PPCE unit 3 (ME8793 – PROCESS PLANNING AND COST ESTIMATION)

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 of manufacture 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 various Standards

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 steps involved 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 steps involved 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 future price 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 This can 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. It can 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 Rate of wage for each worker = Rs. 6 per hour Duration of work = 8 hours per day No. of holidays in the month = 5 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 Total cost, and 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 the asset F = Salvage value of the asset n = Life of the asset B t = Book value of the asset at the end of the period t i D t = Depreciation amount for the time period t D  [ P  F ] STRAIGHT LINE METHOD OF DEPRECIATION Rate of Deprecation = D t / 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 = Rs 8000 F = Rs 8 00 D t = (P-F)/n = (8000 – 800)/10 = 720 Rate of Deprecation = D t / P * 100 = (720/8000)*100 = 9% End of Year Depreciation ( D t ) Book Value B t = B t-1 - D t - 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 1 st December 1985. The installation and erection charges were Rs 1000 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 = Rs 24,000 + Rs 1000 = 25000 F = Rs 5000/- n = 10 years D t = (P-F)/n = (25000-5000)/10 = 2000 B t = 25,000 – 6 × (25000-5000)/10 B t = 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 the asset F = Salvage value of the asset n= Life of the asset B t = Book value of the asset at the end of the period t K = a fixed percentage D t = Depreciation amount at the end of the period “t”

60. FORMULA FOR DECLINING BALANCE METHOD D t  K  B t  1 B t  B t  1  D t B t  (1  K ) B t  1 D t  K  B t  1 B t  B t  1  D t B t  (1  K ) B t  1 D  K (1  K ) t  1  P B  (1  K ) t  P If k = 2 /n t hen it is c alled 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 = Rs 1,50,000. F = Rs 25,000 n= 10 years K = 0.2 Using the formula t t  1 D  K  B B t  B t  1  D t

62. End of year (n) Depreciation ( D t ) Book Value (B t ) - 1,5 , . 00 1 3 , . 00 1,2 , . 00 2 2 4 , . 00 9 6 , . 00 3 1 9 , 2 . 00 7 6 , 8 . 00 4 1 5 , 3 6 . 00 6 1 , 4 4 . 00 5 1 2 , 2 8 8 . 00 4 9 , 1 5 2 . 00 6 9 8 3 .40 3 9 , 3 2 1 . 60 7 7 8 6 4 .32 3 1 , 4 5 7 . 28 8 6 2 9 1 .45 2 5 , 1 6 5 . 83 9 5 3 3 .16 2 , 1 3 2 . 67 10 4 2 6 .53 1 6 , 1 6 . 14

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

64. Using the formula D t = 0.2 (1-0.2) (5-1) × 1,20,000 = Rs 98,430.40/- B t = (1-0.2) 5 × 1,20,000 = Rs 39,321.60 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 D t = Rate × (P-F) B t = B t-1 – D t The formula for D t and B t for a specific year “t” are as follows 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 ( D t ) Book Value (B t ) - 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 5 th year. Using the formula P= Rs. 2,00,000 F = Rs.40,000 n = 8 years 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 B t = Book value of the asset at the end of the period ‘t’ D t = 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 D t = (P-F)(A/F, i ,n)(F/P,i,t-1) B t = 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 = 3 years 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 D t = (P-F)(A/F, i ,n)(F/P,i,t-1) B t = P-(P-F)(A/F, i, n)(F/P,i,t)

73. D t = (P-F)(A/F, i ,n)(F/P,i,t-1) D 2 = at the end of the second year (D 2 ) D 2 = 1,20,840 + 1,20,840 x 0.10 = Rs. 1,32,924. End of the year ‘t’ Fixed D t Net D t Book Value B t 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 D 6 and B 7 , rate of interest 12% compounded annually. Solution : P = Rs. 2,00,000 F = Rs. 40,000 n = 8 years i = 12% D 6 = (P-F)(A/F,12%,8)(F/P,12%,5) ** ** - From interest table D 6 = (2,00,000-40,000) (0.0813)(1.762) = Rs. 22,920 B 7 = P-(P-F) (A/F,12%,8)(F/A,12%,7) ** B 7 = 2,00,000-(2,00,000-40,000)(0.0813)(10.089) = Rs. 68,762.29

75. SERVICE OUTPUT 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 the asset X = Max i mu m c a pa c it y o f s ervice o f the as s e t during its lifetime x = Quantity of service rendered in a period Depreciation/unit of service = (P-F)/X X Depreciation for x unit of service period = ( P  F ) ( x )

76. PROBLEM 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 BC ratio = EquivalentBenefits EquivalentCosts

78. EVALUATION OF PUBLIC ALTERNATIVES P = initial investment C= Early cost of operation and maintenance P A = Annual equivalent of the initial investment P F = Future worth of the initial investment B P = Present worth of total benefits B F = Future worth of total benefits B A = Annual equivalent of total benefits C P = Present worth of yearly cost of operation and maintenance. C F = Future worth of yearly cost of operation   B P B F B A P  C P P F  C F P A  C and maintenance. BCRATIO 

79. PROBLEM Project A 1 and Project A 2 are being considered for investment. Project A 1 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 A 2 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 = 5 Years 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. SOLUTION Alternative 2 : P = Rs 70,00.000 B = Rs. 15,00,000 n = 7 Years 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