1. Equivalence Equations II Infinite number of changes of the interest rate within each year UNLIKE … Equivalence Equations I (where interest rate is constant across year, or changes a finite number of times over the year).

2. Equivalence Equations II Case 1: Description: Finding the compounded amount (F) that would be obtained at the end of a given period, given an initial payment, (P). Interest rate changes an infinite number of times within each year. Example: Jim takes the antique car now, and agrees with dealer for a current price of $20,000, and to pay nothing till 2005. What price will he pay in 2005?

3. Equivalence Equations II CASE 1 (cont’d) Problem Definition: This is a Single Payment Compounded Amount (SPCA) problem Cash Flow Diagram: Computational Formula: 20012005 PF

4. Equivalence Equations II Case 2: Description: Finding the initial amount (P) that would yield a future amount (F)at the end of a given period The interest rate changes an infinite number of times within each year. Example: The Studbaker ’55 car currently under repair. Will be ready by 2005, at a price of $30,000. How much should Jim put away now in order to be able to pay for the car in 2005?

5. Equivalence Equations II

6. Equivalence Equations I CASE 2 (cont’d) Problem Definition: This is a Single Payment Present Worth (SPPW) problem Cash Flow Diagram: Computational Formula: PF 20002005

7. Equivalence Equations II Case 3: Description: Finding the amount of uniform annual payments (A) that would yield a certain future amount (F)at the end of a given period. Interest rate changes at an infinite number of times within each year. Example: The Studbaker ’55 antique car is currently under repair. Will be ready by 2005, at a price of $30,000. Jim agrees to pay 5 yearly amounts until 2005, starting December 2001. How much should he pay every year?

8. Equivalence Equations II CASE 3 (cont’d) Problem Definition: This is a Uniform Series Sinking Fund Deposit (USSFD) problem Cash Flow Diagram: Computational Formula: 2001 2005 AAAAA F

9. Equivalence Equations II Case 4: Description: Finding the final compounded amount (F) at the end of a given period due to uniform annual payments (A). Interest rate changes at an infinite number of times within each year. Example: The Studbaker ’55 antique car is currently under repair, and will be ready by 2005. Jim agrees to pay $5,000 yearly until 2005, starting December 2001. How much will he end up paying for the car by the year 2005?

10. Equivalence Equations II CASE 4 (cont’d) Problem Definition: This is a Uniform Series Compounded Amount (USCA) problem Cash Flow Diagram: Computational Formula: 20012005 F

11. Case 5: Description : Finding the initial amount (P) that would yield specified uniform future amounts (A) over a given period. Interest rate changes at an infinite number of times within each year. Example: Jim takes the car now. He has enough money to pay for it, but rather decides to pay in annual installments over a 5-year period (now till 2005). How much should he set aside now so that he can make such annual payments? Equivalence Equations II

12. CASE 5 (cont’d) Problem Definition: This is a Uniform Series Present Worth (USPW) problem Cash Flow Diagram: Computational Formula: 20012005 P AAAAA

13. Equivalence Equations II Case 6: Description: Finding the amount uniform annual payments (A) over a given period, that would completely recover an initial amount (P). E.g. credit card monthly payments. Interest rate changes at an infinite number of times within each year. Example: Jim receives a loan of $20,000 from PEFCU to pay for the car now. How much will he have to pay to the bank every year?

14. Equivalence Equations II CASE 6 (cont’d) Problem Definition: This is a Capital Recovery (CR) problem. (The bank seeks to recover its capital from Jim.) Cash Flow Diagram: Computational Formula: 2001 2005 P AAAAA