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3 years ago

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Carr Company is considering two capital investment proposals. Estimates regarding each project are provided below:Project Soup P

roject NutsInitial investment $400,000 $600,000 Annual net income 30,000 46,000 Net annual cash inflow 110,000 146,000 Estimated useful life 5 years 6 years Salvage value -0- -0- The company requires a 10% rate of return on all new investments.Present Value of an Annuity of 1Periods 9% 10% 11% 12% 5 3.890 3.791 3.696 3.6056 4.486 4.355 4.231 4.111The annual rate of return for Project Soup is

1 answer:

jeka943 years ago

6 0

Answer:

IRR = 11.65%

Explanation:

Period Cash Flow

0 -400,000

1 110,000

2 110,000

3 110,000

4 110,000

5 110,000

We could list the annual cash inflow and solve using the excel formula

IRR 0.11649

Or if we aren't allowed to do so, use maths

The way to obtain the present value would be:

Annuity x factor = PV

If we know know the rate we can look into the table doing:

factor = PV/annuity

We know Project Soup is 5 year so we look into this row:

400,000/110,000 = 3.6363

The rate wold be between 11% 3.696 and 12% 3.605

Those rate are close to our goal of 3.6363

We could solve an approximation doing excel or working with math:

$\left[\begin{array}{cc}X&Y&\\x_0 = 3.696&y_0 = 0.11&x=3.6363&?&x_1=3.605&y_1 =0.12\end{array}\right]$

$y = y_0 +\frac{y_1-y_0}{x_1-x_0} (x-x_0)$

We got from the table, the 2 rates which are close to our goal, and thetwo factor close to our goal. we post each value into the formula and solve:

$y = 0.11 +\frac{.12-.11}{3.696-3.605} (3.6363-3.605)$

$y = 0.11 +\frac{0.01}{0.091} (0.0313)$

y= 0.1165 = 11.65%

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Answer:

The percentage of variation esplained by the model is given by the determination coefficient, on this case:

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$\hat y = 5.756(8.5) -36.895=12.031$

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Explanation:

Assuming the following dataset:

x y

8.6 9.6

9.3 18.5

10.1 20.9

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Assuming this question: "The data has a correlation coefficient of r = 0.934. Calculate the regression line for this data. What percentage ofvariation is explained by the regression line? Predict the number of doctors per 10,000 residents in a town with a per capita income of $8500."

We want a linear model like this:

$y = mx +b$

Where m represent the slope and b the intercept for the linear model. And we cna find the slope and b with the following formulas:

$m = \frac{n \sum xy - \sum x \sum y}{n \sum x^2 -(\sum x)^2}$

$b = \frac{\sum y}{n} -m \frac{\sum x}{n}$

And from the dataset we have the following values:

$n= 6, \sum x =53, \sum y = 83.7 , \sum xy = 755.89, \sum x^2 = 471.04$

And replacing into the equation for m we got:

$m =\frac{6(755.89) - (53)(83.7)}{6(471.04) -(53)^2}=5.756$

And the intercept:

$b = \frac{83.7}{6}-36.895 5.756 \frac{53}{6}=-36.895$

And then the linear model is given by:

$\hat y = 5.756 x -36.895$

We can find the estimation replacing x = 8.5 into the linear model and we got:

$\hat y = 5.756(8.5) -36.895=12.031$

And we have 12.031 doctors per 10000 residents.

The percentage of variation esplained by the model is given by the determination coefficient, on this case:

$R^2 = 0.934^2 =0.872$

And we have 87.2% of the variation explained by the linear model given.

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