Example 1:
A business purchases software for
₹1,20,000 with an expected useful life of 4 years and no residual value. Calculate the annual amortisation using the straight-line method.
Solution :
- Yahaan
cost = ₹1,20,000
- Residual Value = ₹0
- Useful Life
= 4 saal
- Straight-line
formula: Annual Amortisation=Cost−Residual Value/Useful Life
Amortisation =1,20,000-0/4=₹30,000 per year
Table
Format (Year-wise details):
|
Year
|
Opening
Carrying Value (₹)
|
Amortisation
(₹)
|
Closing
Carrying Value (₹)
|
|
1
|
1,20,000
|
30,000
|
90,000
|
|
2
|
90,000
|
30,000
|
60,000
|
|
3
|
60,000
|
30,000
|
30,000
|
|
4
|
30,000
|
30,000
|
0
|
Example 2:
A company acquires a patent for
₹2,00,000. The expected useful life of the patent is 5 years, and there is no
market for resale, so residual value
is zero. Find the annual amortisation
(straight-line) and the carrying value
after 3 years.
Solution :
1.
Cost =
₹2,00,000
2.
Residual Value =
₹0
3.
Useful Life = 5
saal
Annual amortisation
(straight-line) = (2,00,000 – 0) / 5 = ₹40,000.
3 years ke baad total
amortisation = 3 × ₹40,000 = ₹1,20,000.
Toh Carrying Value after 3 years =
2,00,000 – 1,20,000 = ₹80,000.
Table Format:
|
Year
|
Opening
(₹)
|
Amortisation
(₹)
|
Closing
(₹)
|
|
1
|
2,00,000
|
40,000
|
1,60,000
|
|
2
|
1,60,000
|
40,000
|
1,20,000
|
|
3
|
1,20,000
|
40,000
|
80,000
|
|
4
|
80,000
|
40,000
|
40,000
|
|
5
|
40,000
|
40,000
|
0
|
Example 3:
An intangible asset costs ₹90,000,
has a residual value of ₹10,000,
and a useful life of 4 years.
Determine the annual amortisation
using the straight-line method.
Solution :
- Depreciable Amount = ₹90,000 – ₹10,000 = ₹80,000
- Useful Life
= 4 saal
Annual amortisation = 80,000
/ 4 = ₹20,000.
Table
Format (Year-wise):
|
Year
|
Opening
Value (₹)
|
Amortisation
(₹)
|
Closing
Value (₹)
|
|
1
|
90,000
|
20,000
|
70,000
|
|
2
|
70,000
|
20,000
|
50,000
|
|
3
|
50,000
|
20,000
|
30,000
|
|
4
|
30,000
|
20,000
|
10,000 (Residual)
|
Example 4:
A software license is acquired for ₹1,00,000.
Management initially estimates the useful
life to be 5 years. After 2 years, it is found that the software
might become obsolete sooner, and the remaining
useful life is now revised to 2 more years (total 4 years).
Amortisation is on straight-line.
Show the year-wise amortisation.
Solution :
1.
First 2 years:
Original plan - 5 saal ke hisaab se
Annual amortisation
(years 1 & 2) = (1,00,000 / 5) = ₹20,000
2.
Carrying value
after 2 years = 1,00,000 – (20,000 × 2) = ₹60,000
3.
Revision: Ab
bacha hua life 2 saal (not 3). So year 3 aur year 4 ke liye new annual
amortisation = 60,000 / 2 = ₹30,000.
Table Format:
|
Year
|
Opening
(₹)
|
Amortisation
(₹)
|
Closing
(₹)
|
|
1
|
1,00,000
|
20,000 (based on 5-year
life)
|
80,000
|
|
2
|
80,000
|
20,000 (still old
estimate)
|
60,000
|
|
3
|
60,000
|
30,000 (revised 2-year
life)
|
30,000
|
|
4
|
30,000
|
30,000
|
0
|
Example 5:
A company buys an intangible asset for ₹1,80,000.
Useful life is 3 years. However, the usage pattern is expected to be 40% in Year 1, 30% in Year 2, and 30%
in Year 3. Compute amortisation
accordingly (variable usage).
Solution :
- Cost
= 1,80,000, no residual mentioned → assume 0.
- Usage
pattern ke hisaab se hum cost ko 40%-30%-30% mein allocate kar denge.
So:
- Year
1: 40% of 1,80,000 = ₹72,000
- Year
2: 30% of 1,80,000 = ₹54,000
- Year
3: 30% of 1,80,000 = ₹54,000
Check total: 72 + 54 + 54 =
1,80,000.
Table:
|
Year
|
Percentage
|
Amortisation
(₹)
|
Closing
Value (₹)
|
|
1
|
40%
|
72,000
|
1,80,000 – 72,000 =
1,08,000
|
|
2
|
30%
|
54,000
|
1,08,000 – 54,000 = 54,000
|
|
3
|
30%
|
54,000
|
54,000 – 54,000 = 0
|
Example 6:
A trademark is acquired for
₹60,000 with an expected usage of
10,000 units total. In the first year,
2,000 units are used; second year, 3,500 units; third year, 4,500 units.
Compute the annual amortisation
using the units-of-production
approach (no residual value).
Solution :
- Total
cost = ₹60,000
- Total
expected usage = 10,000 units
- Rate
per unit = 60,000 / 10,000 = ₹6 per unit
So:
- Year
1 (2,000 units): 2,000 × 6 = ₹12,000
- Year
2 (3,500 units): 3,500 × 6 = ₹21,000
- Year
3 (4,500 units): 4,500 × 6 = ₹27,000
Table:
|
Year
|
Units
Used
|
Amortisation
Rate (₹/unit)
|
Amortisation
(₹)
|
|
1
|
2,000
|
6
|
12,000
|
|
2
|
3,500
|
6
|
21,000
|
|
3
|
4,500
|
6
|
27,000
|
|
Total
|
10,000
|
--
|
60,000 (full cost)
|
Example 7:
A logo design is purchased for ₹50,000
on 1st April. It is ready for use
on 1st July. The useful life is 5 years, with no residual
value. Amortisation is straight-line.
Show the amortisation for the
first accounting year ending 31st March
(assume 12-month financial year).
Solution :
- Asset
is purchased on 1st April, but “available
for use” from 1st July.
- So
effectively, amortisation
starts from 1st July.
- For
the year (1st July to 31st March) → 9 months of usage.
Annual amortisation if full year used: (50,000 / 5) = ₹10,000
per year.
But only 9 months in the first year, so pro-rata:
10,000×9/12=₹7,500
Table:
|
Particulars
|
Calculation
|
Amount
(₹)
|
|
Full-year amortisation
(annual)
|
50,000 / 5 = 10,000
|
10,000
|
|
Months used in 1st year
(Jul–Mar = 9 months)
|
10,000 × (9/12)
|
7,500
|
|
Amortisation
for 1st year
|
--
|
7,500
|
Example 8:
On 1st Jan, a company acquires a software for ₹1,80,000, with straight-line amortisation over 6 years.
After 3 years, the software is
found to still have 3 more years
left. No change in overall life.
However, the residual value is
discovered to be ₹30,000 (previously assumed zero). Adjust the remaining amortisation.
Solution :
- Original
plan: cost = 1,80,000, no residual, 6 years →
annual = ₹30,000.
After 3 years:
- Accumulated
amortisation = 30,000 × 3 = ₹90,000
- Carrying
value = 1,80,000 – 90,000 = ₹90,000
Now, new info: residual value = ₹30,000. Baki life = 3
years.
- New
Depreciable Amount = 90,000 (carrying) – 30,000 (residual) = ₹60,000
- New
annual amortisation for next 3 years = 60,000 / 3 = ₹20,000
Table:
|
Period
|
Carrying
Value at Start (₹)
|
Amortisation
(₹)
|
Carrying
Value at End (₹)
|
|
Years 1–3 (old plan)
|
1,80,000 (start)
|
30,000 each year (total
90,000)
|
90,000
|
|
Next 3 years (revised)
|
90,000
|
20,000 each year
|
End with 30,000 (residual)
|
Example 9:
An intangible asset is bought for ₹2,40,000,
with an expected life of 3 years.
Management decides to use the sum-of-the-years’-digits
(SYD) method. Compute the amortisation
for each year, assuming zero
residual value.
Solution :
- SYD
method total of digits for 3 years = 1 + 2 + 3 = 6.
- Cost
= 2,40,000, no salvage.
Year 1 amortisation = 2,40,000 × (3/6) = 2,40,000 × 0.5 =
₹1,20,000
Year 2 amortisation = 2,40,000 ×
(2/6) = 80,000
Year 3 amortisation = 2,40,000 ×
(1/6) = 40,000
Check total: 1,20,000 +
80,000 + 40,000 = 2,40,000
Table:
|
Year
|
SYD
Fraction
|
Amortisation
(₹)
|
Closing
Value
|
|
1
|
3/6 = 0.5
|
1,20,000
|
1,20,000 (2,40,000 –
1,20,000)
|
|
2
|
2/6 = 0.333...
|
80,000
|
40,000 (1,20,000 – 80,000)
|
|
3
|
1/6 = 0.1667
|
40,000
|
0 (40,000 – 40,000)
|
Example 10:
A franchise license is purchased for ₹3,00,000 with a legal right of 10 years. After the 4th
year, the franchise agreement is extended for another 2 years (total 12 years).
Compute the revised amortisation
from year 5 onward.
Solution :
1.
Original plan: 10
years. Straight-line. Annual amortisation = 3,00,000 / 10 = ₹30,000.
2.
After 4 years:
-
Used = 4 × 30,000 =
₹1,20,000
-
Carrying Value =
3,00,000 – 1,20,000 = ₹1,80,000
3.
Now extended to total
12 years. Baki life ab 12 – 4 = 8 years remain.
4.
Revised annual from
5th year = 1,80,000 / 8 = ₹22,500.
Table:
|
Years
|
Amortisation
|
Carrying
Value
|
|
1–4 (original)
|
30,000 each year
(4×30k=1,20k)
|
3,00,000 – 1,20,000 =
1,80,000
|
|
5–12 (extended period)
|
1,80,000 / 8 = 22,500 per
year
|
After 8 more years →
carrying = 0
|
