This page assumes you have seen nothing. We build every symbol the parent Memory bandwidth and latency metrics note leans on, one at a time, each earned before the next.
Figure 1 — one bulb is a bit; a row of eight bulbs is a byte.
Why the topic needs this: bandwidth is measured in bytes per second, so "bytes" is the stuff we are counting. If you do not know a byte is 8 bits, the "64 bits =8 bytes" step in Example 1 looks like magic. It is just 64÷8=8.
Figure 2 — 64 wires carry 8 bytes in one transfer, so W=8; more lanes = more bytes per tick.
Why this symbol? The whole reason a wide bus gives more bandwidth is that more lanes move more bytes per tick. W is the "width" half of "band-width". Look at Figure 2: doubling the lanes doubles the bytes per tick without changing how long any single bit takes to arrive.
Figure 3 — the faster clock (bottom) packs ticks closer, so its period T is shorter.
Why the reciprocal, and why this tool? We use T=1/f because latency is a time but hardware is quoted as a frequency. To turn "16 ticks of waiting" (CAS, see §8) into nanoseconds, we must know how many seconds one tick lasts — that is exactly T. Look at Figure 3: the faster clock packs its ticks closer, so each tick is shorter; that is why the same count of ticks can mean less time.
Why the topic needs this: latency values like 10ns and bandwidths like 51.2GB/s are on wildly different scales; without prefixes the arithmetic in the worked examples is unreadable. Example: a "mega" count of 3200 million transfers per second is 3200×106=3.2×109 transfers/s. (We give the transfers-per-second idea its proper name, MT/s, in §6.)
Why this is the whole bridge: the second form (amount=rate×time) is Little's Law from Example 3: outstanding bytes = bandwidth × latency =B×L. If you understand "distance = speed × time", you already understand it.
Figure 4 — each clock cycle has two edges (up + down); DDR fires one transfer on each, giving 2fclk.
Why this tool, the "edge"? We count edges, not whole cycles, because that is literally when DDR fires a transfer. Look at Figure 4: each full cycle has two edges (up, then down), so a 1600MHz DDR clock delivers 3200MT/s. The name "DDR-3200" is already the doubled number — do not double it again.
Why N multiplies bandwidth but not latency: adding a second highway lets twice as many cars pass per second (bandwidth ×2), but any single car's trip time (latency) is unchanged — it still drives one highway end to end. This is exactly the parent's Forecast answer.
Why we define it here, before the formula: the parent quotes latency as "CL16" or "CL18". Because CL is measured in cycles, and a cycle's length Tclk shrinks as fclk rises, two kits with different CL can still have the same real time. We need CL defined as a cycle-count first, then §9 converts it.
Now every symbol in the parent's formulas is defined. Before we combine them, name the two results we are building toward:
With those two names in hand, re-read the parent's formulas and nothing is unknown:
Here CL (from §8) is how many clock cycles you wait before the first byte — a pure count. Multiply a count of cycles by seconds-per-cycle (Tclk, from §3) and you get seconds, just like the reciprocal rule Tclk=1/fclk.
The figure below shows how each foundation feeds the next, ending in the two headline results.
Figure 5 — prerequisite map: bit → byte → width W, and second → frequency → edges → data rate, all flowing into bandwidth; period + CL flowing into latency.