Information Technology Reference
In-Depth Information
Tabl e 5. 6
The one-way communication time (in seconds) for sending a vector containing n
double-precision numbe
rs on a fast Ethernet network
n (Length of the
T
n
(one-way communi-
vector y)
cation time)
65536
0.048755
131072
0.097074
262144
0.194003
524288
0.386721
1048576
0.771487
only need to do time measurement on one computer, say A. In the table below we
have listed values of n and the communication time T
n
.
(a) Construct a linear least squares model of the form
T.n/
D
˛
C
ˇn:
(b) Try to explain why T.0/
D
˛
¤
0.
(c) In computer terms, ˛ is referred to as the
latency
of the communication network,
and 1=ˇ can be regarded as the practical
bandwidth
, which should be somewhat
lower than the peak-performance bandwidth. Normally, bandwidth is measured
in gigabits (10
9
) per second. Use the fact that one double-precision number has
64 bits and report the bandwidth of the network that has been used for producing
the measurements given in Table
5.6
.
˘
Exercise 5.5.
In the text above we used the fact that
Z
b
.˛
p.t//
2
dt
D
Z
b
a
d
d˛
@
@˛
.˛
p.t//
2
dt
:
(5.141)
a
We will indicate a proof of this. But let us first note that for sufficiently smooth
functions
f
g
"
.t /
g
,wehavethat
Z
b
g
"
.t /
dt
D
Z
b
a
lim
"!0
lim
"!0
g
"
.t /
dt
;
(5.142)
a
see e.g. Royden [25]. Next we consider the problem of computing
Z
b
d
d˛
F.˛;t/
dt
(5.143)
a
for a smooth function F . By definition, we have
"
Z
b
F.˛;t/
dt
#
:
Z
b
F.˛
C
"; t /
dt
Z
b
a
d
d˛
1
"
F.˛;t/
dt
D
lim
"!0
(5.144)
a
a
