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Conceptual Stock Valuation

Conceptual Stock Valuation. Given : Stock price as of Oct 4, 2005 : $105.50/share Earnings growth for next 5 years : 13% Expected cash dividend in 2006 : $2.00/share Expected stock price in 3 years : $230/share Required return on your investment : 10% Find : Current value of stock.

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Conceptual Stock Valuation

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  1. Conceptual Stock Valuation Given: Stock price as of Oct 4, 2005: $105.50/share Earnings growth for next 5 years: 13% Expected cash dividend in 2006: $2.00/share Expected stock price in 3 years: $230/share Required return on your investment: 10% Find: Current value of stock (c) 2001 Contemporary Engineering Economics

  2. Evaluating Business and Engineering Assets Present Worth Analysis (c) 2001 Contemporary Engineering Economics

  3. Common measures on cash flow equivalence • Present Worth (PW) or Present Value (PV) • measure of future cash flow relative to value of • the cash flows at the end of year 0 • PW = CF0+ CF1(P/F,i%, 1)+CF2(P/F, i%,2)+..+CFn(P/F,i%,n) • Future Worth (FW) is a measure of cash flow at • some future planning horizon (at the end of year n.) • FW = CF0 (F/P,i%, n)+CF1(F/P, i%,n-1)+CF2(F/P,i%,n-2)+… • …+CFn-1(F/P,i%)+CFn • Equivalent Annual Worth(AE)- measure of cash flow • in terms of equal payments made on annual basis (c) 2001 Contemporary Engineering Economics

  4. Engineering Economic Decisions • Equipment and Process election • Equipment Replacement • New Product and Product Expansion • Cost Reduction • Service Improvement • investment at the beginning • of the project • stream of cash benefits in • future years (c) 2001 Contemporary Engineering Economics

  5. Loan Customer Bank Repayment Bank Loan vs. Investment Project • Bank Loan Loan cash flow • Investment Project Investment Project Company Return Project cash flow (c) 2001 Contemporary Engineering Economics

  6. Process Plant A used 40% of the time(operating hrs) • to produce X product and 60% of the time to • produce Y product • Annual Production of X product = 30,000kg @$15/kg • The proposed computer cost = $650,000 • Maintenance cost= $53,000/yr • Benefits to product X : • selling price could be increased by $2/kg • production volumes will increase by 4kg/yr • labor cost saving $25/hr • useful life of 8 yrs • Find the Net Cash Flow in each year over the life of the new • System. (c) 2001 Contemporary Engineering Economics

  7. Describing Project Cash Flows (c) 2001 Contemporary Engineering Economics

  8. Mutually Exclusive Alternatives • One of the important functions of financial management and engineering is the creation of “alternatives”. • If there are no alternatives to consider then there really is no problem to solve! • Given a set of “feasible” alternatives, engineering economy attempts to identify the “best” economic approach to a given problem. (c) 2001 Contemporary Engineering Economics

  9. Evaluating Alternatives • Part of Engineering Economy is the selection and execution of the best alternative from among a set of feasible alternatives. • Alternatives must be generated from within the organization. • In part, the role of the engineer. • Feasible alternatives to solve a specific problem/concern • Feasible means “viable” or “do-able” (c) 2001 Contemporary Engineering Economics

  10. Projects to Alternatives Ideas, Data, Experience, Plans, And Estimates • Creation of Alternatives. Generation of Proposals P1 P2 Pn (c) 2001 Contemporary Engineering Economics

  11. Projects to Alternatives • Proposal Assessment. P1 Pj Economic Analysis And Assessment P2 Pk . . Infeasible or Rejected! Pn Viable POK POK (c) 2001 Contemporary Engineering Economics

  12. Assessing Alternatives Economic Analysis And Assessment • Feasible Alternatives . . POK POK Feasible Set Mutually Exclusive Set OR Independent Set (c) 2001 Contemporary Engineering Economics

  13. Mutually Exclusive Set • Mutually Exclusive (ME) Set • Only one of the feasible (viable) projects can be selected from the set. • Once selected, the others in the set are “excluded”. • Each of the identified feasible (viable) projects is (are) considered an “alternative”. • It is assumed the set is comprised of “do-able”, feasible alternatives. • ME alternative compete with each other! (c) 2001 Contemporary Engineering Economics

  14. Initial Project Screening method: Payback Period • Principle: • How fast can I recover my initial investment? • Method: • Based on cumulative cash flow (or accounting profit) • Screening Guideline: • If the payback period is less than or equal to some specified payback period, the project would be considered for further analysis. • Weakness: • Does not consider the time value of money (c) 2001 Contemporary Engineering Economics

  15. Example 7.3 Payback Period N Cash Flow Cum. Flow 0 1 2 3 4 5 6 -$85,000 -$50,000 -$5,000 $45,000 $95,000 $140,000 $175,000 -$105,000+$20,000 $35,000 $45,000 $50,000 $50,000 $45,000 $35,000 Payback period should occurs somewhere between N = 2 and N = 3. (c) 2001 Contemporary Engineering Economics

  16. $45,000 $45,000 $35,000 $35,000 $25,000 $15,000 Annual cash flow 0 1 2 3 4 5 6 Years $85,000 150,000 3.2 years Payback period 100,000 50,000 Cumulative cash flow ($) 0 -50,000 -100,000 0 1 2 4 5 6 3 (c) 2001 Contemporary Engineering Economics Years (n)

  17. Discounted Payback Period Calculation (c) 2001 Contemporary Engineering Economics

  18. Payback Period Analysis-RULE • Never use PBn at the primary means of making an accept-reject decision on an alternative! • Often used as a screening technique or preliminary analysis tool. • Historically, this method was a primary analysis tool and often resulted in incorrect selections. • To apply, the cash flows must have at least one (+) cash flow in the sequence. (c) 2001 Contemporary Engineering Economics

  19. Present Worth Method A process of obtaining the equivalent worth of future cash flows BACK too some point in time. – called the Present Worth Method. At an interest rate usually equal to or greater than the Organization’s established MARR. (c) 2001 Contemporary Engineering Economics

  20. PW(i) > 0 Net Present Worth Measure • Principle: Compute the equivalent net surplus at n = 0 for a given interest rate of i. • Decision Rule: Accept the project if the net surplus is positive. Inflow 0 1 2 3 4 5 Outflow Net surplus PW(i) inflow 0 PW(i) outflow (c) 2001 Contemporary Engineering Economics

  21. Example 7.5 - Tiger Machine Tool Company inflow $55,760 $27,340 $24,400 0 3 1 2 outflow $75,000 (c) 2001 Contemporary Engineering Economics

  22. Present Worth Amounts at Varying Interest Rates (c) 2001 Contemporary Engineering Economics *Break even interest rate

  23. Present Worth Profile 40 Reject Accept 30 20 Break even interest rate (or rate of return) 10 $3553 17.45% 0 PW (i) ($ thousands) -10 -20 -30 0 5 10 15 20 25 30 35 40 i = MARR (%) (c) 2001 Contemporary Engineering Economics

  24. Future Worth Criterion • Given: Cash flows and MARR (i) • Find: The net equivalent worth at the end of project life $55,760 $27,340 $24,400 0 3 1 2 $75,000 Project life (c) 2001 Contemporary Engineering Economics

  25. Future Worth Criterion (c) 2001 Contemporary Engineering Economics

  26. Project Balance Concept N 0 1 2 3 Beginning Balance Interest Payment Project Balance -$75,000 -$11,250 +$24,400 -$61,850 -$61,850 -$9,278 +$27,340 -$43,788 -$43,788 -$6,568 +$55,760 +$5,404 -$75,000 -$75,000 Net future worth, FW(15%) PW(15%) = $5,404 (P/F, 15%, 3) = $3,553 (c) 2001 Contemporary Engineering Economics

  27. Project Balance Diagram 60,000 40,000 20,000 0 -20,000 -40,000 -60,000 -80,000 -100,000 -120,000 Terminal project balance (net future worth, or project surplus) $5,404 Discounted payback period Project balance ($) -$43,788 -$61,850 -$75,000 0 1 2 3 Year(n) (c) 2001 Contemporary Engineering Economics

  28. Capitalized Equivalent Worth • Principle: PW for a project with an annual receipt of A over infinite service life • Equation: • CE(i) = A(P/A, i, ) = A/i A 0 P = CE(i) (c) 2001 Contemporary Engineering Economics

  29. Given: i = 10%, N = Find: P or CE (10%) $2,000 $1,000 0 10 P = CE (10%) = ? (c) 2001 Contemporary Engineering Economics

  30. Example 7.9 Mr. Bracewell’s Investment Problem • Built a hydroelectric plant using his personal savings of $800,000 • Power generating capacity of 6 million kwhs • Estimated annual power sales after taxes - $120,000 • Expected service life of 50 years • Was Bracewell's $800,000 investment a wise one? • How long does he have to wait to recover his initial investment, and will he ever make a profit? (c) 2001 Contemporary Engineering Economics

  31. Mr. Brcewell’s Hydro Project (c) 2001 Contemporary Engineering Economics

  32. Equivalent Worth at Plant Operation • Equivalent lump sum investment • V1 = $50K(F/P, 8%, 9) + $50K(F/P, 8%, 8) + • . . . + $100K(F/P, 8%, 1) + $60K • = $1,101K • Equivalent lump sum benefits • V2 = $120(P/A, 8%, 50) • = $1,460K • Equivalent net worth • FW(8%) = V1 - V2 • = $367K > 0, Good Investment (c) 2001 Contemporary Engineering Economics

  33. With an Infinite Project Life • Equivalent lump sum investment • V1 = $50K(F/P, 8%, 9) + $50K(F/P, 8%, 8) + • . . . + $100K(F/P, 8%, 1) + $60K • = $1,101K • Equivalent lump sum benefits assuming N = • V2 = $120(P/A, 8%,  ) • = $120/0.08 • = $1,500K • Equivalent net worth • FW(8%) = V1 - V2 • = $399K > 0 • Difference = $32,000 (c) 2001 Contemporary Engineering Economics

  34. Problem 7.27 - Bridge Construction • Construction cost = $2,000,000 • Annual Maintenance cost = $50,000 • Renovation cost = $500,000 every 15 years • Planning horizon = infinite period • Interest rate = 5% (c) 2001 Contemporary Engineering Economics

  35. 15 30 45 60 0 $50,000 $500,000 $500,000 $500,000 $500,000 $2,000,000 (c) 2001 Contemporary Engineering Economics

  36. Solution: • Construction Cost • P1 = $2,000,000 • Maintenance Costs • P2 = $50,000/0.05 = $1,000,000 • Renovation Costs • P3 = $500,000(P/F, 5%, 15) • + $500,000(P/F, 5%, 30) • + $500,000(P/F, 5%, 45) • + $500,000(P/F, 5%, 60) • . • = {$500,000(A/F, 5%, 15)}/0.05 • = $463,423 • Total Present Worth • P = P1 + P2 + P3 = $3,463,423 (c) 2001 Contemporary Engineering Economics

  37. 30 45 60 0 15 Alternate way to calculate P3 • Concept: Find the effective interest rate per payment period $500,000 $500,000 $500,000 $500,000 • Effective interest rate for a 15-year cycle • i = (1 + 0.05)15 - 1 = 107.893% • Capitalized equivalent worth • P3 = $500,000/1.07893 • = $463,423 (c) 2001 Contemporary Engineering Economics

  38. Comparing Mutually Exclusive Projects • Mutually Exclusive Projects • Alternative vs. Project • Do-Nothing Alternative (c) 2001 Contemporary Engineering Economics

  39. Revenue Projects • Projects whose revenues depend on the choice of alternatives • Service Projects • Projects whose revenues do not depend on the choice of alternative (c) 2001 Contemporary Engineering Economics

  40. Analysis Period • The time span over which the economic effects of an investment will be evaluated (study period or planning horizon). • Required Service Period • The time span over which the service of an equipment (or investment) will be needed. (c) 2001 Contemporary Engineering Economics

  41. Comparing Mutually Exclusive Projects • Principle: Projects must be compared over an equal time span. • Rule of Thumb: If the required service period is given, the analysis period should be the same as the required service period. (c) 2001 Contemporary Engineering Economics

  42. How to Choose An Analysis Period Analysis period equals project lives Case 1 Analysis period is shorter than project lives Case 2 Analysis = Required period service period Finite Analysis period is longer than project lives Case 3 Case 4 Analysis period is longest of project life in the group Required service period Analysis period is lowest common multiple of project lives Project repeatability likely Infinite Analysis period equals one of the project lives Project repeatability unlikely (c) 2001 Contemporary Engineering Economics

  43. PW (10%) = $283 PW (10%) = $579 A B Case 1: Analysis Period Equals Project Lives Compute the PW for each project over its life $2,110 $2,075 $600 $500 $1,400 $450 0 A B $1,000 $4,000 (c) 2001 Contemporary Engineering Economics

  44. Comparing projects requiring different levels of investment – Assume that the unused funds will be invested at MARR. $600 $450 $500 $2,110 3,993 Project A $2,075 $1,000 $1,400 $600 $450 $500 Project B Modified Project A $4,000 $1,000 $3,000 This portion of investment will earn 10% return on investment. PW(10%)A = $283 PW(10%)B = $579 (c) 2001 Contemporary Engineering Economics

  45. Case 2: Analysis Period Shorter than Project Lives • Estimate the salvage value at the end of • required service period. • Compute the PW for each project over • the required service period. (c) 2001 Contemporary Engineering Economics

  46. Comparison of unequal-lived service projects when the required service period is shorter than the individual project life (c) 2001 Contemporary Engineering Economics

  47. Case 3: Analysis Period Longer than Project Lives • Come up with replacement projects that • match or exceed the required service • period. • Compute the PW for each project over • the required service period. (c) 2001 Contemporary Engineering Economics

  48. Comparison for Service Projects with Unequal Lives when the required service period is longer than the individual project life (c) 2001 Contemporary Engineering Economics

  49. Case 4: Analysis Period is Not Specified • Project Repeatability Unlikely • Use common service (revenue) period. • Project Repeatability Likely • Use the lowest common multiple of project lives. (c) 2001 Contemporary Engineering Economics

  50. Project Repeatability Unlikely PW(15%)drill = $2,208,470 Assume no revenues PW(15%)lease = $2,180,210 (c) 2001 Contemporary Engineering Economics

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