### 金融代写|金融数学代写Financial Mathematics代考|Solving Problems in Interest

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 金融代写|金融数学代写Financial Mathematics代考|Measuring Time Periods: Simple Interest and Fractional Time Periods

In all of our examples so far we have simply stated a value of $t$. In most practical problems, time is expressed in terms of a date of inception and a date of termination. Thus, we might have an investment which is initiated on January 12, 2012, and terminates on September 11, 2014. In order to use our formulas we need to convert these dates into a time period. There are two methods in common use for calculating the value of $t$ corresponding to the time between two specified dates. This method uses the exact number of days and calculates $t$ based on the assumption that there are 365 days in a year.

This method assumes that each month has 30 days, resulting in a year of 360 days. If our two dates are $M_{1}, D_{1}, Y_{1}$ and $M_{2}, D_{2}, Y_{2}$ the formula for computing the number of days between two dates are $M_{1}, D_{1}, Y_{1}$ and $M_{2}, D_{2}, Y_{2}$ days for ordinary simple interest is
$$360\left(Y_{2}-Y_{1}\right)+30\left(M_{2}-M_{1}\right)+\left(D_{2}-D_{1}\right)$$
In both of these two cases, the easiest way to calculate the number of days between two dates is to use the DATE Worksheet on the TI BA II Plus.
We invoke this worksheet by keying in 2ND DATE (it’s above the 1). As for other worksheets, we move from entry to entry using the $\uparrow, \downarrow$. Dates are entered as mm.ddyy. Thus, March 5,2007 , would be entered as $03.0507$. It will display as $3.0507$ until you hit the ENTER key, at which point it will display as 03-05-2007.

There are four variables which are accessed by using the $\uparrow$ or $\downarrow$ keys. DT1 is the first date and is always assumed to be prior to DT2, the second date. DBD is the number of days between the two dates. The final variable is either ACT (for exact number of days) or 360 (using the ordinary simple interest method). This last variable is toggled by using 2ND SET. Here’s how to compute the number of days between March 3,2008 , and October 5,2009 , using both the ordinary simple interest rule and the exact number of days (Table $3.1$ ).
$2 N D$ Date Since there are 581 days between March 3,2014 , and October 5,2015 , using the exact method we obtain $t=\frac{581}{365}=1.592$. For the ordinary simple interest method we have $t=\frac{572}{360}=1.588$.

The equation $F V=P V(1+i)^{t}$ can be evaluated directly for any value of $t$. For example, if we want the value of an investment of $\$ 5,456$at six and a third years earning compound annual interest of$5.6 \%$, we have (Table$3.2$) $$F V=5456(1.056)^{6 \frac{1}{3}}=5456 \cdot 1.412119603=\ 7,704.52$$ This method gives the exact value of the accumulated amount over any time period. In some cases the interest earned over the fractional portion of a time period is computed using simple interest. Historically this practice developed due to the difficulties involved in computing the value of$(1+i)^{t}$if$t$is not an integer. As an example, simple interest is sometimes used to compute the interest due at inception for a home mortgage. Home mortgages are timed to start on the first day of the month subsequent to the date of inception of the loan. If you close on a home loan on a day other than the first of the month, you are assessed simple interest on the loan amount for the number of days remaining in that month at the inception of the loan. After that, interest is computed using compound interest. A loan which closes on May 12 will be assessed interest on the remaining days in May as part of the closing costs. The loan will begin June 1 , and the first regular payment will be due at the end of June. We can calculate the FV of an account using simple interest over the fractional period as follows. Suppose that$t=n+q$where$n$is a whole number and$q$represents the fractional portion of time$(0<q<1)$we would compute the$F V$using simple interest for the fractional portion of time as $$F V=P V(1+i)^{n}(1+i q)$$ ## 金融代写|金融数学代写Financial Mathematics代考|Equations of Value at any Time We will often need to compare two or more amounts of money at different points in time. If the rate of interest is not zero, this is only makes sense if we compute the values of each amount at a common point in time. This common date is called the comparison date. We can choose any date we like. An amount which occurs prior to the comparison date must be accumulated to that date while an amount which occurs after the comparison date must be discounted. It is often helpful to use a number line to place each amount at the appropriate time. This is called a time diagram. The equation which equates two different methods of computing the value of a transaction is called the equation of value. An amount which will be available$n$periods after the comparison date is discounted${ }^{1}$by the factor$v^{n}$Recall that$v=\frac{1}{1+i}$. An amount which will be available$m$periods prior to the comparison date is accumulated by a factor of$(1+i)^{m}$. Example 3.3 In return for the promise of a payment of$\$600$ at the end of eight years, a person agrees to pay $\$ 100$now,$\$200$ at the end of five years, and to make one further payment at the end of ten years. What is the required final payment if the nominal rate of annual interest is $6.2 \%$ compounded semiannually?

Solution. We will measure time in six-month intervals. This gives us $i=\frac{.062}{2}=.031$ and $v=\frac{1}{1.031}$. We will use the inception date as our comparison date. We then have the following situation for each amount of money using $X$ for the (unknown) value of the final payment. See Figure 3.1. $\$ 600$: is not available for eight years or sixteen periods. Its value at inception is thus$600 v^{16}\$100$ : is available now. Its value is thus 100
$\$ 200$: is not available for five years or ten periods. Its value at inception is thus$200 v^{10}\$X$ : is not available for ten years or twenty periods. Its value at inception is thus $x v^{20}$

## 金融代写|金融数学代写Financial Mathematics代考|Measuring Time Periods: Simple Interest and Fractional Time Periods

360(是2−是1)+30(米2−米1)+(D2−D1)

2ñD日期 由于从 2014 年 3 月 3 日到 2015 年 10 月 5 日之间有 581 天，使用我们得到的确切方法吨=581365=1.592. 对于普通的单利方法，我们有吨=572360=1.588.

## 金融代写|金融数学代写Financial Mathematics代考|Fractional Time Periods

F在=磷在(1+一世)n(1+一世q)

## 金融代写|金融数学代写Financial Mathematics代考|Equations of Value at any Time

$100: 现在可以了。因此其值为 100$200: 五年或十期不可用。因此，它在开始时的价值是200在10
\$X: 十年或二十年不可用。因此，它在开始时的价值是X在20

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## MATLAB代写

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