私は長期的なシミュレーションを行っており、ODE 系の解法で可能な限り高い精度を達成しようとしています。4 倍 (128 ビット) 精度の計算と 2 倍 (64 ビット) 精度の計算にかかる時間を調べようとしています。私は少しググったところ、それについていくつかの意見を見ました.4倍長くかかると言う人もいれば、60〜70倍かかると言う人もいます.
program QUAD_TEST
implicit none
integer,parameter :: dp = selected_int_kind(15)
integer,parameter :: qp = selected_int_kind(33)
integer :: cstart_dp,cend_dp,cstart_qp,cend_qp,crate
real :: time_dp,time_qp
real(dp) :: sum_dp,sqrt_dp,pi_dp,mone_dp,zero_dp
real(qp) :: sum_qp,sqrt_qp,pi_qp,mone_qp,zero_qp
integer :: i
! ==============================================================================
! == TEST 1. ELEMENTARY OPERATIONS ==
sum_dp = 1._dp
sum_qp = 1._qp
call SYSTEM_CLOCK(count_rate=crate)
write(*,*) 'Testing elementary operations...'
call SYSTEM_CLOCK(count=cstart_dp)
do i=1,50000000
sum_dp = sum_dp - 1._dp
sum_dp = sum_dp + 1._dp
sum_dp = sum_dp*2._dp
sum_dp = sum_dp/2._dp
end do
call SYSTEM_CLOCK(count=cend_dp)
time_dp = real(cend_dp - cstart_dp)/real(crate)
write(*,*) 'DP sum: ',sum_dp
write(*,*) 'DP time: ',time_dp,' seconds'
call SYSTEM_CLOCK(count=cstart_qp)
do i=1,50000000
sum_qp = sum_qp - 1._qp
sum_qp = sum_qp + 1._qp
sum_qp = sum_qp*2._qp
sum_qp = sum_qp/2._qp
end do
call SYSTEM_CLOCK(count=cend_qp)
time_qp = real(cend_qp - cstart_qp)/real(crate)
write(*,*) 'QP sum: ',sum_qp
write(*,*) 'QP time: ',time_qp,' seconds'
write(*,*)
write(*,*) 'DP is ',time_qp/time_dp,' times faster.'
write(*,*)
! == TEST 2. SQUARE ROOT ==
sqrt_dp = 2._dp
sqrt_qp = 2._qp
write(*,*) 'Testing square root ...'
call SYSTEM_CLOCK(count=cstart_dp)
do i = 1,10000000
sqrt_dp = sqrt(sqrt_dp)
sqrt_dp = 2._dp
end do
call SYSTEM_CLOCK(count=cend_dp)
time_dp = real(cend_dp - cstart_dp)/real(crate)
write(*,*) 'DP sqrt: ',sqrt_dp
write(*,*) 'DP time: ',time_dp,' seconds'
call SYSTEM_CLOCK(count=cstart_qp)
do i = 1,10000000
sqrt_qp = sqrt(sqrt_qp)
sqrt_qp = 2._qp
end do
call SYSTEM_CLOCK(count=cend_qp)
time_qp = real(cend_qp - cstart_qp)/real(crate)
write(*,*) 'QP sqrt: ',sqrt_qp
write(*,*) 'QP time: ',time_qp,' seconds'
write(*,*)
write(*,*) 'DP is ',time_qp/time_dp,' times faster.'
write(*,*)
! == TEST 3. TRIGONOMETRIC FUNCTIONS ==
pi_dp = acos(-1._dp); mone_dp = 1._dp; zero_dp = 0._dp
pi_qp = acos(-1._qp); mone_qp = 1._qp; zero_qp = 0._qp
write(*,*) 'Testing trigonometric functions ...'
call SYSTEM_CLOCK(count=cstart_dp)
do i = 1,10000000
mone_dp = cos(pi_dp)
zero_dp = sin(pi_dp)
end do
call SYSTEM_CLOCK(count=cend_dp)
time_dp = real(cend_dp - cstart_dp)/real(crate)
write(*,*) 'DP cos: ',mone_dp
write(*,*) 'DP sin: ',zero_dp
write(*,*) 'DP time: ',time_dp,' seconds'
call SYSTEM_CLOCK(count=cstart_qp)
do i = 1,10000000
mone_qp = cos(pi_qp)
zero_qp = sin(pi_qp)
end do
call SYSTEM_CLOCK(count=cend_qp)
time_qp = real(cend_qp - cstart_qp)/real(crate)
write(*,*) 'QP cos: ',mone_qp
write(*,*) 'QP sin: ',zero_qp
write(*,*) 'QP time: ',time_qp,' seconds'
write(*,*)
write(*,*) 'DP is ',time_qp/time_dp,' times faster.'
write(*,*)
end program QUAD_TEST
gfortran 4.8.4
最適化フラグなしでコンパイルした後の典型的な実行の結果:
Testing elementary operations...
DP sum: 1.0000000000000000
DP time: 0.572000027 seconds
QP sum: 1.00000000000000000000000000000000000
QP time: 4.32299995 seconds
DP is 7.55769205 times faster.
Testing square root ...
DP sqrt: 2.0000000000000000
DP time: 5.20000011E-02 seconds
QP sqrt: 2.00000000000000000000000000000000000
QP time: 2.60700011 seconds
DP is 50.1346169 times faster.
Testing trigonometric functions ...
DP cos: -1.0000000000000000
DP sin: 1.2246467991473532E-016
DP time: 2.79600000 seconds
QP cos: -1.00000000000000000000000000000000000
QP sin: 8.67181013012378102479704402604335225E-0035
QP time: 5.90199995 seconds
DP is 2.11087275 times faster.
ここで何かが起こっているに違いありません。私の推測では、最適化されたアルゴリズムを使用sqrt
して計算さgfortran
れますが、これはおそらく 4 倍精度の計算には実装されていません。これは と の場合には当てはまらないかもしれませんが、sin
基本cos
演算は 4 倍精度で 7.6 倍遅くなるのに、三角関数では 2 倍しか遅くならないのはなぜですか? 三角関数に使用されるアルゴリズムが 4 倍精度と倍精度で同じである場合、それらの CPU 時間も 7 倍増加すると予想されます。
64 ビットと比較して、128 ビット精度を使用した場合の科学計算の平均速度はどれくらいですか?
これを Intel i7-4771 @ 3.50GHz で実行しています。