Question

我正在做一些长期的模拟，我试图在ODE系统的解决方案中实现最高的准确性。我试图找出四倍（128位）精度计算需要多少时间与双倍（64位）精度。我google了一下，看到了几个意见：有人说它需要4倍的时间，有些则需要60-70倍......所以我决定自己动手，我写了一个简单的Fortran基准程序：

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

使用gfortran 4.8.4进行编译后的典型运行结果，没有任何优化标记：

 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.

这里必须要有一些东西。我的猜测是sqrt通过优化算法用gfortran计算，这可能还没有实现四倍精度计算。这可能不是sin和cos的情况，但为什么基本操作的四倍精度要慢7.6倍而三角函数的速度只减慢2倍？如果用于三角函数的算法对于四次和双精度都是相同的，我预计它们的CPU时间也会增加七倍。

与64位相比，使用128位精度时科学计算的平均减速是多少？

我在Intel i7-4771 @ 3.50GHz上运行它。

Answer 1

更多的是评论而不是答案，但是......

当前CPU为双精度浮点运算提供了大量硬件加速。有些甚至提供扩展精度的设施。除此之外，您仅限于软件实现（正如您所注意到的那样）相当慢。

然而，在一般情况下几乎无法预测这种减速的确切因素。这取决于您的CPU（例如它内置的加速度）以及软件堆栈。对于双精度，您通常使用不同的数学库而不是四倍精度，并且这些算法可能使用不同的算法进行基本操作。

对于使用相同算法的给定硬件上的特定操作/算法，您可能会得到一个数字，但这肯定不会普遍存在。

Answer 2

有趣的是，如果你改变了：

sqrt_qp = sqrt(sqrt_qp)
sqrt_qp = 2._qp

到

sqrt_qp = sqrt(2._qp)

计算会更快！

CPU时间以四倍与双精度

2 个答案: