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3 years ago

7

Consider a no-load mutual fund with $200 million in assets and 10 million shares at the start of the year and with $250 million

in assets and 11 million shares at the end of the year. During the year investors have received income distributions of $2 per share and capital gain distributions of $.25 per share. Assuming that the fund carries no debt, and that the total expense ratio is 1%, what is the rate of return on the fund

1 answer:

frez [133]3 years ago

5 0

Answer:

273.75%

Explanation:

Note: Capital Gain distribution would be $50.25, NOT $.25 (typing mistake)

This is no-load MF. But there are other two types of MF (Mutual Funds).

If FL MF (Front Load Mutual Fund), investors pay something upfront when investing.

In BL MF (Back Load Mutual Fund), investors pay when exiting the MF.

Here, this is no load, so calulations are easier.

Now,

NAV (Net Asset Value) is the total assets divided by number of shares.

NAV beginning of year and NAV end of year. Total expense ratio will be adjusted from NAV, end of year.

NAV, beginning = 200 million / 10 million shares = $20

NAV, end = 250 - (0.01*250) / 11 million shares = $22.5

Now,

Rate of Return of the Fund = (NAV,end - NAV,beginning + Income Distribution + Capital Gain Distribution - Liabilities) / NAV, beginning

We have:

Rate of Return = ($22.5 - $20 + $2 + $50.25 - $0) / $20 = 2.7375

Converting to percentage:

2.7375 * 100 = 273.75%

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Answer:

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Answer:

730 items

Explanation:

The objective of the given information is to determine the number of hamburgers UAHH should order for the following conditions:

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The formula for a given order quantity in a fixed period of time can be expressed as :

$q = \overline d(L+T)+ z \sigma_{L+T}-I$

where;

$q$ = order quantity = ???

$\overline d$ = daily demand average = 600

L = lead time in days = 1

T = time taken = 1

z = no of standard deviation = ???

$\sigma_{L+T}$ = standard deviation of usage in lead time and time taken = ???

I = present inventory level = 800

$\sigma_{L+T}$ = $\sqrt 2$ × standard deviation of daily demand

$\sigma_{L+T}$ = $\sqrt{2} *100$

$\sigma_{L+T}$ = 1.4142 * 100

$\sigma_{L+T}$ = 141.42 items

From the Desired service probability 99% = 0.99; we can deduce the no of standard deviation by using the excel function (=NORMSINV (0.99))

z = 2.33

From $q = \overline d(L+T)+ z \sigma_{L+T}-I$

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Note:

I wasn't able to access the Chester Income Statement but I successfully accessed a similar question Digby.

The Complete Question is as under:

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