — WAF. $WAF = \frac{1}{1 - 0.60} = \frac{1}{0.40} = 2.5$ . Why this step? We need to convert host writes into real flash writes; WAF is that conversion factor.

Step 2 — Total endurance. $512 GB \times 1500 = 768, 000 GB$ of physical writes possible. Why this step? Every GB of cell can be programmed 1500 times before wearing out.

Step 3 — Daily physical wear. $80 GB/day \times 2.5 = 200 GB/day$ actually written. Why this step? The host's 80 GB becomes 200 GB after amplification.

Step 4 — Divide. $\frac{768 , 000}{200} = 3840 days = \frac{3840}{365} \approx 10.5 years$ . Why this step? Total endurance ÷ daily consumption = number of days; divide by 365 for years.

Verify (units): $\frac{GB}{GB/day} = days$ . ✓ Numeric: $200 \times 3840 = 768, 000$ . ✓ About a decade — sensible for a lightly-loaded consumer drive.

Cell G — the payoff of over-provisioning

Worked example Example 7 — Cell G: buy years by adding spare space

Take the drive from Example 6 ( $u = 0.60$ , lifetime 10.5 yr). Adding over-provisioning lets GC wait for emptier blocks, dropping $u$ to 0.40. By what factor does lifetime improve, and what is the new lifetime?

Forecast: Lower $u$ means lower WAF means longer life — but how much?

Step 1 — New WAF. $WAF_{new} = \frac{1}{1 - 0.40} = \frac{1}{0.60} = 1.667$ . Why this step? Emptier victim blocks copy less valid data.

Step 2 — Lifetime scales as $1/ WAF$ . Improvement $= \frac{WAF _{old}}{WAF _{new}} = \frac{2.5}{1.667} = 1.5 \times$ . Why this step? Endurance and daily host writes are unchanged, so only the WAF ratio matters.

Step 3 — New lifetime. $10.5 yr \times 1.5 = 15.75 years$ . Why this step? Multiply the old lifetime by the improvement factor.

Verify: Recompute from scratch: daily physical $= 80 \times 1.667 = 133.3$ GB; $768, 000/133.3 = 5760$ days $= 15.78$ yr. ✓ Rounding-consistent with $15.75$ .

Cell H — the dynamic-only trap (no single plug-in)

Worked example Example 8 — Cell H: uneven wear when static leveling is off

A 100-block drive holds a movie on 80 cold blocks (never rewritten) and cycles all writes through the 20 free blocks. The host performs enough writes to erase 10{,}000 blocks-worth of data over the drive's life. With dynamic-only leveling, how many erases land on each hot block, and is that survivable at a 3000 P/E rating?

Forecast: 10{,}000 erases spread over only 20 blocks… that sounds like a lot each.

Step 1 — Blocks actually sharing the load. Only the $20$ free blocks rotate; the $80$ cold ones are frozen. Why this step? Dynamic leveling only touches blocks it is actively writing — the parent's exact definition.

Step 2 — Erases per hot block. $\frac{10 , 000}{20} = 500$ erases each. Why this step? Total erase work divided evenly over the blocks that share it.

Step 3 — Compare to rating. $500 < 3000$ ? Yes — but check the ideal: if all $100$ blocks shared, each would see $\frac{10 , 000}{100} = 100$ erases. Dynamic-only makes hot blocks age $\frac{500}{100} = 5 \times$ faster. Why this step? The danger is relative — static leveling would give every block the same 100.

Step 4 — Push to failure. If the workload were $5 \times$ heavier ( $50, 000$ erases), hot blocks hit $50, 000/20 = 2500$ while cold blocks sit at $0$ — the drive is $2500/3000 = 83%$ dead on 20 blocks with 80 blocks pristine. Why this step? This is the exact "movie you never touch" failure the parent warned about, now with numbers.

Verify: Total erases conserved: $20 \times 500 = 10, 000$ ✓ (dynamic) and $100 \times 100 = 10, 000$ ✓ (static-ideal). Same total work, wildly different peak wear — that is why static leveling exists.

Cell I — the exam twist: WAF below 1

Worked example Example 9 — Cell I: can a drive write

less than the host asked? A controller with transparent compression takes a 1000 MB stream that compresses to 400 MB, then runs GC at $u = 0.5$ on the compressed data. What is the effective WAF seen from the host?

Forecast: GC alone gives $1/ (1 - 0.5) = 2$ . But compression shrank the data first — could the two effects cancel, or even go below 1?

Step 1 — Physical writes from GC on compressed data. Compressed size $= 400$ MB; GC multiplies by $\frac{1}{1 - 0.5} = 2$ , giving $400 \times 2 = 800$ MB physically written. Why this step? Amplification acts on the stored (compressed) bytes, not the original stream.

Step 2 — Effective WAF vs the host. $\frac{physical}{host requested} = \frac{800}{1000} = 0.8$ . Why this step? WAF is always physical-flash-bytes over host-requested-bytes; the host asked for 1000 MB.

Step 3 — Interpret. $0.8 < 1$ : the drive wrote less to flash than the host handed it. The GC formula's floor of $1$ is not violated — compression is a separate mechanism layered on top, exactly as the parent's mistake box explained. Why this step? Reconciles the " $WAF \geq 1$ from GC" rule with the sub-1 result.

Verify: Without compression, host $= 1000$ MB, physical $= 1000 \times 2 = 2000$ , WAF $= 2$ . Compression cut physical to $800$ , dividing WAF by $2000/800 = 2.5$ : $2/2.5 = 0.8$ . ✓

Recall Self-check questions

WAF when victim blocks are 80% valid ::: $1/ (1 - 0.8) = 5$ Physical meaning of $u = 1$ giving $WAF = \infty$ ::: GC copies every page and frees none — a forbidden, useless erase Given $WAF = 2$ , what is $u$ ? ::: $u = 1 - 1/2 = 0.5$ Why does dynamic-only leveling let hot blocks die 5× early in Ex 8? ::: writes rotate over only 20 of 100 blocks; static leveling would share all 100 How can end-to-end WAF be 0.8 despite GC? ::: transparent compression shrinks data before GC, lowering physical bytes below host bytes