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Assignment - 2
Question 1:-Common data to question (i) to (iv):-
The initial cost of a piece of construction equipment is Rs.30,00,000 having a useful life of 10 years. The estimated salvage value of the equipment at the end of the useful life is Rs.450,000.
[8 marks]
Question (i):- The book value of the construction equipment at the end of 5th^ year using Straight - line method is:
a) Rs. 17,25, b) Rs. 11,81, c) Rs. 11,45, d) None of these
Question (ii):- The book value of the construction equipment at the end of 5th^ year (BV 5 ) and depreciation (d 5 ) for 5th^ year using Double-declining balance method are:
a) BV 5 = Rs. 11,81,712.19;d 5 = Rs. 2,78,181. b) BV 5 = Rs. 11, 81,712.19; d 5 = Rs. 2,42,037. c) BV 5 = Rs. 11,45,454.54 ;d 5 = Rs. 2,78,181. d) BV 5 = Rs. 17,25,000.00 ; d 5 = Rs. 1,27,341.
Question (iii):- Determine the book value (BV 5 ) of the construction equipment at the end of 5th year and depreciation (d 5 ) for 5th^ year using Sum-of-the-years-digits method?
a) BV 5 = Rs. 11,81,712.19 ;d 5 = Rs. 2,78,181. b) BV 5 = Rs. 11, 81,712.19; d 5 = Rs. 2,42,037. c) BV 5 = Rs. 11,45,454.54 ; d 5 = Rs. 2,78,181. d) BV 5 = Rs. 17,25,000.00 ; d 5 = Rs. 1,27,341.
Question (iv):- Determine accumulated depreciation at the end of 5th^ year using Sinking fund method, if interest rate is 8.2 % per year?
a) Rs.10,27,341. b) Rs. 13,81,712. c) Rs. 13,45,487. d) Rs. 12,56,800.
Solution:
Given: Original cost of equipment (V) = Rs. 30, 00, Salvage value of equipment (Vs) = Rs. 4, 50, Useful life (n) = 10 years Calculate: Annual depreciation and book value of the construction equipment at the end of 5thyear i.e. d 5 and BV 5 respectively. Ans: (i) Using straight line method- As in this method annual depreciation d 1 = d 2 =d 3 =…=d. Therefore, Annual depreciation (d) =d 5 = (Original value of equipment (V)-Salvage value of equipment (Vs))/service life (n) d= d 5 = (3000000-450000)/10 = Rs.2,55, Book Value (V 5 ) after 5 years =V-da = 3000000-2550005 = Rs.1,72,
(ii) Using Double-declining balance method
Amount to be depreciated = (V-Vs) = 3000000-450000= Rs. Using straight line depreciation method Annual depreciation = 2550000/10= Rs.255000 per year Annual depreciation (in terms of fraction of original cost) = 255000/3000000 =0. Thus for double decline method annual depreciation will be =20.085 =0. Thus for 1st year depreciation amount will be = 30000000.17= Rs. 510000 Book value at the end of 1st^ year (V 1 ) = (3000000-510000) = Rs. 2490000 For 2nd year depreciation amount will be= Recent Book value0.17 =24900000.17 = Rs. Book value at the end of 2nd^ year (V 2 ) = (2490000-423300) = Rs. 2066700 For 3rd year depreciation amount will be= Recent Book value0.17 =20667000.17= Rs. Book value at the end of 3rd^ year (V 3 ) = (2066700-351339) = Rs. 1715361 For 4th year depreciation amount will be= Recent Book value0.17 =17153610.17= Rs.291611. Book value at the end of 4th^ year (V 4 ) = (1715361-291611.37) = Rs. 1423749. For 5th year depreciation amount will be= Recent Book value0. =1423749.630.17= Rs.242037. Book value at the end of 5th^ year (V 5 ) = (1423749.63-242037.44) = Rs. 1181712.
Alternate Method d 5 = VB4f =V(1-f)^4 *f = 3000000(1-0.17)^4 *0.17 = Rs.2,42,037.
Increased in fund value for 3rd^ year= (174420+29777.68) = Rs. 204197. Accumulated depreciation at the end of 3rd^ year = (363142.44+204197.68) = Rs. 567340. Book value at the end of 3rd^ year = (2636857.56-204197.68) = Rs. 2432659. Interest earned at the end of 4th^ year = 567340.12.082 = Rs. 46521. Increased in fund value for 4th^ year= (174420+46521.89) = Rs. 220941. Accumulated depreciation at the end of 4th^ year = (567340.12+220941.89) = Rs. 788282. Book value at the end of 4th^ year = (2432659.88-220941.89) = Rs. 2211717. Interest earned at the end of 5th^ year = 788282.01.082 = Rs.64639. Increased in fund value for 5th^ year= (174420+64639.12) = Rs. 239059. Accumulated depreciation at the end of 5th^ year = (788282.01+239059.12) = Rs. 1027341. Book value at the end of 5th^ year = (2211717.09-239059.12) = Rs. 1972657.
Year Annual Dep. Computed(Rs.)
Interest earned (Rs.)
Increase in fund value (Rs.)
Accumulated Depreciation (Rs.)
Book Value (Rs.)
0 3000000 1 174420 - 174420 174420 2825580 2 174420 14302.44 188722.44 363142.44 2636857. 3 174420 29777.68 204197.68 567340.12 2432659. 4 174420 46521.89 220941.89 788282.01 2211717. 5 174420 64639.12 239059.12 1027341.13 1972657.
Question 2:- The original cost of a heat exchanger is Rs.1, 00,000. It has a useful life of 10 years. The estimated salvage value of the heat exchanger at the end of useful life is zero. Calculate the book value at the end of 3rd^ year, using repair provision method, if the repairs and maintenance charges together were estimated to be Rs. 18, during the lifetime of the equipment. And also determine the annual depreciation to be provided? [2 marks]
a) BV 5 = Rs. 81,000 ;d= Rs. 18, b) BV 5 = Rs. 81,000; d= Rs. 12, c) BV 5 = Rs. 64,600; d= Rs. 11, d) BV 5 = Rs. 38,000; d= Rs. 27,
Ans. Given: Original cost of heat exchanger (V) = Rs. 1,00,
Salvage value of equipment (Vs) = 0 Useful life (n) = 10 years Estimated total cost of repair= Rs. Calculate: Annual depreciation and book value of the heat exchanger at the end of 3rd^ year Solution:
Annual amount to be provided for depreciation (d): = [(original cost-salvage value) +Estimated total cost of repair]/expected useful life = [(100000-0) +18000]/10 = Rs. Book value at the end of 3rd^ year = V-ad = 100000-311800 = Rs. 64600
Question 3:- There are two plans for a new godown construction for storage. We can either go in for a new concrete building (Plan-1) or have an extension to the existing building (Plan-2). The new concrete building is estimated to cost Rs. 60,000 with a permanent life. Its annual maintenance, insurance and tax cost is expected to be Rs. 500.The extended building will cost Rs. 20,000 and annual maintenance, insurance, and tax cost being Rs. 800. Both plans have life spans of 20 year. Assuming 10% as an attractive return, and using the annual cost method, choose the correct statement- [4 marks]
a) We should go for the concrete construction (Plan-1), which has a total annual cost of Rs.3149. b) We should go for the extended construction (Plan-2), which has total annual cost Rs.7547.58. c) We should go for extended construction (Plan-2), which has total annual cost Rs.3149. d) Both are equally economical.
Solution: Given: Plan- 1 Plan- 2 Capital investment(Rs.) 60000 20000 maintenance, insurance, and tax cost per year(Rs.)
Useful life (years),n^20 Rate of return (%), i 10 10
Annual cost of the capital recovery is the annuity based on time value of money that one has to pay throughout the useful life, which will be equal to the capital investment at the start of the 1st^ year. Annual cost of capital recovery = Capital investment*i / [1-(1+i)-n]
Interest rate (i) = 9 % per year =0. V-Vs=Rs. 18500, is the depreciable cost which should be accumulated at the end of 4th year or it can be called the future value at the end of 4th^ year which needs to be generated yearly investment of Rs. R. R is the annual equal amount of depreciation.
Annual depreciation=R = 185000.2187 = Rs. 4045. Book value at the end of 1st^ year = (20000-4045.37) = Rs. 15954. Interest earned at the end of 2nd^ year = 4045.370.09= Rs.364. Increased in fund value for 2nd^ year= (4045.37+364.08) = Rs. 4409. Accumulated depreciation at the end of 2nd^ year = (4045.37+4409.45) = Rs. 8454. Book value at the end of 2nd^ year = (15954.63-4409.45) = Rs. 11545. Interest earned at the end of 3rd^ year =8454.82*.09 =Rs. 760. Increased in fund value for 3rd^ year = Rs. 4806. Accumulated depreciation at the end of 3rd^ year = (8454.82+4806.3) = Rs. 13261. Book value at the end of 3rd^ year = (11545.18-4806.30) = Rs.6738.
Year Annual Dep. Computed(Rs.)
Interest earned (Rs.)
Increase in fund value (Rs.)
Accumulated Depreciation (Rs.)
Book Value (Rs.)
0 20000 1 4045.37 - 4045.37 4045.37 15954. 2 4045.37 364.08 4409.45 8454.82 11545. 3 4045.37 760.93 4806.30 13261.12 6738. 4 4045.37 1193.51 5238.88 18499.99 1500.
Question 5:- 12 years ago M/s Z. Limited purchased a piece of equipment for Rs. 40,000. At the time the equipment was put into use the service life estimated was 20 years and salvage value was zero. On this basis a straight line method was set up. The equipment can now (after 12 years) be sold for Rs. 10,000. The total replacement cost of the equipment is Rs. 55,000. Assuming depreciation fund is available for purchase, compute how much new capital must be made available for the purchase of the equipment? [3 marks]
a) Rs. 45000 b) Rs. 34000 c) Rs. 26000
^ *[^ ]
s (1 ) n 1
i
R V V
i
d) None of these
Solution: Given: V= Rs. 40, n = 20 years; Vs= zero (after 20 years) Present value of equipment = Rs. 10,000 (according to question after 12 years) Total replacement cost of equipment= Rs. 55, 000 Straight line method is used for depreciation calculation, therefore Annual depreciation (d) = (Original value of equipment (V)-Salvage value of equipment (Vs))/service life (n) d= (40000-0)/20 = Rs. Total depreciation accumulated at the end of 12th^ year =2000*12 = Rs. 24, Rs.24, 000 is available as a fund for purchase (according to question) So total available fund after selling the equipment = (10,000+24,000) = Rs. 34, Therefore, New capital needed for the purchase of the new equipment is = (total cost of new equipment-total available fund after selling the equipment) = (55,000-34,000) = Rs. 21,
Question6:- Two pumps under consideration for installation at a plant have the following capital investments, salvage values and annual interest.
Capital investment (Rs.) Salvage value (Rs.) Interest rate per annum (%) Pump A 40,000 3900 10 Pump B 50,000 20,000 10
If annual cost of capital recovery is same for both the pumps. Then determine what should be the common life of the pumps. Maintenance and operational costs are negligible. [4 marks]
a) 8 years b) 5 years c) 9 years d) 6 years Ans.
Given:
Capital investment (Rs.),P Salvage value (Rs.) ,SV
Interest rate per annum (%),i Pump A 40,000 3900 10
(-n)*log (1.1) =log (0.6211)
n= 4.996 year ≈ 5 years
Question 7 :- A choice has to be made by Mr. Abhishek between a new bike and a second-hand bike. The relevant data are:
New bike Second-hand bike Initial investment(Rs.) 15,000 5, Salvage value after 10 years of use(Rs.) 5,000 zero Annual fuel cost for average 10,000 miles run(Rs.)
Annual repair cost(Rs.) 1,000 3,
If Abhishek invest Rs. 15,000 or Rs. 5000 elsewhere, he except a return of 10%. Using the annual cost method, find out whether Abhishek should go for new bike or the second-hand bike? [5 marks]
a) The New bike with total annual cost of Rs. 5313.72. b) The Second-hand bike with total annual cost Rs. 5313.72. c) Data insufficient to make a selection. d) None of these
Solution: Given: New bike Second-hand bike Capital investment (Rs.) 15,000 5, Estimated useful life, year 10 10 Salvage value (Rs.) 5000 zero Annual fuel cost for average 10,000 miles run(Rs.)
Annual repair cost(Rs.) 1,000 3,
Solution:
In this problem the salvage value (which is a receipt) is at a different time line than the capital investment. Hence, the Present value of the salvage value is to be calculated and then
it should be deducted from the capital investment for calculating annual cost of capital recovery.
Investment at the start of 1st year (for New bike) = Rs.15, Salvage value = Rs.5000 (at the end of 10th year) To bring it to the time line of the investment, the Present worth of Rs.5000 is computed Present worth of Rs.5000= 5000/ (1+0.1)^10 = Rs.1927. Hence the capital expenditure at the start of 1st year = (Rs.15000-Rs.1927.72)= Rs.13072. Annual cost of capital recovery = Capital investmenti/[1-(1+i)-n] For “new bike” Annual cost of Capital recovery =13072.280.1/[1-(1+0.1)-10] = Rs.2127. For “second hand bike” Investment at the start of 1st year = Rs.5, Salvage value = zero (at the end of 10th year) Hence the capital expenditure at the start of 1st year = Rs. Annual cost of Capital recovery =5000*0.1/[1-(1+0.1)-10] = Rs.813.
New bike Second hand bike Capital investment Rs.15,000 Rs.5, Estimated useful life^10 Salvage Value, Rs. Rs.5000^ zero Annual fuel cost for average 10, miles run(Rs.)
Annual repair cost(Rs.) 1,000 3, Rate of return 10% 10% Solution given below Annual cost of capital recovery Rs. 2127.45 Rs.813. Total annual cost(annual cost of capital recovery + annual operating cost+ annual repair cost)
Rs.4127.45 Rs.5313.
Decision: He should purchase “new bike” in comparison to “second hand-bike” as its total annual cost is low.
Question 8:- A firm purchased a heat exchanger for Rs. ‘V’. The salvage value of the heat exchanger after 9 years of service life is expected to be Rs. ‘VS’. In which year, annual depreciation amount by sum-of-years-digits method will be equal to the annual depreciation by straight line method for the above available information?
If sum of all values (i.e. depreciation as well as book values) calculated above in (i),(ii) and (iii) is ‘S’. Then the value of S will be - [5 marks]
a) Rs. 55023 b) Rs. 65245 c) Rs. 53023 d) Rs. 57045
Solution: Given: Original cost of automated packaging equipment = Rs.30, 000 Salvage value of equipment (Vs) = Rs. 3000 Useful life (n) = 5 years
(i) Using Straight line method As in this method annual depreciation d 1 = d 2 =d 3 =…=d. Therefore, Annual depreciation (d) =d 2 = (Original value of equipment (V)-Salvage value of equipment (Vs))/service life (n) d=d 2 = (30000-3000)/5 = Rs. Book Value (V 2 ) after 2 years =V-da = 30000-54002 = Rs.
(ii) Using Sum of the years digits method Depreciable cost = Rs.30000-Rs.3000=Rs. Sum of the years' digits for n years = 1 + 2 + 3 + ...... + (n-1) + n = (n+1) x (n / 2) = (5+1)5/2= 15 or Sum of the years' digits = 1+2+3+4+5 = 15 Depreciation for 1st^ year = (27000) x 5/15 = 51800=Rs. Book value at the end of 1st year= (30000-9000) = Rs. 21000 Depreciation for 2nd^ year= (27000) x4/15 =4*1800= Rs. Book value at the end of 2nd^ year= (21000-7200) = Rs. 13800
(iii) Given: Initial investment = Rs.30, 000 Salvage value = Rs.3000 (at the end of 5th^ year) Annual repairs and maintenance cost = Rs. 2000 Ans: To bring it to the same time line of the investment, the Present worth of Rs.3000 is computed. Present worth of Rs.3000= 3000/ (1+0.1)^5 = Rs.1862.
Hence the capital expenditure at the start of 1st year = (Rs.30000-Rs.1862.76) = Rs.28137. Annual cost of capital recovery = Capital investmenti/[1-(1+i)-n] Annual cost of Capital recovery =28137.240.1/[1-(1+0.1)-5] = Rs.7422. Hence total annual cost of equipment is = (annual capital recovery+ annual repairs cost=Rs. (7422.53+2000) = Rs. 9422.
Thus, the value of ‘S’ = (5400+19200+7200+13800+9422.53) = Rs. 55022.
Question 10:- M/s ABC Ltd. purchased a new brick making machine for Rs. 8,40,000. The salvage value of machine is anticipated to be zero at the end of it’s five-year life. Compute the book value of the machine at the end of 4th^ year, using the Modified accelerated cost recovery system method. Assuming the half-year convention is relevant. [4 marks]
a) Rs. 2,41, b) Rs. 1,72, c) Rs. 1,45, d) None of these Solution:
By modified accelerated cost recovery system (MACRS) method:
Year Depreciation rate (%)
Calculation using formula Depreciation (Rs.)
Book value(Rs.) DDB SLM 0 - - - 8,40, 1 20 [1]^ 20 840000x(1/5)x2(200%)x0. (Half-year convention)
2 32 17.7 (6,72,000)x(1/5)x(2) 2,68,800 4,03, 3 19.20 13.7 (4,03,200) x (1/5) x 2 1,61,280 2,41, 4 11.52 11.52 (2,41,920)x(1/5)x(2) 96,768 1,45, 5 6.912 11.52 [2]^ (1,45,152)(1/1.5) 96,768 48384 6 5.76 [3]^ 96,7680.5 48384 0
[1] The double-declining-balance (DDB) method allows a depreciation of 2(1/5) = 0.4, but due to the half-year convention it reduces to 0.4/2=0.2 or 20% [2] During computation of depreciation value using MACRS, double declining balance method (DDBM) changes to straight-line method (SLM) when the later method provides greater depreciation than DDBM.. Deductions under 200% declining balance MACRS for 5th^ year
