UNIT III - INTRODUCTION TO COST ESTIMATION Importance of costing and estimation –methods of costing-elements of cost estimation –Types of estimates – Estimating procedure- Estimation labor cost, material cost- allocation of over head charges- Calculation of depreciation cost

1. WELCOME
UNIT III - INTRODUCTION TO COST ESTIMATION
ME8793 – PROCESS PLANNING AND
COST ESTIMATION
Mr. TAMIL SELVAN M, A/P, MECH, KIT.

2. UNIT III INTRODUCTION TO COST
ESTIMATION
Importance of costing and estimation –methods of costing-
elements of cost estimation –Types of estimates – Estimating
procedure- Estimation labor cost, material cost- allocation of
over head charges- Calculation of depreciation cost

3. COST ESTIMATION
Define
Cost estimating may be defined as the process of determining
the probable cost of the product before the start ofmanufacture
Purpose
The purpose of cost estimating is to find the cost of the
manufacturing operations and to assist in setting the price for the
product

4. FEATURES OF COST ESTIMATION
 Important activity in engineering design and production
 It is forecasting the future cost
 Cost estimating & process planning are prominent activities in the
manufacturing system
 It considers all the expenditures involved like engineering,
administration, etc
 It requires high technical knowledge

5. IMPORTANCE OF COST ESTIMATION
The only accurate estimating can enable the leaders to make
vital decisions such as manufacturing, selling policies
Case1
 If job is over estimated,
The firm will not be able to compete with its competitors
 If the job is under estimated,
The firm will face huge financial loss

6. AIMS OF COST ESTIMATION
To establish the selling price of a product, so as to ensure
reasonable profit to the company
 To determine the most economical process
 To make/buy decisions
 To evaluate the alternate designs
 To prepare production budget
 To initiate the cost reduction in existing facilities

7. COSTING
Define
Costing is the determination of an actual cost of a component after
adding different expenses incurred in various departments
Costing may be defined as a systematic procedure for recording
accurately every item of expenditure incurred on the manufacture of a
product by different sections of any manufacturing concern

8. METHOD OF COSTING
1.Unit Costing
2. Job Costing
3. Contract Costing
4. Batch Costing
5. Operating Costing
6. Process Costing.
7. Multiple Costing
8. Uniform Costing.

9. 1.Unit Costing
 This method also called 'Single output Costing’.
 This method of costing is used for products which can be expressed in
identical quantitative units and is suitable for products which are
manufactured by continuous manufacturing activity.
 Costs are ascertained for convenient units of output.
Examples: Brick making, mining, cement manufacturing, dairy, flour
mills etc.

10. 2. JOB COSTING
 Under this method costs are ascertained for each work order
separately as each job has its own specifications and scope.
Examples: Painting, Car repair, Decoration, Repair of building
etc.

11. 3. CONTRACT COSTING
 Under this method costing is done for big jobs which involves
heavy expenditure and stretches over a long period and often it is
undertaken at different sites.
 Each contract is treated as a separate unit for costing.
 This is also known as Terminal Costing.
Construction of bridges, roads, buildings, etc. comes under
contract costing.

12. 4. BATCH COSTING
 This methods of costing is used where the units produced in a
batch are uniform in nature and design.
 For the purpose of costing each batch is treated as a job or
separate unit.
Industries like Bakery, Pharmaceuticals etc. usually use
batch costing method.

13. 5. OPERATING COSTING OR SERVICE
COSTING:
 Where the cost of operating a service such as nursing home, Bus,
railway or chartered bus etc. this method of costing is used to
ascertain the cost of such particular service.
 Each particular service is treated as separate units in operating
costing.
In the case of a Nursing Home, a unit is treated as the cost of a
bed per day and for buses operating cost for a kilometer is treated as
a unit.

14. 6. PROCESS COSTING
 This kind of costing is used for the products which go through different
processes.
For example, manufacturing cloths goes through different process.
 Fist process is spinning.
 The second step is the weaving process.
 The third process is converting cloth in to finished product such as shirt
or trouser etc.

15. 7. MULTIPLE COSTING
 When the output comprises many assembled parts or components such
as in television, motor Car or electronics gadgets, costs have to be
ascertained or each component as well as the finished product.
 Such costing may involve different methods of costing for different
components.
Therefore this type of costing is known as composite costing or
multiple costing.

16. 8. UNIFORM COSTING
 This is not a separate method of costing.
 This is a system of using the same method of costing by a number of
firms in the same industry.
 It is treated as a common system of using agreed principles and
standard accounting practices in the identical firms or industry.
This helps in fixation of price of the product and inter-firm
comparisons.

17. ELEMENTS OF COST ESTIMATION

18. TYPES OF ESTIMATES
 To fix the selling price of the product
 To help the contractors to submit the accurate tenders
 To forecast the progress of production and cost
 To set the variousStandards

19. COST ESTIMATING PROCEDURE
Step 1
 Study the cost estimation request thoroughly and understand it completely
Step 2
 Analyze the product and decide the requirements and specifications of
the product
Step 3
 Prepare the list of all the parts of the product and their bill of Materials
Step 4
 Take make or buy decisions and prepare a separate list of parts to be
purchased &manufactured

20. Step 5
 Estimate the materials cost for the parts to be manufactured in the plant
Step 6
 Determine the cost of the parts to be purchased from outside
Step 7
 Make a manufacturing process plan for the parts to be manufactured in
the plant
Step 8
 Estimate the machining time for each operations listed in the
manufacturing process plan
Step 9
 Determine the direct labour cost

21. Step 10
 Determine the prime cost by adding direct expenses, direct material cost,
and direct labour cost
prime cost =direct expenses +direct material cost +direct labour cost
Step 11
 Estimate the factory overheads, which include all indirect expenditure
incurred during production such as indirect material cost, indirect labour
cost, depreciation and expenditure on maintenance of the plant,
machinery, power, etc.
Step 12
 Estimate the administrative expenses
Step 13
 Estimate the selling and distribution expenses, which include packing
and delivery charges, advertisement charges, etc.

22. Step 14
 Now calculate the total cost of the product
Total cost = Prime cost + Factory overheads + Administrative
expenses + Selling and distribution expenses
Step 15
 Decide the profit and add the profit to the total cost to fix the selling price
of the part
Selling price = Total cost + Profit
Step 16
 Finally estimate the time of delivery in consultation with the production
and sales department

23. ESTIMATION OF LABOUR COST
Following stepsinvolved
 To estimate the labour cost, the estimator should have the
knowledge of various operations, machines, sequence, tools &
labour to be used.
Labour cost =Estimated labour time needed to the product
x Cost of the labour per hour

24. ESTIMATION OF MATERIAL COST
 Following stepsinvolved
 Prepare the list of all materials required to manufacture the product
 Estimate the weight of all the materials expected (the allowance for the
material wastage, spoilage and scrap are also added )
Estimated materials cost =Estimated weight of each part x Estimated futureprice
 Finally, the estimated cost of all the parts is added to get the total
estimated material cost of the product

25. ALLOCATION OF OVER HEAD CHARGES
 Thiscan not be charged directly
 All expenses other than direct cost are known as overhead cost or
Indirect expenses, ex:- Administrative expenses, selling and distribution
expenses,etc.
 Itcan be estimated by referring the previous records

26. FORMULA
 Prime cost = Direct material cost + Direct labour cost + Direct expenses
 Factory cost = Prime cost + Factory expenses
 Production cost = Factory cost + Administrative expenses
 Total or Ultimate cost = Production cost + Selling and distribution
expenses.
 Selling price = Total cost + Profit

27.  Example 1: Calculate prime cost, factory cost, production cost, total cost and
selling price per item from the data given below for the year 2019-2020.
Rs.
Cost of raw material in stock as on 1-04-2019 25,000
Raw material purchased 40,000
Direct labour cost 14,000
Direct expenses 1,000
Factory/Works overhead 9,750
Administrative expenditure 6,500
Selling and distribution expenses 3,250
No. of items produced 650
Cost of raw material in stock as on 31-03-2020 15,000
Net profit/item is 10 percent of total cost of the product.

28. Solution : For 650 units produced during 2019-2020
(i) Direct material used = Stock of raw material on 1-04-2019 + raw
material purchased – stock of raw material on 31-03-2020
= 25,000 + 40,000 – 15,000
= Rs. 50,000
(ii) Direct labour = Rs. 14,000
(iii) Direct expenses = Rs. 1,000
Prime cost = 50,000 + 14,000 + 1,000
= Rs. 65,000
Factory cost = Prime cost + Factory expenses
= 65,000 + 9,750
= Rs. 74,750

29. Production cost = Factory cost +
Administrative expenses
= 74,750 + 6,500
= Rs. 81,250
Total cost = Production cost +
Selling expenses
= 81,250 + 3,250
= Rs. 84,500
Selling price = 84,500 + 10 percent
of 84,500
= 84,500 × 1.10 = Rs. 92,950
Prime cost/item =65,000/650
= Rs. 100
Factory cost/item =74,750/650
= Rs. 115
Production cost/item
=81,250/650
= Rs. 125
Total cost/item =84,500/650
= Rs. 130
Selling price/item =92,950/650
= Rs. 143

30. Example 2 : From the following data for a sewing machine
manufacturer, prepare a statement showing prime cost, Works/factory
cost, production cost, total cost and profit.
Description Rs.
Value of stock of material as on 1-04-2003 26,000
Material purchased 2, 2,74,000
Wages to labour 1,20,000
Depreciation of plant and machinery 8,000
Depreciation of office equipment 2,000
Rent, taxes and insurance of factory 16,000
General administrative expenses 3,400

31. Water, power and telephone bills of factory 9,600
Water, lighting and telephone bills of office 2,500
Material transportation in factory 2,000
Insurance and rent of office building 2,000
Direct expenses 5,000
Commission and pay of salesman 10,500
Repair and maintenance of plant 1,000
Works Manager salary 30,000
Salary of office staff 60,000
Value of stock of material as on 31-03-2004 36,000
Sale of products 6,36,000

32. Solution :
(i) Material cost = Opening stock value + Material purchases – Closing
balance
= 26,000 + 2,74,000 – 36,000
= Rs. 2,64,000
Prime cost = Direct material cost + Direct labour cost + Direct expenses
= 2,64,000 + 1,20,000 + 5,000
= Rs. 3,89,000
(ii) Factory overheads are :Rs.
Rent, taxes and insurance of factory= Rs 16,000
Depreciation of plant and machinery = Rs 8,000
Water, power and telephone bill of factory= Rs 9,600
Material transportation in factory = Rs 2,000

33. Material transportation in factory = Rs 2,000
Repair and maintenance of plant = Rs 1,000
Work Manager salary = Rs 30,000
Factory overheads or Factory expenses= Rs 66,600
Factory cost = Prime cost + Factory expenses
= 3,89,000 + 66,600
= Rs. 4,55,600
(iii) Administrative/office expenses are :
Depreciation of office equipment= Rs 2,000
General administrative expenses= Rs 3,400
Water, lighting and telephone bills of office = Rs 2,500
Rent, insurance and taxes on office building= Rs 2,000
Salary of office staff = Rs 60,000
Total 69,900

34. Production cost = Factory cost + Office expenses
= Rs. 4,55,600 + Rs. 69,900
= Rs. 5,25,500
(iv) Selling overheads are :
Commission and pay to salesmen = Rs. 10,500
Total cost = Production cost + Selling expenses
= 5,25,500 + 10,500
= Rs. 5,36,000
(v) Profit = Sales – Total cost
= 6,36,000 – 5,36,000
= Rs. 1,00,000

35. Example 3 : Calculate the selling price per unit from the following data :
Direct material cost = Rs. 8,000
Direct labour cost = 60 percent of direct material cost
Direct expenses = 5 percent of direct labour cost
Factory expenses = 120 percent of direct labour cost
Administrative expenses = 80 percent direct labour cost
Sales and distribution expenses = 10 percent of direct labour cost
Profit = 8 percent of total cost
No. of pieces produced = 200

36. Solution :
Direct material cost = Rs. 8,000
Direct labour cost = 60 percent of direct material cost
=60 × 8,000/100
= Rs. 4,800
Direct expenses = 5 percent of direct labour cost
=4,800 × 5/100
= Rs. 240
Prime cost = 8,000 + 4,800 + 240
= Rs. 13,040
Factory expenses = 120 percent of direct labour cost
=4,800 × 120/100
= Rs. 5,760

37. Administration Expenses = 80 percent of direct labour cost
=4,800 × 80/100
= Rs. 3,840
Sales and distribution expenses = 10 percent of direct labour cost
=10 × 4,800/100
= Rs. 480
Total cost = Prime cost + Factory expenses + Office expenses + Sales and
distribution expenses
= 13,040 + 5,760 + 3,840 + 480
= Rs. 23,120

38. Profit = 8 percent of Total cost
=23,120 × 8/100
= Rs. 1,849.60
= Rs. 1,850 (say)
Selling price = Total cost + Profit
= 23,120 + 1,850
= Rs. 24,970
Selling price 1 unit =24,970/200 = Rs. 124.85
= Rs. 125

39. Example 4 : A factory is producing 1000 high tensile fasteners per hour on
a machine. The material cost is Rs. 375, labour cost is Rs. 245 and direct
expense is Rs. 80. The factory on cost is 150 percent of the total labour
cost and office on cost is 30 percent of the factory cost. If the selling price
of each fastener is Rs. 1.30, calculate whether there is loss or gain and
by what amount ?

40. Solution : For 1000 fasteners
Material cost = Rs. 375.00
Labour cost = Rs. 245.00
Direct expenses = Rs. 80.00
Factory on cost = 150 percent of labour cost
= 245 × 1.5
= Rs. 367.50
Factory cost = 375 + 245 + 80 + 367.50
= Rs. 1,067.50
Office on cost = 30 percent of factory cost
=1,067.50 × 30/100 = Rs. 320.25

41. Total cost for 1000 fasteners = 1,067.50 + 320.25
= Rs. 1387.75
Cost per fastener =1387.75/1000
= Rs. 1.387 = Rs. 1.39
Selling price = Rs. 1.30
As selling price is lower than total cost per fastener, the management will
suffer a loss.
Loss per fastener = (1.39 – 1.30) = Rs. 0.09 Loss per
1000 fastener = 0.09 × 1000 = Rs. 90

42. Example 5 : A certain product is manufactured in batches of 100. The
direct material cost is Rs. 50, direct labour cost in Rs. 80 and factory
overhead charges are Rs. 65. If the selling expenses are 45 percent
of factory cost, what should be selling price of each product so that
the profit is 10 percent of the total cost ?

43. Solution : Batch size = 100
Direct material cost = Rs. 50
Direct labour cost = Rs. 80
Factory overheads = Rs. 65
Factory cost = 50 + 80+ 65 = Rs. 195
Selling expenses = 45 percent of factory cost
=45 × 195/100
= Rs. 87.75
Total cost = 195 + 87.75
= Rs. 282.75

44. Profit = 10 percent of total cost
=282.75 × 10/100
= Rs. 28.28
Selling price of 100 components = 282.75 + 28.28
= Rs. 311
Selling price per component =311/100
= Rs. 3.11 = Rs. 3.15

45. Example 6 : A factory owner employed 50 workers during the month of
November 2004, whosedetailed expenditure is given below :
Material cost = Rs. 30,000
(i)Rate of wage for each worker = Rs. 6 per hour
(ii)Duration of work = 8 hours per day
(iii)No. of holidays in the month = 5
(iv)Total overhead expenses = Rs. 15,000
 If the workers were paid over time of 400 hours at the rate of Rs. 12 per
hour, calculate
(a)Total cost, and
(b)Man hour rate of overheads.

46. Solution : (a) Material cost = Rs. 30,000 No. of workers= 50
Duration of work = 8 hrs/day
No. of working days = 30 – 5 = 25
Total no. of work hours for the month of November 2004
= 25 × 8 × 50 = 10,000 hrs
Wage rate = Rs. 6 per hour
Labour cost = 10,000 × 6 = Rs. 60,000

47. Overtime paid = No. of overtime hours × hourly rate
= 400 × 12 = Rs. 4800
Total labour cost = 60,000 + 4800 = Rs. 64,800
Overhead expenses = Rs 15,000
Total cost = 30,000 + 64,800 + 15,000 = Rs. 1,09,800
(b) Total no. of man hours = 10,000 + 400 = 10,400
Man hour rate of overheads =Total overheads/Total no. of man hours
=15,000/10,400 = Rs. 1.44

48. Example 7 : The catalogue price of a certain gadget is Rs. 1,050, the
discount allowed to distributors being 20 percent. Data collected for a
certain period shows that the selling price and factory cost are equal. The
relation between material cost, labour cost and factory oncost (overhead
expenses) are in the ratio 1 : 2 : 3. If the labour cost is Rs. 200, what profit
is being made on the gadget ?

49. Solution :
Catalogue Price = Rs. 1,050
Distributors discount = 20% =1,050 × 20/100 = Rs. 210
Selling price = 1,050 – 210 = Rs. 840
Labour cost = Rs. 200
Material cost = 200 ×1/2 = Rs. 100
Factory oncost = 200 ×3/2 = Rs. 300

50. Factory cost = 200 + 100 + 300 = Rs. 600
It is given that selling price = Factory cost
= Rs. 600
Selling price = Total cost + Profit 840 = 600 + Profit
Profit = 840 – 600
= Rs. 240
Profit = Rs. 240 per gadget

51. Definition :
Carter defines : Depreciation is the gradual and permanent
decrease in the value of an asset from any cause
Accounting point of view : Depreciation is an annual charge
reflecting the decline in value of an asset due to causes such as
wear and tear action of elements obsolescence.
DEPRECIATION

52. • Straight line method of depreciation
• Declining balance method
• Sum of years digits method
• Sinking fund method
• Service output method
METHODS OF DEPRECIATION

53. • In this method of depreciation a fixed sum is charged as
depreciation amount throughout the life time.
• At the end of the life of an asset the accumulated sum of
the asset is exactly equal to the purchase value of the
asset.
• Assumption : Inflation is absent
STRAIGHT LINE METHOD OF DEPRECIATION

54. t t
n
i1
B  B  D  P t[P  F ]
t
n
P= First cost of theasset
F= Salvage value of the asset
n = Life of the asset
Bt = Book value of the asset at the end of the periodti
Dt = Depreciation amount for the time period t
D 
[P  F]
STRAIGHT LINE METHOD OF DEPRECIATION
Rate of Deprecation = Dt/ P* 100

55. • A company purchased a machinery for Rs 8000, the useful life of
the machinery is 10 years and the estimated salvage value of the
machinery at the end of the lifetime is Rs 800. Determine the
depreciation charge and book value at the end of the various
years using the straight line method of depreciation.

56. P= Rs8000
F= Rs 800
Dt= (P-F)/n
= (8000 – 800)/10
= 720
Rate of Deprecation
= Dt/ P* 100
= (720/8000)*100
= 9%
End of Year Depreciation ( Dt ) Book Value Bt = Bt-1 - Dt
0 - 8000
1 720 7280
2 720 6560
3 720 5840
4 720 5120
5 720 4400
6 720 3680
7 720 2960
8 720 2240
9 720 1520
10 720 800

57. • A machine costing Rs 24,000 was purchased on 1st December
1985. The installation and erection charges were Rs1000 and its
useful life is expected to be 10 years. The scrap value of the
machine at the end of the useful life is Rs 5000. Compute the
depreciation and the book value for the period 6

58. P= Rs24,000 + Rs1000 = 25000
F= Rs5000/-
n = 10years
Dt = (P-F)/n = (25000-5000)/10 = 2000
Bt = 25,000 – 6 × (25000-5000)/10
Bt= Rs.13,000/-
n
t i  1 t
B  B  D  P  t [ P  F ]

59. A constant % of book value of the previous period of the asset will be charged
as the independent amount for the current period.
The book value at the end of the life of the asset may not be exactly equal to
the salvage value of the asset.
P= First cost of theasset
F= Salvage value of the asset
n= Life of the asset
Bt = Book value of the asset at the end of the periodt
K= a fixedpercentage
Dt = Depreciation amount at the end of the period “t”

60. FORMULA FORDECLINING BALANCE
METHOD
Dt  K  Bt1
Bt  Bt1 Dt
Bt  (1 K)Bt1
Dt  K Bt1
Bt  Bt1 Dt
Bt  (1 K)Bt1
D  K(1 K)t1
 P
B  (1 K)t
 P
•If k = 2/n then it is called
as double declining balance
method

61. Glaxo company has purchased a machine for Rs 1,50,000. The plant
engineer estimates that the machine has a useful life of 10 years and a
salvage value of Rs 25,000 at the end of the useful life. Demonstrate the
calculations of the declining balance method of depreciation by
assuming 0.2 for K
P= Rs1,50,000.
F= Rs25,000
n= 10 years
K= 0.2
Using the
formula
t t1
D  K B
Bt  Bt1 Dt

62. End of year (n) Depreciation ( Dt) Book Value (Bt)
0 - 1,50,000.00
1 30,000.00 1,20,000.00
2 24,000.00 96,000.00
3 19,200.00 76,800.00
4 15,360.00 61,440.00
5 12,288.00 49,152.00
6 9830.40 39,321.60
7 7864.32 31,457.28
8 6291.45 25,165.83
9 5033.16 20,132.67
10 4026.53 16,106.14

63. Calculate the depreciation and book value for the period 5 using the
declining balance method of depreciation by assuming 0.2 for Kand Rs
1,20,000 for P and salvage value Rs10,000. The useful life of the
machinery is 10 years.
P= Rs1,20,000
F= Rs10,000
n= 10 years
K= 0.2

64. • Using the formula
• Dt = 0.2 (1-0.2)(5-1) × 1,20,000 = Rs98,430.40/-
• Bt = (1-0.2)5 × 1,20,000 = Rs39,321.60
t
D  K(1 K)t1
 P
t
B  (1 K)t
 P

65. The scrap value of the asset is deducted from its original cost and it is assumed that the
book value of the asset decreases at a decreasing rate.
Sum of the years = n (n+1)/2
Dt = Rate × (P-F)
Bt = Bt-1 – Dt
The formula for Dt and Bt for a specific year “t” are asfollows
t
n t 1
D  (P  F)
n(n 1) / 2
t
B  (P  F)
n t (n t 1)
 F
n (n 1)

66. ABC company has purchased an equipment whose first cost is
Rs 2,00,000 with an estimated life of eight years. The estimated scrap
value of the equipment at the end of the lifetime is Rs 40,000/-.
Determine the depreciation charge and book value at the end of
various years using sum of the years digits method of depreciation.

67. P= Rs.2,00,000
F= Rs.40,000
n = 8 years
Sum = (8 × 9)/2 = 36
End of Year (n) Depreciation (
Dt)
Book Value (Bt)
0 - 2,00,000
1 35,555.55 1,64,444.44
2 31,111.11 1,33,333.33
3 26,666.67 1,06,666.66
4 22,222.22 84,444.44
5 17,777.77 66,666.67
6 13,333.33 53,333.34
7 8888.88 44,444.46
8 4444.44 40,000.02

68. Consider problem 1 and find the depreciation and book value for
the 5th year.
Using the formula
P= Rs. 2,00,000
F= Rs.40,000
n = 8years
t
n  t 1
D  (P  F)
n(n 1) / 2
t
B  (P  F )
n  t (n  t 1)
 F
n (n 1)
t
D 
8 5 1
(2,00,000  40,000)  Rs.17,777.78
8(81) / 2
t
8 81
B  (2,00,00040,000)85 851
 40,000  Rs.66,666.67

69. • In this method of depreciation a depreciation fund equal to actual
loss in the value of the asset is estimated for each year.
• This amount is invested outside the business in a separate account
sinking fund investment account and interest will be earned on
the fund.
• The sinking fund will rise year after year.

70. P= first cost of the asset
F= salvage value of the asset
n= life of the asset
i = Rate of return compounded annually
A = Annual equivalent amount
Bt = Book value of the asset at the end of the period ‘t’
Dt = Depreciation amount at the end of the period ‘t’.
The loss of value of the asset (P-F) is made available in the form of
cumulative depreciation amount
A = (P-F)[A/F, i, n]
The fixed sum depreciated at the end of every time period earns an interest at
the rate of i% compounded annually
Dt = (P-F)(A/F, i ,n)(F/P,i,t-1)
Bt = P-(P-F)(A/F, i,n)(F/P,i,t-1)

71. Find the depreciation annuity by annuity method after three years
when the initial cost of the machine is Rs.8,00,000 and the salvage
value at the end of three years is Rs. 4,00,000. Rate of interest is
10%.

72. P= Rs. 8,00,000
F = Rs.4,00,000
n = 3years
i = 10%
A = (P-F)[A/F, i, n]
A = (8,00,000-4,00,000) [A/F,10%,3] value from interest table is substituted
A = (8,00,000-4,00,000) x 0.3021 = Rs.1,20,840
Dt = (P-F)(A/F, i,n)(F/P,i,t-1)
Bt = P-(P-F)(A/F, i,n)(F/P,i,t)

73. Dt = (P-F)(A/F, i ,n)(F/P,i,t-1)
D2 =at the end of the second year (D2)
D2 = 1,20,840 + 1,20,840 x 0.10 = Rs.1,32,924.
End of the year ‘t’ Fixed Dt Net Dt Book Value Bt
0 1,20,840 - 8,00,000.00
1 1,20,840 1,20,840.00 6,79,160.00
2 1,20,840 1,32,924.00 5,46,236.00
3 1,20,840 1,46,216.40 4,00,019.60

74. ABC & Co has purchased a machinery and its first cost is Rs. 2,00,000 with an
estimated standard life of 8 years. The salvage value is Rs. 40,000 find D6 and
B7, rate of interest 12% compounded annually.
Solution :
P= Rs.2,00,000 F= Rs.40,000 n = 8 years i = 12%
D6= (P-F)(A/F,12%,8)(F/P,12%,5)**
** - From interest table
D6= (2,00,000-40,000) (0.0813)(1.762) = Rs.22,920
B7= P-(P-F)(A/F,12%,8)(F/A,12%,7)**
B7= 2,00,000-(2,00,000-40,000)(0.0813)(10.089)
= Rs.68,762.29

75. METHOD
• In this method the life of the machine is
expressed in terms of number of units that a
machine is expected to produce over the
estimated life
• P= first cost of the asset
• F= Salvage value of theasset
• X = Maximum capacity of service of the asset
during its lifetime
• x = Quantity of service rendered in aperiod
• Depreciation/unit of service = (P-F)/X
X
• Depreciation for x unit of service period = (P  F)
(x)

76. M
• The first cost of a road laying machine is Rs 60,00,000/-.
Its salvage value after 5 years of service is Rs. 40,000/-.
The length of road that can be laid by the machine
during the lifetime is 55,000km. In its third year of
operation the length of road laid is 1500 km. Find the
depreciation of the equipment for that year
Solution:
• P = 60,00,000 F= 40,000 X = 55,000 km, x = 1500 km
• Depreciation for x unit of service in year 3 =
3
55,000
D 
60,00,000  40,000
 (1500)  Rs.1,62,545.45

77. EVALUATION OF PUBLIC ALTERNATIVES
• Evaluation of public alternatives is selection of
best alternative from the available alternatives.
• The factor considered in selection is profit
maximization.
• For the evaluation of public alternatives the
benefit – cost ratio is used
• BCratio = EquivalentBenefits
EquivalentCosts

78. EVALUATION OF PUBLIC ALTERNATIVES
• P= initialinvestment
• C= Early cost of operation and maintenance
• PA = Annual equivalent of the initial investment
• PF = Future worth of the initial investment
• BP = Present worth of total benefits
• BF = Future worth of total benefits
• BA= Annual equivalent of total benefits
• CP= Present worth of yearly cost of operation
and maintenance.
• CF = Future worth of yearly cost of operation
 
BP BF BA
P  CP PF  CF PAC
and maintenance.
BCRATIO 

79. PROBLE
M
• Project A1 and Project A2 are being considered
for investment. Project A1 requires an initial
investment of Rs 40,00,00 and net receipts
estimated as Rs10,00,000 per year for the next 5
years. The initial overlay for the A2 is Rs
70,00,000 and the net receipts have been
estimated as Rs. 16,00,000 per year for the next
seven years. There is no salvage value associated
with either of the projects. Using the benefit to
cost ratio which project would you select?
Interest rate = 10%

80. SOLUTION
Alternative 1 :
• P = Rs 40,00.000 B= Rs. 10,00,000 n = 5Years
i= 10%
BCRATIO  AnnualequivalentBenefits
AnnualequivalentCosts
• Annual equivalent of initial cost = P(A/P,10%,5)
=40,00,000 x 0.2638
= Rs.10,55,200
10,55,200
BCRATIO  10,00,000
 0.9476

81. N
Alternative 2 :
• P = Rs 70,00.000 B= Rs. 15,00,000 n = 7Years
i= 10%
BCRATIO  AnnualequivalentBenefits
AnnualequivalentCosts
• Annual equivalent of initial cost = P(A/P,10%,7)
=70,00,000 x 0.2054 = Rs.14,37,800
14,37,800
• The benefit cost ratio of alternative 2 (i.e 1.0342 >1) is
more than alternative 1. Hence alternative 2 is selected
BCRATIO  15,00,000
1.0432