Mortgage%20Calculation

1
Mortgage Calculation

• Using z-transforms

2
Setup the Original Equation

• Let kmonth number 0,1,2,3 etc.
• Let y(k)loan balance in month k
• Let u(k)payment during month k
• Let iinterest rate as a proportion
• Let mmonthly payment and Pamount of loan
• The loan balance then at month k1 is
• Y(k1)y(k)(1i/12) u(k)
• If a(1i/12) then
• Y(k1)ay(k) u(k) (no z transforms yet!!)
  

3
Z-transform Each Term

• Z(y(k1))aYz Uz but u(k) was a steady
  unit amount each month, so to find Uz look it up
  in the table Uzmz/(z-1)Z(u(k))
• And using the shift theorem
• Z(y(k1))zYz zy(0)
• Note initial values like y(0) are in most cases
  ignored but cant be in this example (my bad).
• Final z-transformed equation
• zYz zy(0) aYz -mz/(z-1)

4
Next Solve for Yz Algebraically

• zYz zy(0) aYz -mz/(z-1)
• zYz - aYz zy(0) -mz/(z-1)
• Yz(z a) zy(0) -mz/(z-1)
• Yz zy(0) /(z a) -mz/((z-1) (z a) )

5
Begin the inverse z-transform

• Yz zy(0) /(z a) - mz/((z-1) (z a) )
• The term zy(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
• K1z/(z-1) k2z/(z-a)-mz/((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
• Yzzy(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)

• Yzzy(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
• 0y(0)ak m(1- ak )/(a-1)
• m(1- ak )/(a-1) -y(0) ak but if Py(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 )
• mP(1-a)/(a-k 1) but recall a1i/12
• mP(-i/12)/((1i/12)-k 1) or
• mP(i/12)/(1 - (1i/12)-k) or
• mP(i/12)(1i/12)k/((1i/12)k - 1)
• In C and assuming Ji/12 you could write
• M P J( (pow(1.0 J,n))/(pow(1.0J,n) - 1.0))

8
Example

• 20 year mortgage, 11 interest, 30000 house
• mP(-i/12)/((1i/12)-k 1)
• m30000(-.11/12)/((1.11/12)-240 1)
• m30000(-.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.