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From: nisha kannookadan (nishak44_at_[hidden])
Date: 2008-01-26 06:51:29
Hi
I slowly understand,
the first version isnt that straight forward to me
>> 127^5 * 5 doubles need
>> 2^(log2(127)*5 + log2(5 columns) + log2(8 bytes/double)) bytes = 1.2 Tera
>> Bytes
but the second one, I almost understand
> Or simply:
> ( (127^5)*5)*8 / (1024^3)
> =
> 1230.78 GByte
((127^5)*5)*8 is the size in byte, which is used for all these entries,
but why do we divide by 1024^3 and not by 10^9 to receive the memory in
GB?
I was able to allocate a matrix of size 140000*140000:
matrix a(140000,140000),
how is that possible?
This would need 146 GB, but I acctually have only 32GB RAM.
Thanks
Nisha K
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