Financial Planning- Home Mortgage management (ii)

In this section you will need excel. Also you will have to go through part (i) as prerequisite to understand the formulation of system equations. 

In part (i) we assumed that annual income of client is constant, i.e Y(bar) = $80,000. However in reality income of an individual is not constant in nature (because it might decrease or increase due unforeseeable circumstances). In this regard let us assume that the client's income is growing at rate 'g'= 2% and so we incorporate this dynamics as continuous compounding growth of Y:

Further, we take the differential equation describing debt (D(dot))(we have already discussed about this equation in part i) and include the value of Y(t). Then we transform the equation by transferring Exponential value on right hand side.


Now we multiply both the sides by integrating factor e^(rt) and integrate both the sides to obtain 'D' :

By simplifying above equation ( transfer e^(-tr) on the right hand side) we derive the expression for D(t) which is known as general solution of the system of equation. 

As we have a general solution for the system of equation, we embed initial condition and determine value of 'C' in terms of Initial Debt, Initial Income, growth rate of income, interest rate on debt; and then formulate analytical solution for D(t). This will enable us to calculate the maximum loan, that we can approve with the help of given information.

 We know: Value of e^(0) =1



Make 't' =0 , as we are considering initial debt condition.



Transfer 'C' on the left hand side and express the relationship.


Include C in  D(t) as

Now you have readymade decision making equation that can be used for mortgage/auto loan financing.  When you embed D(t) equation in excel along with decision making parameters/factors you will get the visual dynamics of debt.

# Techniques  covered in blog posts labelled AED are taught in courses Mathematics for Applied Economics, and Applied Economics Dynamics offered at Crawford School of Public Policy @ The Australian National University. 

Financial Planning- Home Mortgage management (i)

#If you are new to compounding interest formula and continuous time series please refer my post on Financial Planning (basics) before reading this post

Consider that you a lending officer of a bank (e.g Standard Chartered or HSBC) are calculating a home loan contract with a client. The client wishes to borrow some debt amount (D), which can be denoted as Do > 0  (Here o means at time zero or at the time of burrowing). The annual interest rate is quote to be 'r', and since it is daily compounded, we will treat it as continuously compounding rate. The annual income of the client is Y(t)>0, which is a yearly income in some positive numbers. A fixed proportion (S) of the income (Y) is saved to pay the outstanding debt.

The relationship can be exhibited with the help of differential equation:

D(dot) is the change in debt, which is the function of saving (S) from Income (Y) deducted from incurred interest rate from debt  D(t)

In this first section we considered that Y (income) remains constant over time, so we change Y(t) as Y(bar). The solution method we use in solving the equation relies on the nature of the equation. So with the assumption of Y(t) as constant, the equation becomes autonomous, non-homogeneous and first order equation. It is of the first-order because it the higher-order derivative in the equation and it is autonomous because the coefficients doesn't vary with time. So first step to solve fist order, non-homogeneous and autonomous equation is to identify the steady state. If a system is in steady state, then the recently observed behavior of the system will continue into the future, which  means system remains unchanged for ever. So in-order to determine steady state we will have to solve by making change in Debt as zero or D(dot)=0 . We get steady state D(bar) and then incorporate it in solution for non-homogeneous equation.

I will include the process of deriving a particular solution for given nature of equation in next post or update in this post by 7th June 2015. 

For now just understand that the particular solution looks like:

A) Now first  question client asks you is "If i get home loan (D)o  in how many years i will pay up the debt  (remember here we assumed that whatever client save (S) from constant income (Y) is used to service the debt) 

So for particular solution we make D(t) = 0  (Client pays the debt in time  't' )

 





We solve the equation by keeping e^rt on left side and rest on the right. Further taking log on the both sides simplifies the equation,and we will have expression without exponential form. Now client reveals us that his/her income is Y= $80,000, and saves (s) 30% out of the yearly income, and wants to take $500,000  (given that valuation of property is sufficient enough provide the required sum). By feeding the value in the formula we can easily calculate the number of years it will take client to pay the loan, by servicing debt from his/her savings. 

                                                                        
 B) Now if the client could choose the length of the contract (e.g. 40 years or 60 years), what would be the maximum loan you should approve? If client wants to pay up the debt in 30 years

We took particular solution and embedded 30 in-place of 't', since we want to calculate the maximum loan we can approve for the given income and savings we calculate D(o) such that D(30)=0 i.e debt at 30th year is zero. 

Knowing this process enables us to understand the dynamics that goes inside system of equation. Further it also enables us to calculate specific value without use of complex excel formulas and software.



# Techniques  covered in blog posts labelled AED are taught in courses Mathematics for Applied Economics, and Applied Economics Dynamics offered at Crawford School of Public Policy @ The Australian National University. 

Financial Planning (Basics) Continuous Time

The Time value of money in Continuous Time

Calculating continuous time series in real life scenario means that we might be dealing with infinite frequency and time horizon. The continuous time series is applicable in finance when we deal with continuously changing  Foreign Exchange rate,  Interest rate (continuously compounding) and price of the perpetual bonds. Normally, to calculate the value of perpetual and high frequency instruments is very difficult, but with the help of mathematics and calculus we can estimate there value within few steps, that also without use of software and spread sheets.

Before we began the calculations of perpetual bonds let's go through a important concept that we will use in this section. It is called continuously compounding.

The relationship above illustrates that in continuously compounding formula assuming interest rate is equal to '1%' as value of 'n' (number of times compounding) approaches to infinity the output approaches to 'e'  ( Euler's number) which is equal to 2.71828~. The progression is demonstrated  by plotting the values in excel and feeding the formula. Further to aid the claim we plot it graphically to illustrate the convergence towards 'Euler's number':

.
This concept is now helpful to understand the continuously compounding interest rate formula. Let us take an example where we have to invest $1 dollar at compounding rate 'r'  and compounding 'n' number of times in one (1) year, when 'n' tends to infinity.

The continuous compounding formula looks as follows:  

                  

Further in relationship we take 'r' to the denominator and represent (n/r) as N to make it ready for transition into 'Euler's number'

 
Finally the formula for compounding interest rate at rate 'r'can be written as: 'e'raised to the power 'r'= e^r. Now if we want to calculate for 't' years instead of '1' year we write it as 'e' raised to the power 'rt or  e^rt.

For example:

 A is the amount that will be received for the principle 'P' with 'r' continuously compounding interest rate in 't' years when number of times compounding 'n' tends to infinity. 

Calculating the present value when the time step is very small and we are continuously compounding for 't' year(s).

As mention above section, in the financial market we usually come across situations when the time step is very small and the frequency of compounding gets closer to infinity. In such case inorder to sum the continuously compounding value we take help of Integration formula.

Example :

The interest is 5% per annum. A bond pays $1 coupon in one year (from the beginning of year 0 to the beginning of year 1), but the coupon payment spreads evenly throughout 365 days of the year. 
Calculate present value of the bond!

(1/365) means that $1 is divided among 365 days.  

Then present value of $ 1 couple paid every day (i.e time step 1/365) within the period of time 't' years is given by :





Please note that the exponential has negative sign, as we are dealing with present value. 





However the challenge is to sum all the them, and if we try to do it in excel we will have to drag it all the way down to 365 days. So in order to sum at time step (1/365) we take integral  from 0 to 1 years.

Here the time step is very small (e.g 1/365)





We further solve the integration (sum of many small components) in-order to get:

Please click here to know the process (believe me it is quite simple)


So the Present  value of bond with the interest rate 5% per annum which pays $1 coupon  in one year (from the beginning of year 0 to the beginning of year 1),whose  coupon payment spreads evenly throughout 365 days of the year (very small time step) is $0.9754

Please note that this is just the illustration of the use of integration in continuous time series. In real life small differences in decimal when multiplied with millions of dollar can show huge gap in the result



# Techniques  covered in blog posts labelled AED are taught in courses Mathematics for Applied Economics, and Applied Economics Dynamics offered at Crawford School of Public Policy @ The Australian National University. 

Financial Planning (Basics)

Time value of money in Discrete Time series 

Ever wondered how Banks calculate Annual Payments?  (Without considering Annual Income of an Individual)

This section illustrates the method to calculate annual equal payments for n number of years required to pay $A sum at annual interest rate R. Without using excel files and heavy calculations.

For example: You are borrowing $300,000 to buy a house (at year zero) and you sign a loan contract which fixes the interest rate at 5% per annum and you pay 30 annual equal payments (from year 1 to year 30). What is the amount of money you have to pay each year?

Let's first  illustrate the annual payments, with assumption that the equal annual payment/installment is '$X'.

= $300,000

We know that sum (S) of discrete series (geometric) is given by the formula :

S= (1-B^T+1)/ (1-B)    (Proof is shown at end of this section)

In this case B= 1/(1+r)= 1/(1+0.05) = 1/1.05

Now taking 'X' and '1/(1.05)' out the bracket, the series becomes:

x/(1.05)[1+]=$300,000

(X/1.05)[1+(1/1.05)+ (1/1.05)²)+ (1/1.05)³+ (1/1.05)⁴+ (1/1.05)⁵……………+(1/1.05)²⁹]= $300,000

After converting the sum inside the bracket as per the formula we can write:

[X/ (1.05)]*[ (1-(1/1.05)³⁰)/(1-(1/1.05))]  = $300,000

So value of annual installment ($X) =   ($300,000/[ (1-(1/1.05)³⁰)/(1-(1/1.05))])*1.05 = $19,515.4

This means you will have to pay $19,515.4 as annual installments for $300,000 to be paid in 30 years at 5% annual interest rate. 

Proof of the Summation formula:

Let, 'S' be the sum of discrete series 'B'
S= 1+B+B²+ B³+B⁴+B⁵+B⁶   …………………….B^{T      (i)
Now multiply both the sides by B, so the series becomes}
SB= B+B²+ B³+B⁴+B⁵+B⁶   …………………….B^T+1                 (ii)
^{Now Subtract (ii) from (i)}
S-SB= 1- B^T+1

Now, take ‘S’ common and transfer (1-B) to right hand side. 

Finally, the Sum of discrete series 'B' is given by the formula.
S= (1-B^T+1)/ (1-B)

(It is also possible to calculate in excel, but if you remember this formula you can calculate summation of discrete number upto any length without having to drag the excel cell) 

# Techniques  covered in blog posts labelled AED are taught in courses Mathematics for Applied Economics, and Applied Economics Dynamics offered at Crawford School of Public Policy @ The Australian National University.