INDE 301 Midterm (2), Fall 2016

al Ff Pi La Fa Jif d 7 - INDE. 30) Midterm Vaan ones Nome Aey Problem | (15 pointe) (Highway) AWC [ 50 IOP, 10%, 1) 2400 (PIP Los ro APIA, 10% SCPE, LOM, DO) | CAMP, "LA AWE = [ 65(P/A IM PIP, Lote, 2) + 7SCPA, I wir a ge + BOCPLA, PPP, 10%, 10) OAL, Des, Po} fy - 7 Py # “ -/ | Ai by é = 7 7 gif & AWE <AWC, the investment is not atiracthve AW =e / F nie Problem 2 (15 points) (Car) ( | 12000 A * (PA, 196,18) # A, then : \ Ot ge S A= $680 f Finally / ( Pf Pavoll = 689.70 * (PIA, 1%, 7) + O89.70 = § 5929.96, - Problem 3 (15 points) (Oi well) (a) The expected revenue for the first year of production A, = $30° 100,000 = $3,000 000) gradient series havinit sing peometr The revenue stream follows a dec # 10%. Therefore, the expected pr P =$3,000,000(P|A,, —10%,1 nt worth of the revenues for the next 7] year ye 3,000,0003, 5620) $10,686,037 41 1 pre LL while th (b) In this case, the oil price increases at the annual rate decreases at the annual rate of 10%. Therefore ihe annual revenue in , ani ti expressed as = x uy cit A, = Ay, (1+ 0.05)(1 = 0.10) = A, (0.945 This revenue series is equivalent to a decreasing geometric gradient with § - The present worth of this new series 1s; P= $3,000,000(P|A),-5.5%, 12%, 7)°$3 000, 00006 9746) = $11,923,946 (c) A fair price wouldequal the present worth of the remaining en (/A4, A 6, / : } end of Year 3, / P =$2,$34,.725.88(P/A1,-5 5%, 12%,4) = $2,531,725.88(2.81 82) oP $7,134,615.9 v/ (~c Ais = Ps j | f ¥ as = ——— “) A Je , Finding OE of ¢ Ny Deseo Ah ique FOR sobaticn. Surtees OU eal } ‘Obtaining with 4@ 200+) 0088 — 325 f oe (PAG «) = 30769 => Im 11% 7 \ i - Ate 11%, 100 007 P|A FA LOOCPIF LAY 4,39, i. 1 At de 10%, —1000+300(P/A,(,4)+ 1 OOCP(F fA y9 19.29 a SofOk, <= 11% < 15% . Discard A Finding ROR of B a Oe ys re By Deacartes' rule, B has a unique ROR solution yf 3 fy L Solwe: —f00+60(PIA. 1. 2)-S00P PF £,2 0 { Obtaining an initial guess with 4 & 60+ SO0/2 a5 “ —~{O0+H(G0+2S)(PIA.42) = O = (P/A,I,2) = 1.1764 => is 40% 6 — Z As = 40%, —1000+ 300(P|A,/,4)+ 100(P|F, 1,4 yn 1 02 F wy fay | aad of Ata = 35%, —10004+-300(P|A.i/.4)¢ 10O(P(F,/,4 ye4.8 RORg > 39% > 15%, Keep B Finding ROR of C By Descartes’ rule, C has a unique RC IR solution. Solve —300+200(P/A,i,2)+40(P|F,1,2) = 0 Obtaining an initial guess, A = 200 + 40/2 = 220 -300+220(P/A,i,2)—0 => (P/A,/,.2)" 1.3636 = f= 30% At / = 30%, —300+200(P|A,/,2)+40(P|F i.2)= — 4.4 At i= 25%, —300+200(P/A.i,2)+-40(P/F 1,2)" 13.6 => ROR, > 25%>15%. Keep C ; to Yn \? Do incremental analysis of C-B The incremental cumulative cash flows in Years 1-4, are —200, 48s 4g), 920-70- Therefore, by Norstrom's criterion, a unique ROR solution exists. — ol Life, years | 196 } [_ Project | Initial cost | Annual revenue | Salvage Value | [| _C-B =200 | 140 —10 { : th Fi,2) = 0, I} / ( Solve —200+140(P|A.i,2) —10(P Obtaining an initial guess with A = 140 — 10/2 = 135. Ds d ~200+135(PIA,i,2) = 0 => (P/A,i,2) = 1.48148 = ix 22% ( are - At é=22%, —200+140(P/A,i,2) —10(P/F 1,2) = 2.1 At i= 24%, —200+140(P/A,i,2) —10(P/F,i,2)= -2.55 =>RORe-9 > 22% > 15%. Choose Project C.