Contemporary Engineering Economics, 4th edition, © 2007

Book Depreciation

Lecture No. 33Chapter 9Contemporary Engineering EconomicsCopyright © 2006

Contemporary Engineering Economics, 4th edition, © 2007

Book Depreciation Methods Purpose: Used to report net income to

stockholders/investors Types of Depreciation Methods:

Straight-Line Method Declining Balance Method Unit Production Method

Contemporary Engineering Economics, 4th edition, © 2007

Straight – Line (SL) Method

• Principle A fixed asset as providing its service in a uniform fashion over its life

• Formula•Annual Depreciation

Dn = (I – S) / N, and constant for all n.•Book Value

Bn = I – n (D)where I = cost basis

S = Salvage valueN = depreciable life

Contemporary Engineering Economics, 4th edition, © 2007

Example 9.3 – Straight-Line Method

n Dn Bn 1 1,600 8,400 2 1,600 6,800 3 1,600 5,200 4 1,600 3,600 5 1,600 2,000

I = $10,000N = 5 YearsS = $2,000D = (I - S)/N

n

Contemporary Engineering Economics, 4th edition, © 2007

Declining Balance Method• Principle: A fixed asset as providing its service in a decreasing fashion• Formula

• Annual Depreciation

• Book Value

1 nn BD 1)1( nI

nn IB )1( where 0 << 2(1/N)

Note: if is chosen to be the upper bound, = 2(1/N),

we call it a 200% DB or double declining balance (DDB) method.

Contemporary Engineering Economics, 4th edition, © 2007

Example 9.4 – Declining Balance Method

n012345

Dn

$4,0002,4001,440

864518

Bn$10,000

6,0003,6002,1601,296

778

I

N

S

D B

I

B I

n n

n

nn

= $10,

= years

= $778

=

= ( -

000

5

1

1

1

1

( )

Contemporary Engineering Economics, 4th edition, © 2007

• SL Dep. Rate = 1/5• (DDB rate) = (200%) (SL rate)

= 0.40

Asset: Invoice Price $9,000Freight 500Installation 500

Depreciation Base $10,000Salvage Value 0Depreciation 200% DBDepreciable life 5 years

Example 9.5 DB Switching to SL

Contemporary Engineering Economics, 4th edition, © 2007

Adjustments to the DB Method

Switch from DB to SL after n’

No further depreciation allowances are availableafter n”

Contemporary Engineering Economics, 4th edition, © 2007

n Depreciation

Book

Value

12345

10,000(0.4) = 4,000 6,000(0.4) = 2,400 3,600(0.4) = 1,440 2,160(0.4) = 864 1,296(0.4) = 518

$6,0003,6002,1601,296

778

n

Book

Depreciation Value

12345

4,000 $6,0006,000/4 = 1,500 < 2,400 3,6003,600/3 = 1,200 < 1,440 2,1602,160/2 = 1,080 > 864 1,0801,080/1 = 1,080 > 518 0

(a) Without switching (b) With switching to SL

Note: Without switching, we have not depreciated the entirecost of the asset and thus have not taken full advantage of depreciation’s tax deferring benefits.

Case 1: S = 0

Contemporary Engineering Economics, 4th edition, © 2007

End of Year

Depreciation Book Value

1 0.4($10,000) = $4,000 $10,000 - $4,000 = $6,000

2 0.4(6,000) = 2,400 6,000 – 2,400 = 3,600

3 0.4(3,600) = 1,440 3,600 –1,440 = 2,160

4 0.4(2,160) = 864 > 160 2,60 – 160 = 2,000

5 0 2,000 – 0 = 2,000

Note: Tax law does not permit us to depreciate assets belowtheir salvage values.

Case 2: S = $2,000

Contemporary Engineering Economics, 4th edition, © 2007

Units-of-Production Method

• PrincipleService units will be consumed in a non

time-phased fashion

• Formula•Annual Depreciation

Dn = Service units consumed for yeartotal service units

(I - S)

Contemporary Engineering Economics, 4th edition, © 2007

Example 9.7 Units-of-Production Given: I = $55,000, S = $5,000, Total service

units = 250,000 miles, usage for this year = 30,000 miles

Solution:

30,000($55,000 $5,000)

250,000

3($50,000)

25

$6,000

Dep