Math

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

Metes and Bounds Examples W 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

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

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

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

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

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

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 + 20,000 = 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

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)]% = 29,940 600,000 + 29,940 = 629,940 ASP = $629,940

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%

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 + 6,000 + 3,000 = 103,000 The price today would be $103,000

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

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 (5000 + 6000 - 2000 + 5000 - 2000) = 2000 (4500- 4000- 3000) = -11500 (-10000 + 5000 - 2000- 4000 + 3000)= - 8000 Total Adjustment Sale Price 307,500 + 2000 = $309,500 315,000 - 11,500 = $303,500 315,000 - 8000 = $307,000

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

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

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 $104.000 Depreciated Cost of Improvement + 60,304 Indication of Value (rounded to $164,300) $164,304

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

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 0.0929 x $ p/ Square Meter (SqMt) = RCN

Continuation Meter x Meter = Square Meter (SqMt) SqMt x $ p/ Square Meter = RCN / 0.0929 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

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

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

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

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

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

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

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

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

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, 170.73