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PRESENTATION 9 Chapter 15
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## PRESENTATION 9 Chapter 15

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1. PRESENTATION 9Chapter 15 Real Estate Finance Mathematics

2. I. Approach to Solving Math Problems

3. Approach to Solving Math Problems • Solving math problems is simplified by using a step-by-step approach. • The most important step is to thoroughly understand the problem. • You must know what answer you want before you can successfully work any math problem. TQ • Once you have determined what it is you are to find (for example, interest rate, loan-to-value ratio, amount, or profit), you will know what formula to use.

4. (cont.) • The next step is to substitute the numbers you know into the formula. TQ • It may be necessary to take one or more preliminary steps, for instance, converting fractions to decimals.

5. (cont.) • Once you have substituted the numbers into the formula you will have to do some computations to find the unknown. • Most of the formulas have the same basic form: A=B x C • You will need two of the numbers (or the information that enables you to find two of the numbers) and then you will either have to divide or multiply them to find the third number—the answer you are seeking.

6. (cont.) • Whether you will need to multiply or divide is determined by which quantity (number) in the formula you are trying to discover. • For example, the formula A=B x C may be converted into three different formulas. All three formulas are equivalent, but are put into different forms, depending upon the quantity (number) to be discovered. • If the quantity A is unknown, then the following formula is used: A = B x C • The number B is multiplied by C; the product of B times C is A.

7. (cont.) • If the quantity B is unknown, the following formula is used: B = A ÷ C • The number A is divided by B; the quotient of A divided by B is C. C = A ÷ B • Notice that in all these instances, the unknown quantity is always by itself on one side of the “equal” sign.

8. II. Converting Fractions to Decimals

9. Converting Fractions to Decimals • There will be many times when you will want to convert a fraction into a decimal. • Most people find it much easier to work with decimals than fractions. • Also, hand calculators can multiply and divide by decimals. • To convert a fraction into a decimal, you simply divide the top number of the fraction (the “numerator”) by the bottom number of the fraction (the “denominator”).

10. Example: • To change 3/4 into a decimal, you must divide 3 (the top number) by 4 (the bottom number). 3 ÷ 4 = .75 • To change 1/5 into a decimal, divide 1 by 5. 1 ÷ 5 = .20 • If you are using a hand calculator, it will automatically give you the right answer with the decimals in the correct place.

11. To add or subtract by decimals,think “MONEY” \$ 101.18 line the decimals up by decimal point and add or subtract. • Example:

12. To multiply by decimals, do the multiplication. The answer should have as many decimal places as the total number of decimal places in the multiplying numbers. • Just add up the decimal places in the numbers you are multiplying and put the decimal point the same number of places to the left. • Example: 57.999 x 23.7 1374.5763

13. To divide by decimals, move the decimal point in the outside number all the way to the right and then move the decimal point in the inside number the same number of places to the right. • Example: 44.6 ÷ 5.889 44600 ÷ 5889 = 7.57

14. III. Percentage Problems

15. Percentage Problems • You will often be working with percentages in real estate finance problems. For example, loan-to-value ratios and interest rates are stated as percentages. • It is necessary to convert the percentages into decimals and vice versa, so that the arithmetic in a percentage problem can be done in decimals.

16. To convert a percentage to a decimal, remove the percentage sign and move the decimal point two places to the left. This may require adding zeros. • Example: 80% becomes .80 9% becomes .09 75.5% becomes .755 8.75% becomes .0875

17. To convert a decimal to percentage, do just the opposite. Move the decimal two places to the right and add a percentage sign. • Example: .88 becomes 88% .015 becomes 1.5% .09 becomes 9%

18. The word “of” means to multiply. • Whenever something is expressed as a percent of something, it means MULTIPLY. • Example: If a lender requires a loan-to-value ratio of 75% and a house is worth \$89,000, what will be the maximum loan amount? (What is 75% of \$89,000?) .75 x \$89,000 = \$66,750 maximum loan amount • Percentage problems are usually similar to the above example. You have to find a part of something, or a percentage of the total.

19. A general formula is: • A percentage of the total equals the part, or part = percent x total P = % x T

20. Example: • Smith spends 24% of her monthly salary on her house payment. Her monthly salary is \$2,750. What is the amount of her house payment? 1. Find amount of house payment. 2. Write down formula: P = % x T. 3. Substitute numbers into formula. P = 24% x \$2,750 • Before you can perform the necessary calculations, you must convert the 24% into a decimal. Move the decimal two places to the left: 24% = .24 P = .24 x \$2,750 4. Calculate: multiply the percentage by the total. .24 x \$2,750 = \$660 • Smith’s house payment is \$660.

21. IV. Problems Involving Measures of Central Tendency

22. Problems Involving Measures of Central Tendency • Appraisers, finance officers, and lenders evaluate data and information by the use of averages. • These average figures are known as the mean, the median, or the mode. It is useful to know how they are derived.

23. MEAN • The average figure that is called a MEAN is derived by taking a set of numbers and adding them up. The result is then divided by the numbers in the set. • Example: • A group of houses have the following monthly rental prices: Rental #1 – \$1,300 Rental #2 – \$900 Rental #3 – \$1,200 Rental #4 – \$1,100 Rental #5 – \$1,150 \$5,650 Total • To determine the Mean, the total rentals of \$5,650.00 are divided by the number of rental houses. Thus, \$5650 = \$ 1130 Mean monthly rental. 5

24. MEDIAN – Odd Number Example • The average figure that is described as a MEDIAN is derived by simply selecting the middle number in a set of numbers. The numbers need to be in ascending order first. • Example: Rental #1 – \$1,100 Rental #2 – \$1,150 Rental #3 – \$1,200 Rental #4 – \$1,200 = Median Rental #5 – \$1,250 Rental #6 – \$1,250 Rental #7 – \$1,300 • There are three numbers before Rental #4 and three numbers after. Therefore, rental #4 is the median number and the median rental price is \$1,200.

25. MEDIAN – Even Number Example • The example above had an odd number of rentals. The following example shows how to determine a mean with an even set of numbers. • Example: Rental #1 – \$1,100 Rental #2 – \$1,150 Rental #3 – \$1,200 Rental #4 – \$1,200 Rental #5 – \$1,250 Rental #6 – \$1,250 Rental #7 – \$1,275 Rental #8 – \$1,275 • The median is determined by adding the two middle rental numbers in the set and dividing the result by 2. • In this set the middle numbers are rentals number 4 and 5. \$1,200 \$2,450 = \$1,225 +\$1,250 Therefore: 2 \$2,450 • Thus, the median rent in this even number example is \$1,225.

26. MODE • A MODE measures the most frequently occurring number in a series of numbers. Appraisers and lenders often use this benchmark to determine the predominant value of housing in a neighborhood. • Example: Sale #1 – \$325,000 Sale #2 – \$328,000 Sale #4 – \$332,000 Sale #6 – \$335,000 Sale #7 – \$335,000 Sale #8 – \$340,000 Sale #9 – \$345,000 • There are two sales at \$335,000. This would be the Mode. It would also be the predominant value of this set of sales. • The value range states from the lowest number to the highest number, in a set of numbers. Thus, the range in the example above would be expressed as from \$325,000 to \$345,000.

27. V. Interest Problems

28. Interest Problems • Interest can be viewed as the “rent” paid by a borrower to a lender for the use of money (the loan amount, or principal). • INTEREST is the cost of borrowing money. • SIMPLE INTEREST • COMPOUND INTEREST

29. SIMPLE INTEREST • Simple interest problems are worked in basically the same manner as percentage problems, except that the simple interest formula has four components rather than three: interest, principal, rate, and time. Interest = Principal x Rate x Time I = P x R x T • Interest: The cost of borrowing expressed in dollars; money paid for the use of money. • Principal: The amount of the loan in dollars on which the interest is paid. • Rate: The cost of borrowing expressed as a percentage of the principal paid in interest for one year. • Time: The length of time of the loan, usually expressed in years.

30. One must know the number values of three of the four components in order to compute the fourth (unknown) component. • a. Interest unknown Interest = Principal x Rate x Time Example: Find the interest on \$3,500 for six years at 11%. I = P x R x T I = (\$3,500 x .11) x 6 I = \$385 x 6 I = \$2,310

31. b. Principal unknown Principal = Interest ÷ Rate x Time P= I ÷ (R x T) • Example: How much money must be loaned to receive \$2,310 interest at 11% if the money is loaned for six years? P = I ÷ (R x T) P = \$2,310 ÷ (.11 x 6) P = \$2,310 ÷ .66 P = \$3,500

32. c. Rate unknown Rate = Interest ÷ Principal x Time R = I ÷ (P x T) • Example: In six years \$3,500 earns \$2,310 interest. What is the rate of interest? R = I ÷ (P x T) R = \$2,310 ÷ (\$3,500 x 6) R = \$2,310 ÷ \$21,000 R = .11 or 11%

33. d. Time unknown Time = Interest ÷ Rate x Principal T = I ÷ (R x P) • Example: How long will it take \$3,500 to return \$2,310 at an annual rate of 11%? T = I ÷ (R x P) T = \$2,310 ÷ (\$3,500 x .11) T = \$2,310 ÷ \$385 T = 6 years

34. A. COMPOUND INTEREST • Compound interest is more common in advanced real estate subjects, such as appraisal and annuities. • As previously stated, compound interest is interest on the total of the principal plus its accrued interest. • For each time period (called the “conversion period”), interest is added to the principal to make a new principal amount. Therefore, each succeeding time period has an increased principal amount on which to compute interest. Conversion periods may be monthly, quarterly, semi-annual, or annual. • The compound interest rate is usually stated as an annual rate and must be changed to the appropriate “interest rate per conversion period” or “periodic interest rate.” To do this, you must divide the annual interest rate by the number of conversion periods per year. This periodic interest rate is called “i.”

35. COMPOUND INTEREST EXAMPLE: • The formula used for compound interest problems is: Interest = principal x periodic interest rate I = P x i

36. Example: • A \$5,000 investment at 9% interest compounded annually for three years earns how much interest at maturity? I = P x i I = \$5,000 x (.09 ÷ 1) • First year’s I = \$5,000 x .09 or \$450. Add to \$5,000. • Second year’s I = \$5,450 x .09 or 490.50. Add to \$5,450. • Third year’s I = \$5,940.50 x .09 or \$534.65. Add to \$5,940.50 • At maturity, the borrower will owe \$6,475.15. • The \$5,000 loan has earned interest of \$1,475.15 in three years.

37. Example: • How much interest will a \$1,000 investment earn over two years at 12% interest compounded semi-annually? • Since the conversion period is semi-annual, the interest is computed every six months. Thus, the periodic interest rate “i” is divided by two conversion periods: i = 6%. I = P x i 1. Original principal amount = \$1,000.00 2. Interest for 1st period (\$1,000 x .06) = \$60.00 3. Balance beginning 2nd period = \$1,060.00 4. Interest for 2nd period (\$1,060 x .06) = \$63.60 5. Balance beginning 3rd period = \$1,123.60 6. Interest for 3rd period (\$1,123.60 x .06) = \$67.42 7. Balance beginning 4th period = \$1,191.02 8. Interest for 4th period (\$1,191.02 x .06) = \$71.46 9. Compound principal balance = \$1,262.48 i for 2 years = \$1,262.48 - \$1,000 or \$262.48

38. B. EFFECTIVE INTEREST RATE • The NOMINAL (“NAMED”) INTEREST RATE is the rate of interest stated in the loan documents. • The EFFECTIVE INTEREST RATE/APR/ANNUAL PERCENTAGE RATE is the rate the borrower is actually paying. TQ • In other words, the loan papers may say one thing when the end result is another, depending upon how many times a year the actual earnings rate is compounded. • The effective interest rate equals the annual rate, which will produce the same interest in a year as the nominal rate converted a certain number of times. • For example, 6% converted semi-annually produces \$6.09 per \$100; therefore, 6% is the nominal rate and 6.09% is the effective rate. A rate of 6% converted semi-annually yields the same interest as a rate of 6.09% on an annual basis.

39. C. DISCOUNTS • In alternative methods of financing, the loan proceeds disbursed by the lender are often less than the face value of the note. • This occurs when the borrower (or a third party) pays discount points. Generally when a borrower wants a lower rate they “buy down” the rate. TQ • The lender is paid points up front as compensation for making the loan on the agreed terms, at a lower rate. • When a discount is paid, the interest costs to the borrower (and the yield to the lender) are higher than the contract interest rate.

40. DISCOUNT EXAMPLE: • When more accurate yield and interest tables are unavailable, it is possible to approximate the effective interest cost to the borrower and the yield rate to the lender when discounted loans are involved. • The formula for doing so is as follows: i = (r + (d/n)) ÷ (P - d) • i: approximate effective interest rate (expressed as a decimal) • r: contract interest rate (expressed as a decimal) • d: discount rate, or points deducted (expressed as a decimal) • P: principal of loan (expressed as the whole number 1 for all dollar amounts) • n: term (years, periods, or a fraction thereof)

41. Example: • What is the estimated effective interest rate on a \$60,000 mortgage loan, with a 20-year term, contract rate, if interest being 10% per annum, discounted 3%, so that only \$58,200 is disbursed to the borrower? i = .10 + (.03/20) = .10 + .0015 = .10150 = .10463 or 10.46% 1 - .03 .97 .97 • The effective interest rate (or yield) on the loan is 10.46%.

42. VI. Profit and Loss Problems

43. Profit and Loss Problems • Every time a homeowner sells a house, a profit or loss is made. • Many times you will want to be able to calculate the amount of profit or loss. • Profit and loss problems are solved with a formula that is a variation of the percent formula: value after = percentage x value before. VA = % x VB • The VALUE AFTER is the value of the property after the profit or loss is taken. • The VALUE BEFORE is the value of the property before the profit or loss is taken. • The PERCENT is 100% plus the percent of profit or minus the percent of loss.

44. Example: • Green bought a house ten years ago for \$50,000 and sold it last month for 45% more than she paid for it. What was the selling price of the house? VA = % x VB VA = 145% x VB (To get the percent, you must add the percent of profit to or subtract the percent of loss from 100%). VA = 1.45 x \$50,000 VA = \$72,500 was the selling price

45. Profit means TAXES (Rentals) • Taking a profit can be even more expensive than what your tax rate is on your profit • What about recaptured depreciation? • Can be very expensive • More the longer you hold the property • Just be well informed when investing and doing tax writeoffs • PERSONAL RESIDENCE (Still has a non-recaptured interest deduction) • The government want’s that money though!!

46. STILL NO TAX ON PERSONAL RESIDENCE PROFITS • Be aware, if that deduction is taken, the incentive for people to buy will be greatly decreased! • If a bill is introduced in congress you might want to let your senator and representative know you are against it!

47. Example: • Now we will use the profit and loss formula to calculate another one of the components. • Green sold her house last week for \$117,000. She paid \$121,000 for it five years ago. What was the percent of loss? VA = % x VB \$117,000 = % x \$121,000(Because the percent is the unknown, you must divide the value after by the value before.) % = \$117,000 ÷ \$121,000 % = .9669 or 97% (rounded) Now subtract 97% from 100% to find the percent of loss. % = 100% - 97% = 3% loss

48. Example: • Your customer just sold a house for 17% more than was paid for it. The seller’s portion of the closing costs came to \$4,677. The seller received \$72,500 in cash at closing. What did the seller originally pay for the house? VA = % x VB \$72,500 + 4,677 = 117% x VB VB = (\$72,500 + 4,677) ÷ 117% (Since the value before is 117% unknown, you must divide the value after [the total of the closing costs and the escrow proceeds] by the percent of profit.) VB = \$77,177 ÷ 1.17 VB = \$65,963.25 was the original price

49. VII. Prorations

50. Prorations - Intro • There are some expenses connected with owning real estate that are often either paid for in advance or in arrears. • For example, fire insurance premiums are normally paid for in advance. Landlords usually collect rents in advance, too. • On the other hand, mortgage interest accrues in arrears. • When expenses are paid in advance and the owner then sells the property, part of these expenses have already been used up by the seller and are rightfully the seller’s expense. • Often, however, a portion of the expenses of ownership still remain unused and when title to the property transfers to the buyer, the benefit of these advances will accrue to the buyer. • It is only fair that the buyer, therefore, reimburse the seller for the unused portions of these homeownership expenses.