提交 9eb2d627 编写于 作者: S Stephen Hemminger 提交者: David S. Miller

[TCP] cubic: use Newton-Raphson

Replace cube root algorithim with a faster version using Newton-Raphson.
Surprisingly, doing the scaled div64_64 is faster than a true 64 bit
division on 64 bit CPU's.
Signed-off-by: NStephen Hemminger <shemminger@osdl.org>
Signed-off-by: NDavid S. Miller <davem@davemloft.net>
上级 89b3d9aa
......@@ -52,6 +52,7 @@ MODULE_PARM_DESC(bic_scale, "scale (scaled by 1024) value for bic function (bic_
module_param(tcp_friendliness, int, 0644);
MODULE_PARM_DESC(tcp_friendliness, "turn on/off tcp friendliness");
#include <asm/div64.h>
/* BIC TCP Parameters */
struct bictcp {
......@@ -93,67 +94,51 @@ static void bictcp_init(struct sock *sk)
tcp_sk(sk)->snd_ssthresh = initial_ssthresh;
}
/* 65536 times the cubic root */
static const u64 cubic_table[8]
= {0, 65536, 82570, 94519, 104030, 112063, 119087, 125367};
/* 64bit divisor, dividend and result. dynamic precision */
static inline u_int64_t div64_64(u_int64_t dividend, u_int64_t divisor)
{
u_int32_t d = divisor;
if (divisor > 0xffffffffULL) {
unsigned int shift = fls(divisor >> 32);
d = divisor >> shift;
dividend >>= shift;
}
/* avoid 64 bit division if possible */
if (dividend >> 32)
do_div(dividend, d);
else
dividend = (uint32_t) dividend / d;
return dividend;
}
/*
* calculate the cubic root of x
* the basic idea is that x can be expressed as i*8^j
* so cubic_root(x) = cubic_root(i)*2^j
* in the following code, x is i, and y is 2^j
* because of integer calculation, there are errors in calculation
* so finally use binary search to find out the exact solution
* calculate the cubic root of x using Newton-Raphson
*/
static u32 cubic_root(u64 x)
static u32 cubic_root(u64 a)
{
u64 y, app, target, start, end, mid, start_diff, end_diff;
if (x == 0)
return 0;
u32 x, x1;
target = x;
/* first estimate lower and upper bound */
y = 1;
while (x >= 8){
x = (x >> 3);
y = (y << 1);
}
start = (y*cubic_table[x])>>16;
if (x==7)
end = (y<<1);
else
end = (y*cubic_table[x+1]+65535)>>16;
/* binary search for more accurate one */
while (start < end-1) {
mid = (start+end) >> 1;
app = mid*mid*mid;
if (app < target)
start = mid;
else if (app > target)
end = mid;
else
return mid;
}
/* Initial estimate is based on:
* cbrt(x) = exp(log(x) / 3)
*/
x = 1u << (fls64(a)/3);
/* find the most accurate one from start and end */
app = start*start*start;
if (app < target)
start_diff = target - app;
else
start_diff = app - target;
app = end*end*end;
if (app < target)
end_diff = target - app;
else
end_diff = app - target;
/*
* Iteration based on:
* 2
* x = ( 2 * x + a / x ) / 3
* k+1 k k
*/
do {
x1 = x;
x = (2 * x + (uint32_t) div64_64(a, x*x)) / 3;
} while (abs(x1 - x) > 1);
if (start_diff < end_diff)
return (u32)start;
else
return (u32)end;
return x;
}
/*
......
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