a) Determine the annual depreciation cost by the straight line method.
b) Determine the annual depreciation cost by the sinking fund method.
Assume interest at
6
1
2
% compounded annually.
Solution:
C
O
= P53,000 + P1,500 = P54,500
C
L
= P5,000
a)
? =
?
?
−?
?
?
=
?54,500−?5,000
10
= P4,950
b)
? =
?
?
−?
?
?[?,𝑖%,?]
=
?54,500−?5,000
(1+.065)
10
−1
.065
=
?49,500
13.49
=
P3,668.18
Example
A firm bought an equipment for P56,000. Other expenses including
installation amounted to P4,000. The equipment is expected to have a life of
16years with a salvage value of 10% of the original cost. Determine the book

Engineering Economy
Page 134
value at the end of 12years by (a) the straight line method and (b) sinking fund
method at 12% interest.
Solution:
C
O
= P65,000 + P4,000 = P60,000
C
L
= P60,000(0.10) =P6,000
L
= 16
n=12
i-12%
a) Straight line method
? =
?
?
−?
𝐿
?
=
?60,000−?6,000
16
= P3,375
D
12
= (d) (n) = P3,375 (12) = P40,500
C
12
= C
O
–
D
12
= P60,000
–
P40,500 = P19,500
b) Sinking fund method
? =
?
?
−?
?
?[?,𝑖%,?]
=
?60,000−?6,000
(1+0.12)
16
−1
0.12
=
?54,000
42.7533
= P1,263.06
D
12
=
?
(1+0.12)
12
−1
0.12
= P1,263.06(14.1331) = P30,481.60
C
12
= C
O
–
D
12
= P60,000
–
P30,481.60 = P29,518.40
Problem
Solve the above problem using sinking fund method at i =5%
Given:
C
O
= ₱ 4,000
SV = 0
L = 10 years

Engineering Economy
Page 135
Declining Balance Method
In this method, sometimes called the constant percentage method or the
Matheson Formula, it is assumed that the annual cost of depreciation is a fixed
percentage of the salvage value at the beginning of the year. The ratio of the
depreciation in any year to the book value at the beginning of that year is
constant throughout the life of the property and is designed by k, the rate of
depreciation
C
0
C
1
C
2
C
3
C
n-1
C
n
C
L-1
C
L
0
1
2
3
n-1
n-1
L-1
L
d
1
d
2
d
3
d
n
d
L
d
n
= depreciation during the n
th
year
Year
Book Value at
beginning of
year
Depreciation during the
year
Book Value at the end of the
year
1
C
0
d
1
= kC
0
C
1
= C
0
–
d
1
= C
0
(1-k)
2
C
0
(1-k)
d
2
= kC
1
C
2
= C
1
–
d
2
= C
0
(1-k)
2
3
C
0
(1-k)
2
d
3
= kC
2
C
3
= C
2
–
d
3
= C
0
(1-k)
3
n
C
0
(1-k)
n-1
d
n
= kC
n-1
C
n
= C
n-1
–
d
n
= C
0
(1-k)
n
L
C
0
(1-k)
L-1
d
L-1
= kC
L
C
L
= C
L-1
–
d
L
= C
0
(1-k)
L
d
n
= C
0
(1-k)
n-1
k
(3 - 7)
C
n
= C
0
(1-k)
n
=
?
0
[
?
?
?
0
]
?
?
(3 - 8)
C
L
= C
0
(1-k)
L
(3 - 9)
k = 1 -
√
?
𝑛
?
0
𝑛
= 1 −
√
?
𝐿
?
0
𝐿
(3 - 10)

Engineering Economy
Page 136
This method does not apply , if the salvage value is zero, because k will
be equal to one and d
1
will be equal to C
0
.
Example
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 the book value at the end of each year for 5years.
Solution:
Year
Book value at
beginning of
year
Depreciation
during the year
10%
Total
depreciation at
end of year
Book value at
end of year
1
P2,000.00
P200.00
P200.00
P1,800.00
2
P1,800.00
P180.00
P380.00
P1,620.00
3
P1,620.00
P162.00
P542.00
P1,458.00
4
P1,458.00
P145.80
P687.80
P1,312.20
5
P1,312.20
P131.22
P819.02
P1,180.98
Double Declining Balance(DDB) Method
This method is very similar to the declining balance method except that
thw rate of depreciation k is replaced by 2/L.
?
?=
?
0
(1 −
2
?
)
?−1
2
?
(3 - 11)
?
?=
?
0
(1 −
2
?
)
?
(3 - 12)
?
?=
?
0
(1 −
2
?
)
?
(3 - 13)
When the DDB method is used, the salvage value should not be
subtracted from the first cost when calculating the depreciation charge.