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Mortgage Calculation

Mortgage Calculation. Using z-transforms. Setup the Original Equation. Let k=month number: 0,1,2,3 etc. Let y(k)=loan balance in month k Let u(k)=payment during month k Let i=interest rate as a proportion Let m=monthly payment and P=amount of loan

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Mortgage Calculation

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  1. Mortgage Calculation Using z-transforms

  2. Setup the Original Equation • Let k=month number: 0,1,2,3 etc. • Let y(k)=loan balance in month k • Let u(k)=payment during month k • Let i=interest rate as a proportion • Let m=monthly payment and P=amount of loan • The loan balance then at month k+1 is: Y(k+1)=y(k)*(1+i/12) – u(k) If a=(1+i/12) then Y(k+1)=a*y(k) – u(k) (no z transforms yet!!)

  3. Z-transform Each Term… • Z(y(k+1))=a*Yz – Uz but u(k) was a steady unit amount each month, so to find Uz look it up in the table: Uz=m*z/(z-1)=Z(u(k)) • And using the shift theorem: Z(y(k+1))=z*Yz – z*y(0) Note initial values like y(0) are in most cases ignored but can’t be in this example (my bad). • ========================== Final z-transformed equation: • z*Yz – z*y(0) = a*Yz -m*z/(z-1)

  4. Next: Solve for Yz Algebraically • z*Yz – z*y(0) = a*Yz -m*z/(z-1) • z*Yz - a*Yz = + z*y(0) -m*z/(z-1) • Yz*(z – a) = + z*y(0) -m*z/(z-1) • Yz= + z*y(0) /(z – a) -m*z/((z-1) *(z – a) )

  5. Begin the inverse z-transform • Yz= + z*y(0) /(z – a) - m*z/((z-1) *(z – a) ) The term z*y(0) /(z – a) can be looked up in the table but the second half must be expanded using partial fractions: ==============Notice a trick, multiplying by z======== K1*z/(z-1) +k2*z/(z-a)=-m*z/((z-1) *(z – a) ) Or K1/(z-1) +k2/(z-a)=-m/((z-1) *(z – a) ) Solving gives: K1=-m/(1-a)=m/(a-1) k2=-m/(a-1) ==============Substituting back gives============= Yz=z*y(0)/(z – a) + m/(a-1) *(z/(z-1)) -m/(a-1) *(z/(z – a) ) Ready, get set, magic happens………

  6. Solving for Y(k) Yz=z*y(0)/(z – a) + m/(a-1) *(z/(z-1)) -m/(a-1) *(z/(z – a) ) • Y(k)=y(0)*ak +m/(a-1) –m/(a-1 )*ak Or factoring: • Y(k)=y(0)*ak +m(1- ak )/(a-1) If at end of k months the loan is paid off, Y(k)=0 then solve for needed monthly payment m: • 0=y(0)*ak +m(1- ak )/(a-1) • +m(1- ak )/(a-1)= -y(0)* ak but if P=y(0) then: • m= -y(0)* ak *(a-1)/(1- ak )=-P*(a-1)/(a-k – 1) or:

  7. Final answer: • Final Answer (fanfare: “do to do to do to do” ) m=P*(1-a)/(a-k – 1) but recall: a=1+i/12 m=P*(-i/12)/((1+i/12)-k – 1) or m=P*(i/12)/(1 - (1+i/12)-k) or m=P*(i/12)*(1+i/12)k/((1+i/12)k - 1) In C++ and assuming J=i/12 you could write: • M = P *J*( (pow(1.0+ J,n))/(pow(1.0+J,n) - 1.0));

  8. Example: • 20 year mortgage, 11% interest, $30000 house: m=P*(-i/12)/((1+i/12)-k – 1) m=30000*(-.11/12)/((1+.11/12)-240 – 1) m=30000*(-.0091666)/(-.88808)=$309.65 Rent it for 600$/month + utilities and begin the joys of landlord/lady ownership……….

  9. Banking Mania 2008 This formula was used in the mortgage program that is included here. However, the reason for using this formula was never obvious to me. I had simply copied it from a book and I was amazed, when I started to learn about z-transforms, at their power to do this calculation. I had found other non transform methods on the internet for accomplishing this but they too were complex.

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