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

C

E W

B

S

AN

D

• 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 CD-South Forty-Five Degrees WestThe 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 BA has an

approximate bearing of North Forty-Five Degrees West • BA 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

1. Direct Comparison Approach

2. Cost Approach

3. 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

1. Time Adjustment

2. Location Adjustment

3. Lot Size Adjustment

4. 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

1. 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)]% = 10,5003500,000 – 10,500 = 339,500ASP = $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)]% = 29,940600,000 + 29,940 = 629,940ASP = $629,940

Percentage Increase or DecreaseMarket 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 – 400,000 400,000

% / 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

Item Subject Comparable 1 Comparable 2 Comparable 3

Address 5646 Ellie Cres 7227 Victorian Ave 1234 Kristof Blvd 9907 Yukan DrDistance 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 AgoDays on MarketTime Adjustment +7500 +5,000 0$Time Adjusted Price $307,500 $315,000 $315,000

Location Normal Normal Normal CornerLot size 50 Ft 55 Ft -$5,000 50 Ft 0 45 Ft +$5,000

House StyleAge of HouseTotal Sq Footage 1205 1145 +$6,000 1250 -$4,500 1205 0$Family RoomBedroomsBathrooms 4 Pc 4pc + 2 Pc -$2,000 4 Pc 0$ 4 Pc + 2 Pc -$2,000

Basement/ % Furnished

Recreational Room Yes No +$5,000 Yes 0$ Yes 0$Garage/Parking 1 1 0$ 2 -$4,000 2 -$4,000

Interior ConditionExterior Condition

Fireplace No Yes -$2,000 No 0$ No 0$Air Conditioning No No 0$ Yes -$3,000 No 0$Deck Yes Yes 0$ Yes 0$ No +$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

Sales Analysis

Cost Approach Valuation

1. Estimate the value of the site

2. Estimate Replacement Cost New (RCN)

3. Estimate Accrued Depreciation (Age)

4. Add Replacement costs to the value and subtract depreciation costs

5. Find the total value

Example- Cost ApproachAn 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 #3Sale Date 6 Months Ago 2 Months Ago 1 Month AgoSale Price $103,900 $101,750 $103,275Time 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 $103,668Adjust Sale Price psf $12.80 $13.20 $13.00Site Valuation $104,000

Replacement Cost Main Building Storage Measurement 30 x 66 18 x 32Total Square Footage 1,980 576Replacement Cost (psf) $34.50 $21.00Replacement Cost $68,310 $12,096Total Accrued Depreciation

$80,406

Accrued Depreciation Estimation

Main Building Storage

Effective Age/Economic Life

10/40 10/40

Replacement Cost $68,310 $12,096Depreciation $17,078 $3,024Total Accrued Depreciation

$20,102

Depreciated Cost Of Improvements Total Replacement Cost $80,406Accrued Depreciation - 20,102Depreciated Cost of Improvements $60,304

Indication Of Value Site Value $104.000Depreciated Cost of Improvement + 60,304Indication 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 newStraightforward Mostly adequate for basic cost estimatesLacks 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 CostMain Building 26 x 42 ft. 1,092 $180.00/SqFt 196,560Addition 14 x 26 ft. 364 $120.00/SqFt 43,680Garage 16 x 21 ft. 336 $47.00/SqFt 15,792

Other Improvements:

Storage ShedTotal Replacement Cost

$265,032

$9,000 $9,000

Accrued Depreciation (Depreciation)Two Methods:

1.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

Actual Age

Effective Age

Economic Life

Remaining Economic Life

0 5 10 15 20 25 30 35 40

Number of years2Number of years3Number of years

ACUTALAGE

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:

320 x 50,000

= 7500

The Current Value = 50,000 – 7,500 = 42,500Current Value $42,500

Example: Accrued DepreciationFlat 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 DepreciationEconomic 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

1. Find Total Gross Potential Income

2. 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