29/09/2010. Outline Module 4. Selection of Alternatives. Proposals for Investment Alternatives. Module 4: Present Worth Analysis

Transcription

1 Outline Module 4 Proposals for nvestment Alternatives Selection of Alternatives Future Worth Analysis Capitalized-cost calculation Module 4: Present Worth Analysis S-4251 konomi Teknik 4-2 S-4251 konomi Teknik Proposals for nvestment Alternatives An investment is usually established from (engineering) proposal. very proposal can be considered as an investment alternatives, but An investment alternative can consist of a group or set of proposals, which in turn may include option to do nothing ndependent Proposal The condition at which the acceptance of a proposal from a set of proposals has not effect on the acceptance of any of the other proposals in the set. Dependent Proposal The condition at when the acceptance of a proposal from the set will influence the acceptance of the others Mutually exclusive the acceptance of a proposal from the set precludes the acceptance of any others Selection of Alternatives Decision criteria in an economic analysis can be done comparing mutually exclusive alternatives: The differences between alternatives highest inflow or lowest outflow The minimum attractive rate of return highest rate of return Payback period shortest payback period Do nothing 4-3 S-4251 konomi Teknik 4-4 S-4251 konomi Teknik 1

2 Selection of Alternatives Selection of two or more (investment) alternatives by comparing their economic values. Method for comparison of alternatives: present values, future worth, capitalized cost, annual values, rate of returns, or payback period Conditions of comparison: equal lives different lives MARR Minimum Acceptable Rate of Return Minimum amount the investor is willing to accept for the use of money Different from lending rates More akin to Opportunity Cost of the money Determined by Company Policy All proposals must offer at Least MARR to be considered Rate of Return xpected Rate of Return on New Proposal Range for ROR On accepted proposals MARR ROR on Safe investment 4-5 S-4251 konomi Teknik 6 S-4251 konomi Teknik The comparison of alternatives is made by transforming all future receipts and expenditures into equivalent today s rupiah xample: Two types of production systems are being considered based on MARR of 12% per year and the following characteristics: system A system B nitial cost Rp ,- Rp. 57..,- Monthly expenses Rp ,- Rp ,- Receipts Rp ,- / quarter Rp. 14..,- / 4 months Salvage value Rp ,- Rp. 19..,- Life 3 tahun 3 tahun System A: R R R R R R R R R R R R 3? ffective monthly interest rate, i A = i / 12 = 1% P A = - - (P/A, i A, 36) + R(P/F, i A, 3) + R(P/F, i A, 6) + + R(P/F, i A, 36) + (P/F, i A, 36) 4-7 S-4251 konomi Teknik 4-8 S-4251 konomi Teknik 2

3 System B: R R R R R R R R R 3 P B = - - (P/A, i B, 36) + R(P/F, i B, 4) + R(P/F, i B, 8) + + R(P/F, i B, 36) + (P/F, i B, 36) Select System A if P A > P B To have a fair comparison of alternatives with different lives, the time span over must made equal: a) The time period of comparison is made equal to the least common multiple (LCM) for their lives. Cash flows of the shorter period will be extended up to the remaining time period of comparison b) At any time span to be considered (the study period approach or planning horizon approach), when LCM is impossible to perform. Only cash flows up to the time span is to be considered; 4-9 S-4251 konomi Teknik 4-1 S-4251 konomi Teknik xample: Two types of production systems are being considered based on MARR of 12% per year and the following characteristics: System A: R system A system B nitial cost Rp ,- Rp. 77..,- Monthly expenses Rp ,- Rp ,- Monthly receipts Rp. 32..,- Rp. 4..,- Salvage value Rp ,- Rp. 11..,- Life 2 tahun 4 tahun System B: R S-4251 konomi Teknik 4-12 S-4251 konomi Teknik 3

4 (LCM) (Study Periods: 2 years) System A: 1 R1 2 2 R2 2 System A: R 3 4 System B: R System B: R stimated S-4251 konomi Teknik 4-14 S-4251 konomi Teknik xample 1: Three Alternatives xample 1: Cash Flow Diagrams Assume i = 1% per year A 1 : lectric F = 2 A1 lectric Power First Cost: -25 Ann. Op. Cost: -9 Sal. Value: +2 Life: 5 years A2 Gas Power First Cost: -35 Ann. Op. Cost: -7 Sal. Value: +35 Life: 5 years A3 Solar Power First Cost: -6 Ann. Op. Cost: -5 Sal. Value: +1 Life: 5 years A = -9/Yr. F = 35 A 2 : Gas A = -7/Yr. -35 A 3 :Solar F = 1 Which Alternative if any, should be selected based upon a present worth analysis? 4-15 S-4251 konomi Teknik A = -5/Yr. -6 i = 1%/yr and n = S-4251 konomi Teknik 4

5 Calculate the Present Worth's xample 2: Present Worth's are: 1. PW lec. = -25-9(P/A,1%,5) + 2(P/F,1%,5) = $ PW Gas = -35-7(P/A,1%,5) + 35(P/F,1%,5) = $ PW Solar = -6-5(P/A,1%,5) + 1(P/F,1%,5) = $-6127 Select lectric which has the min. PW Cost! Two Location Alternatives, A and B where one can lease one of two locations. Which option is preferred if the interest rate is 15%/year? Location A Location B First cost, $ -15, -18, Annual lease cost, $ per year -3,5-3,1 Deposit return,$ 1, 2, Lease term, years S-4251 konomi Teknik 4-18 S-4251 konomi Teknik Use LCM, where n = 18 yrs. Unequal Lives: 2 Alternatives The Cash Flow Diagrams are: A 6 years 6 years 6 years Cycle 1 for A Cycle 2 for A Cycle 3 for A B 9 years 9 years Cycle 1 for B Cycle 2 for B 18 years i = 15% per year 4-19 S-4251 konomi Teknik LCM(6,9) = 18 year study period will apply for present worth 4-2 S-4251 konomi Teknik 5

6 LCM Present Worth's PW Calculation for A and B -18 yrs Since the leases have different terms (lives), compare them over the LCM of 18 years. For life cycles after the first, the first cost is repeated in year of the new cycle, which is the last year of the previous cycle. These are years 6 and 12 for location A and year 9 for B. Calculate PW at 15% over 18 years. PW A = -15, - 15,(P/F,15%,6) + 1(P/F,15%,6)- 15,(P/F,15%,12) + 1(P/F,15%,12) + 1(P/F,15%,18) - 35(P/A,15%,18)= $-45,36 PW B = -18, - 18,(P/F,15%,9) + 2(P/F,15%,9) + 2(P/F,15%,18) - 31(P/A,15 %,18) = $-41,384 Select B : Lowest PW 15% 4-21 S-4251 konomi Teknik 22 S-4251 konomi Teknik Use The Study Period Approach An alternative method; mpress a study period (SP) on all of the alternatives; A time horizon is selected in advance; Only the cash flows occurring within that time span are considered relevant; May require assumptions concerning some of the cash flows. Common approach and simplifies the analysis somewhat S-4251 konomi Teknik xample Problem with a 5-yr SP Assume a 5- year Study Period for both options: For a 5-year study period no cycle repeats are necessary. PW A = -15, - 35(P/A,15%,5) + 1(P/F,15%,5) = $-26,236 PW B = -18,- 31(P/A,15%,5) + 2(P/F,15%,5) = $-27,397 Location A is now the better choice. Note: The assumptions made for the A and B alternatives! Do not expect the same result with a study period approach vs. the LCM approach! 4-24 S-4251 konomi Teknik 6

7 Future Worth Analysis Future Worth Approach (FW) n some applications, management may prefer a future worth analysis; Analysis is straight forward: Find P of each alternative: Then compute F n at the same interest rate used to find P of each alternative. For a study period approach, use the appropriate value of n to take forward. Applications for the FW approach: Wealth maximization approaches; Projects that do not come on line until the end of the investment (construction) period: Power Generation Facilities Toll Roads Large building projects tc S-4251 konomi Teknik 4-26 S-4251 konomi Teknik Capitalized Cost Calculations Derivation of Capitalized Cost CAPTALZD COST- the present worth of a project which lasts forever. Government Projects; Roads, Dams, Bridges, project that possess perpetual life; nfinite analysis period; n in the problem is either very long, indefinite, or infinity. We start with the relationship: P = A[P/A,i%,n] Next, what happens to the P/A factor when we let n approach infinity. Some math follows S-4251 konomi Teknik 4-28 S-4251 konomi Teknik 7

8 P/A where n goes to infinity CC Derivation The P/A factor is: n (1 i) 1 P A i n (1 i ) On the right hand side, divide both numerator and denominator by (1+i) n 1 1 (1 ) n i P A i 4-29 S-4251 konomi Teknik 1 (1 ) n i P A i Repeating: 1 f n approaches the above reduces to: P A i 4-3 S-4251 konomi Teknik CC xplained For this class of problems, we can use the term CC in place of P. Restate: Or, AW: annual worth A CC i CC AW i CC Problem: Public Works xample Problem Parameters The suspension bridge will cost $5 million with annual inspection and maintenance - costs of $35,. n addition, the concrete deck would have to be resurfaced every 1 years at a cost of $1,. The truss bridge and 'approach roads' are expected to cost $25 million and have annual maintenance costs of $2,. The bridge would have to be painted every 3 years at a cost of $4,. n addition, the bridge would have to be sandblasted every 1 years at a cost of $19,. The cost of purchasing right-of-way is expected to be $2 million for the suspension bridge and $15 million for the truss bridge. Compare the alternatives on the basis of their capitalized cost if the interest rate is 6% per, year. Two, Mutually xclusive Alternatives: Select the best alternative based upon a CC analysis 4-31 S-4251 konomi Teknik 4-32 S-4251 konomi Teknik 8

9 Bridge Alternatives: Suspension Cash Flow Diagrams Suspension Bridge Alternative $35,/yr $5 Million $1, $2 Million Suspension Bridge Analysis CC 1 = -52 million at t =. A $35, 1 A 1,( A/ F,6%,1) $7,587 2 CC A A 35, ( 7,587) i.6 2 $79, 783. Total CC suspension bridge is: -52 million + (-79,783) = -$52.71 million 4-33 S-4251 konomi Teknik 4-34 S-4251 konomi Teknik Truss Bridge Alternative Truss Bridge Alternative For the Truss Bridge Alternative: Cash Flow Diagram: 1. CC 1 nitial Cost: -$25M + (-15M) = -$4M Truss Design: / / / / / / A. Maint. = $2,/yr Truss Design: / / / / / / A. Maint. = $2,/yr n = n = Paint: -4, Paint: -4, Paint: -4, Paint: -4, Paint: -4, Paint: -4, -25M +(-15M) Sandblast: - 19, -25M +(-15M) Sandblast: - 19, 4-35 S-4251 konomi Teknik 4-36 S-4251 konomi Teknik 9

10 Truss Bridge Alternative 2. Annual Maintenance is already an A amount so: A 1 = -$2,/year Truss Design: / / / / / / A. Maint. = $2,/yr -25M +(-15M) Paint: -4, Paint: -4, Paint: -4, Sandblast: - 19, n = Truss Bridge Alternative 3, A 2 : Annual Cost of Painting Truss Design: / / / / / / A. Maint. = $2,/yr -25M +(-15M) Use A/F,6%,3 Paint: -4, Paint: -4, Paint: -4, Sandblast: - 19, For any given cycle of painting compute: A 2 = -$4,(A/F,6%,3) = -$12,564/year n = 4-37 S-4251 konomi Teknik 4-38 S-4251 konomi Teknik Truss Bridge Alternative 3, A 3 Annual Cost of Sandblasting Truss Design: / / / / / / A. Maint. = $2,/yr -25M +(-15M) Use The A/F,6%,1 to convert to an equivalent $/year amount Paint: -4, Paint: -4, Paint: -4, Sandblast: - 19, For any given cycle of Sandblasting Compute A 3 = -$19,(A/F,6%,1) =-$14,421 n = Bridge Summary for CC(6%) CC 2 = (A 1 +A 2 +A 3 )/i CC 2 = -(2,+12,564+14,421)/.6 CC 2 $783,83 CC Total = CC 1 + CC 2 = million CC Suspension = -$52.71 million CCT russ million Select the Truss Design! 4-39 S-4251 konomi Teknik 4-4 S-4251 konomi Teknik 1

11 xercise: The municipal government of Bandung is considering two proposals to build new toll highway system. The first alternative calls for upgrading the existing system, which would cost Rp. 225,75 B for construction and additional Rp 21 M and Rp 25 M per year for maintenance and operation cost The other option is to build a new elevated highway that is estimated to cost Rp. 885 B for construction and annual maintenance cost of Rp 325 M f either alternatives would yield Rp 815 M revenue per year, and the interest rate is set at 8% p.a., which alternative should the government go about realizing the toll highway system? Homework #4 1. A manufacturing company is trying to decide among three different pieces of equipment that have the following characteristics: equipment A equipment B equipment C First cost Rp ,- Rp ,- Rp ,- Annual M&O cost Rp ,- Rp. 95..,- Rp. 75..,- Salvage value Rp ,- Rp. 25..,- Rp ,- Overhaul cost Rp ,- / 2 years Rp ,- / 3 years Rp ,- / 3 years useful life = 6 years and interest rate of 12% per year. 2. Which of these two machines that have the following costs is to be selected for a continuous production process, if the i = 15% p.a: machine X machine Y First cost Rp ,- Rp ,- Annual operating cost Rp ,- Rp ,- Salvage value Rp ,- Rp. 25..,- Life 5 years 3 years 3. Ganesha consulting firm is considering to build or lease an office space. For interest rate of 6% compounded semiannualy compare and select alternative. build own lease Construction cost Rp ,- - Lease cost - Rp ,- / 3 years Maintenance cost Rp. 11..,- / year Rp. 75..,- / year Period 3 years 4-41 S-4251 konomi Teknik 4-42 S-4251 konomi Teknik 11