#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.
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:
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:
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
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.
# 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
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