Is due to the lessening of the physical ability of the property to produce results.

Engineering Economy Depreciation

1

Depreciation This module deals with depreciation and the different methods of solving depreciation charges. At the end of this module, the learner should be able to: 1. Calculate the depreciation charge for any particular year using the different methods. 2. Compute for the book value for any particular year using any depreciation method. Definition of Terms 1. Depreciation – is the decrease in the value of physical property with the passage of time. 2. Value – present worth of all future profits that are to be received through ownership of a particular property. 3. Market Value- the amount which a willing buyer will pay to willing seller for the property each has equal advantage and is under no compulsion to buy or sell. 4. Utility/Use Value – is what the property is worth to the owner as an operating unit. 5. Fair Value- is the value which is usually determined by a disinterested third party in order to establish a price that is fair to both seller and buyer. 6. Book Value – sometimes called depreciated book value, it is worth of a property as shown on the accounting records of an enterprise. 7. Salvage or Resale Value- is the price that can be obtained from the sale of the property after it has been used. 8. Scrap Value- is the amount of property would sell for if disposed off as junk. 9. Straight Line Method – this method assumes that the loss in value is directly proportional to the age of property. 10. Sinking Fund Method – this method assumes that a sinking fund is established in which funds will accumulate for replacement. 11. Declining Balance method (Matheson Formula) – also called the constant percentage method – it is assumed that the annual cost of depreciation is a fixed percentage of the salvage value of the beginning of the year.

Engineering Economy Depreciation

2

12. Rate of Depreciation, k – is the ration of the depreciation in any year to the book value at the beginning of that year is constant throughout the life of the property. 13. Double Declining Balance (DDB) Method – similar to the declining balance method except that the rate of depreciation, k, is replaced by2/L. Types of Depreciation 1. Normal depreciation a) Physical depreciation – is due to the lessening of the physical ability of a property to produce results. Its common causes are wear and deterioration. b) Functional depreciation – is due to the lessening in the demand for the function which the property was designed to render. Its common causes are inadequacy, changes in styles, population centers shift, saturation of markets or more efficient machines are produced. 2. Depreciation due to changes in price levels – is almost impossible to predict and therefore is not considered in economy studies. 3. Depletion – refers to the decrease in the value of a property due to the gradual extraction of its contents. Physical life of a property – is the length of time during which it is capable of performing the function for which it was designed and manufactured. Economic life – is the length of time during which the property may be operated at a profit. Depreciation Methods: Where: 𝐿= useful life of the property in years 𝐶𝑜 = the original cost 𝐶𝐿 = the value at the end of the life, the scrap value/ salvage value 𝑑 = the annual cost of depreciation 𝐶𝑛 = the book value at the end of n years

3

Engineering Economy Depreciation

𝐷𝑛 = total deprecation up to n years 1. Straight Line Method (SLM) 𝐶𝑜 − 𝐶𝐿 𝐿 𝑛(𝐶𝑜 − 𝐶𝐿 ) 𝐷𝑛 = 𝐿 𝐶𝑛 = 𝐶𝑜 − 𝐷𝑛 𝑑=

2. Sinking Fund Method (SFM) 𝐶𝑜 − 𝐶𝐿 𝐹 𝐴 , 𝑖%, 𝐿 𝐶𝑜 − 𝐶𝐿 𝑑= (1 + 𝑖)𝐿 − 1 𝑖 𝐹 𝐷𝑛 = 𝑑( , 𝑖%, 𝑛) 𝐴 (1 + 𝑖)𝑛 − 1 𝐷𝑛 = 𝑑( ) 𝑖 𝐶𝑛 = 𝐶𝑜 − 𝐷𝑛 𝑑=

3. Declining Balance Method/ Matheson Formula (DBM) 𝑑𝑛 = 𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑑𝑢𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝑛𝑡ℎ 𝑦𝑒𝑎𝑟 𝑑𝑛 = 𝐶𝑜 (1 − 𝑘)𝑛−1 𝑘 𝐶𝑛 = 𝐶𝑜 (1 − 𝑘)𝑛 𝐶𝐿 = 𝐶𝑜 (1 − 𝑘)𝐿 𝐿 𝐶𝐿 𝑘 =1− √ 𝐶𝑜

This method does not apply if the salvage value is zero, because k will be equal to one and 𝑑1 will be equal to 𝐶𝑜 . 4. Double Declining Balance Method (DDBM) 2 𝑛−1 2 𝑑𝑛 = 𝐶𝑜 (1 − ) 𝐿 𝐿

4

Engineering Economy Depreciation

2 𝐶𝑛 = 𝐶𝑜 (1 − )𝑛 𝐿 2 𝐶𝐿 = 𝐶𝑜 (1 − )𝐿 𝐿 When the DDB method is used, the salvage value should not be subtracted from the first cost when calculating the depreciation charge. 5) Sum of the Year’s Digit Method (SYD) Let 𝑑𝑛 = depreciation charge during the nth year 𝑑𝑛 = (depreciation factor)(total depreciation) 𝑑𝑛 =

𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑑𝑖𝑔𝑖𝑡 (𝐶 −𝐶 ) 𝑠𝑢𝑚 𝑜𝑓 𝑑𝑖𝑔𝑖𝑡𝑠 𝑂 𝐿

𝐿(𝐿 + 1) 2 For example, for a property whose life is 5 years. ∑ 𝑜𝑓 𝑑𝑖𝑔𝑖𝑡𝑠 =

Year

Year in reverse order

Depreciation factor

Depreciation during the year

1

5

2

4

3

3

4

2

5 15 4 15 3 15 2 15

5 (𝐶 −𝐶 ) 15 𝑂 𝐿 4 (𝐶 −𝐶 ) 15 𝑂 𝐿 3 (𝐶 −𝐶 ) 15 𝑂 𝐿 2 (𝐶 −𝐶 ) 15 𝑂 𝐿

5

1

1 15

1 (𝐶 −𝐶 ) 15 𝑂 𝐿

∑ of digits=15 Let 𝑑𝑛 = 𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑢𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝑛𝑡ℎ 𝑦𝑒𝑎𝑟 𝑑𝑛 = (𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟)(𝑡𝑜𝑡𝑎𝑙 𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛) 𝑑𝑛 =

𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑑𝑖𝑔𝑖𝑡 (𝐶 − 𝐶𝐿 ) 𝑠𝑢𝑚 𝑜𝑓 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜

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Engineering Economy Depreciation

∑ 𝑜𝑓 𝑑𝑖𝑔𝑖𝑡𝑠 =

𝑛(𝑛 + 1) 2

Illustrative Example 1: A firm bought an equipment for P56,000. Other expenses including installation mounted to P4,000. The equipment is expected to have a life of 16 years with a salvage value of 10% of the original cost. Determine the book value at the end of 12 years by a) the straight line method and b) sinking fund method at 12% interest. Solution 𝐶𝑜 = P56,000 + P4,000 = P60,000 𝐶𝐿 = P60,000(0.10) = P6,000 𝐿 = 16

𝑛 = 12

𝑖 = 12%

a) Straight Line Method 𝐶𝑜 − C𝐿 𝐿 = 𝑛(𝑑)

𝑑 = 𝐷12

𝑃60,000 − 𝑃6,000 = 𝑃3,375 16 = 12(𝑃3,375) = 𝑃40,500

=

𝐶12 = 𝐶𝑜 − 𝐷12

= 𝑃60,000 − 𝑃40,500 = 𝑃19,500

b) Sinking Fund method 𝐶 −C

𝐿 𝑑 = 𝐹⁄𝐴𝑜,12%,16 =

P60,000−P6,000 42.7533

= 𝑃1,263

𝐷12 = 𝑑(𝐹 ⁄𝐴 , 12%, 12) = 𝑃1,263(24.1331) = 𝑃30,480 𝐶12 = 𝐶𝑜 − 𝐷12 = 𝑃60,000 − 𝑃30,480 = 𝑃29,520 Illustrative Example 2: A certain type of machine loses 10% of its value each year. The machine costs P2,000 originally. Make out a schedule showing the yearly depreciation, the total depreciation and book value at the end of each year for 5 years.

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Engineering Economy Depreciation

Solution: year

Book value at beginning of the year

Depreciation during the year

Book value at the end of year

1

P2,000.00

P200.00

P1,800.00

2

P1,800.00

P180.00

P1,620.00

3

P1,620.00

P162.00

P1,458.00

4

P1,458.00

P145.80

P1,312.20

5

P1,312.20

P131.22

P1,180.98

Illustrative Example 3: Determine the rate of depreciation, the total depreciation up to the end of the 8th year and the book value at the end of 8 years for an asset that costs P15,000 new and has an estimated scrap value of P2,000 at the end of 10 years by a) the declining balance method and b) the double declining balance method. Solution 𝐶𝑜 = 𝑃15,000

𝐶𝐿 = 𝑃2,000

𝐿 = 10

a) Declining balance method 𝐿

𝑘 = 1− √

10 𝑃2,000 𝐶𝐿 = 1− √ = 0.1825 𝑜𝑟 18.25% 𝐶𝑜 𝑃15,000

𝐶8 = 𝐶𝑜 (1 − 𝑘)𝑛 = 𝑃15,000(1 − 0.1825)8 = 𝑃2,992 𝐷8 = 𝐶𝑜 − 𝐶8 = 𝑃15,000 − 𝑃2,992 = 𝑃12,008 b) Double declining balance method 2

2

Rate of depreciation = 𝐿 = 20 = 0.20 𝑜𝑟 20%

𝑛=8

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Engineering Economy Depreciation

2 2 8 𝐶8 = 𝐶𝑜 (1 − )8 = 𝑃15,000(1 − ) = 𝑃2,517 𝐿 10 𝐷8 = 𝐶𝑜 − 𝐶8 = 𝑃15,000 − 𝑃2,517 = 𝑃12,483 Illustrative Example 4: A plant bought a calciner for P220,000 and used it for 10 years, the life span of the equipment. What is the book value of the calciner at the end of 5 years if the salvage value isP20,000 for straight line method; P22,000 for declining balance method and P20,000 for the double declining balance method. Solution: 𝐶𝑜 = 𝑃220,000

𝐿 = 10

a) Straight line method

𝑛=5 𝐶𝐿 = 𝑃20,000

𝑛(𝐶𝑜 − C𝐿 ) 5(𝑃220,000 − 𝑃20,000) = = 𝑃100,000 𝐿 10 𝐶5 = 𝐶𝑜 − 𝐷5 = 𝑃220,000 − 𝑃100,000 = 𝑃120,000 𝐷5 =

b) Declining balance method 𝐶𝐿 = 𝑃22,000 𝐶

𝑛

𝑃22,000

5

𝐶5 = 𝐶𝑜 (𝐶𝐿 ) 𝐿 = 𝑃220,000(𝑃220,000)10 = 𝑃69,570 𝑜

c) Double declining balance method 𝐶𝐿 = 𝑃20,000 2 2 𝐶5 = 𝐶𝑜 (1 − )5 = 𝑃220,000(1 − )5 = 𝑃72,090 𝐿 10 Illustrative Example 5: (3-9) A machine costs P7,000 , last 8 years and has a salvage value at the end of life of P350. Determine the depreciation charge during the 4 th year and the book value at the end of 4 years by the a) straight line method. b) Declining balance method, c) SYD method, and d) sinking fund method with interest at 12%. Solution 𝐶𝑜 = 𝑃7,000

𝐶𝐿 = 𝑃350

𝐿=8

𝑛=4

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Engineering Economy Depreciation

a) Straight line method 𝐶𝑜 − C𝐿 𝐿 𝐶4 = 𝐶𝑜 − 𝐷4 𝑑 =

𝑃7,000 − 𝑃350 = 𝑃831 8 = 𝑃7,000 − 𝑃831(4) = 𝑃3,676

=

b) Declining balance method 𝐿

𝑘 = 1− √

8 𝑃350 𝐶𝐿 = 1− √ = 0.3123 𝐶𝑜 𝑃7,000

𝑑4 = 𝐶𝑜 (1 − 𝑘)4−1 𝑘 = 𝑃7,000(1 − 0.3123)3 (0.3123) = 𝑃711 𝐶4 = 𝐶𝑜 (1 − 𝑘)4 = 𝑃7,000(1 − 0.3123)4 = 𝑃1566 c) SYD method 1

8

2

7

3

6

4

5

5

26

𝑆𝑢𝑚 𝑜𝑓 𝑑𝑖𝑔𝑖𝑡𝑠 =

𝑑4 =

5 (𝑃7,000 − 𝑃350) = 𝑃924 36

𝐷4 =

26 (𝑃7,000 − 𝑃350) = 𝑃4,803 36

6 7

8(8 + 1) = 36 2

8 36

𝐶4 = 𝐶𝑂 − 𝐷4 = 𝑃7,000 − 𝑃4,803 = 𝑃2,197

d) Sinking fund method 𝑑 =

𝐶𝑜 − C𝐿 P7,000 − P350 = = 𝑃541 𝐹 ⁄𝐴 , 12%, 8 12.2997 𝐷4 = 𝑑(𝐹 ⁄𝐴 , 12%, 4) = 𝑃541(4.7793) = 𝑃2,586 𝐶4 = 𝐶𝑜 − 𝐷4 = 𝑃7,000 − 𝑃2,586 = 𝑃4,414

Illustrative Example6: A contractor imported a bulldozer for his job, paying P250,000 to the manufacturer. Freight and insurance charges amounted to P18000; custom’s, and broker’s fees, P8500; taxes, permits, and other expenses, P25000. If the

9

Engineering Economy Depreciation

contractor estimates the life of the bulldozer to be 10 years with a salvage value of P20,000; determine the book value at the end of 6 years, using the a) straight-line formula, b) sinking fund formula at 8%, c) Matheson formula, d) double declining balance method and e) SYD method. Solution: a) Straight-line formula: 𝐶𝑜 = 250000 + 18000 + 8500 + 25000 = 𝑃301500 𝐶𝐿 = 20000; 𝐿 = 10 𝑦𝑒𝑎𝑟𝑠; 𝑛 = 6𝑦𝑒𝑎𝑟𝑠 𝐶𝑜 − C𝐿 301500 − 20000 = = 𝑃28150 𝐿 10 6(301500 − 20000) 𝐷6 = 𝑛(𝑑) = = 𝑃168900 10 𝐶𝑛 = 𝐶𝑜 − 𝐷𝑛 ; 𝐶6 = 301500 − 168900 𝑑 =

𝑪𝟔 = 𝑷𝟏𝟑𝟐𝟔𝟎𝟎 𝑨𝒏𝒔. b) Sinking fund formula at 8%: 𝐶𝑜 − 𝐶𝐿 𝐹 𝐴 , 𝑖%, 𝐿 301500 − 20000 𝑑= = 𝑃19431.80 (1 + 0.08)10 − 1 0.08 𝐹 𝐷𝑛 = 𝑑( , 𝑖%, 𝑛) 𝐴 (1 + 0.08)6 − 1 𝐷6 = 19431.80 [ ] = 𝑃142550.31 0.08 𝐶𝑛 = 𝐶𝑜 − 𝐷𝑛 𝐶6 = 301500 − 142550.31 𝑪𝟔 = 𝑷𝟏𝟓𝟖𝟗𝟗. 𝟔𝟗 𝑨𝒏𝒔. 𝑑=

c) Declining Balance Method: 𝐿 𝐶𝐿 𝑘 =1− √ 𝐶0 10

20000

𝑘 = 1− √

301500

=0.2376

𝐶6 = 301500(1 − 0.2376)6

1 0

Engineering Economy Depreciation

𝑪𝟔 = 𝑷𝟓𝟗𝟐𝟎𝟏. 𝟓𝟑 𝑨𝒏𝒔.

d) Double Declining Balance Method: 2 6 𝐶6 = 𝐶0 (1 − ) 𝐿 2 6 𝐶6 = 301500 (1 − ) 10 𝑪𝟔 = 𝑷𝟕𝟗𝟎𝟑𝟔. 𝟒𝟐 𝑨𝒏𝒔. e) SYD:

∑ 𝑜𝑓 𝑑𝑖𝑔𝑖𝑡𝑠 =

𝐷6 =

𝐿(𝐿+1) 2

=

1(10+1) 2

= 55

10 + 9 + 8 + 7 + 6 + 5 (301500 − 20000) = 𝑃230381.18 55 𝐶𝑛 = 𝐶𝑜 − 𝐷𝑛

𝐶6 = 301500 − 230381.18 𝑪𝟔 = 𝑷𝟕𝟏𝟏𝟖𝟏. 𝟖𝟐 𝑨𝒏𝒔.

Is due to the lessening of the physical ability of a property to produce results?

What do you call to the decrease in value of a physical property due to the passage time?

What causes the decrease in value of physical properties with the passage of time and use?

Is the amount the property would Sale For if disposed off as junk?