Example :

 It is a load bearing structure. Age is 8 years. Life is 60

years.

i) What is the percentage depreciation by straight line

method assuming a salvage value of 10%.
ii)What is the depreciation by constant percentage
method if the depreciation rate is1.5%.
Exercise 2 :
It is a load bearing structure of 20 years old. Plinth
area : 1275 sq.ft.. Replacement rate = Rs. 1,650/sq.ft.
What is the depreciated value of the building (Life : 60
years, salvage value = 10%) by adopting straight line
method?

Plinth area = 1,275 sq.ft.

Replacement rate Replacement
value
= Rs. 1,650/sq.ft.
= Rs. 21,03,750/-
Age of the = 20 Years
building
Life of the building
= 60 years
Salvage value = 10%
=20 x (100 - 10) =
Depreciation percentage 30%
60

Depreciated value or = 0.7 x 21,03,720

(Net Present Value)
 = Rs.
14,72,625/-
Exercise 3 :

The built up area of a GF building is 5,000 sq.ft. and

the carpet area is 4,000 sq.ft. Plot size is 10,000 sq.ft.
What is the FSI? What is plot coverage?
FSI =
Plot coverage =
Exercise 4 :
A building of 8,000 sq.ft (GF & FF - 4,000 sq.ft
each) is existing in a plot of 8,000 sq.ft.
What is the plot coverage?
Exercise 5 :
20 years factory building of 5,000 sq.ft. is situated in 1 acre of industrial
land. The unit replacement rate of building is Rs. 1,000/-. Assuming the
life as 40 years and a salvage value of 30%, find the depreciated value
and salvage value of the building.
Plinth area = 5,000 sq.ft.
Replacement rate = Rs. 1,000/sq.ft.
Replacement value = Rs. 50,00,000
Age of the building = 20 years
Life assumed = 40 years
Salvage value assumed = 30%
Depreciation
percentage = 20/40 x (100 - 30) = 35%

= Rs.
Depreciated value = 0.65 x 50,00,000 32,50,000
 = Rs.
Salvage value = 0.3 x 50,00,000 15,00,000
Exercise 6 :

Building area = 1,200 m2 ; Age = 25 years ; Life = 50 years ; Salvage

value = Nil ; Plot area= 2,000 m2 ; Land rate = Rs. 8,000/m2 ;
Replacement cost of building = Rs. 25,000/m2

What is the value?

= Rs. 1,60,00,000
Land value = 2,000 x 8,000
Depreciation percentage
= 25/50 x 100
= 50%

Depreciated value of the =

building 1,200x25,000x0.5
= Rs. 1,50,00,000
Total value = Rs. 3,10,00,000
Exercise 7 :

The plinth area of a RCC roofed load bearing residential building (16
years old) is1,000 sq.ft. The life of the building as 60 years and a salvage
value of 10%,

Questions :
1) Calculate the depreciated value if the unit replacement cost is Rs.
1,800/-.
2) For the above building, if the age of the first floor is 10 years, what
will be the depreciated value of first floor of built up area 1,200 sq.ft.
assuming the unit rate of construction as Rs. 1,400/-.
Data :
Type of structure = Load bearing
Plinth area = 1,000 sq.ft.
Life
= 60 years
Age of the building = 16 years
Salvage value
= 10%

Calculations :
GF
Plinth area = 1,000 sq.ft.
Replacement rate
= Rs. 1,800/sq.ft.
Replacement value = Rs. 18,00,000
Age
= 16 years
Life = 60 years
Salvage value Depreciation percentage
= 10%
= 16/60 x 90
= 24%
Depreciated value = 0.76 x 18,00,000
= Rs. 13,68,000/- (1)
FF
Plinth area = 1,200 sq.ft.
Age
= 10 years
Life = 60 Years

Depreciation (as of GF)

= 24%
Replacement rate = Rs. 1,400/sq.ft.
Replacement value
= 1,200 x 1,400
= 16,80,000
Depreciated value = 0.76 x 16,80,000
= Rs. 12,76,800/- (2)

Answers :
1) Rs. 13,68,000/- 2) Rs. 12,76,800/-
Exercise 8 :
A RCC framed structure building consists of front portion (1,500 sq.ft. -
24 years age) and rear portion (1,200 sq.ft. - 16 years). The replacement
unit rate of construction is Rs. 1,600 per sq.ft. Life 80 years. Salvage
value - 10%.
Questions :
1) What is the depreciated value of rear portion?
2)What is the depreciated value of the front
portion? Data :
Number of portion =2
Type of structure
= RCC framed
Area of front portion = 1,500 sq.ft.
Age of front portion
= 24 years
Area of rear portion = 1,200 sq.ft.
Age of rear portion
= 16 years
Replacement rate of = Rs. 1,600/sq.ft.
construction
Life
= 80 years
Salvage value = 10%
Calculations :
Rear portion :
Plinth area of rear
portion = 1,200 sq.ft.
= Rs.
Replacement rate 1,600/sq.ft.
Replacement value = 1,200 x 1,600
= Rs. 19,20,000
Age of the building = 16 years
Life of the building = 80 years
Salvage value assumed = 10%
Depreciation percentage = 16/80 x 90
= 18%

Depreciation value = 0.18 x 19,20,000

=Rs. 3,45,600
= 19,20,000 -
Depreciated value 3,45,600
= Rs. 15,74,400/-
Front portion :
Plinth area of front
portion = 1,500 sq.ft.
Replacement rate = Rs. 1,600/sq.ft.
Replacement value = 1,500 x 1,600
= Rs. 24,00,000
Age = 24 years
Life = 80 years
Salvage = 10%
Depreciation = 24/80 x 90
= 27%
Depreciation value = 0.27 x 24,00,000
= Rs. 6,48,000
= 24,00,000 -
Depreciated value 6,48,000
= Rs. 17,52,000/- (2)
Answers :
1) Rs. 15,74,400/- 2) Rs. 17,52,000/-
Exercise 9 :
It is a residential building of GF & FF. The age of GF is 16 years and
FF is 8 years. Plinth area of each floor is 1,200 sq.ft. Replacement unit
rate of GF & FF is Rs. 1,600 & 1,200 respectively. Assume life as 60
years and salvage value as 10%.
Questions :
1.What is the depreciated value of GF?
2.What is the depreciated value of
FF? Data :
Number of floors =GF&FF
Plinth area of ground floor
= 1,200 sq.ft.
Age of ground floor = 16 years
Replacement rate of ground floor
= Rs. 1,600/-
Plinth area of first floor = 1,200 sq.ft.
Age of first floor
= 8 years
Replacement rate of first floor = Rs. 1,200/-
Life
= 60 years
Salvage value = 10%
 GF FF
Plinth area = 1,200 sq.ft. 1,200 sq.ft.
Replacement
rate = Rs.1,600/sq.ft. Rs.1,200/sq.ft.
Replacement value = Rs.19,20,000 Rs.14,40,000
Age = 16 years 8 years
Life = 60 years 60 years
Salvage = 10% 10%
Depreciation =16 x 90 = 24% 8x 90 = 12%
60 60
 But, 24% is adopted
14,40,000 x
 Depreciation value = 19,20,000 x 0.24 0.24
Rs. 4,60,800 Rs. 3,45,600
Rs.
 Depreciated value = Rs. 14,59,200/- 10,94,400/-
 Answers :
 1) Rs. 14,59,200/- 2) Rs. 10,94,400/-
Exercise 10 :
A load bearing building (1,500 sq.ft.) of 20 years old is existing in a
plot of 2,400 sq.ft. The unit land rate of plot is Rs. 2,000 and
replacement unit rate of construction is Rs. 1,700 sq.ft. It is a collateral
security. Salvage value = 10%.
Questions :
1) Determine the market value?
2) Determine the forced value (assume a reduction factor as
15%)? Data :
Type of structure = Load bearing
Plinth area
 =
1,500
sq.ft.
Age = 20 years
Plot area
= 2,400 sq.ft.
Land rate = Rs. 2,000/-
Replacement rate of construction
= Rs. 1,700/sq.ft.
Salvage value = 10%
Purpose
= Collateral security to bank
Calculations :
Land value = 2,400 x 2,000
= Rs. 48,00,000
Building area = 1,500 sq.ft.
= Rs.
Replacement rate 1,700/sq.ft.
Age of the building = 20 years
Life of the building = 60 years
Salvage value = 10%
Depreciation
percentage = 20/60 x 90
= 30%
Depreciated value of building = 0.7 x 1,500 x
1,700 = Rs.
17,85,000
Total market value = Rs. 65,85,000/- (1)
48,00,000 + 17,85,000

Forced sale value 0.85x65,85,000= Rs. 55,97,250/- (2)

Answers :
1) Rs. 65,85,000/- 2) Rs. 55,97,250/-
Exercise 11 :
Plinth area is 1,000 sq.ft. Replacement rate of construction is Rs.
2,000/sq.ft.
Age is20 years. Life is 60 years. Salvage value 10%.
Questions :
1) What is replacement value?
2) What is depreciation percentage by straight line method?
3) What is the net present value?
4) What is the depreciation percentage by constant percentage
method assuming a rate of depreciation as 1.5%.
5) What is the balance economic life?
Data :
Plinth area = 1,000 sq.ft.
Replacement rate of = Rs. 2,000/sq.ft.
construction
Age of the building
= 20 years
Life of the building = 60 years
Salvage value
= 10%
Plinth area = 1,000 sq.ft.
Replacement rate = Rs. 2,000/sq.ft.
Replacement value = Rs. 20,00,000/- (1)
Age of the building = 20 Years
Life of the building = 60 Years Salvage value = 10%
Depreciation = (20/60) x 90
= 30% (2)
Depreciation value = 0.3 x 20,00,000
= 6,00,000
Net present value = 20,00,000 - 6,00,000
= Rs. 14,00,000/- (3)
n
Depreciation percentage = 1 - ( 1 – r/100 )
20
by constant %age method = 1 - ( 1 - 1.5/100 )
20
= 1 - (0.985)
Depreciation factor = 0.26087
Depreciation percentage = 0.26087 x 100
= 26.09% (4)
Balance economic life =60-20
= 40 Years (5)
Answers :
5) 40
1) Rs. 20,00,000/- 3) Rs. 14,00,000/- Years
2) 30% 4) 26.09%
Exercise 12 :

Ground floor of a residential bungalow was

constructed in 1985 at a cost of Rs. 3,50,000.First floor
was constructed in 1990 at the cost of Rs. 6,00,000/-.
Work out Replacement cost of bungalow for the year
2003 by Book value method. The building cost
multiplier factor with 1960 as base year for year 1985,
1990 and 2003 were 14.16, 27.08 and 87.50
respectively.
Questions :
1.What is the Replacement cost of ground floor by book value
method?
2.What is the Replacement cost of first floor by book value method?
3.What is the Total replacement cost of the building by book value
method? Data :
Year of construction of GF = 1985
Cost invested for GF
= Rs. 3,50,000/-
Year of construction of FF = 1990
Cost invested for FF
= Rs. 6,00,000/-
Cost multiplier factor (1960 as base = 14.16 year) for the year 1985
Cost multiplier factor (1960 as base = 27.08 year) for the year 1990
Cost multiplier factor (1960 as base = 87.50 year) for the year 2003
3. Total Replacement
= 21,62,782 + 19,38,700
cost
= Rs. 41,01,482/- (3)

Answers :

1. Rs. 21,62,782/-
2. Rs. 19,38,700/-
3. Rs. 41,01,482/-