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庄鑫 · 2022年09月23日

请教

  1. CFA I
  2. Portfolio

NO.PZ2017092702000030

问题如下：

At the beginning of Year 1, a fund has $10 million under management; it earns a return of 14% for the year. The fund attracts another $100 million at the start of Year 2 and earns a return of 8% for that year. The money-weighted rate of return is most likely:

选项：

A.

less than the time-weighted rate of return.

B.

the same as the time-weighted rate of return.

C.

greater than the time-weighted rate of return.

解释：

A is correct.

The money-weighted rate of return is found by setting the present value (PV) of investments into the fund equal to the PV of the fund’s terminal value. Because most of the investment came during Year 2, the measure will be biased toward the performance of Year 2. Set the PV of investments equal to the PV of the fund’s terminal value: 10+1001+r=10×1.14×1.08+100×1.08(1+r)210+\frac{100}{1+r}=\frac{10\times1.14\times1.08+100\times1.08}{{(1+r)}^2}   Solving for r results in r = 8.53%. The time-weighted return of the fund is =(1.14)(1.08)2−1=10.96\sqrt[2]{{(1.14)}{(1.08)}}-1=10.96

Q1：求解MWRR的解题方法就是：把期间发生的现金流全部按照一个折现率折现到期初，使得NPV为零，算出折现率，是吗？

Q2：求解TWRR的解题方法就是：用根号下（1+%）（1+%）（1+%）等于（1+TWRR）的那个公式，是吗？

Q3：题目说的期初有10，为什么要视为现金流流出？

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1 个答案

Kiko_品职助教 · 2022年09月26日

嗨，爱思考的PZer你好：


Q1：是。

Q2：是。

Q3：因为相当于我投进去了10.所以对我来说是现金流的流出，所以是负号

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NO.PZ2017092702000030问题如下the beginning of Ye1, a funh$10 million unr management; it earns a return of 14% for the year. The funattracts another $100 million the start of Ye2 anearns a return of 8% for thyear. The money-weighterate of return is most likely:A.less ththe time-weighterate of return. B.the same the time-weighterate of return. C.greater ththe time-weighterate of return.A is correct. The money-weighterate of return is founsetting the present value (PV) of investments into the funequto the PV of the funs terminvalue. Because most of the investment came ring Ye2, the measure will biasetowarthe performanof Ye2. Set the PV of investments equto the PV of the funs terminvalue: 10+1001+r=10×1.14×1.08+100×1.08(1+r)210+\frac{100}{1+r}=\frac{10\times1.14\times1.08+100\times1.08}{{(1+r)}^2}10+1+r100​=(1+r)210×1.14×1.08+100×1.08​ Solving for r results in r = 8.53%. The time-weightereturn of the funis =(1.14)(1.08)2−1=10.96\sqrt[2]{{(1.14)}{(1.08)}}-1=10.962(1.14)(1.08)​−1=10.96CF0 -10CF1 1.4-100CF2 100*0.08+10*0.08

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