1. Math

2. Metes and Bounds Metes & Bounds = Distances & Directions
Do not replace surveys
Metes- Limit or limiting mark (distance)
Bounds- Boundary lines (directions)
This is a process that consists of minutes, degrees and seconds and are set out in compass directions
A system of written land descriptions where all boundary lines are described using terminal points and angles
All metes and bounds as recorded in land registry

3. Metes and Bounds ExamplesW E C B S
The line that travels in the direction from C  D has an approximate bearing of North Forty-Five Degrees East
A circle contains 360 degrees
Four quadrants each with 90 degrees
The line traveling from C  D proceeds into the northeast quadrant at approximately the midpoint
The bearing will describe a northerly direction measured from North-South line that represents 0 degrees and towards the East is 45 degrees

4. The reverse (reciprocal) bearing of line CD-
South Forty-Five Degrees West
The reverse bearing merely travels in the opposite direction through the southwest quadrant at approximately the midpoint
The bearing will describe a southerly direction measured from the North-South line representing 0 degrees toward the west – 45 Degrees
The Line travelling from BA has an approximate bearing of North Forty-Five Degrees West
BA goes into the Northwest Quadrant at the mid-point
The bearing will describe a Northwest direction measured from the North-South line that represents 0 degrees toward the West at 45 degrees

5. Methods of Finding Value of Property
Direct Comparison Approach
Cost Approach
Income Approach

6. 1. Direct Comparison Approach
Obtain the information
Find the differences by analyzing the data that exists between two subject properties and comparable properties
Compare with the subject property with comparing it to the subject property and make any adjustments.
Less is More and More is Less
When the comparable property has More features than the subject property, Subtract less from the comparable
When the comparable property has less features than the subject property, add more from the comparable
Reconcile/ Final estimate which is finding the best estimate
Check all calculations
Give more preference to the most recent sale with less number of adjustments
Do not make adjustments for properties that were sold less than one month ago

7. Four Types of Adjustments
Time Adjustment
Location Adjustment
Lot Size Adjustment
Physical Characteristics Adjustments

8. Selecting Comparables
Market Value: Comparable property has to have a reasonable value that is compared to the market
Time & Market Conditions: Something that was sold closer to today’s date
Proximity: Comparable property should be located on same street or close to subject property
Similarity: Comparable property should similar to subject property so the less adjustments are needed to be made
Adjusted Sales Price = ASP
Prices of the comparable after all the necessary adjustments

9. Adjustment Calculations
TA – Time Adjustment
The Equation-
Time Adjustment is Sold price X Increase of Decrease % X number of month(s)
ASP = Original price (+) or (-) Time adjustment
Examples:
A subject property was sold 5 months ago for $400,000. The property value has been increasing by 1% every month. The adjusted sales price is :
400,000 x [(1x5)]% = 20,000
400, ,000 = 420,000
ASP = $420,000
Another subject property was sold 9 months ago for $500,000/ The property value has decreased by 2% every month. The adjusted sale price is:
500,000 x [(2 x 9)]% = 90,000
500,000 – 90,000 = 410,000
ASP = $410,000

10. Another subject property was sold 5 months ago for $350,000 and the property values have increased by 6% in the last 10 months. The adjusted sales price is:
350,000 x [(6/10 x5)]% = 10,500
3500,000 – 10,500 = 339,500
ASP = $339,500
Another subject property was sold 6 months ago for $600,000 and the property values have increased by 10% in the last 12 months. The adjusted sales price is:
600,000 x [(10/12 x6)]% = 29,940
600, ,940 = 629,940
ASP = $629,940

11. To find % increase or decrease the formula to use is =
Percentage Increase or Decrease Market Trend Analysis and Time Resale Analysis
To find % increase or decrease the formula to use is =
New Price – Old Price
Old Price
% / # of Months
Examples:
Property was sold 9 months ago for $400,000 and was resold again the last month for $450,000. The percentage increase per month is:
450,000 – 400,000
400,000
How about if it was sold for $350,000, is the percentage now a decrease:
350,000 – 400,000
% / 8 Months = 1.56%
% / 8 Months = 1.56%

12. A property was sold 6 months ago for $100,000 and the prices are going up about 1% monthly and the property is superior to the subject property by 3% because of the location, the adjustments required are:
Time Adjustment – 100,000 x 1% x 6 – 6000
You would ADD 6000 to the adjustment
Location Adjustment- 100,000 x 3% = 3000
You would SUBTRACT 3000 to the adjustment
Total Adjustment- 100, , ,000 = 103,000
The price today would be $103,000

13. Question Example What you need to know: Prices are going up 10% a year
Corner lots are valued at $10,000 more than a normal lot
Value per SqFt of interior space is $100 SqFt
Value per SqFt of lot is $1,000 p/ Foot
Recreational rooms add $5,000 to its value
A 2 Pc Washroom adds $4,000 to its value
1 Car garage adds $6,000 to its value
2 Car garage adds $10,000 to its value
Fireplaces add $2,000 to its value
Air conditioning adds $3,000 to its value
Decks add $3,000 to its value

14. Sales Analysis Item Subject Comparable 1 Comparable 2 Comparable 3
Address
5646 Ellie Cres
7227 Victorian Ave
1234 Kristof Blvd
9907 Yukan Dr
Distance To Subject
Sale Price
$300,000 (300,00 x 10%)/12 x 5
$310,000 (310,000 x 10%)/12 x 2
$315,000
Date sold
3 Months Ago
2 Months Ago
2 Weeks Ago
Days on Market
Time Adjustment
+7500
+5,000 0$
Time Adjusted Price
$307,500
Location
Normal
Corner
Lot size
50 Ft
55 Ft
-$5,000
45 Ft
+$5,000
House Style
Age of House
Total Sq Footage
1205
1145
+$6,000
1250
-$4,500
Family Room
Bedrooms
Bathrooms
4 Pc
4pc + 2 Pc
-$2,000
4 Pc + 2 Pc
Basement/ % Furnished
Recreational Room
Yes No
Garage/Parking 1 2
-$4,000
Interior Condition
Exterior Condition
Fireplace
Air Conditioning
-$3,000
Deck
+$3,000
Total Adjustments
( ) = 2000
( ) =
( )=
Total Adjustment Sale Price
307, = $309,500
315, ,500 = $303,500
315, = $307,000

15. Cost Approach Valuation
Estimate the value of the site
Estimate Replacement Cost New (RCN)
Estimate Accrued Depreciation (Age)
Add Replacement costs to the value and subtract depreciation costs
Find the total value

16. Example- Cost Approach An Industrial Property
Industrial site is being valued
Improvements are a small main building and it is attached to a storage area
Three comparable sites have been found
Prices for properties have been increasing at 4% p/month
Depreciation- both structures have an effective age of 10 years and economic remaining life of 30 years
SqFt for each site is –
Sale 1-8, 312 SqFt
Sale 2-7, 7,770 SqFt
Sale 3-7,976 SqFt
Subject site has 8,000 SqFt

17. Site Valuation
Sale #1
Sale #2
Sale #3
Sale Date
6 Months Ago
2 Months Ago
1 Month Ago
Sale Price
$103,900
$101,750
$103,275
Time Adjustment
2.4%
0.8%
0.4%
Time Adjustment Calculation
0.024 x 103,900
.008 x 101,750
.004 x 103,275
Adjusted Sale Price
$106,394
$102,564
$103,668
Adjust Sale Price psf
$12.80
$13.20
$13.00
$104,000
Replacement Cost
Main Building
Storage
Measurement
30 x 66
18 x 32
Total Square Footage
1,980
576
Replacement Cost (psf)
$34.50
$21.00
$68,310
$12,096
Total Accrued Depreciation
$80,406
Accrued Depreciation Estimation
Main Building
Storage
Effective Age/Economic Life
10/40
Replacement Cost
$68,310
$12,096
Depreciation
$17,078
$3,024
Total Accrued Depreciation
$20,102
Depreciated Cost Of Improvements
Total Replacement Cost
$80,406
Accrued Depreciation
- 20,102
Depreciated Cost of Improvements
$60,304
Indication Of Value
Site Value $
Depreciated Cost of Improvement
+ 60,304
Indication of Value (rounded to $164,300)
$164,304

18. Site Value Estimation Methods
Comparative Sales Method: like done previously
Abstraction Method: To find value of the empty lot
Site value- Total purchase price – Value of improvements (or structure)
Land Residual Method:
Estimate Gross Income
Find Net Income
Apply Capitalization Rate Formula
Land Development Method: limited/not important

19. Replacement Cost New (RCN)
The cost of recently buildings that are made by comparing square metre/foot methods and divide it by the number of square metres/feet
This approach-
Gives a price p/square Metre/Foot to build structure from new
Straightforward
Mostly adequate for basic cost estimates
Lacks accuracy when considering complex buildings especially with major structural differences
Cost Services Method-
Various firms give costing manuals outlining basic unit costs for residential/commercial structures
Variations exist but unit costs are categorized to structure type
Appraiser selects most comparable and relative component costs are fully detailed
Feet x Feet = Square Feet (SqFt)
SqFt x $ p/ SqFt = RCN
X x $ p/ Square Meter (SqMt) = RCN

20. Continuation Meter x Meter = Square Meter (SqMt)
SqMt x $ p/ Square Meter = RCN
/ x $ p/ Square Foot = RCN
10’6” x 9’9” Means 10.5 x 9.75
If asked to calculate total RCN then you add all $RCN

21. Example 1- Structure Measurement Total Sq.Ft Replacement Cost
Total Cost
Main Building
26 x 42 ft.
1,092
$180.00/SqFt
196,560
Addition
14 x 26 ft.
364
$120.00/SqFt
43,680
Garage
16 x 21 ft.
336
$47.00/SqFt
15,792
Other Improvements:
Storage Shed
Total Replacement Cost $265,032
$9,000

22. Accrued Depreciation (Depreciation)
Two Methods:
Flat Depreciation Methods(Widely Used)-
A flat depreciation % and the time period will be given in:
Accrued Depreciation = % x Number of/Years x Replacement Cost
Effective Age
Economic Life
2. Economic-Age Life Depreciation Method-
Accrued Depreciation =
X RCN
ACUTAL
AGE

23. Example: Main Building-
Actual Age = 5 years (We DON’T care about this)
Effective Age is 3 years
Remaining Economic Life is 12 years
RCN is $200,000
Depreciation is: 3
(12+3)
X 200,000 = 40,000
The current value = 200,000 – 40,000 = 160,000
Current Value = $160,000

24. Addition: Actual Age is 3 years Effective Age is 3 years
Economic Life is 20 years
RCN- $50,000
Depreciation is: 3
20 x 50,000
Addition:
= 7500
The Current Value = 50,000 – 7,500 = 42,500
Current Value $42,500

25. Example: Accrued Depreciation Flat Depreciation Method
Costing service companies provide flat annual depreciation rate for various structures. A rule of thumb for residential structures is 1% a year during the first 25 years of economic life. A single-family residence has an actual age of 20 years and a replacement cost of $147,300
The flat depreciation based on 1% rate is .01 x 20 years x $147,300 = 29,460

26. Example: Accrued Depreciation Economic Age- Life Depreciation Method
The approach takes into account a structures effective age and its remaining economic life
Ex. Main Building
A single family residence has an actual age of 15 years and has estimated effective age of 10 years and remaining economic life of 30 years. The replacement cost is $83,500
Structure
Total Replacement Cost
Depreciation Rate Eff Age + Total Economic Life
Depreciation
Main Building
$83,500
10/40 or 25%
$20,875

27. Estimated Value of the Property
Step 1
Site Value $975,000
Step 2
Replacement Cost
1,347,000
Step 3
Depreciation
268,700
Step 4
Total Depreciation Cost
1,078,300
Step 5
Estimated of Value by the Cost Approach
$2.053,300

28. Income Approach
The Income Approach is an appraisal procedure consisting of 6 steps
Find Total Gross Potential Income
Find Effective Gross Income
a. Effective Income = Gross Income + Other Income – Vacancy Loss
3. Find a gross operating expenses (usually given in the question)
4. Find Net Operating Income (NOI)
a. Net Operating Income = Effective Income – Expenses or
b. Net Operating Income = Total Gross Income – All the losses
5. Then find the Capitalization rate
6. Find the VALUE of Property
Gross Potential Income – Vacancy Loss + Other Income = Effective Income
Effective Income – Expenses = NOI
NOI/R = Value
R= I/V

29. Example
An investor owns an industrial property that contains several rental units with an effective gross income $393,000 and its operating expenses are about $303,500. Based on the Capitalization rate of 10.25%, the estimated value of the property is: NOI= 393,000 – 303,500 = 89,500 VALUE= NOI/R = 89,500/10.25% = 873,