Example
1:
A customer list is purchased for
₹1,50,000 (marketing intangible), with an expected
life of 3 years. However, usage is heavily skewed in the first year (50%), then 30% in second year,
and 20% in third. Show the amortisation.
No salvage.
Solution :
- Cost
= 1,50,000. No salvage.
- Usage
pattern: 50%, 30%, 20%.
So,
- Year
1 = 1,50,000 × 50% = ₹75,000
- Year
2 = 1,50,000 × 30% = ₹45,000
- Year
3 = 1,50,000 × 20% = ₹30,000
- Total
= 75k + 45k + 30k = 1,50,000.
Table:
|
Year
|
Usage%
|
Amortisation
(₹)
|
Remaining
Value (₹)
|
|
1
|
50%
|
75,000
|
75,000 (1,50,000 – 75,000)
|
|
2
|
30%
|
45,000
|
30,000 (75,000 – 45,000)
|
|
3
|
20%
|
30,000
|
0 (30,000 – 30,000)
|
Example
2:
A brand name is purchased for
₹4,00,000 with an initial 8-year
expected useful life. It has no
active market at the end of 8 years, so residual
value = 0. Using straight-line
amortisation, find (a) the annual
expense and (b) the carrying
amount after 5 years.
Solution :
1.
Cost =
4,00,000
2.
Life = 8
years
3.
Residual = 0
(no active market)
(a) Annual amortisation =
(4,00,000 / 8) = ₹50,000.
(b) After 5 years, total amortisation = 5 × 50,000 = ₹2,50,000. So carrying value = 4,00,000 – 2,50,000 =
₹1,50,000.
Table:
|
Year
|
Opening
(₹)
|
Amortisation
(₹)
|
Closing
(₹)
|
|
1
|
4,00,000
|
50,000
|
3,50,000
|
|
2
|
3,50,000
|
50,000
|
3,00,000
|
|
3
|
3,00,000
|
50,000
|
2,50,000
|
|
4
|
2,50,000
|
50,000
|
2,00,000
|
|
5
|
2,00,000
|
50,000
|
1,50,000
|
|
…
|
…
|
…
|
…
|
|
8
|
50,000
|
50,000
|
0
|
Example 3
On 1st October, ABC Ltd. acquired an exclusive license for ₹1,20,000. The
license is valid for 4 years, and
is available for use immediately on the same date (1st October). The company’s
financial year ends on 31st March.
1.
Calculate the annual
straight-line amortisation if it were a full year.
2.
Calculate the
amortisation for the first financial year
(i.e., from 1st Oct to 31st Mar).
Solution
1.
Pehla step:
Full-year (12 months) ke liye agar hum amortise karein, toh
Annual Amortisation=1,20,000/4 years=₹30,000 har saal.
Iska matlab agar poore 12
mahine use karte toh 30k expense hota.
2.
First financial year
(1st Oct se 31st Mar) tak sirf 6 mahine hi ho rahe hain. Toh hum pro-rata basis
pe karenge:
30,000×6/12=₹15,000
Matlab pehle saal me 15k hi
amortisation charge hoga kyunki use sirf 6 mahine hua hai.
Table
|
Particulars
|
Calculation
|
Amount
(₹)
|
|
Annual
Amortisation (full 12 months)
|
1,20,000 / 4 = 30,000
|
30,000
|
|
First 6
months (Oct–Mar)
|
30,000 × (6/12)
|
15,000
|
Example 4
On 1st April Year 1, a company purchased Brand X for ₹5,00,000 with an expected useful life of 10 years. After 2 years, management decides the brand can actually last for 12 years total. Show how the amortisation is revised from the 3rd year
onward (straight-line, zero residual).
Solution (Hinglish)
- Shuru
mein hum assume kar rahe the 10 saal. Toh annual amortisation = 5,00,000/10=₹50,000 per saal.
- 2
saal ho gaye, toh total amortise = 50,000 × 2 = 1,00,000. Bacha hua
carrying value = 5,00,000 – 1,00,000 = ₹4,00,000.
- Ab
management ko lagta hai brand 12 saal chal sakta hai total. Par 2 saal
already use ho chuke. Matlab ab 10 saal remain (12 – 2).
- Revised
amortisation from 3rd year = 4,00,000 / 10 = ₹40,000 har saal.
- Is
tarah se aap 3rd year se lekar 12th year tak ₹40k charge karoge.
Table
|
Period
|
Amort.
Method
|
Amort.
(₹)
|
Carrying
Value (₹)
|
|
Years 1–2
|
50k per year
|
Total 1,00,000
|
5,00,000 – 1,00,000 =
4,00,000
|
|
Years 3–12 (revised)
|
4,00,000 / 10 = 40k per
year
|
40k per year
|
Finally reduces to 0 after
year 12
|
Example 5
XYZ Ltd. acquired a software package for ₹3,60,000 on 1st January with a useful life of 4 years. However, they estimate that usage is 30%
in the first year, 25% in the second, 25% in the third, and 20% in the fourth. Compute the year-wise amortisation (no residual value).
Solution
- Cost
= 3,60,000. Residual = 0 assume kar lo.
- Usage
pattern diya hua hai: 30%, 25%, 25%, 20%.
- Aap
cost ke hisaab se har year utna percent apply kar do.
Calculation:
- 1st
year: 3,60,000 × 30% = ₹1,08,000
- 2nd
year: 3,60,000 × 25% = ₹90,000
- 3rd
year: 3,60,000 × 25% = ₹90,000
- 4th
year: 3,60,000 × 20% = ₹72,000
- Total
= 1,08,000 + 90,000 + 90,000 + 72,000 = 3,60,000 (puri cost).
Table
|
Year
|
Usage %
|
Amortisation
(₹)
|
Remaining
(₹)
|
|
1
|
30%
|
1,08,000
|
2,52,000 (3,60,000 –
1,08,000)
|
|
2
|
25%
|
90,000
|
1,62,000
|
|
3
|
25%
|
90,000
|
72,000
|
|
4
|
20%
|
72,000
|
0
|
Example 6
On 1st April, PQR Ltd. purchased a patent for ₹2,00,000 with an expected life of 5 years. The annual straight-line amortisation is
₹40,000. After 3 years, the patent
is found to be impaired due to a
competitor’s technology. The new recoverable
amount at the end of Year 3 is ₹30,000.
Show the amortisation plus impairment effect.
Solution
1.
Shuru se hum 2,00,000
/ 5 = ₹40,000 har saal charge kar rahe the.
2.
3 saal ke baad total
amortise = 40k × 3 = 1,20,000. Carrying value = 80,000.
3.
Par ek competitor ki
wajah se patent ki value gir gayi. Ab recoverable sirf 30,000 hai. Toh
impairment = 80,000 – 30,000 = ₹50,000.
4.
Year 3 ke end me hum
impairment record kar denge.
5.
Bacha hua carrying =
₹30,000. Baaki 2 saal me hum normal amortisation kar sakte hain (assuming abhi
2 aur saal use hoga). Toh 30,000 ÷ 2 = ₹15,000 each year for year 4 & 5.
Table
|
Year
|
Opening
CV (₹)
|
Amort
(₹)
|
Impairment
(₹)
|
Closing
CV (₹)
|
|
1
|
2,00,000
|
40,000
|
0
|
1,60,000
|
|
2
|
1,60,000
|
40,000
|
0
|
1,20,000
|
|
3
|
1,20,000
|
40,000
|
50,000 (80k →
30k)
|
30,000
|
|
4
|
30,000
|
15,000
|
0
|
15,000
|
|
5
|
15,000
|
15,000
|
0
|
0
|
Example 7
A license costing ₹4,20,000 has a useful life of 3 years. The company decides to use the double-declining balance (DDB) method for amortisation (no salvage). Calculate the amortisation for each year.
Hint: DDB
rate for 3 years is typically (100% ÷ 3) × 2 = 66.67% per year on the opening carrying amount.
Solution
1.
Rate nikaalo:
1/3 = 33.33%. Double = 66.67%.
2.
Year 1:
Opening 4,20,000. 66.67% of 4,20,000 = 2,80,000. Closing = 1,40,000.
3.
Year 2:
Opening 1,40,000. 66.67% of 1,40,000 = 93,333. Closing 46,667.
4.
Year 3:
Opening 46,667. 66.67% of that = 31,111 approx. Closing 15,556.
-
DDB method me aap end
tak pura zero ho sakta hai ya ek chota leftover rahta hai jo final me adjust
kar sakte ho.
Table
|
Year
|
Opening
CV (₹)
|
Amort
(66.67%) (₹)
|
Closing
CV (₹)
|
|
1
|
4,20,000
|
2,80,000
|
1,40,000
|
|
2
|
1,40,000
|
93,333
|
46,667
|
|
3
|
46,667
|
31,111
|
15,556
|
(Rounding
ke karan final figure thoda differ ho sakta hai. Agar company chaahe toh last
year me balancing figure le sakti hai to bring it to zero.)
Software License Amortization
Example
8:
ABC Ltd. purchases a software license for $50,000. The useful life of the
software is 5 years. Calculate the annual amortization expense using the
straight-line method.
Solution:
Amortization Expense=50,000/5=10,000 per year
Hinglish Explanation:
Company ne ek software license purchase kiya jo 5 saal tak kaam karega.
Straight-line method ka matlab hai ki har saal ek equal amount amortize hoga.
Toh $50,000 ko 5 saal ke beech divide karenge toh har saal ka amortization
expense = $10,000 hoga.
Copyright Amortization
Example
9:
LMN Ltd. purchases copyrights for $60,000, and the useful life is 6 years.
Calculate the annual amortization expense.
Solution:
Amortization Expense=60,000/6=10,000 per year
Hinglish
Explanation:
Copyright movies, books, music compositions ke intellectual property rights ko
protect karta hai. Yahaan 6 saal ka useful life diya gaya hai, toh har saal
$10,000 amortize kiya jayega.
Customer Contracts
Amortization
Example
10:
A company buys customer contracts for $120,000, and the useful life is 12
years. What is the annual amortization expense?
Solution:
Amortization Expense=120,000/12=10,000 per year
Hinglish
Explanation:
Company ne customer contracts $120,000 me purchase kiye, jinka useful life 12
saal hai. Annual amortization ka matlab hai ki har saal equal amount amortize
hoga.
Toh $120,000 ko 12 saal ke beech divide karenge, toh har saal $10,000 amortize
hoga.
Summary Table of
Amortization Examples
|
Asset
Type
|
Cost of
Asset ($)
|
Useful
Life (Years)
|
Annual
Amortization ($)
|
|
Software
License
|
50,000
|
5
|
10,000
|
|
Patent
|
100,000
|
10
|
10,000
|
|
Trademark
|
80,000
|
8
|
10,000
|
|
Copyright
|
60,000
|
6
|
10,000
|
|
Customer Contracts
|
120,000
|
12
|
10,000
|
Overall
Hinglish Recap
- Amortisation
intangible assets pe hota hai, similar to Depreciation par yeh intangible ke liye use karte hain.
- Agar
partial year ho, toh hum
months ka ratio leke pro-rata charge kar dete hain.
- Usage-based
methods (jaise percentage usage, units-of-production) tab use karte hain
jab actual usage pattern pata ho.
- Double-declining ya other accelerated methods aap tab use karte ho
jab aapko lagta hai pehle saalon me zyada benefit use ho raha hai.
- Agar
impairment ho jaye, carrying
value ko reduce karke difference impair record karna hota hai.
