Parallel Performance Theory
CS 441 Lecture, Dr. Lawlor
Network Performance
Most network cards require some (fairly large) fixed amount of time per message, plus some (smaller) amount of time per byte of the message:
tnet = a + b * L;
Where
- tnet: network time. The total time, in seconds, spent on the network.
- a: network latency. The time, in seconds, to send a
zero-length message. For MPICH running on gigabit Ethernet, this
is something like 50us/message, which is absurd. Fancier, more
expensive networks such as Myrinet or Infiniband have latencies as low as 5us/message (in 2005; Wikipedia now claims 1.3us/message). (Network latency is often written as alpha or α).
- b: 1/(network bandwidth). The time, in seconds, to send
each byte of the message. For 100MB/s gigabit ethernet (1000Mb/s
is megabits), this is 10ns/byte. For 4x Infiniband, which sends 1000MB/s, this is 1ns/byte. (Network 1/bandwidth is often written as beta or β).
- L: number of bytes in message.
The bottom line is that shorter messages are always faster. *But*
due to the per-message overhead, they may not be *appreciably*
faster. For example, for Ethernet a zero-length message takes
50us. A 100-byte message takes 51us
(50us/message+100*10ns/byte). So you might as well send 100 bytes
if you're going to bother sending anything!
In general, you'll hit 50% of the network's achievable bandwidth when sending messages so
a = b * L
or
L = a / b
For Ethernet, this breakeven point is at L = 5000 bytes/message, which
is amazingly long! Shorter messages are "alpha dominated", which
means you're waiting for the network latency (speed of light,
acknowledgements, OS overhead). Longer messages are "beta
dominated", which means you're waiting for the network bandwidth
(signalling rate). Smaller messages don't help much if you're
alpha dominated!
The large per-message cost of most networks means many applications cannot
get good performance with an "obvious" parallel strategy, like sending
individual floats when they're needed. Instead, you have to
figure out beforehand what data needs to be sent, and send everything at once.
Parallel CPU Performance
OK, that's the network's performance. What about the CPU? A typical model is:
tcpu = k * n / p
Where
- tcpu is the total time spent computing, in seconds.
- k is the (average) time spent per element. For example,
Mandelbrot set rendering takes about 300ns/pixel. Some folks
break k down into the (average) number of floating-point operations per
element (flops/element), multiplied by the time taken per floating-point operation (seconds/flop, typically around 1ns (CPU) to 0.1ns (GPU)).
- n is the number of elements in the computation. For example, a 1000x1000 image has 1m pixels.
- p is the number of processors. Dividing the total work by p
implicitly assumes the problem's load is (more or less) evenly balanced!
Parallel Problem Performance
Say we're doing an n=1mpixel Mandelbrot rendering in parallel on p=10 processors. Step one is to render all the pixels:
tcpu = k * n / p
tcpu = 300 ns/pixel * 1m pixels / 10 processors = 30ms / processor
After rendering the pixels, we've got to send them to one place for
output. So each processor sends a message of size L = (3
bytes/pixel * n / p) = 300Kbytes/processor, and thus:
tnet = a + b * L
tnet = a + b * (3 bytes/pixel * 1m bytes / 10 processors) = 3.05 ms/processor
(Careful: this assumes all the messages sent can be received at one place in parallel, which often isn't the case!)
Hence in this case we spend almost 10 times longer computing the data as we do sending it, which is pretty good.
Unlike the network, the CPU spends its time actually solving the
problem. So CPU time is usually seen as "good" useful work, while
network time is "bad" parallel overhead. So more generally, you'd
like to make sure the "computation to communication ratio", tcpu/tnet
is as high as possible. For the mandelbrot problem:
tcpu / tnet = k * n / p / ( a + b * n / p * 3)
Or, simplifying:
tcpu / tnet = k / ( a * p / n + b * 3)
Remember, a higher CPU/net ratio means more time spent computing, and
hence higher efficiency. So we can immediately see that:
- As the amount of work k per element grows, the efficiency goes
up. (This makes sense: more time spent computing is "better", at least from the point of view of CPU/net ratio!)
- As the network's per-message cost a grows, efficiency goes down. (Slower network is worse)
- As you add processors to a fixed-size problem, the per-message cost goes up, and efficiency goes down. (More processors means more messages.)
- As the problem size n grows, the per-message cost gets less
important, and efficiency goes up. (Messages get longer with
bigger problems, which uses the network more efficiently.)
- As the network's time per byte b goes down, efficiency goes up. (Faster network is better)
Or, to spend more time computing *relative* to the time spent communicating:
- Get a faster network.
- Find a bigger problem.
- Do *more* work per element (or get a *slower* CPU!).
- Use *fewer* processors.
Of course, what users really care about is the *total* time spent, tcpu + tnet:
tcpu + tnet = k * n / p + a + b * n / p * 3
To spend less time overall:
- Get a faster network.
- Find a *smaller* problem.
- Do *less* work per element (or get a *faster* CPU).
- Use more processors.
Note that even if you do all these things, you can *never* solve the
mandelbrot problem in parallel in less than the per-message time
a. For Ethernet, this means parallel problems are always going to
take 50us. In particular, if the problem can be solved serially
in less than 50us, going parallel can't possibly help! Luckily,
it's rare you care about a problem that takes less than 50us
total. And if that sub-50us computation is just a part of a
bigger problem, you can parallelize the *bigger* problem.
In general, the per-message cost means you want to parallelize the *coarsest* pieces of the computation you can!