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# Fires

Fires. Floods. Hurricanes. Don’t worry…. Where do I start ?. I hope the contractor can do the measuring ?. Is this my new claim ?. How am I going to do this ?. What are the formulas ?. Message from our Vice President.

## Fires

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### Presentation Transcript

1. Fires Floods Hurricanes

2. Don’t worry… Where do I start ? I hope the contractor can do the measuring ? Is this my new claim ? How am I going to do this? What are the formulas ?

3. Message from our Vice President In the world of business, the Insurance Industry is part of the Financial sector. Numbers are a critical focus in the financial sector and the adjustment of claims is an example of that focus. Having the basic math fundamentals is essential in the day to day handling of claims . Master the basic math skills and feel confident about your work product. Wendy Hillier Vice President, Property Claims

4. And Now a Message from our Technical Trainer Dollars & Sense is a practical refresher for our experienced adjusters, advisors and examiners. Property Advisors, Field Adjusters and Claims Examiners will. Revisit the concepts of Common Units of Measurements, Converting Inches into Decimals, Area & Perimeter Formulas; determining measurements such as rooms, roofs, lumber; calculating Coinsurance, Actual Cash Value, Additional Living Expenses, Deductibles, Depreciation, Replacement, Average Distribution. Apply the lessons to calculation exercises that an adjuster needs to do on a daily basis. Prepare for the Advanced Math course, which will be released in Q1 of 2006. Good luck and I look forward to your comments! Stan Bodal, FCIP. Technical Property Trainer, Claims Training

6. Introduction Menu • Introduction This module discusses the need for the course and objectives are introduced. • Math Fundamentals This module refreshes concepts of Common Units of Measurements, Converting Inches into Decimals, Area & Perimeter Formulas. • Measurement This module addresses perimeter, area, circumference. • Measuring on the Job This module demonstrates common practical property measurements such as rooms, roofs, lumber and trim. • Calculating Coinsurance This module explains the Coinsurance Claim, the Coinsurance Penalty, a Waiver of Coinsurance, the Formula for Coinsurance, and is illustrates with interactive Scenarios. • Other Calculations This module explores calculating Actual Cash Value, Additional Living Expenses, Deductibles, Depreciation, Replacement, Average Distribution Clause and each concept is supported by interactive Scenarios. Math Fundamentals Measurement Measuring on the Job Calculating Coinsurance Other Calculations

7. Introduction This course is a supplement to the Residential Construction and Property Estimating Workshops. Identified gaps in the course results and on-the-job application have determined the need for a refresher in arithmetic skills. Addition, subtraction, multiplication, and division are absolutely essential. Adjusters lacking in these arithmetic skills will not be able to write a proper building damage repair estimate. Arithmetic accuracy is a must. One of the most common estimating mistakes is measuring. Page 1 of 2

8. Learning & Development Strategy A robust and interactive combination of a 2-hour online tutorial, a subsequent online Mastery Test and a follow up leader-led hands-on workshop: Online Tutorial ·Online Tutorial with Immediate Feedback of Comprehension ·Online Arithmetic Activities ·Online Mastery Test ·Application Workshop ·In-Class Question/Answer Period ·In-Class Scenario/File Clinic ·Online/In-Class Course Test ·Online/In-Class Evaluation Who is this course designed for: Participants who have successfully completed the Residential Construction & Estimating Workshops Upon completing this course, you will be able to: IDENTIFY the mathematical formulas for measurement of area APPLY the formulas to measure various geometric shapes IMPLEMENTthe applications to common property scenarios the end Course Objective Page 2 of 2

9. Math Fundamentals Common Units of Measurements Converting Inches into Decimals Area & Perimeter Formulas Formula Exercise Page 1 of 4

10. 1 Foot = 1 Yard = 1 Yard = 1 Square Yard = 1 Square = 12 Inches 3 Feet 36 Inches 9 Square Feet 100 Square Feet Common Units of Measurements Page 2 of 4

11. Fraction Decimal 1/16"1/8"3/16"1/4"5/16"3/8"7/16"1/2"9/16"5/8"11/16"3/4"13/16"7/8"15/16" .0625.125.1875.25.3125.375.4375.5.5625.625.6875.75.8125.875.9375 Converting Inches into Decimals Estimating Guide Table 1” = .08’ 2”= .17’ 3”= .25’ 4”= .33’ 5”= .42’ 6”= .50’ 7”= .58’ 8”= .67’ 9”= .75’ 10”= .83’ 11”= .92’ 12”= 1’ Page 3 of 4

12. A B Area of triangle base X height  2 Area of Circle r2 Circumference of Circle d Perimeter of Square a+b+c+d Area of Square s2 Perimeter of triangle a+b+c Area of Rectangle length X width Perimeter of a Rectangle 2 (length X width) Area of Parallelogram base X height Area of a Trapezoid height (b1+b2) 2 the end Area & Perimeter Formulas Page 4 of 4

13. Measurement • The proper use of geometry and algebra is vital to provide a competent estimate. • Buildings are comprised of a variety of shapes. Surface areas commonly conform to the shape of parallelograms (e.g., rectangles and squares), triangles, circles, and trapezoids. • Surface area, linear dimensions, and cubic volume are the end products of the adjuster's measurements and calculations. • To measure areas, you must use geometric formulas based on the shape of the area. You can measure complex areas by dividing the whole area into parts that fits the basic shapes of rectangles, squares, triangles, trapezoids, parallelograms, and sometimes circles. Page 1 of 10

14. Perimeter The perimeter is the linear distance around a partially or fully enclosed area. One type of estimate that requires perimeter measurement is a fence. • The image represents a typical fence scope around a house. To find out how much material you need for fence construction, you normally determine the length of the fence. Since fences usually are used to enclose an area, you can find out the length of the fence by measuring the perimeter. If the fence borders a structure, such as the back of a house, you can subtract the length of the structure from the perimeter. • The perimeter is the total distance around a given area. Most of the time, perimeter can be measured linearly or by using geometric formulas. • Let's look at few simple formulas. • The perimeter of a rectangle is two times the length plus two times the width. • The perimeter of a square is 4 times the length of one side. • If the shape of the fence is irregular, you can simply measure each side and then add together the measurements to get the estimate. (Courtesy of IMACC.net) Page 2 of 10

15. Perimeter of a Circle The perimeter of a circle is the same as the circumference of the circle. You find the circumference by first determining the diameter. The diameter is the width of the circle. The distance from the center of the circle to the circumference is called the radius. The radius is exactly half the diameter. To find the circumference multiply the diameter by the number 3.14. This number is called pi. The circumference is 3.14 (or Pi as in the Greek Letter) times the length of the diameter e.g. d = 5, so c = 5 x (3.14) = 15.7, so the circumference = 15.7 The diameter of a circle is twice the radius, d = 2r (Courtesy of IMACC.net) Page 3 of 10

16. Area of Reactangles and Squares When estimating for flooring materials you usually will be measuring rectangular areas. To find an area of a rectangle, multiple the length by the width. This gives you the square footage. To calculate the area multiply the base (or length) by the width (or height). This rectangle could represent a ceiling, wall or a floor. A square differs from a rectangle in that all 4 sides are of equal length. A square is a rectangle with four equal sides. So to estimate an area that is perfectly square, multiply one side by itself. (Courtesy of IMACC.net) Page 4 of 10

17. Area of Triangles This triangle could be a roof gable end or part of a cut up ceiling. To calculate the area multiply the 1/2 Base x Height To find the area of a triangle, first measure the base (any one side of the triangle), then the height. Multiply the base by the height, and then divide the result by 2. (Courtesy of IMACC.net) Page 5 of 10

18. Area of Triangles So if one side of a triangle is 4 feet, and the height of the triangle is 8 feet, multiply 4 by 8, and then divide it in half. The result is 16. So the area of the triangle is 16 square feet. The two triangles above, R and H, though different in looks and shape, have the exact same area. (20' × 20' )/2 = 200 square feet = 2 squares (Remember, one square is equal to 100 sq. ft.) (Courtesy of IMACC.net) Page 6 of 10

19. Area of Trapezoids The Trapezoid is a figure with two opposite sides that are parallel, and two opposite sides that are not parallel. Knowing how to recognize and calculate a trapezoid will aid the adjuster in scoping floors and roofs. To calculate the area add 1/2 B1 and 1/2 B2 then multiply by the height A trapezoid is 4-sided figure with exactly one pair of parallel sides. The other two sides are not parallel. The two parallel sides are the bases: we will call them B1 (base one) and B2 (base two). To find the area of a trapezoid, you must add base1 and base2 and divide by two. This number is multiplied by the height. (Courtesy of IMACC.net) Page 7 of 10

20. Area of Parallelograms This is identical to the rectangle formula because a parallelogram is a rectangle at a slant. Be careful not to measure the slanted side for the height. A parallelogram is any four-sided figure that has two pairs of opposite sides that are parallel. To calculate the area of a parallelogram, multiply the base by the height. (Courtesy of IMACC.net) Page 8 of 10

21. Area of Circles Adjusters will run into estimating circle areas in all manner of ceiling, floor and exterior losses. You must be able to define the circumference, diameter, and radius of a circle before you can work any problems using circles The circumference (c) of a circle is the distance around the circle. The diameter (d) of a circle is a straight line that passes through the center of a circle from one side to the other. The radius (r) is a straight line from the center to any point on the circumference. The circumference is 3.14 (or Pi as in the Greek Letter) times the length of the diameter (Courtesy of IMACC.net) Page 9 of 10

22. the end Area of Circles (continued) The diameter of a circle is twice the radius, d = 2r The area of a circle = r2 e.g. Assume radius is 8‘ x r2 3.14 x 82 3.14 x 64 = 200.96‘ Area of circle = 200.96 sq. ft. For example you may need to measure a circular floor. First you measure the diameter, which is the length across the circle. The diameter is 8 feet. What is the area of the circle? First, find the radius by dividing the diameter by two. This gives you 4 feet. Now multiply 4 feet by itself. This gives you 16 feet, the number for the radius squared. Finally, multiply the radius squared by pi, which is about 3.14. The result is 50.24 square feet. (Courtesy of IMACC.net) Page 10 of 10

23. Measuring on the Job The following categories highlight some of the main issues to address when scoping these type of losses. Please note there are innumerable possible situations each unique and too varied for this treatise. The following is meant as a base to build on with personal experience and further learning. We will cover: Room Measurement Roof Measurement Calculating Lumber Calculating Trim Page 1 of 7

24. Height 10’ 12’ 15’ Room Measurements Using the measurements you will be able to calculate: (i) gross area • Height 10’ • 15’ • 12’ (ii) gross perimeter (iii) gross wall area When floor measurements must be taken: the width, the length and the height of each room must be calculated. Page 2 of 7

25. Height 8’ 15”’ 12’ Room Measurements This room (pictured) has: (i) Gross Floor Area (GFA) Length x Width or 15' x 12' = 180 sq.ft (ii) Gross Perimeter (GP) (Length + Width) x 2 or (15' + 12') x 2 or 15' + 12' + 15' + 12' = 54 Lineal Feet (iii) Gross Wall Area (GWA) Gross Perimeter x Height or 54' x 10' = 540 sq.ft Page 3 of 7

26. interactive activity Roof Measurements The picture shown is an aerial view of a roof with both a hip end and gable ends. It is strongly recommended that you make a rough sketch of your roof. In order to make it easier for measuring, the roof will be broken up into sections A through F. Click each letter to view a measurement breakdown C C B D A E E F F When you have completed each letter click here to see the final sum Page 4 of 7 (Courtesy of IMACC.net)

27. Roof Measurements - Section E s = (15' × 15')/2 = 112.5 square feet t = 40' × 15' = 600 square feet Remember to check both small triangles in the center part to make sure that they are the same size. x = (10' 6" × 10' 6")/2 = 55 square feet y = 21' × 4.5' = 94.5 square feet The actual answer of x is 55-1/8 square feet, but we rounded off for ease of measuring. (Note: in this example we are rounding off, but adjusters should follow their company guidelines.) z = 34' × 15' = 510 square feet Be careful adding all of these up. Remember that there are two different x sections so we'll need to add it twice. Section E = 112.5 + 600 + 55 + 55 + 94.5 + 510 = 1,427 square feet. This section has several different subsections so we'll have to be careful and make sure we do it right. If your roof has a section similar in shape, double check your sketch to make sure that every piece is either rectangular or triangular. (Courtesy of IMACC.net)

28. Roof Measurements – The Sum If we take all sections and add them up… Section A =225.0Section B =1050.0Section C = 1125.0 Section D = 262.5 Section E =1427.0Section F = 530.0 we get a total square footage that is equal to 4,619.5 square feet

29. r h b 10’ Alternative Roof Measurements Pythagorean Theory In a right angle triangle the square of the length of the hypotenuse (rafter) is equal to the sum of the squares of the lengths of the other two sides. b2 +h2 = r2 10x10 +8x8= r² 164= r² √164=r r=12.8 Add for overhang Page 5 of 7

30. Measuring and Calculating Lumber • Measure the distance between the studs in the center i.e from the middle of one stud to the adjacent stud. 16”, 24”etc. • If length of the wall is 30’ multiply by the factor .75 = 22.5 or 23 studs. • Add one for the end and two for every opening. • Refer to the Estimating Guide for the Tables. Page 6 of 7

31. a d b c the end Measuring & Calculating Trim • The quantity of baseboard/quarter round/ chair rail etc can be determined by calculating the perimeter of the room. • Trim around the windows and doors can be physically measured and agreed upon. • Most trim is pine or mdf. However take a sample in to a hardware store if not sure. Page 7 of 7

32. interactive activity Calculating Coinsurance The Coinsurance Claim Click each category to learn more The Coinsurance Penalty Waiver of Coinsurance Page 1 of 6

33. Formula for Coinsurance Amount of Insurance Carried X Amount of Loss Amount of Insurance Required = Amount Paid Or better referred to as DID X Amount of Loss SHOULD = Amount Paid Page 2 of 6

34. Coinsurance Scenario #1 Scenario Value of Building \$156,250 Amount of Loss \$ 30,000 Amount of Insurance \$100,000 Coinsurance on building 80% Calculation \$100,000 X \$30,000 \$156,250 X 80% = \$100,000 X \$30,000 \$125,000 = \$24,000 Page 3 of 6

35. Coinsurance Scenario #2 Scenario Value of Building \$500,000 Amount of Loss \$ 15,000 Amount of Insurance \$300,000 Coinsurance on building 80% Calculation \$300,000 X \$15,000 \$500,000 X 80% = \$300,000 X \$15,000 \$400,000 = \$11,250 Page 4 of 6

36. interactive activity Coinsurance Activity Scenario Value of Building \$200,000 Amount of Loss \$ 90,000 Amount of Insurance \$180,000 Coinsurance on building 100% Click the correct answer \$18,000 \$81,000 \$81,500 Page 5 of 6

37. interactive activity the end Coinsurance Activity (Optional Loss Settlement Clause) Scenario Value of Building \$300,000 Amount of Loss \$ 90,000 Amount of Insurance \$200,000 Is the settlement on ACV or Replacement cost? Note If the amount of insurance carried on the building is less than 80%, then the settlement will only be on actual cash value. Click the correct answer ACV Replacement Page 6 of 6

38. interactive activity Other Calculations Actual Cash Value Click each category to learn more Additional Living Expenses Deductibles Depreciation Average Distribution Clause Replacement

39. Actual Cash Value • Actual Cash Value Settlements are made for contents as well as for building claims. • We dealt with the contents ACV calculations under the SOL presentation. • We now need to address calculations for ACV value for Buildings or their components such as roofs,floors,walls etc. • This becomes more important when the basis of settlement is ACV such as for subrogation or some commercial policies. Page 1 of 5

40. Actual Cash Value Calculations of ACV for Buildings The factors to be considered would be: • Life Span • Age of the item in question. • Current replacement cost. Page 2 of 5

41. Actual Cash Value • Lets’s take an example of a 10yr old roof with a 20yr lifespan which needs to be replaced. • If the replacement cost of the roof in the above example is \$6000 what is the Actual Cash Value? • 10/20=1/2 or 10/20x100 =50%(to arrive at proportion used) 1/2 or 50% of the roof’s lifespan has been consumed. Balance life span is 50% (100 - 50) Therefore the ACV is 1/2 x \$6,000 = \$3,000 or (\$6,000 - \$3,000) Page 3 of 5

42. EXAMPLE 1:High Fashion Cocktail Dress interactive activity Actual Cash Value Activity #1 Calculate the ACV of a hardwood floor with a 20-year life span. The floor is 5 years old and has a replacement value of \$10,000. Click the correct answer \$6,500 \$7,000 \$7,500 Page 4 of 5

43. EXAMPLE 1:High Fashion Cocktail Dress interactive activity Actual Cash Value Activity #2 Calculate the ACV of a roof with a 20-year life span. The roof is 5 years old and has a replacement value of \$6,000. Click the correct answer \$3,500 \$4,500 \$5,500 Page 5 of 5

44. interactive activity Additional Living Expenses Activity #1 Additional Living expense is incurred when an insured is required to spend more money than they usually do for their day-to-day living expenses. Example, if the insured’s home is unlivable as a result of an insured peril then they are eligible for Additional Living Expenses (ALE). Activity Sammy’s house has had a bad fire. The house is unlivable. Does Sammy have a claim under the Additional Living Expense coverage for staying at a hotel until his house is repaired? Click the correct answer Yes No Page 1 of 2

45. interactive activity Additional Living Expenses Activity #2 Chuck normally spends \$200 on weekly groceries. As a result of a fire, he has moved out and spends \$700 on food at a hotel. How much Additional Living Expense would you allow? Click the correct answer \$900 \$700 \$500 Page 2 of 2

46. interactive activity Deductibles Activity Deductibles are amounts which the insured has agreed to self-insure for or are stipulated in the policy as being the amount by which a claim will be reduced. Activity When a fire damages his building and contents, Dan claims \$5,000 for the building and \$1,000 for the contents. The deductible on the policy is \$500. He will be paid: \$5,000 + \$1,000 = \$6,000 \$6,000 - \$500 = \$5,500 Dan has another fire loss two weeks later. The adjuster applies the deductible of \$500 to this claim as well. Later in the year, Dan has a burglary. His coin collection, valued at \$5,000, is stolen. Coverage for numismatic property is limited to \$200 under the policy. The adjuster first applies the deductible reducing the claim to \$4,500 and then applies the limitation of \$200. If Dan’s coin collection were valued at \$600, how much much would he collect under his policy? \$100 \$200 \$175 Click the correct answer Page 1 of 1

47. EXAMPLE 1:High Fashion Cocktail Dress Replacement Cost \$5000 Life Expectancy 10 years Actual Age 5 years Calculation 5 10 X \$5000 ACV \$2500 Replacement Cost \$8000 Life Expectancy 25 years Actual Age 10 years Calculation • 15 • X \$8000 ACV \$4800 Depreciation EXAMPLE 2:Broadloom Flooring EXAMPLE 1:Roofing Shingles Depreciation Depreciation means loss of value.  Depreciation in personal property may be physical (wear and tear) or obsolescence (change in style or function). Page 1 of 1

48. Average Distribution Clause Let us assume that we have 2 detached structures on a piece of property. The value of the detached property A is \$20,000 and B is \$30,000. The available insurance is \$30,000. How would you apportion the available insurance between the two structures? The approach to this would be as follows: • Total coverage available is \$30,000 • Total value of the property is \$50,000 Property A: \$30,000/\$50,000 X \$20,000 = \$12,000 Property B: \$30,000/\$50,000 X \$30,000 = \$18,000 If Property A was to be destroyed then only \$12,000 of coverage is available. Page 1 of 2

49. interactive activity Average Distribution Clause Activity A property has two detached structures Structure C & Structure D C=\$20,000 D=\$40,000 Insurance available is \$30,000. What is the allocation of insurance for C & D? Click the correct answer C=\$15,000 D=\$15,000 C=\$20,000 D=\$10,000 C=\$10,000 D=\$20,000 Page 2 of 2

50. Replacement Replacement cost insurance lets you replace a damaged item with a similar item, even if it's more expensive than the original. Example of replacement costAmount paid for a refrigerator ten years ago\$800 Current cost of a refrigerator with the same features\$1,000 Total compensation\$1,000 Page 1 of 1

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