[Beowulf] bandwidth: who needs it?

Philippe Blaise philippe.blaise at cea.fr
Fri Oct 22 03:41:58 EDT 2004


The naive formula

T_size = T_0 + size / max_bandwidth

for
size1/2 = T_0 * max_bandwith
gives

T_size1/2 = 2 * T0

which is a characteristic message length : you reach (more or less) half 
the bandwith,
and it takes 2 *  latency seconds to send / recv the message.

For example, with T_0 = 5 usec and max_bandwith = 400 or 800 MB/s you obtain

T_size1/2(5, 400) = 5 * 400 = 2 kB
T_size1/2(5, 800) = 5 * 800 = 4 kB

Phil.

Richard Walsh wrote:

>Greg Lindahl wrote:
>
>  
>
>>>do you have applications that are pushing the limits of MPI bandwidth?
>>>for instance, code that actually comes close to using the 8-900 MB/s
>>>that current high-end interconnect provides?
>>>      
>>>
>>Bandwidth is important not only for huge messages that hit 900 MB/s,
>>but also for medium sized messages. A naive formula for how long it
>>takes to send a message is:
>>
>>T_size = T_0 + size / max_bandwidth
>>
>>For example, for a 4k message with T_0 = 5 usec and either 400 MB/s or
>>800 MB/s,
>>
>>T_4k_400M = 5 + 4k/400M = 5 + 10 = 15 usec
>>T_4k_800M = 5 + 4k/800M = 5 +  5 = 10 usec
>>
>>A big difference. But you're only getting 266 MB/s and 400 MB/s
>>bandwidth, respectively.
>>
>>Of course performance is usually a bit less than this naive model. But
>>the effect is real, becoming unimportant for packets smaller than ~ 2k
>>in this example. The size at which this effect becomes unimportant
>>depends on T_0 and the bandwidth.
>>    
>>
>
>The above also makes a point about a mid-range regime of message sizes 
>whose transfer times are affected ~equally by bandwidth and latency
>changes.  Halving the latency in the 4K/800M case above is equivalent
>to doubling the bandwidth for a message of this size: 
>
>  T_4k_800M. = 2.5 + 4k/800M  = 2.5 +  5.0 =  7.5 usec
>  T_4k_800M  = 5.0 + 4k/800M  = 5.0 +  5.0 = 10.0 usec
>  T_4k_1600M = 5.0 + 4k/1600M = 5.0 +  2.5 =  7.5 usec
>
>For a given interconnect with a known latency and bandwidth there is
>a "characteristic" message size whose transfer time is equally sensitive
>to perturbations in bandwidth and latency (latency and bandwidth piece
>of the transfer time are equal).  So, for an "Elan-4-like" interconnect 
>characteristic message length would be 1.6k:
>
>  T_4k_800M   = 1.0 + 1.6k/800M   = 1.0 + 2.0 = 3.0 usec
>  T_4k_800M   = 2.0 + 1.6k/800M   = 2.0 + 2.0 = 4.0 usec
>  T_4k_1600M  = 2.0 + 1.6k/1600M  = 2.0 + 1.0 = 3.0  usec
>
>Messages sizes in the vicinity of the characteristic length will 
>respond approximately equally to improvements in either factor.
>Messages much larger in size will be more sensitive to bandwidth 
>improvements in an interconnect upgrade while message sizes much
>smaller will be more sensitive to latency improvements in an upgrade. 
>
>One might argue that bandwidth actually matters more because message
>sizes (along with problem sizes) can in theory grow indefinitely (drop 
>in some more memory and double you array sizes) while they can be made 
>only be so small -- this is a position supported by the rate of storage 
>growth, but undermined by slower bandwidth growth and processor count 
>increases.  
>
>I think I will keep my bandwidth though ... and take any off of the 
>hands of those who ... don't need it ... ;-) ...
>
>rbw
>
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