Midterm Solution Part 2

Midterm Solution Part 2 - UW ID Number Problem 4. A...

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Unformatted text preview: UW ID Number Problem 4. A proposed new product would have sales revenue of $500,000 next year, $470,000 in the second year of sales, $440,000 in the third year, and so on, decreasing by $30,000 each year compared to the previous, for altogether 6 years of sales. Operating costs would be $200,000 per year, each year for 6 years, and in addition, there would be a special maintenance cost of $150,000 at the end of the third year from now. The first cost would be $800,000 now, and the equipment would have a salvage value of $75,000 in 6 years. The company has a MARR=10%, and a 4-year payback policy. a) [8 marks] Use the present worth method to recommend whether this is a good investment. , t) _ PWl: 400,000 + [Wt°°°"3°«°°°(4"'l'W)‘ WWW/AM!) "mm(P/rgoxqvtwea/gmzt) : 400,000 ‘t [(001600 -—3o,ooo(a. a236)~acoloooj(‘h {$33) - ld'olooo (a, 35m) +4571”; 0 ( o. “q q q) i aims, at: _ WWW?) *) 54ft PVV?O \ (\c .fl ' l M J6 i .,) {cwfi “mm/(ch: (9,» b) [8 marks] Use the payback method to recommend whether this is a good investment. 1‘ l : mmt #1; ‘ Wag-W “4M4 FMMWC): $300,090 i = (.—)/IcrwJ°¥+ 8 I .. p g . _ FC w__~--- ‘1“: MW(%‘.) ma “#44 1W- , ., h w ' $300,000 $200, we % 7‘ 5 93°19“) $570,002 L 3 quoflfldtfotoowi‘io‘eoc g 660‘000 .. g. yum it “Amoco $340.000 -* Y va> 2 N N ‘4' ‘ é .H ~ , . WM .‘ ism-02, {31000 ("x t' . (é ' t o N V , C Q ‘ 1" m - me‘, % quw' m “00.000 IP+M1 Jib; Mi 1M4 < ‘t yum: . 4“ rut may}; Mi. 1%“:5 4a WWW 4%» real Wk yam 1mg,“ < 4 7W1. Ha WM Man/r UW ID Number Problem 5. A proposed investment in a new product would have a first cost now of $850,000. Annual operating and repair expenses during the 10 years of production would be $42,000 in the first year from now, and would increase by $2000 per year for each of the following 9 years. In 5 years from now, a special overhaul is necessary, at a cost of $50,000. At the end of production, i.e 10 years from now, the equipment would be sold for a salvage value of $100,000. It is expected that 13,000 units of the new product will be sold in the first year, and that sales would then decrease by 500 units per year (each year, compared to the previous year), with 10 years of sales altogether. The MARR is 10%. (a) [10 marks] Write an expression for the levelized cost of the new product. Use the above data, and interest factor symbols in your expression. (See page 1 for the list of allowed interest factor symbols.) Do not evaluate the expression, as it would take too much time. l M ' -L L e - . *\\ ha 7 _ i P I l0 L, _ 7 “'0? (A/a"°vfilla)]( P/A,loz’im)\:fiS-GPOO '02,, f) ifteo'ow '07., W \[fflooc j {WU/6. ,'°%,Io)](i’/A .67» to) l '7‘; I ft. 2 Sfio‘ooow‘li‘iamo +31‘oocig- ¥1S{)]( inl't‘lb) 1331000 (0. 5;! on) -9100‘000 (0‘ 3351-9) mm [(3.000— 5‘00 ( ztlass)](g.mqs) am :4»; ' 34,» wt in M3 *9 film: ’0 pi; (b) [1 mark] If the company can sell the product at a price that is greater than the levelized cost, then is this a good investment ? Ya fiat \ UW ID Number Problem 6. [11 marks] Cash flows for a proposed investment in a new product are shown in the table below. The first cost would be spent now, at “year zero,” the revenue and annual operating costs would happen in years 1 through 8, inclusive, the salvage value would be received at the end of year 8, and the special overhaul of some equipment would happen at the end of year 5 from now. $ 890 000 First cost now Salv e Value in 8 ears $ 45,000 Sales Revenue $ 760 000 er ear for 8 cars Annual operating cost $ 320,000 in the first year, then increasing by $ 10 000 each ear over the revious ear $ 80,000 at the end of ear 5 S ecial Overhaul cost Calculate the annual worth, with a MARR of 10%, and state whether the proposed investment is a good one. "7 I O‘M'L ,, l” i {J L ~ _: K. \‘\. , x . ' - “ .- ‘ _ , \ ‘ ziol 00 G (Am '°7°.',§) “QC: 00°(NF.“33)6730‘06117320‘W‘Ql O‘ecc (A/G,{0% '3') -‘ lr‘l/ i 'L 80.000U’lFitov/HS) (A/P’ rev/0‘ .3) r 2 4557331.“ 39mm -+ “0.060 ~ 30m:— $210.? 5 313 "l. "+3111 5 Mar/(I M UW ID Number A L \M t» 0' "’l Problem 7. A power tool is needed for a factory. There are two mutually exclusive alternatives, made by Acme or by Best: _ $350 $500 $100 er ear Annual maintenance Useful life Both‘brand’s tools would have a zero salvage value. The MARR is 10%. a) [7 marks] Using the repeated lives assumption and the present worth method, determine which brand is preferred. W half Can/(min Mu/fip/z 0/ 2&3 2'! 6. 50 .' 3mm: PW”) = -350-350 (P4zzl07. ,z) —3_YO(P/F1IOZ)4)—([00+50)(’% » I02 ’4) I / {id { _ _3sa, 350X0.6264I- BSOXO-KJ’SOI— Mia/r4355 =-—/6€2v27" mafv’e" Alre/nafmly: Pun/(A); (350+(IW+3°)(P/,4 ; IOZ>Z)(I+(t},;)loz;zj+(llzllox 24)) 3ma_~"‘£S Fw (3) = -500, 500 (P4,, (07, )3) —- ( WNW—NZ I ’07- J 6) 1,13" — ref/WU 2 = -Soo -5-00 x 07513! - I65 x q ’35)“; — —HLE?_4M§E amjwe/ Nfernafh/ely: Pw(B) s (goo +(70+7i)( 2, 10215)) (1+ {ID/F , /0/, ,3» l monk (grand W/v‘h lawgy aif U Pkg/("101‘ b) [7 marks] Using the study period method, determine which brand is preferred. For this case, assume a salvage value of $ 85 at the end of year 2 for the Best brand. 5moul<5 M IF” flirted al/IJVGV )Z) : -359 _/go;<ln7’35f: .. 662137 ___—___,____. Pw(,4) = _ 350 — {400+80)(/>/A)/oz Pw(B) ; —-500 - (90f7f)(P/A M07~U+ 6502:; I07. )2) I?“ ans Nix" I _J’00 , 15:» I-7355+ 8&(2'33‘45’ = —7’5‘ ” I mark Acme brand wrlh lower [well/'7‘ Warfl' C4” ’1 [Ni/“Ved' UW ID Number Problem 8. A proposed oil sands plant would take 2 years to build, at costs of $ 2 X 109 in year 0, and $ 4 >< 109 in year 1. There would be 35 years of production altogether, with the first oil produced in year 2 from now, at a rate of 0.025 X 109 barrels in year 2; in later years, production would increase by 1% per year, each year compared to the previous year. The last year of production would be year 36 from now (for 35 years of production altogether). An oil price of $100 per barrel is expected, on average in each of the 35 years of production. Annual operating costs would be steady at $1.50 X 109 per year for each of the 35 years of production. Salvage value on the equipment at the end of production would be $0.2 >< 109. Some special additional equipment costing $ 0.4 X 109 would be installed at the end of the 3rd year of production. The following cash flow diagram illustrates the timing assumptions for the cash flows described above. (a) [12 marks] Write out an equation which, if solved, would give the internal rate of return of the proposed investment. Use the above data, and interest factor symbols in your equation. (See page 1 for t - ' factor symbols.) Do not solve the equation, as it would take too much time. f. may/<5 q 4 ,r 7' ”' 4‘12" " Fm}; —leO - than (1%! I I U + you. 0.9m mi (fiw' _. w . . . ‘ 1+ 0~ /"’ 1 A: “.V‘\ ' 4 4t 7 ‘ e. 1.6m (l4 71,3?)(fi‘21192 + oLx/Oifl/u ,35) + (—on/o/{3 xx,“ I; f / F f [7?" ‘0 =1> L z a r «was, ~—) (0.: E.‘_I Lol Cm‘fm/j luau/L “1 4’1"” ‘ (b) [1 mark] Briefly state what you would do with the solution to your equation of part (a), in order to use the internal rate of return method to judge whether the proposed new product is economically acceptable. f’ 25% ) MflK ,fl’wv f/a/Wc/ Phjeof [5 filly/awe 10 (A lax [wHam V 1! may \\ \ 79 l / UW ID Number Problem 9. [10 marks] There are five mutually exclusive proposed projects, labelled A, B, C, D and E, and the “do- nothing” alternative is possible. Their internal rates of return, are displayed in the table on the left below, along with their first costs. The tables on the right give the incremental IRRs when comparing one proposal to another. The MARR is 11.5%. Which is the best project? Explain, using the algorithm from the course. -- First Cost Overall Investment $3800 I133 ' Memo Tit 1m 4 cm a?“ (Mme—r MARK .14%>n.rx.T/ngwtl 04‘, Wit/Kt MVMWAAM‘ 9. KW 1“ 4 8&0 Aim MARQ'IQ'K‘21LIZ @ WWW 9” £44 MW Qflzmb 14%?“ W, 11 ~> S'x vadg :flf’f-EVB .Molleé‘vx 8A2. mwwqm, ...
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This note was uploaded on 04/01/2012 for the course MSCI 261 taught by Professor Bonkoo during the Winter '09 term at Waterloo.

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Midterm Solution Part 2 - UW ID Number Problem 4. A...

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