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Regulation Z – 12 CFR 1026

Alan Dombrow CFPB, Retired FRBNY, retired OCC, retired Orlando FL, just tired. Regulation Z – 12 CFR 1026. Section 1026.22(a) Appendix J. May 20, 2014.

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Regulation Z – 12 CFR 1026

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  1. Alan Dombrow CFPB, Retired FRBNY, retired OCC, retired Orlando FL, just tired Regulation Z – 12 CFR 1026 Section 1026.22(a) Appendix J May 20, 2014

  2. This slide show was created by Alan Dombrow. Alan was an operations officer with Bank of America (1968-2003) and worked in the consumer examination related departments of the Office of the Comptroller of the Currency (1973-1997), Laser Pro/Harland Financial Solutions (1997-2001), the Federal Reserve Bank of New York (2002-2011) and the Consumer Financial Protection Bureau (2011-2013). He is now retired but, as evidenced by this slide show, not inactive.The views expressed in this presentation are those of Alan Dombrow and do not necessarily represent the views of any regulator responsible for interpreting or enforcing Regulation Z. I am not an attorney and the contents of the slide show should not be considered or relied upon as legal or tax advice. For legal or tax advice, please contact a qualified attorney or tax adviser. This slide show does not include an electronic signature for E-SIGN. Although Alan relied on his understanding of Regulation Z (12 CFR 1026), Appendix J to the regulation in particular, the Commentary to Regulation Z and on his own experience writing annual percentage rate programs and calculations, he makes no warranty of complete or legal accuracy.Appendix J is complicated and perhaps mostly of interest to programmers who write code for creditors to calculate annual percentage rates. Alan hopes this slide show will be useful to programmers and anyone else interested in understanding the underlying rules of Regulation Z that are associated with computing accurate annual percentage rates.Anyone may use or copy this slide show at no cost. Alan wishes this work to be in the public domain. Caveat

  3. What’s in this slide show? • Definition of APR (Regulation Z) • Reference to Appendix J to Regulation Z • Description of U.S. Rule method • Description and rules applicable to actuarial method • Time intervals • Unit periods • Fractional unit periods • Definition of “term of the transaction” • General equation (b)(8) for all closed-end consumer loans • How to compute the actuarial APR for closed-end credit

  4. What’s an Annual Percentage Rate? § 1026.22 Determination of annual percentage rate (a) Accuracy of annual percentage rate (1) The annual percentage rateis a measure of the cost* of credit, expressed as a yearly rate, that relates the amount and timing of value received by the consumer to the amount and timing of payments made. The annual percentage rate shall be determined in accordance with either theactuarial method or the United States Rule method . . . * Annual percentage ratemeasures the total finance charge relative to theamount financed, as defined by federal law (12 CFR 1026.4(a) and 12 CFR 1026.18(b)(2), respectively). Simple interest ratemeasures the interest relative to the principal loan amount, as defined and regulated by individual states.

  5. Which $50,000 loan below is the best bargain? Loan A $50,000 Loan amount 8% Annual simple interest rate $1,013.82 Monthly payment amount 60 Number of payments $2,000 Loan origination fee $10,829.20 Total interest Loan B $50,000 Loan amount 7.8% Annual simple interest rate $1,009.41 Monthly payment amount 60 Number of payments $3,000 Loan origination fee $10,564.60 Total interest Loan C $50,000 Loan amount 8% Annual simple interest rate $366.88 Monthly payment amount 360 Number of payments $2,000 Loan origination fee $82,076.80 Total interest Loan D $50,000 Loan amount 7.8% Annual simple interest rate $359.94 Monthly payment amount 360 Number of payments $3,000 Loan origination fee $79,578.40 Total interest

  6. Which $50,000 loan below is the best bargain? Can the APR help you decide? Loan A $50,000 Loan amount 8% Annual simple interest rate $1,013.82 Monthly payment amount 60 Number of payments $2,000 Loan origination fee $10,829.20 Total interest 9.74% APR Loan B $50,000 Loan amount 7.8% Annual simple interest rate $1,009.41 Monthly payment amount 60 Number of payments $3,000 Loan origination fee $10,564.60 Total interest 10.47% APR Loan C $50,000 Loan amount 8% Annual simple interest rate $366.88 Monthly payment amount 360 Number of payments $2,000 Loan origination fee $82,076.80 Total interest 8.44% APR Loan D $50,000 Loan amount 7.8% Annual simple interest rate $359.94 Monthly payment amount 360 Number of payments $3,000 Loan origination fee $79,578.40 Total interest 8.46% APR

  7. Which $50,000 loan below is the best bargain? Can the APR help you decide? Loan A $50,000 Loan amount 8% Annual simple interest rate $1,013.82 Monthly payment amount 60 Number of payments $2,000 Loan origination fee $10,829.20 Total interest 9.74% APR Loan B $50,000 Loan amount 7.8% Annual simple interest rate $1,009.41 Monthly payment amount 60 Number of payments $3,000 Loan origination fee $10,564.60 Total interest 10.47% APR Loan C $50,000 Loan amount 8% Annual simple interest rate $366.88 Monthly payment amount 360 Number of payments $2,000 Loan origination fee $82,076.80 Total interest 8.44% APR Loan D $50,000 Loan amount 7.8% Annual simple interest rate $359.94 Monthly payment amount 360 Number of payments $3,000 Loan origination fee $79,578.40 Total interest 8.46% APR THE WINNER!!

  8. Loan A Interest rate = 8% APR = 9.74% Interest: $10,829.20 Loan B Interest rate = 7.8% APR = 10.47% Interest: $10,564.60 Loan C Interest rate = 8% APR = 8.44% Interest: $82,076.80 Loan D Interest rate = 7.8% APR = 8.46% Interest: $79,578.40 Why is Loan C the best bargain? Loan C $50,000 Loan amount 8% Annual simple interest rate $366.88 Monthly payment amount 360 Number of payments $2,000 Loan origination fee $82,076.80 Total interest 8.44%APR THE WINNER!! • Loan C is the best bargain because the APR takes into account: • The total cost of credit (interest + loan fee in this case) • The amount of the advance and how long the consumer has use of that advance • The amounts and timing of payments • If the loan is repaid as agreed, the lower the APR, the lower the total cost of credit.Loan C has the lowest APR so Loan C is the best bargain, even though the interest rate on Loan C is higher than on Loans B & D, and the amount of interest on Loan C is higher than on Loans A, B & D.

  9. So how do we calculate an APR? • Regulation Z provides a uniform method and formulas to calculate APRs under the actuarial method • Uniformity is important if the cost of any one loan can be compared effectively to another loan • The uniform* rules for calculating APRs on closed-end credit are set forth in Regulation Z and Appendix J to Regulation Z * Complete uniformity does not exactly exist, due to exceptions and tolerances in Regulation Z, but there is significant uniformity that did not exist before the Truth in Lending Act was enacted in 1968.

  10. Appendix J • Provides a brief description of APRs under the U.S. Rule method • Explains APRs under the actuarial method • Contains equations for calculating the actuarial APR • Includes instructions, with examples, on how the actuarial method is applied to single advance and multiple advance transactions

  11. Appendix J: U.S. Rule Method • At the end of each payment period, the unpaid balance of the amount financed is increased by the finance charge earned during that payment period and is decreased by the payment made at the end of that payment period. • If the payment is less than the finance charge earned, the adjustment of the unpaid balance of the amount financed is postponed until the end of the next payment period. • If at that time the sum of the two payments is still less than the total earned finance charge for the two payment periods, the adjustment of the unpaid balance of the amount financed is postponed still another payment period, and so forth.

  12. Appendix J: U.S. Rule Method • Example 1For this example, the actuarialAPR is also 6% (5.996% = 6%) $1,000.00 Amount financed $ 336.67 Monthly payment 3 Number of payments 6.00% U.S. Rule APR Month 1: 1,000 x (.06/12) = 5.00 Finance Charge 1,000 + 5 – 336.67 = 668.33 Balance Month 2: 668.33 x (.06/12) = 3.34 Finance Charge 668.33 + 3.34 – 336.67 = 335.00 Balance Month 3: 335 x (.06/12) = 1.68 Finance Charge 335 + 1.68 – 336.68 = 0.00 Balance (months are equal) 1,000 336.67 336.67 336.68 Total FC = $10.02 5 + 3.34 + 1.68 = $10.02

  13. Appendix J: U.S. Rule Method • Example 2 $1,000.00 Amount financed $ 3.00 Two monthly payments $1,009.00 Final payment 6.00% U.S. Rule APR Month 1: 1,000 x (.06/12) = 5.00 Finance Charge 1,000 + 3 – 3 = 1,000.00 Balance (postpone $2) Month 2: 1,000 x (.06/12) = 5.00 Finance Charge 1,000 + 3 – 3 = 1,000.00 Balance (postpone $2 more) Month 3: 1,000 x (.06/12) = 5.00 Finance Charge 1,000 + (5 + 2 + 2) – 1,009 = 0.00 Balance 1,000 (months are equal) 3 3 1,009 Total FC = $15.00 3 + 3 + 9 = $15.00 5 + 5 + 5 = $15.00 (Unpaid FC from prior periods)

  14. Appendix J: U.S. Rule Method • Example 3 (For this example, the actuarial APR is 11.36% - see slide 22) (months are equal) $1,000.00 Amount financed $1,240.00 One payment due in 2 years 12.00% U.S. Rule APR 1,000 x .12 x 2 = 240 Finance Charge 1,000 + 240 – 1,240 = 0.00 Balance 1,000 Year 1 Year 2 1,240

  15. Appendix J: Actuarial Method1 • At the end of each unit period2 (or fractional unit period) the unpaid balance of the amount financed is increased by the finance charge earned during that period and is decreased by the total payment (if any) made at the end of that period. 1Most federal financial institution regulators use the actuarial method to calculate and verify APRs. 2 Understanding the term “unit period” is essential to understanding the actuarial method under Appendix J to Regulation Z.

  16. Appendix J: Actuarial Method • Time intervals • A period is the interval of time between advances or between payments • A common period is any period that occurs more than once in a transaction • A standard interval of time is a day, week, semi-month, month, or a multiple of a week or a month up to, but not exceeding, 1 year • All monthsshall be considered equal

  17. Appendix J: Actuarial Method • Unit period • The unit period is the common period, not to exceed 1 year, that occurs most frequently in the transaction • Exceptions • If 2 or more common periods occur with equal frequency, the smaller common period is the unit period

  18. Appendix J: Actuarial Method • Unit period (cont’d) • Exceptions (cont’d) • If there is no common period, the unit-period is the average of all periods rounded to the nearest whole standard interval of time. If the average is equally near 2 standard intervals of time, the lower shall be the unit-period. • In a single advance, single payment transaction, the unit-period is the term of the transaction, but shall not exceed 1 year.

  19. Appendix J: Actuarial Method • Unit period examples • Single advance, monthly payments • Unit period = one month • Single advance, quarterly payments • Unit period = three months • Single advance, three monthly payments and three quarterly payments • Unit period = one month

  20. Appendix J: Actuarial Method • Unit period examples (Cont’d) • Single advance, with first payment due in one month, second payment due two months after the first payment and the last payment due three months after the second payment • Unit period = 2 months Single advance 1 + 2 + 3 = 6 6/3 = 2 months Two months One month Three months Pmt 1 Pmt 3 Pmt 2

  21. Appendix J: Actuarial Method • Unit period examples (Cont’d) • Single advance, single payment • Payment due in 8 months • Unit period = 8 months Annual simple interest rate = 12% If loan amount = amount financed and there areno prepaid finance charges, then U.S. Rule andactuarial APR = 12% Single advance $1,000 (AF) Eight months Eight months Principal x periodic interest rate x time = interest $1,000 x .12/12 x 8 = $80 Payment = principal + interest = $1,080 Amount financed x periodic APR x time = finance charge $1,000 x .12/12 x 8 = $80 Payment = amount financed + finance charge = $1,080 $1,080 Pmt 1

  22. Appendix J: Actuarial Method • Unit period examples (Cont’d) • Single advance, single payment • Payment due in 24 months • Unit period = 1 Year (there are 2 full unit periods from loan date to date of single payment) Year 1 1,000 x .113553 = 113.55Finance Charge 1,000 + 113.55 = 1,113.55 Balance Year 2 1,113.55 x .113553 = 126.45Finance Charge 1,113.55 + 126.45 = 1,240.00 1240 - 1,240 = 0.00 Balance $1,000 (Loan amount and AF) 1,113.55 (bal.) 1,240 (bal.) Year 2 Year 1 $1,240 (single payment) Annual simple interest rate = 12% APR = 11.3553 = 11.36% (See slide 14) Interest = $240 (1,000 x .12 x 2) Finance charge = $240 ($113.55 + 126.45)

  23. Appendix J: Actuarial Method • Fractional unit periods • If unit period = a month, 12 unit periods per year • Number of full unit periods between 2 dates is the number of months measured back from the later date. • The remaining fraction of a unit period is the number of days measured forward from the earlier date to the beginning of the first full unit period, divided by 30. February 12 April 1 – March 1 = one unit period (one month) February 12 – March 1 = 17 days = 17/30 fractional unit period May 1 June 1 July 1 April 1 March 1 Pmt Pmt Pmt Pmt 17 days No Pmt

  24. Appendix J: Actuarial Method • Fractional unit periods (Cont’d) • If unit period = a semi-month or a multiple of a month not exceeding 11 months, the number of days between 2 dates shall be 30 times the number of full months measured back from the later date, plus the number of remaining days Loan date = February 12 Payments due quarterly, beginning July 1 July 1 – March 1 = 4 full months = 4 x 30 = 120 days February 12 – March 1 = 17 days Loan date to first payment date = 120 + 17 = 137 days to the first payment February 12 July 1 October 1 March 1 Pmt 2 Pmt 1 4 months (120 days) 17 days 137 days

  25. Appendix J: Actuarial Method • Fractional unit periods (Cont’d) • If unit period = a semi-month or a multiple of a month not exceeding 11 months, the number of full unit periods and the remaining fraction of a unit period shall be determined by dividing the number of days between two dates by 15 in the case of a semimonthly unit period or by the appropriate multiple of 30 in the case of a multi-monthly unit period February 12 Loan date = February 12 Payments due quarterly, beginning July 1 July 1 – March 1 = 4 full months = 4 x 30 = 120 days February 12 – March 1 = 17 days 120 + 17 = 137 days; 137/90 = one unit period (90 days)+ 47 days to the first payment = one unit period plus 47/90 fractional unit period July 1 March 1 October 1 Pmt 2 Pmt 1 30 days + 1 unit period (120 days/30 = 3 months + 30 days) 17 days 17 days + 30 days = 47 days = 47/90 fractional unit period

  26. Appendix J: Actuarial Method • Fractional unit periods (Cont’d) • If the unit period is a semi-month, the number of unit periods per year shall be 24 • Example: Payments are due on the 1st and 16th of each month or on the 15th and last day of each month • If the number of unit periods is a multiple of a month, the number of unit periods per year shall be 12 divided by the number of months per unit period • Example: If payments are quarterly, the unit period is 3 months and the number of unit periods per year is 4 (12/3 = 4)

  27. Appendix J: Actuarial Method • Fractional unit periods (Cont’d) • If the unit period is a day, a week, or a multiple of a week, the number of full unit periods and the remaining fractions of a unit period shall be determined by dividing the number of days between the 2 given dates by the number of days per unit period • Example: Payments are due bi-weekly; unit period = two weeks February 12 19 Days March 3 March 17 Loan date = February 12 Payments due biweekly, beginning March 3 March 3 – February 12 = 19 days 19 – 14 = 5 days = one unit period (14 days)+ 5 days to the first payment = one unit period plus 5/14 fractional unit period to the first payment Pmt 2 Pmt 1 1 unit period (14 days) 5 days

  28. Appendix J: Actuarial Method • Fractional unit periods (Cont’d) • If the unit period is a day, the number of unit periods per year shall be 365. If the unit period is a week or a multiple of a week, the number of unit periods per year shall be 52 divided by the number of weeks per unit period. • Example: Payments are due bi-weekly; unit period = two weeks; Number of unit periods per year = 26 (52/2)

  29. Appendix J: Actuarial Method • If the unit period is a year, the number of full unit-periods between 2 dates shall be the number of full years (each equal to 12 months) measured back from the later date • The remaining fraction of a unit-period shall be • The remaining number of months divided by 12 if the remaining interval is equal to a whole number of months*, or • The remaining number of days divided by 365 if the remaining interval is not equal to a whole number of months * Under the U.S. Rule method, the remaining number of days may be divided by 365, even if the remaining interval is equal to a whole number of months

  30. Appendix J: Actuarial Method • Single advance, single payment transaction • If the term is less than a yearand equalto whole number of months, the number of unit periods in the term is 1, and the number of unit periods per year is 12 divided by the number of months in the term or 365 divided by the number of days in the term. • Example: Payment is due in 6 months (182 days); unit period = 1 Number of unit periods per year = 2 (12/6), or Number of unit periods per year = 2.0055 (365/182)

  31. Appendix J: Actuarial Method • Single advance, single payment transaction (Cont’d) • If the term is less than a yearand not equal to a whole number of months, the number of unit periods in the term is 1, and the number of unit periods per year shall be 365 divided by the number of days in the term. • Example: Payment is due in 200 days; unit period = 1 Number of unit periods per year = 1.825 (365/200)

  32. Term of the Transaction • The term of the transaction begins on the date of its consummation, except: • If the finance charge or any portion of it is earned beginning on a later date, the term begins on the later date. • The term ends on the date the last payment is due, except: • If an advance is scheduled after the date of the last payment, the term ends on the later date. • For computation purposes: • The length of the term shall be equal to the time interval*between any point in time on the beginning date to the same point in time on the ending date. * Day, week, semi-month, month, or a multiple of a week or a month

  33. Term of the Transaction • Examples • Single advance, multiple payments Installment Loan 5/1/14 Date of consummation 5/1/14 Date advance made available for consumer’s use 5/1/14 Date FC begins to be earned (term begins) 6/1/14 Date of first P&I payment 5/1/19 Date of final payment (term ends) Term of loan 5/1/14 – 5/1/19 = 5 years

  34. Term of the Transaction • Examples (Cont’d) • Single advance, multiple payments Rescindable Transaction 5/1/14 Date of consummation 5/7/14 Date advance made available for consumer’s use 5/7/14 Date FC begins to be earned (term begins) 6/1/14 Date of first P&I payment 5/1/19 Date of final payment (term ends) Term of loan 5/7/14 – 5/1/19 = 4 years, 11 months and 25 days

  35. Term of the Transaction • Examples (Cont’d) • Multiple advances, multiple payments Multiple-advance Loan 5/1/14 Date of consummation 5/1/14 Date first advance made available for consumer’s use 5/1/14 Date FC begins to be earned (term begins) 6/1/14 Date of first payment (interest only) 12/1/14 Date of first P&I payment 5/1/19 Date of final payment (term ends) Term of loan 5/1/14 – 5/1/19 = 5 years

  36. Term of the Transaction • Examples (Cont’d) • Multiple advances, multiple payments Multiple-advance Loan 5/1/14 Date of consummation 5/1/14 Date first advance made available for consumer’s use 5/1/14 Date FC begins to be earned (term begins) 6/1/14 Date of first payment (interest only) 12/1/14 Date of first P&I payment 5/1/19 Date of final payment 6/1/19 Date of final advance(term ends) Term of loan 5/1/14 – 6/1/19 = 5 years and 1 month

  37. General Equation (b)(8) Simplified • Formula • Example • $1,000 loan with 3 monthly payments at $340.02 each • If you know amount financed and payments, you can solve for i • Multiply i by the number of unit periods per year (w) = APR 1,000 = 340.02 340.02 340.02 (1+.00999685359)1 (1+.00999685359)2 (1+.00999685359)3 A = 1,000 (amount financed); P = 340.02; w= 12; i= 0.00999685359 APR = w i x 100 APR= 12 x 0.00999685359 x 100 =12.00% + + One advance Three payments A1 = P 1 P2 P3 Pn (1+i)t1 (1+i)t2 (1+i)t3(1+i)tn . . . + . . . + +

  38. Appendix J: General Equation (b)(8)(the hard way ) • General equation (b)(8) semi-simplified A1 = P 1 P2 P3 Pn (1+i)t1 (1+i)t2 (1+i)t3(1+i)tn . . . + . . . + + 1,000 = 33.61 33.61 33.61 33.6136 (1+.010688)1 (1+.010688)2 (1+.010688)3 (1+.010688)36 i = 0.010688 APR= i x 12 = 0.010688 x 12 x 100 =12.83% . + . + +

  39. Appendix J: Calculate APR • How to solve for APR • GUESS! • AKA “ITERATION” • Appendix J begins with: • w = no. unit periods per year • i = periodic APR (decimal equivalent) • I = wi x 100 = estimated APR • Monthly payment loan • w = 12 • First guess “i” = .010416667 • I = 12.50% (12 x .010416667x100) • First guess APR: 12.50% • Last guess APR: 12.83% APR (based on interpolation) • 12.83 ÷ 12÷100 = .010688= i (12 x .010688 x 100 = 12.83% = I) 1,000 = 33.61 33.61 33.61 33.6136 (1+.010688)1 (1+.010688)2 (1+.010688)3 (1+.010688)36 . + . + +

  40. Appendix J: Calculate APR • Example (c)(i) t = 1 (full unit periods from loan date to first payment date) f = 0 (i.e., no odd days); W = 12 i = periodic rate = .0080714 wi = 12 x .0080714 = .0969 APR = I = .0969% x 100 = 9.69% Single advance /// 5,000 = 230 230 230 23024 (1+.0080714)1 (1+.0080714)2 (1+.0080714)3 (1+.0080714)24 A = P P P P24 (1+i)1 (1+i)2 (1+i)3 (1+i)24 . + . . + . + + + + One month One month One month Pmt 2 Pmt 1 Pmt 24

  41. What if there are odd days? • Let’s go back to the simplified formula on slide 37: • This formula assumes a full unit period between the loan date and the first payment, as well as between monthly payments Unit period = 1 month In the denominator, the exponentials, 1, 2 and 3, represent 1 whole unit period (1 whole month) to the first payment, 2 whole unit periods (2 whole months) to the second payment and 3 whole unit periods (3 whole months) to the third payment. A1 = P 1 P2 P3 Pn (1+i)t1 (1+i)t2 (1+i)t3(1+i)tn . . . + . . . + + 1,000 = 340.02 340.02 340.02 (1+.00999685359)1 (1+.00999685359)2 (1+.00999685359)3 A = 1,000 (amount financed); P = 340.02; w= 12; i= 0.00999685359 APR = w i x 100 APR= 12 x 0.00999685359 x 100 =12.00% + +

  42. What if there are odd days? • Let’s now assume that there are 15 days and one month to the first payment: Loan date: 4/16/14 First payment date: 6/1/14 6/1/14 back to 5/1/14 = one month (1 unit period) 4/16/14 forward to 5/1/14 = 15 days (long odd days) 15 days = fraction of unit period = 15/30 = ½ unit period April 16 15 days + 1 month to first payment 15 days + 2 months to second payment 15 days + 3 months to third payment August 1 May 1 June 1 July 1 Pmt3 Pmt2 Pmt1 1 unit period (1 month) 15 days

  43. What if there are odd days? • General equation (b)(8) with odd days: • Simplified equation (b)(8) with long odd days (3 monthly payments): A1 = P 1 P2 P3 Pn (1+f1i)(1+i)t1 (1+f2i)(1+i)t2 (1+f3i) (1+i)t3(1+fni) (1+i)tn A1 = P 1 P2 P3 (1+f1i)(1+i)t1 (1+f2i)(1+i)t2 (1+f3i) (1+i)t3 . . . + . . . + + + + Note the introduction of the variable “f”; f = .5 (or ½) when there are 15 long odd days on a monthly payment loan (15/30 = .5)

  44. What if there are odd days? • Assume there are 15 days and one month to the first payment: Loan date: 4/16/14 First payment date: 6/1/14 6/1/14 back to 5/1/14 = one month (1 unit period) 4/16/14 forward to 5/1/14 = 15 days (long odd days – fractional unit period) f = .5 (15/30 = ½ unit period ) In the denominator, the exponentials, 1, 2 and 3, represent 1 whole unit period (1 whole month) to the first payment, 2 whole unit periods (2 whole months) to the second payment and 3 whole unit periods (3 whole months) to the third payment. In each case, however, there remain 15 extra odd days (1/2 unit period) between the loan date and the payment date. • 1,000 = 340.02 340.02 340.02 • [1+(.5)(.008)](1+.008)1[1+(.5)(.008)](1+.008) 2 [1+(.5)(.008)](1+.008)3 • A = 1,000 (amount financed); P = 340.02; w= 12; i= 0.00798153021* • f = .5 (one-half unit period); APR= w i x 100 • APR= 12 x 0.00798153021 x 100 =9.577836248% = 9.58% + + * To simplify and shorten the example for fractional unit periods above, the periodic rate in the denominator (i = 0.0079815302) is rounded to .008; actual calculations should use all decimals available until the final APR is obtained and rounded to the nearest 100th of one percentage point (e.g. 9.577836248% = 9.58% APR).

  45. What if there are odd days? • Example (c)(ii) t = 1 (full unit periods from loan date to first payment date) f = 19/30 (i.e., 19 odd days); W = 12 i = periodic rate = .00984709 (simplified below to .0098) wi = 12 x .00984709 = .11816508 APR = .1182% x 100 = 11.82% Single advance (2-10-78) 6,000 = 200 200 20036 [1+(19/30)(.0098)] (1+.0098) 1 [1+(19/30)(.0098)] (1+.0098)2 [1+(19/30)(.0098)] (1+.0098) 36 A = P P P P36 (1+fi1)(1+i)1(1+fi2)(1+i)2(1+fi3)(1+i)3 (1+fi36)(1+i)36 . . . + . . . . + . + + + /// One month One month One month Pmt 2 Pmt 3 Pmt 1 (4-1-78) Pmt 36 19 days

  46. Appendix J 10 REM: APR PROGRAM (Actuarial Method) - BASIC 20 CLS : DEFINT I-N: DEFDBL A-H, O-Z 30 DIM D(400, 4) 35 REM: The number 400 means 400 payment streams are possible 40 INPUT "Enter number of days in unit period................: ", U 50 IF INT(U) = U AND U > 0 AND U <= 365 THEN GOTO 70 60 GOTO 40 70 INPUT "Enter number of days (360-364-365) in year...........: ", YR 80 IF INT(YR) = YR AND YR = 360 OR YR = 364 OR YR = 365 THEN 100 90 GOTO 70 100 INPUT "Enter AMOUNT FINANCED............: ", P 110 INPUT "Enter disclosed (or estimated) APR............: ", W 120 U1 = U / YR 130 FOR N = 1 TO 400 135 REM: The number 400 means 400 payment streams are possible 140 PRINT 145 IF N = 1 THEN INPUT "Enter payment amount (must not be zero)...............: ", D(N, 0) IF D(N, 0) <> 0 THEN GOTO 200 ELSE GOTO 145 END IF 150 INPUT "Enter payment amount (or 0 to compute APR)............: ", D(N, 0) 160 IF D(N, 0) = 0 AND N > 1 THEN 310 170 IF D(N, 0) <> 0 THEN 200 180 IF N = 1 THEN 140 190 GOTO 150 200 INPUT "Enter No. of payments (in this stream)...........: ", D(N, 1) 210 NP = D(N, 1) 215 IF D(N, 1) AND D(N, 1) > 0 AND INT(D(N, 1)) = D(N, 1) THEN 230 220 GOTO 200 230 IF D(N, 0) < 0 THEN PMT = ABS(D(N, 0)) : Q = PMT * NP: S = S + Q 240 INPUT "Enter whole unit periods to first payment in this stream: ", D(N, 2) 250 IF INT(D(N, 2)) = D(N, 2) AND INT(D(N, 2)) >= 0 THEN 270 260 GOTO 240 270 INPUT "Enter No. of 'odd' days..............: ", D(N, 3) 280 IF INT(D(N, 3)) = D(N, 3) AND D(N, 3) >= 0 THEN 300 290 GOTO 270 300 NEXT N 310 PRINT 315 PRINT USING " AMOUNT FINANCED = #########,.##"; P + S 320 T = 0: N = N - 1 325 FOR J = 1 TO N: D(J, 4) = D(J, 0): T = T + D(J, 4) * D(J, 1): NEXT J 330 PRINT USING " FINANCE CHARGE = ##########,.##": T - P 340 PRINT USING " TOTAL OF PAYMENTS = ##########,.##": T + S 350 R = W * U1 / 100! 355 REM: Lines 360-380 are used to iterate for the annual percentage rate 360 X = .0001: IF R <> 0! THEN X = R 370 GOSUB 500: P1 = P0: X = R + .0001: GOSUB 500 375 APR = R + (P - P1) / (PO - P1) * .0001 380 A = ABS(APR - R): IF A > .0000001 THEN R = APR: GOTO 360 390 APU = (100! / U1) * APR: REM: APU = the annual percentage rate 400 PRINT 405 PRINT USING " ANNUAL PERCENTAGE RATE = ####.####": APU 410 PRINT : PRINT : INPUT "Do you wish to do another (Y/N)? ", ANS$ 420 ANS$ = LEFT$(ANS$, 1): IF ANS$ = "y" OR ANS$ = "Y" THEN 40 430 IF ANS$ = "n" OR ANS$ = "N" THEN END 440 GOTO 410 450 REM SUBROUTINE FOR AGGREGATE PRESENT VALUE OF ALL PAYMENTS (P0) 500 P0 = 0! : FOR J = 1 TO N: X1 = 1! + X: Y = 1! + D(J, 3) * X / U 510 V1 = (1! / X1 " D(J, 2)) / Y: V2 = 1! / X1 " D(J, 1) : V3 = 1! - V2 520 Z = V1 * X1 * (D(J, 0) * V3 / X): P0 = P0 + Z: NEXT J: RETURN • General equation (b)(8) • How can I convert general equation (b)(8) above into language a computer would understand? • The BASIC programming code on the right: • Calculates the APR • Follows general equation (b)(8) • Does not include input error-proofing, construction loans, reimbursement or bells and whistles; it’s core programming only • Must be “compiled” for greater accuracy BASIC Language

  47. OCC’s APRWin Program • First version written and compiled with MS-DOS operating system in GW-BASIC (1984) • Subsequently, the APR program was expanded and written for newer DOS operating systems in: • Quick BASIC • Visual BASIC • Visual BASIC Professional Development System • In 1997, the OCC converted the APR program into a Windows programming language (v 5.0) • The current version is 6.2 • This program may be downloaded for free on the OCC’s website: http://www.occ.gov/tools-forms/tools/compliance-bsa/aprwin-software.html

  48. Questions?

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