‘EXERCISE NO. 7 Capitalized Cost Bruan, Kimberly C._2A Engr. Lilian A. Lerios Name, Year & Section Instructor ECOM 11 June 30,2020 ‘Subject. Date 1. To maintain a bridge, P25,000 will be required at the end of 3 years and annually thereafter. If money is worth 8%, determine the capitalized cost of all future maintenance. Given Az P 25 000 is O% nes = 64427-4248 i waaay: 42de PF 64,427.42 2. A manufacturing plant installed a new boiler at a total cost of P750,000 and is estimated to have a useful life of 10 years. It is estimated to have a scrap value at the end of its useful life of P25,000. If interest is 12% compounded annually, determine its capitalized cost. Capitalized Cost (Cc) = First Cost (FC) + Cost of Perpetual Maintenance Capitalized Cost = FC + % Where: % = the amount of principal invested at it per period, the interest on which will amount to S every k periods S = the amount needed to replace or maintain the property every k periods xi interest on the amount % for each period If the property or structure costs PS to obtain and it will have to be replaced every k periods for the same amount, then Capitalized Cost = S + x = —S— “aCost of replacement = FC or original cost Scrap Value or Salvage Value (Depend sometimes on what is given in the problem) Given Solvtion? Fe" 15 000 nt fe 7 Fe -s Gy® 28000 = cot hos Ter vt 2% as0000- 16060 = 750.000 TP 000 + Goa 1 wena = 1044279325 ce? ce * P1,044,279-33| 3. The capitalized cost of a piece of equipment was found to be P710,000. The rate of interest used in the computations was 12%, with a salvage value of P50,000 at the end of a service life of 8 years. Assuming that the cost of perpetual replacement remains constant, determine the original cost of the equipment. Given? Req'd: ce 0 998 ree? s1= $0 000 ne ® yrs TE hed oe Solvtion: Pe _ cr ke © aa Fee qus4 97 -064s [re=fa 48,437.07]4. Compare the capitalized costs of the following road pavements: (a) An asphalt pavement costing P500,000 which would last for 5 years with negligible repairs. At the end of 5 years, P25,000 would be spent to remove the old surface before P500,000 is spent again for a new surface. (b) A thick concrete pavement costing P1,250,000 which would last indefinitely, with a cost of P100,000 for minor repairs at the end of every 3 years. Money is worth 8% compounded annually. Solution: © sx0 20 A750 900+ rng 4 BSCS. 7 198) 5-1 Hy = F 1, 61%, 620.4% Ci g> 2-50 000 + 190.00 Tro8T Cem Pl USS 1041. 29 5.A research foundation wishes to set up a trust fund earning 10% compounded annually to (1) Provide P10,000,000 for the lot and building and 5,000,000 for the initial equipment of a Structural Engineering and Materials Laboratory: (2) Pay P2,000,000 for the annual operating costs every year, and (3) Pay P2,500,000 for the purchase of new equipment and replacement of some equipment every 5 years beginning 5 years from now. How much money should be paid into the fund for the building and equipment and to pay for perpetual operation and equipment replacement? @® Solution: Fo + Me ge a ‘tad tae Te e00 000 # § W0oo90 + 2 LCP CO, 2 sv0.wW ele Teer BAOGdABT-OR Ke » P 49, 094,937. 02]6.A heat exchanger is needed in a chemical process. If interest is 10% compounded annually, determine which of the following heat exchangers is cheaper by comparing the capitalized costs: (a) Exchanger A costs P110,000 with a scrap value of P5,000 and a useful life of 6 year: (b) Exchanger B costs P140,000 with a scrap value of P7,500 and a useful life of 10 years. Sdykon' © Me Stas Fe RE Cts> Fe + Tr wd + Move +52, = 40 000 +S Tei Trond 1G, 490-39 cogs P (44,706.90 Heat Exchanger BLS economically cheaper than B duc > ie uch life years: 7. certain type of untreated pole to be used for an electrical distribution system costs P3,120 in place, and it is expected to last 10 years before replacement - If this pole is treated with a wood preservative it is expected to last 20 years before replacement, but will cost P3,600 in piace. Assuming the cost of renewal to be the same as the cost for each type of pole, compare the capitalized costs of the poles using an interest rate of 14% compounded annually. e TUF > PL 1st.47 Re __ 3400 : cee THM > Tray +P 282 Se se The pole treated with wood preservative would Pave a lesser capitalized coct Eran that of an vatreated pole:8. Corrosive liquids are transported through pipes in a factory. Ordinary pipes will have an installed cost of P180,000 and their useful life is 4 years. Stainless steel pipes are highly resistant to the corrosive effect of the liquids and are being considered as an alternative. These pipes are estimated to have an installed cost of P330,000. Scrap value is zero in each case. If money is worth 10% compounded annually and assuming replacement costs to be the same as the original prices, what should be the useful life of the stainless steel pipes to have equal capitalized cost as the ordinary pipes? solution: Cla > CoB [g0_ 080 330 Ge GFR ne 4ST yrs, 9. Compare the capitalized costs of the following penstocks for a hydro-electric plant with interest at 10% compounded annually: zimber First Cost 300,000 Estimated life 12 years Scrap Value P12, 000 Annual Maintenance P7,200 Solvtion® umber * 200 | 12090 ce = 200 000 ¥ “OTT (ipyT Co P B77, GUI GO) steel? io ce = 4B 000* “Gy fe = P 442,000) © ‘oe Steel 480, 000 30 years None P1,20010. A financial analysis of two types of bridges based on capitalized cost and on the following data is to be made: Bridge A Bridge B Initial Cost 2,000,000 2, 400, 000 Cost of renewal 2,000,000 — P2, 400,000 Salvage value 0 200,000 Annual maintenance P10, 000 None Repairs every 5 years P100,000 P50,000 Life 30 years 40 years If the rate of interest is 10% compounded annually, determine the capitalized cost of each type of bridge. Solution: Bridge A? : 10.900 , 100 00 , 2000000 CC 2 000 exo + EEO + 10 OT Tey 23S ae 2 44h CU> P 2,395,382. 45 Bridge B: ce 50.002 4 2400 908 2400 000 Eye * Getoad USBVITA GRY co = Pz, 5 3b, (24. 4]