1. Paired analysis Provides market evidence for amount and direction of a particular adjustment Pair only two sales for the adjustment and check them against other sales Appraiser has to use some judgment “but judgment without market evidence is simply not acceptable appraisal practice.”

2. Impact of a paved road; Sale A on paved and Sale B on dirt Sale ASale B Selling price$800$750 Acres 160 160 FinancingCashCash Market conditions CurrentCurrent Size00 Location?? Land00 Buildings00 Adjusted price$800$750

3. Paired Comparison $50 per acre impact from the paved vs dirt road Appraiser will keep checking this amount as they find sales that let them make the comparison Once one adjustment has been ‘proved’ in the market it can be used for other comparisons Assume two similar sales except location and size

4. Size Comparison Sale CSale D Selling price$700$750 Acres320160 Financing CashCash MarketCurrentCurrent Size ?? Location-$500 Land00 Buildings00 Adjusted price$650$750

5. Size comparison Adjustment is $100 per acre assuming that the adjustment of $50 for location holds Next is time adjustment Remember with time adjustment that the percentage adjustment is a reflection of compounding too Sale two years ago at $1000, next year at $1,100 and this year at $1,200 What’s the percent change due to time?

6. Market conditions (time) Sale ESale F Selling Price$780$950 Acres320150 FinancingCashCash Market conditionsCurrent1.5 yrs. Size$1000 Location00 Land00 Buildings00 Adjusted Price$880$950

7. Market conditions (time) What is the percentage change due to time? – $880 - $950 = -$70 -$70/950 = -7.4%/1.5 = -4.9% Sale 1 yr. ago for $600,000 Sale this yr. for $742,000 but property had $65,500 in improvements

8. Comparison Sale A: 160 Ac. on Hyw. 6, two miles from Growthville, all Class II soils, no improvements, sold 1 year ago for $1,200, Cash

9. Sale A Sale B Sale C Sale D Subject Price $1,200 $1,000 $860 $905 Acres 160 140 180 150 160 Financing Cash Cash Loan Cash Cash Market 1 yr. Now Now 1yr. Now Location 2mi. 4 mi 6 mi. 7 mi. 6 mi. Land 0 0 0 0 0 Improvements 0 0 0 0 0 5% decrease in land values over past year

10. Steps First thing to do is to find a control sale. This is the sale as similar to the subject property as possible. Comparisons will be made to this sale Assume that we are interested in finding the market contribution of location using these sales. Time adjustments have to be made first Which is the most like the subject?

11. Pairs for Location Sale CSale A Selling Price$860$1,200 Acres180160 FinancingCashCash Market conditionsCurrent Size 00 Location6 miles2 miles Land00 Buildings00 Adjusted Price$$ Difference due to location

12. Pairs for Location Sale CSale B Selling Price$860$1,000 Acres180140 FinancingCashCash Market conditionsCurrentCurrent Size 00 Location6 miles4 miles Land00 Buildings00 Adjusted Price$$ Difference due to location

13. Pairs for Location Sale CSale D Selling Price$860$905 Acres180150 FinancingCashCash Market conditionsCurrent Size 00 Location6 miles7 miles Land00 Buildings00 Adjusted Price$$ Difference due to location

14. Improved land Appraisal of improved land starts with determining the value for each land class through evaluating sales of unimproved land These land classes will then be used to determine the value of the land to the sale and the residual will be the value of the improvements The value of the improvements will then be allocated among the buildings, etc.

15. Value of the land classes Assume that the appraiser knows that unimproved sales indicate this division for each land class – Class I100% – Class II 60% – Class III 40% What is the value of each land class if we had a sale for $260,000 with 100 acres of Class I, 200 acres of Class II and 100 acres of Class III.

16. Land class example Class I 100%, Class II 60% and Class III 40% 100 ac. Class I, 200 ac. Class II, 100 ac. Class III $260,000 sale price 100 * 100% = 100 200 * 60% = 120 100 * 40% = 40 260 $260,000/260 = $1,000 Class I $1,000 *.60 = 600 Class II $1,000 *.40 = 400 Class III