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KiRa [710]
2 years ago
15

A pension fund that begins with $500,000 earns 15% the first year and 10% the second year. At the beginning of the second year,

the sponsor contributes another $300,000. The dollar-weighted and time-weighted rates of return, respectively, were
Business
1 answer:
Margaret [11]2 years ago
8 0

Answer:

time-weighted   12.5%

dollar-weighted 12%

Explanation:

<em><u>time-weighted:</u></em>

<em><u></u></em>\sqrt{1.15 + 1.10}  - 1 = 0.124722188 = 12.5<em><u></u></em>

<em>Dollar-value</em>

<em>Return on dollars </em>

<em></em>500,000 \times 1.15 \times 1.10\\ = 632,500\\<em></em>

<em></em>

<em></em>300,000 \times  1.1 = 330,000<em></em>

<u>Present value of the return</u>

(632,500 + 330,000) \div (1+r)^2

<u>Present value of the investment:</u>

<em></em>500,000  + 300,000 \div (1+r)<em></em>

<em></em>

<u>Both this concepts should match as they work with the same weighted rate</u>

<u />

<u />500,000 + 300,000 /(1+r)  \times (1+r)^2 = 962,500\\500,000 (1+r)^2 + 300,000 (1+r) - 962,500 = 0<u />

<u />

We use cuadratic equation and solve getting a rate of 12%

<u />

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Answer:

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4 0
3 years ago
Variable costs A. are fixed per unit and vary in total as production levels change. B. are fixed in total as production levels c
Bumek [7]

Answer: Option A

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8 0
2 years ago
Why should policies that attempt to create a ladder of opportunity be carried out in the spirit of open-mindedness? Select all t
OverLord2011 [107]

Answer:

b. Such policies have the potential to equalize society.

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4 0
3 years ago
120*45+120*54*120+0+0+0+0+0+0+0=
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783000

Explanation:

3 0
2 years ago
Read 2 more answers
A researcher wants to test the order of integration of some time series data. He decides to use the DF test. He estimates a regr
pav-90 [236]

Answer:

a) H0: u = presence of a unit root

   HA: u ≠ presence of a unit root  ( i.e. stationary series )

b) t stat = -0.064

c) We will reject the Null hypothesis and the next step will be to accept the alternative hypothesis

d) It is not valid to compare the estimated t stat with the corresponding critical value because a random walk is non-stationary while the difference is stationary because it is white noise

Explanation:

<u>a) stating the null and alternative hypothesis</u>

H0: u = presence of a unit root

HA: u ≠ presence of a unit root  ( i.e. stationary series )

<u>b) performing the test </u>

critical value = -2.88

T stat = coefficient / std error

          = -0.02 / 0.31  = -0.064

c) From the test, the value of T stat > critical value we will reject the Null hypothesis hence the next step will be to accept the alternative hypothesis

d) It is not valid to compare the estimated t stat with the corresponding critical value because a random walk is non-stationary while the difference is stationary because it is white noise

   

5 0
3 years ago
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