Engineering Economics

Depreciation
– is the decrease in the value of physical property
with the passage of time.

Requirements of a Depreciation
Method
•It should be simple.
•It should recover capital.
•The book value will be reasonably close to the
market value at any time.
•The method should be accepted by the Bureau
of International Revenue.

Depreciation Methods
•Straight Line Method
•Sinking Fund Method

The following symbols are used for the different
depreciation methods:
L = useful life of the property (in years)
Co = the original cost
CL = the value at the end of the life (scrap value)
d = the annual cost of depreciation
Cn = the book value at the end of (n) years
Dn = depreciation up to age (n) years

Straight Line Method
-This method assumes that the loss in value is directly
proportional to the age of the property.
d =
Co −CL
𝐿
Dn = n (
Co −CL
𝐿
) or n(d)
Cn = Co - Dn

Example 1
An electronic balance costs P90,000 and has an estimated
salvage value of P8,000 at the end of its 10 years life time. What
would be the book value after three years, using the Straight
Line Method in solving for the depreciation?
Given:
L = 10 years
Co = P90,000.00
CL = P8,000.00
n = 3 years
Cn = ? P90,000
P8,000
1y 2y 3y 4y 5y 6y 7y 8y 9y
A AA AAAA AA

Solution:
d =
Co −CL
𝐿
d =
90,000 − 8,000
10
d = P8,200
Dn = n (
Co −CL
𝐿
) or n(d)
D3 = 3(8,200)
D3 = P24,600
Cn = Co - Dn
C3 = 90,000 - P24,600
C3 = P65,400

Example 2
An equipment costs P10,000 with a salvage value of P500
at the end of 10 years. Calculate the annual depreciation
by Straight Line Method.
Given:
L = 10 years
Co = P10,000.00
CL = P500.00
d = ? P10,000
P500
1y 2y 3y 4y 5y 6y 7y 8y 9y
A AA A AA A A AA

Solution:
d =
Co −CL
𝐿
d =
10,000 − 500
10
d = P950.00

Example 3
A machine has an initial cost of P50,000.00 and a salvage
value of P10,000.00 after 10 years. What is the book
value after five years using straight line method
depreciation?
Given:
L = 10 years
Co = P50,000.00
CL = P10,000.00
n = 5 years
Cn = ? P50,000
P10,000
AA A A A A A AA A
4y 5y 6y 7y 8y 9y 10y1y 2y 3y

Solution:
d =
Co −CL
𝐿
d =
50,000 − 10,000
10
d = P4,000.00
Dn = n (
Co −CL
𝐿
) or n(d)
D5 = 5(4,000)
D5 = P20,000.00
Cn = Co - Dn
C5 = 50,000 – 20,000
C5 = P30,000.00

Example 4
A machine has a first cost of P13,000, an estimated
life of 15 years, and an estimated salvage value of
P1,000. Using the Straight Line Method, find:
a) The annual depreciation.
b) The book value at the end of 9 years.
Given:
L = 15 years
Co = P13,000.00
CL = P1,000.00
n = 9 years
1y 2y 3y 4y 5y 6y 7y 8y 9y 10y
15y
P1,000
P13,000
A A A A A A A A A A
A

Solution:
d =
Co −CL
𝐿
d =
13,000 − 1,000
15
d = P800.00
Cn = Co - Dn
C9 = 13,000 - 9(800)
C9 = P5,800.00

Example 5
An asset is purchased for P500,000. The salvage value in
25 years is P100,000. What are the depreciations in the
first three years?
Given:
L = 25 years
Co = P500,000.00
CL = P100,000.00
n = 3 years
Dn = ?
P500,000
P100,000
1y 2y 3y 4y
25yA A AA

Solution:
d =
Co −CL
𝐿
d =
500,000 − 100,000
25
d = P16,000.00
Dn = n (
Co −CL
𝐿
) or n(d)
D3 = 3(16,000)
D3 = P48,000.00

Sinking Fund Method
-This method assumes that a sinking fund is established
in which funds will accumulate for replacement.
d =
Co − CL
(1+𝑖) 𝐿−1
𝑖
or
(Co −CL) 𝑖
(1+𝑖) 𝐿−1
Dn = (
(Co −CL) 𝑖
(1+𝑖) 𝐿−1
) (
(Co −CL) 𝑖
(1+𝑖) 𝑛−1
) or (d)(
(Co −CL) 𝑖
(1+𝑖) 𝑛−1
)
Cn = Co - Dn

Example 1
A broadcasting corporation purchased equipment for P53,000
and paid P1,500 for freight and delivery charges to the job site.
The equipment has a normal life of 10 years with a trade-in
value of P5,000 against the purchase of a new equipment at the
end of the life. Determine the annual depreciation cost and
assume interest at 6.5% compounded annually.
Given:
L = 10 years
Co = P53,000 + P1,500
CL = P5000.00
i = 6.5%
d = ? P50,000
P5,000
1y 2y 3y 4y 5y 6y 7y 8y 9y
A AA AAAA AA 10y

Solution:
d =
Co − CL
(1+𝑖) 𝐿−1
𝑖
or
(Co −CL) 𝑖
(1+𝑖) 𝐿−1
d =
(54,500 − 5,000)(.065)
(1+.065)10−1
d = P3,668.18

Example 2
An equipment costs P10,000 with a salvage value of
P500 at the end of 10 years. Calculate the annual
depreciation by Sinking Fund method at 40% interest.
Given:
L = 10 years
Co = P10,000.00
CL = P500.00
i = 40%
d = ? P10,000
P500
1y 2y 3y 4y 5y 6y 7y 8y 9y
A AA AAAA AA 10y

Solution:
d =
Co − CL
(1+𝑖) 𝐿−1
𝑖
or
(Co −CL) 𝑖
(1+𝑖) 𝐿−1
d =
(10,000 − 500)(.40)
(1+.40)10−1
d = P136.08

Example 3
A P110,000 chemical plant had an estimated life of 6 years and a
projected scrap value of P10,000. After 3 years of operation an
explosion made it a total loss. How much money would have to be
raised to put up a new plant costing P150,000, if a depreciation
reserved have been maintained during its 3 years of operation by
Sinking Fund Method. Assume that the interest is 6%.
Given:
L = 6 years
Co = P110,000.00
CL = P10,00.00
n = 3
i = 6% P110,000
P10,000
1y 2y 3y
6yA A A

Solution:
d =
Co −CL
(1+𝑖) 𝐿−1
𝑖
𝑜𝑟
(Co −CL) 𝑖
(1+𝑖) 𝐿−1
d =
(110,000 − 10,000)(.06)
(1+.06)6−1
d = P14,336.26
Dn = (d)(
(1+𝑖) 𝑛−1
𝑖
)
D3 = (P14,336.26) (
(1+.06)3−1
.06
)
D3 = P45,640.92
Amount to be raised:
= P150,000 – P45,640.92
= P104,359.08

Example 4
A unit of welding machine cost P45,000 with an estimated life of
5 years. Its salvage value is P2,500 find its depreciation rate of
sinking fund method. Assuming that will deposit the money to a
bank giving 8.5%. Solve the depreciation.
Given:
L = 5 years
Co = P45,000.00
CL = P2,500.00
i = 8.5%
d = ?
P45,000
AAA
P2,500
1y 3y 4y
5yA
2y

Solution:
d =
Co −CL
(1+𝑖) 𝐿−1
𝑖
𝑜𝑟
(Co −CL) 𝑖
(1+𝑖) 𝐿−1
d =
(45,000 − 2,500)(.085)
(1+.085)5−1
d = P7,172.54

Example 5
A certain machinery costs P50,000, last 12 years with a salvage value of
P5,000. Money is worth 5%. If the owner decides to sell it after using it
for 5 years, what should his price be so that he will not lose or gain
financially in the transaction?
Given:
L = 12 years
Co = P50,000.00
CL = P5,000.00
i = 5%
n = 5 years
Cn = ? P50,000
P5,000
1y 2y 3y 4y 5y 6y
12yA A AAAA

Solution:
d =
Co −CL
(1+𝑖) 𝐿−1
𝑖
𝑜𝑟
(Co −CL) 𝑖
(1+𝑖) 𝐿−1
d =
(50,000 − 5,000)(.05)
(1+.05)12−1
d = P2,827.14
Dn = (d)(
(1+𝑖) 𝑛−1
𝑖
)
D5 = (2,827.14)(
(1+.05)5−1
.05
)
D5 = P15,621.75
Cn = Co - Dn
C5 = 50,000 - 15,621.75
C5 = P34,378.25

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