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