Consider a stipulated synchronous <shared computer bus>, rather than a timing claim about any named hardware protocol. Assume a 32-<bit> <data bus>, a 100 MHz clock, and one four-<byte> payload transfer per <bus cycle> during the data phase. Both initiators $A$ and $B$ request at time zero. $A$ has one four-transfer write <bus burst> and $B$ one two-transfer write <bus burst>, to an immediately ready target in <computer memory>. Each <bus burst> pays one <bus cycle> for <bus arbitration> and one for its starting address and command; subsequent addresses advance implicitly. The arbiter chooses $A$ first, owners finish their entire <bus bursts>, and no phases overlap. There are no <bus wait states>, retries, extra completion cycles, direction changes or intervening requests. Payload is counted when each data phase completes; the observation interval starts at the first <bus arbitration> cycle and ends at the last payload completion.

\Image[/computer-bus-arbitration-payload-timing.png]
{title=Stipulated <bus arbitration> and payload timing for two competing <bus bursts>}

A <bus cycle> lasts
$$
T=\frac{1}{100\times10^6\ \mathrm{s}^{-1}}=10\ \mathrm{ns}.
$$
The timeline uses half-open intervals of <bus cycles>: $A$ has <bus arbitration> on $[0,1)$, address and command on $[1,2)$, and four payload transfers on $[2,6)$. Then $B$ has <bus arbitration> on $[6,7)$, address and command on $[7,8)$, and two payload transfers on $[8,10)$. Their final payload completions are at 60 ns and 100 ns respectively. $B$ waits 60 ns before its own selection cycle even though both requests arrived together.

The six payload transfers contain 24 <bytes>. Ten <bus cycles> take 100 ns, giving finite-interval <bus throughput>
$$
R=\frac{24\ \mathrm{bytes}}{100\ \mathrm{ns}}=240\times10^6\ \mathrm{bytes/s}=240\ \mathrm{MB/s},
$$
where MB means $10^6$ <bytes>. The raw <data bus> rate is $4\ \mathrm{bytes}\times100\times10^6\ \mathrm{s}^{-1}=400\ \mathrm{MB/s}$. Payload occupies $6/10=60\%$ of the interval. If this same pair repeats with no gaps, its long-run <bus throughput> equals the finite-interval value; isolated observations need not predict a later workload.

Reverse the <bus arbitration> order, keeping these assumptions. $B$ finishes after four <bus cycles>, at 40 ns; $A$ finishes at 100 ns. Aggregate <bus throughput> stays 240 MB/s, but $B$'s request-to-completion time falls from 100 ns to 40 ns and $A$'s rises from 60 ns to 100 ns. Scheduling changes who waits without changing the available payload capacity in this example.

For an independent perturbation, retain $A$ first and add two <bus wait states> before $B$'s final payload transfer. The interval grows to 120 ns with the same 24 <bytes>, so <bus throughput> falls to 200 MB/s. For another workload, one initiator sends all six transfers in a single <bus burst>: the stipulated overhead is paid once, giving eight <bus cycles>, 80 ns and 300 MB/s. This latter comparison assumes one combined request is valid; combining different initiators' operations is not automatically permitted.


 Shared-bus timing example (source code)