Number of Levels Cumulative Annual operating Income per year construction costs costs ($) ($) ($) 1 600,000 35, , ,200,000 60, ,000


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1 Problem Set 5 Solutons 1 MIT s consderng buldng a new car park near Kendall Square. o unversty funds are avalable (overhead rates are under pressure and the new faclty would have to pay for tself from parkng fees over a 15 year perod. mnmum rate of return of 1% (before taxes s requred for the project. The archtect has developed four alternatve desgns for dfferent numbers of parkng levels n the structure. Based on the ncome and cost data n the Table below, how many levels should be bult? (OTE: Prepare your analyss on a beforetax bass. umber of Levels Cumulatve nnual operatng Income per year constructon costs costs ($ ($ ($ Soluton: 1 6, 35, 1, 2 2,2, 6, 35, 3 3,6, 8, 57, 4 4,8, 15, 75, Snce the lfetme of each faclty s 15 years, we can use the PW crteron to compare the alternatves. We can construct the followng generc cash flow dagram
2 I j Etc 15 C j I We can calculate the present worth based on the followng formula PW = I o ( I C ( P,1%,15 j j The present worth for each opton s calculated below PW (1 = $6, ($1, $35,( P,1%,15 = $15,65 PW (2 = $2,2, ($35, $6,( P,1%,15 = $5,763 PW (3 = 3,6, ($57, $8,( P,1%,15 = $126,979 PW (4 = 4,8, ($75, $15,( P,1%,15 = $15,921 Snce PW(3 s the greatest t s to our advantage to buld the 3 level parkng garage.
3 2 muncpalty s tryng to establsh a polcy on how long to keep economy passenger cars that are used for offcal busness. The purchase cost of a car at fleet dscount s $12,. nnual costs for nsurance, ol, and repars for 2, mles of annual use are $2, for a new car and ncrease by $1year wth the age of the car. Fuel consumpton s not perceved to change wth age and s therefore excluded from the analyss. It s recognzed that as a vehcle becomes older t suffers more breakdowns, causng employees to become less productve. The average loss of employee productvty s estmated to be 4 daysyear for a new car and to ncrease by 2 daysyear wth the age of the car. The average payroll cost of any employee lkely to use the car s $2day. fter 8 years spare parts become dffcult to obtan, and the downtmes of the vehcle become unacceptably long and frequent. Salvage values as a functon of age up to 8 years are gven below. Salvage Value Table S Usng = 1%, fnd the economc lfe of such a car. Ignore ncome taxes, nflaton and technologcal mprovements. Soluton: To calculate the economc lfetme of the car we must use the levelzed annual cost (LC crteron. From the problem we are gven the followng nformaton I = $12, M = M M n ( n 1 = $2, ( n 1$1 L = 4($2 ( n 1(2($2 = L L n ( n 1 = $8 ( n 1$4 We can now calculate the levelzed annual cost based on the nvestment, the operatng costs, and the productvty loss. For the gradent terms we use the annuty factor for a gradent ncrease. Let us frst lst the annuty factors we wll need to use to complete ths problem
4 = = = = = P G P P G P F (1 1 (1 1 (1 (1 (1 1 (1 %,, ( %,, ( %,, ( %,, ( 1 (1 (1 %,, ( 1 (1 %,, ( 2 2 The levelzed cost can be calculated as thus,1%, $5( $2,8,1%, (,1%, $12,(,1%, (,1%, (,1%, (,1%, ( G F I P LC G L L G M M F I P I LC n n = = The followng page shows a table whch calculates the LC gven that the car operates for any number of years between 1 and 8
5 (P,1%, (F,1%, (G,1%, I I G C LC $12,. $9,. $5. $2,8. $7, $12,. $6,75. $5. $2,8. $6, $12,. $5,6. $5. $2,8. $6, $12,. $3,8. $5. $2,8. $6, $12,. $2,85. $5. $2,8. $6, $12,. $2,14. $5. $2,8. $6, $12,. $1,6. $5. $2,8. $6, $12,. $1,2. $5. $2,8. $6, The Economc Lfetme s gven by the opton where the LC s the smallest, n ths case the economc lfetme s 6 years.
6 3 new automoble wth a lst prce of $1, s beng consdered as a replacement for a used automoble presently owned by the company. The new automoble can be acqured by tradng n the old one plus $6 cash. lternatvely, the company can sell outrght the present auto for $2 and purchase the new car outrght for $8 cash. Current remanng book value (unallocated cost of the present auto s $3. a. If EC of the present auto s based on ts remanng book value, the resultng EC tends to be (a overstated, (b understated, or (c correctly stated b. If EC of the new automoble s based on the $1, lst prce and f EC of the used auto s based on the traden allowance of $4, the study tends to be based n favor of (a the proposed, (b the present, or (cnether automoble. c. If EC of the new automoble s based on the $6 cash pad and f EC of the used auto s based on the fact that because already owned t requres zero outlay and therefore has zero present bass, the study tends to be based n favor of (a the proposed, (b the present, or (c nether auto. Soluton: ssumng that EC stands for the nnual Equvalent Captal Cost, we can evaluate the optons gven the nterest rates and tax rates, however, we are only asked to evaluate the trends n the EC based on the methods of calculaton. The EC s calculated usng the followng formula for an ntal nvestment of I and a salvage value of I over an economc lfetme of, wthout accountng for deprecaton allowances towards tax deductons. [ I I ( P F, %, ] EC = ( P, %, (a From the outsder s pont of vew to decde between nvestng n the new auto, one would have to consder purchasng the defendng auto at the book value and calculatng the EC. Ths s compared to the outsder purchasng a new auto at the lst prce and computng the EC. If we assume that the new and old automobles are effectvely the same, where no new technology has been employed, usng the BV of the automoble as the market prce does not bas the decson n ether drecton. (b Snce the traden allowance of the old automoble s hgher than the BV, t s effectvely makng the defender more expensve to the outsder relatve to the case n (a thus basng the decson towards purchasng the new automoble.
7 Equvalently, ths opton can be thought of from the nsder s perspectve of makng the new automoble cheaper. (c If t s assumed that because the vehcle s already owned, t has zero present bass, then usng the EC as a decson crteron s an nvald technque. It wll bas the decson maker n favor of the present, because t assumes that the EC of the present opton of the automoble does not nclude the nvestment cost, whereas the EC of the proposed s based on a $6 purchase prce. It s also noteworthy that the $6 purchase prce s obtaned from combnng the salvage value of the defender wth the nvestment cost of the defender, thus shftng part of the worth of the defender to the worth of the challenger.
8 4 You have calculated the economc lfe of a faclty. ow you are asked to ndcate for each factor below whether an ncrease n that factor would shorten or lengthen that economc lfe. a. The amount of ncrease n the postve gradent pattern of yearbyyear costs b. Rate of return requrement c. Estmated future salvage values d. Frst cost of the faclty e. Rate of technologcal progress f. pplcable ncome tax rate g. Emphass on prestge and mage of the frm
9 Soluton: We wll use problem 2 as a prototype to answer these questons by changng the values n the spread sheet and seeng the effect on the economc lfetme. The followng s the LC equaton as shown n problem 2 LC = I ( P,1%, I ( F,1%, M M n ( G,1%, L Ln ( G,1%, LC = $12,( P,1%, I ( F,1%, $2,8 $5( G,1%, (a The gradent pattern s n the Mn and Ln terms. If these are ncreased the economc lfetme changes as follows mathematcally (P,1%, (F,1%, (G,1%, I I G C LC $12,. $9,. $1,. $2,8. $7, $12,. $6,75. $1,. $2,8. $6, $12,. $5,6. $1,. $2,8. $7, $12,. $3,8. $1,. $2,8. $7, $12,. $2,85. $1,. $2,8. $7, $12,. $2,14. $1,. $2,8. $7, $12,. $1,6. $1,. $2,8. $7, $12,. $1,2. $1,. $2,8. $7, We ve ncreased the gradent from 5 to 1 and we can see that the economc lfetme shortens from 6 to 2 years. Ths s because the LC ncreases as the hgher costs n subsequent years are added.
10 (b The rate of return requrement. The rate of return s 1%, f we ncrease the RR to 15%, the LC table changes as follows. (P,15%, (F,15%, (G,15%, I I G C LC $12,. $9,. $5. $2,8. $7, $12,. $6,75. $5. $2,8. $7, $12,. $5,6. $5. $2,8. $7, $12,. $3,8. $5. $2,8. $6, $12,. $2,85. $5. $2,8. $6, $12,. $2,14. $5. $2,8. $6, $12,. $1,6. $5. $2,8. $6, $12,. $1,2. $5. $2,8. $6, The economc lfetme n ths case s 7 years. Increasng the RR lengthens the economc lfetme. The ntal nvestment s always greater than the salvage value, gnorng the effects of the gradent term, one can see that ncreasng the RR effectvely ncreases the contrbuton of the ntal nvestment and salvage value to the LC. The ntal nvestment contrbuton wll outwegh the salvage value, and the economc lfetme ncreases because you must earn more ncome from the asset to counter ths effectve ncrease n the nvestment cost of the asset.
11 (c Estmated Future Salvage values were ncreased by 25% and the economc lfe compared to the case n problem 2 (P,1%, (F,1%, (G,1%, I I G C LC $12,. $11,25. $5. $2,8. $4, $12,. $8,437.5 $5. $2,8. $5, $12,. $6,325. $5. $2,8. $6, $12,. $4,75. $5. $2,8. $6, $12,. $3,562.5 $5. $2,8. $6, $12,. $2,675. $5. $2,8. $6, $12,. $2,. $5. $2,8. $6, $12,. $1,5. $5. $2,8. $6,42.4 s one can see the economc lfetme s dramatcally reduced (1 year, ths s because f you can get more money for the asset at the end of lfe, sellng t sooner s advantageous.
12 (d Frst Cost of the Faclty, the ntal nvestment was ncreased by 5%. (P,1%, (F,1%, (G,1%, I I G C LC $18,. $9,. $5. $2,8. $13, $18,. $6,75. $5. $2,8. $1, $18,. $5,6. $5. $2,8. $8, $18,. $3,8. $5. $2,8. $8, $18,. $2,85. $5. $2,8. $7, $18,. $2,14. $5. $2,8. $7, $18,. $1,6. $5. $2,8. $7, $18,. $1,2. $5. $2,8. $7,571.3 The economc lfetme s ncreased (8 years. Ths s because t takes more tme earnng money from the asset to make up for the larger ntal nvestment.
13 (e Rate of technologcal progress The economc lfetme s reduced, because the asset wll lkely become obsolete compared wth the challengng technologes avalable. The LC of the newer technologes wll be less due to the progress, as the rate of progress ncreases the challenger becomes economcally vable as a replacement sooner. (f pplcable Income Tax Rate The deprecaton allowance per year ncreases as the lfetme of the asset decreases typcally. Lkewse as shown wth the example wth no taxes, the LC wthout tax ncreases as the economc lfetme s shortened. The frm wll receve a tax deducton for the costs of operaton and the deprecaton allowance. The true cost to the frm for the asset can be thought of as τd τlc LC = ( 1 τ LC τd From the standpont of the frm t s worthwhle to mnmze ths quantty, where LC s the levelzed annual cost wthout tax. If the LC and D are larger n earler years, and the tax rate s large, then t s advantageous for the frm to shorten the lfetme of the asset. (g Emphass on Prestge and Image of the Frm ewer assets ncrease the prestge of the frm; therefore the frm that emphaszes ths mage wll replace assets frequently, thus shortenng the economc lfetme f a dollar value can be attached to mage.
14 5 (PSB defender has a current salvage value of $15, a remanng lfe of 5 years wth zero salvage thereafter, and O&M costs of $26, per year. The challenger costs $5,, has a lfe of 12 years wth $5, salvage value thereafter, and O&M costs of $2,year. MRR = 15% a. Use E wth the outsder pont of vew to make a decson. b. Use E wth the cash flow approach to make a decson. c. Use PV wth the outsder pont of vew to make a decson. d. Use PV wth the cash flow approach to make a decson. e. Reconcle any conflctng answers. (Hnt: What s the study perod? Soluton: We wll examne ths problem by regroupng the problem segments; parts (a and (c share smlar cash flow dagrams, whereas parts (b and (d share smlar dagrams that are both dfferent from (a and (c.
15 Outsder Pont of Vew (a and (c Defender: The outsder buys the defender for the book value of the defender and operates the defender for the rest of ts lfe. Therefore the book value appears as an expense. 5 $15k Etc $26k The mathematcal expresson for the annual expendture s: E D = I ( P,15%,5 OC = $15,( P,15%,5 $26, = $3,474 and the present value of ths opton s PV D = I OC( P,15%,5 = $15, $26,( P,15%,5 = $12,156
16 Challenger: The aspects of the defender are not ncluded from the outsder s pont of vew. $5k 12 $5k Etc $2k The analogous expressons for annual expendture and present value, respectvely: E C = I ( P,15%,12 OC I ( F,15%,12 = $5,( P,15%,12 $2, $5,( F,15%,12 = $29,32 PV C = I OC( P,15%,12 $5,( P F,15%,12 = $5, $2,( P,15%,12 $5,( P F,15%,12 = $157,478 So from the pont of vew of the Outsder usng the E method, the challenger (E=$29,32 s a better opton than the defender (E=$3,474. From the pont of vew of the Outsder usng the PV method, the defender (PV=$12,156 s a better opton than the challenger (PV=$157,478
17 Cash Flow Pont of Vew (b and (d Defender: There s no cost assocated wth keepng the defender from the cash flow pont of vew. 5 Etc $26k The mathematcal expresson for the annual expendture s: E D = OC = $26, = $26, and the present value of ths opton s PV D = OC( P,15%,5 = $26,( P,15%,5 = $87,156
18 Challenger: from the cash flow perspectve, the frm sells the defender for the book value and then purchases the challenger. $15k $5k 12 $5k Etc $2k The analogous expressons for annual expendture and present value, respectvely: E C = I, Defender ( P,15%,12 I ( P,15%,12 OC I, Challenger ( F,15%,12 = $15,( P,15%,12 $5,( P,15%,12 $2, $5,( F,15%,12 = $26,284 PV C = I I OC P,15%,12 I ( P F,15%,12 = $15, $5, $2,( P,15%,12 $5,( P F,15%,12 = $142,478, Defender (, Challenger
19 So accordng to the cash flow approach usng the E method, the defender (E=$26, s a better opton than the challenger (E=$26,284. ccordng to the cash flow approach usng the PV method, the defender (PV=$87,156 s stll a better opton than the challenger (PV=$142,478. The nvestgaton has been summarzed n the followng table: pproach Method Defender Cost Challenger Cost Decson (a E, Outsder $3,474 $29,32 Challenger (b E, Cash Flow $26, $26,284 Defender (c PV, Outsder $12,156 $157,478 Defender (d PV, Cash Flow $87,156 $142,478 Defender (e There s a dscrepancy for the outsder between the E and PV evaluatons. Ths dscrepancy s due to the fact that the PV evaluaton compares a 5year project to a 12year project. lthough the PV of the 5year project s much less, t s only runnng for fve years, whereas the other project dsplays costs that run for 12 years. The PV s thus an nvald comparson, and from the outsder s pont of vew, the challenger should be the preferred opton. The dfference from cash flow to outsder can be attrbuted to the levelzaton of the salvage value of the defender; for the defender, t s levelzed over fve years. The challenger, by contrast, levelzes the salvage value over twelve years.
20 6 (Sullvan et al, Q 136 small dam s beng planned for a rver trbutary that s subject to frequent floodng. From past experence, the probabltes that water flow wll exceed the desgn capacty of the dam durng a year, plus relevant cost nformaton, are as follows: Desgn Probablty of greater flow durng a year Captal nvestment.118, B.5195, C.2528, D , E.6224, Estmated annual damages that occur f water flows exceed desgn capacty are $15, $16, $175, $19,, and $21 for desgn, B, C, D, and E, respectvely. The lfe of the dam s expected to be 5 years, wth neglgble salvage value. For an nterest rate of 8% per year, determne whch desgn should be mplemented. What nonmonetary consderatons mght be mportant to the selecton? The desgn to be chosen should have the mnmum PW. The present worth of the total cost ncludes the ntal nvestment and the annual damages resultng from the probablty of the overflow. The present worth can be calculated accordng to the followng formula PW = I p( P, %, Where % s the nterest rate, s the lfetme of the project, s the annual repar cost, p s the probablty of an overflow and Io s the ntal nvestment. PW ( = I p ( P, %, = $18,.1($15,( P,8%,5 = $363,52 PW ( B = $195,.5($16,( P,8%,5 = $292,868 PW ( C = $28,.25($175,( P,8%,5 = $261,521 PW ( D = $214,.15($19,( P,8%,5 = $248,865 PW ( E = $224,.6($21,( P,8%,5 = $239,414
21 Snce the PW of the E desgn s the smallest, ths s the desgn opton that should be pursued assumng that each generates the same annual revenue. Dams dsrupt the local ecosystem and ths s not taken nto account through the monetary analyss. We should also consder potental loss of human lfe or qualty thereof due to water flow excess f t s not already ncluded n the annual damage cost estmates.
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