Excel是否创建一系列相同的值？
我想知道是否有办法使内部收益率的计算更简单？例如，我每月有500美元的稳定收入，我希望看到6个月、12个月、18个月等的内部收益率 目前，我必须创建三列，一列有6行$500，一列有12行$500，等等 有没有办法简化计算 另外，实际上有一个不同的上次付款值，因此我无法创建12行并选择相同值的子集。编辑 在阅读了lori-m的评论之后，我在回复中添加了更多内容，以说明由Excel RATE函数内部执行的费率计算Excel是否创建一系列相同的值？,excel,Excel,我想知道是否有办法使内部收益率的计算更简单？例如，我每月有500美元的稳定收入，我希望看到6个月、12个月、18个月等的内部收益率 目前，我必须创建三列，一列有6行$500，一列有12行$500，等等 有没有办法简化计算 另外，实际上有一个不同的上次付款值，因此我无法创建12行并选择相同值的子集。编辑 在阅读了lori-m的评论之后，我在回复中添加了更多内容，以说明由Excel RATE函数内部执行的费率计算 原始答复 我假设最后一个值是负数，因为您至少需要一个负数才能找到 当年金支付金额固定
原始答复 我假设最后一个值是负数，因为您至少需要一个负数才能找到 当年金支付金额固定时，您可以选择使用计算内部收益率 RATE函数接受以下值 =费率（NPER、PMT、PV、[FV]类型） NPER将是案例6、12、18中的周期数 在这种情况下，PMT是定期付款 PV是年金的现值如果它是一个流出的现金流，那么您需要将其指定为负值 FV是残值 使用速率函数要求以下值中至少一个或最多两个必须为负FV、PV或PMT 然而，从你所写的内容来看，你甚至可能没有负现金流，因此无法使用内部收益率或利率函数
下面的文本显示了RATE函数如何使用
TVM方程=0的Newton-Raphson法IRR计算
TVM等式1:PV（1+i）^N+PMT（1+i*类型）[（1+i）^N-1]/i+FV=0
f（i）=0+500*（1+i*0）[（1+i）^6-1）]/i+-2500*（1+i）^6
f’（i）=（500*（6*i*（1+i）^（5+0）-（1+i）^6+1）/（i*i））+6*-2500*（1+0.1）^5
i0=0.1
f（i1）=-571.0975
f'（i1）=-14420.4
i1=0.1--571.0975/-14420.4=0.0603965562675
错误界限=0.0603965562675-0.1=0.039603>0.000001
i1=0.0603965562675
f（i2）=-63.1212
f'（i2）=-11318.2776
i2=0.0603965562675--63.1212//-11318.2776=0.0548196309075
错误界限=0.0548196309075-0.0603965562675=0.005577>0.000001
i2=0.0548196309075
f（i3）=-1.1104
f'（i3）=-10921.6385
i3=0.0548196309075--1.1104//-10921.6385=0.0547179582964
错误界限=0.0547179582964-0.0548196309075=0.000102>0.000001
i3=0.0547179582964
f（i4）=-0.0004
f'（i4）=-10914.4953
i4=0.0547179582964--0.0004//-10914.4953=0.0547179250235
错误界限=0.0547179250235-0.0547179582964=0<0.000001
内部收益率=5.47%
TVM方程=0的牛顿-拉斐逊法IRR计算
TVM等式2:PV+PMT（1+i*type）[1-{（1+i）^-N}]/i+FV（1+i）^-N=0
f（i）=-2500+500*（1+i*0）[1-（1+i）^-6]/i+0*（1+i）^-6
(i)=(500*(1+i)^-6*(1+i)^-6*(1+i)/(i*i)+(0*-6*)(1+i)^-6*(-6-1
i0=0.1
f（i1）=-322.3697
f'（i1）=-4842.0856
i1=0.1--322.3697//-4842.0856=0.0334233887655
错误界限=0.0334233887655-0.1=0.066577>0.000001
i1=0.0334233887655
f（i2）=178.1297
f'（i2）=-6438.6863
i2=0.0334233887655-178.1297//-6438.6863=0.0610889228796
错误界限=0.0610889228796-0.0334233887655=0.027666>0.000001
i2=0.0610889228796
f（i3）=-49.7272
f'（i3）=-5702.834
i3=0.0610889228796---49.7272//-5702.834=0.0523691914645
错误界限=0.0523691914645-0.0610889228796=0.00872>0.000001
i3=0.0523691914645
f（i4）=18.7303
f'（i4）=-5922.4426
i4=0.0523691914645-18.7303//-5922.4426=0.055531790412
错误界限=0.055531790412-0.0523691914645=0.003163>0.000001
i4=0.055531790412
f（i5）=-6.4397
f'（i5）=-5841.5461
i5=0.055531790412--6.4397//-5841.5461=0.054429394433
错误界限=0.054429394433-0.055531790412=0.001102>0.000001
i5=0.054429394433
f（i6）=2.2892
f'（i6）=-5869.581
i6=0.054429394433-2.2892//-5869.581=0.0548194083235
错误界限=0.0548194083235-0.054429394433=0.00039>0.000001
i6=0.0548194083235
f（i7）=-0.8044
f'（i7）=-5859.6427
i7=0.0548194083235--0.8044//-5859.6427=0.0546821305725
误差范围=0.0546821305725-0.0548194083235=0.000137>0.000001
i7=0.0546821305725
f（i8）=0.2838
f'（i8）=-5863.1383
i8=0.0546821305725-0.2838//-5863.1383=0.0547305377303
误差范围=0.0547305377303-0.0546821305725=4.8E-5>0.000001
i8=0.0547305377303
f（i9）=-0.1
f'（i9）=-5861.9054
i9=0.0547305377303--0.1//-5861.9054=0.0547134792018
错误界限=0.0547134792018-0.0547305377303=1.7E-5>0.000001
i9=0.0547134792018
f（i10）=0.0352
f'（i10）=-5862.3398
i10=0.0547134792018-0.0352//-5862.3398=0.054719491928
错误界限=0.054719491928-0.0547134792018=6.0E-6>0.000001
i10=0.054719491928
f（i11）=-0.0124
f'（i11）=-5862.1867
i11=0.054719491928--0.0124//-5862.1867=0.0547173727531
错误界限=0.0547173727531-0.054719491928=2.0E-6>0.000001
i11=0.0547173727531
f（i12）=0.0044
f'（i12）=-5862.2407
i12=0.0547173727531-0.0044//-5862.2407=0.0547181196735
误差范围=0.0547181196735-0.0547173727531=1.0E-6<0.000001
内部收益率=5.47%
原始答复 我假设最后一个值是负数，因为您至少需要一个负数才能找到 当年金支付金额固定时，您可以选择使用计算内部收益率 RATE函数接受以下值 =费率（NPER、PMT、PV、[FV]类型） NPER将是案例6、12、18中的周期数 在这种情况下，PMT是定期付款 PV是年金的现值如果它是一个流出的现金流，那么您需要将其指定为负值 FV是残值 使用速率函数要求至少
Newton Raphson Method IRR Calculation with TVM equation = 0
TVM Eq. 1: PV(1+i)^N + PMT(1+i*type)[(1+i)^N -1]/i + FV = 0
f(i) = 0 + 500 * (1 + i * 0) [(1+i)^6 - 1)]/i + -2500 * (1+i)^6
f'(i) = (500 * ( 6 * i * (1 + i)^(5+0) - (1 + i)^6) + 1) / (i * i)) + 6 * -2500 * (1+0.1)^5
i0 = 0.1
f(i1) = -571.0975
f'(i1) = -14420.4
i1 = 0.1 - -571.0975/-14420.4 = 0.0603965562675
Error Bound = 0.0603965562675 - 0.1 = 0.039603 > 0.000001
i1 = 0.0603965562675
f(i2) = -63.1212
f'(i2) = -11318.2776
i2 = 0.0603965562675 - -63.1212/-11318.2776 = 0.0548196309075
Error Bound = 0.0548196309075 - 0.0603965562675 = 0.005577 > 0.000001
i2 = 0.0548196309075
f(i3) = -1.1104
f'(i3) = -10921.6385
i3 = 0.0548196309075 - -1.1104/-10921.6385 = 0.0547179582964
Error Bound = 0.0547179582964 - 0.0548196309075 = 0.000102 > 0.000001
i3 = 0.0547179582964
f(i4) = -0.0004
f'(i4) = -10914.4953
i4 = 0.0547179582964 - -0.0004/-10914.4953 = 0.0547179250235
Error Bound = 0.0547179250235 - 0.0547179582964 = 0 < 0.000001
IRR = 5.47%
Newton Raphson Method IRR Calculation with TVM equation = 0
TVM Eq. 2: PV + PMT(1+i*type)[1-{(1+i)^-N}]/i + FV(1+i)^-N = 0
f(i) = -2500 + 500 * (1 + i * 0) [1 - (1+i)^-6)]/i + 0 * (1+i)^-6
f'(i) = (-500 * (1+i)^-6 * ((1+i)^6 - 6 * i - 1) /(i*i)) + (0 * -6 * (1+i)^(-6-1))
i0 = 0.1
f(i1) = -322.3697
f'(i1) = -4842.0856
i1 = 0.1 - -322.3697/-4842.0856 = 0.0334233887655
Error Bound = 0.0334233887655 - 0.1 = 0.066577 > 0.000001
i1 = 0.0334233887655
f(i2) = 178.1297
f'(i2) = -6438.6863
i2 = 0.0334233887655 - 178.1297/-6438.6863 = 0.0610889228796
Error Bound = 0.0610889228796 - 0.0334233887655 = 0.027666 > 0.000001
i2 = 0.0610889228796
f(i3) = -49.7272
f'(i3) = -5702.834
i3 = 0.0610889228796 - -49.7272/-5702.834 = 0.0523691914645
Error Bound = 0.0523691914645 - 0.0610889228796 = 0.00872 > 0.000001
i3 = 0.0523691914645
f(i4) = 18.7303
f'(i4) = -5922.4426
i4 = 0.0523691914645 - 18.7303/-5922.4426 = 0.055531790412
Error Bound = 0.055531790412 - 0.0523691914645 = 0.003163 > 0.000001
i4 = 0.055531790412
f(i5) = -6.4397
f'(i5) = -5841.5461
i5 = 0.055531790412 - -6.4397/-5841.5461 = 0.054429394433
Error Bound = 0.054429394433 - 0.055531790412 = 0.001102 > 0.000001
i5 = 0.054429394433
f(i6) = 2.2892
f'(i6) = -5869.581
i6 = 0.054429394433 - 2.2892/-5869.581 = 0.0548194083235
Error Bound = 0.0548194083235 - 0.054429394433 = 0.00039 > 0.000001
i6 = 0.0548194083235
f(i7) = -0.8044
f'(i7) = -5859.6427
i7 = 0.0548194083235 - -0.8044/-5859.6427 = 0.0546821305725
Error Bound = 0.0546821305725 - 0.0548194083235 = 0.000137 > 0.000001
i7 = 0.0546821305725
f(i8) = 0.2838
f'(i8) = -5863.1383
i8 = 0.0546821305725 - 0.2838/-5863.1383 = 0.0547305377303
Error Bound = 0.0547305377303 - 0.0546821305725 = 4.8E-5 > 0.000001
i8 = 0.0547305377303
f(i9) = -0.1
f'(i9) = -5861.9054
i9 = 0.0547305377303 - -0.1/-5861.9054 = 0.0547134792018
Error Bound = 0.0547134792018 - 0.0547305377303 = 1.7E-5 > 0.000001
i9 = 0.0547134792018
f(i10) = 0.0352
f'(i10) = -5862.3398
i10 = 0.0547134792018 - 0.0352/-5862.3398 = 0.054719491928
Error Bound = 0.054719491928 - 0.0547134792018 = 6.0E-6 > 0.000001
i10 = 0.054719491928
f(i11) = -0.0124
f'(i11) = -5862.1867
i11 = 0.054719491928 - -0.0124/-5862.1867 = 0.0547173727531
Error Bound = 0.0547173727531 - 0.054719491928 = 2.0E-6 > 0.000001
i11 = 0.0547173727531
f(i12) = 0.0044
f'(i12) = -5862.2407
i12 = 0.0547173727531 - 0.0044/-5862.2407 = 0.0547181196735
Error Bound = 0.0547181196735 - 0.0547173727531 = 1.0E-6 < 0.000001
IRR = 5.47%