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EML4550 - Engineering Design Methods

EML4550 - Engineering Design Methods. Engineering-Economics Introduction Economic Decision Rules. Hyman: Chapter 8. What are we missing from previous economical analysis?. Time value of money. Criterion 5: Present value analysis (NPV).

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EML4550 - Engineering Design Methods

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  1. EML4550 - Engineering Design Methods Engineering-Economics Introduction Economic Decision Rules Hyman: Chapter 8

  2. What are we missing from previous economical analysis? Time value of money

  3. Criterion 5: Present value analysis (NPV) • How do we account for the fact all expenditures/incomes occur at different times? • Time value of money • ‘Bring to the present’ of all costs • Assume a hurdle rate of i = 20% (high?) and rework example

  4. Present Worth Factor • Convert a future transaction F into an equivalent present value P by considering the accumulation of annual rate of return (interest rate) over a certain time period. • For n years (or n periods): F=P(1+i)n=P*[F/P]=P*[TP F,i,n] • Present value can also be expressed knowing the future transaction: P=F(1+i)-n. That is, how much investment (P) should be made now in order to achieve the future value (F) in a period of n years given an annual rate of return of i. • Present worth factor: [P/F]=(1+i)-n=[TFP i,n] Convert future transaction value (F) into the present value (P).

  5. Uniform Series Present Worth Factor • An “equal” amount (U) is invested periodically to the product, therefore, both the capital and interest accumulated contribute to the total expenses of the product. Examples: maintenance and productivity benefits. • U: uniform investment each period: first period U, 2nd period U+U(1+i) and so on.. • Fn: future worth after n periods

  6. [P/F] [P/U] [P/A] [P/G] [U/F] [U/G] Transformation Table

  7. Summary of NPV Ajax i = 20% Blaylock

  8. NPV calculations • Initial cost • Already ‘present’ amounts, no need to account for interest • Rebuilding • P/F (20%, 3 yrs) = 1/(1+0.2)3 = 0.58 • PV-Rebuild = 3x0.58 = 1.73 • Salvage • P/F (20%, 5 yrs) = 1/(1+0.2)5 = 0.402 • PV-Salvage = 4x0.402 = 1.61 • Maintenance • P/U (20%, 5 yrs) = 2.99 • 1x2.99 and 2x2.99=5.98 • Benefits • Same P/U as maintenance

  9. Summary of NPV Ajax i = 20% Blaylock

  10. NPV Calculation (Geometric series present worth factor) • Electricity • Cannot use P/U because amounts are not uniform (growing with inflation by a given percentage e)

  11. NPV of geometric series of amounts • Electricity • i=20%, e=5% inflation • Ajax: $3k*3.25=$9.75k; Blaylock: $3.5k*3.25=$11.375k

  12. Summary of NPV Ajax i = 20% Blaylock

  13. Conclusion of NPV analysis • For i = 0%, A is a better option (no time value of money) • For i = 20% B is better (high time value) but very close • For what “i” are A and B indistinct? (internal rate of return (IRR), that is the discount rate by setting the overall project gains/losses at zero)

  14. Criterion 6: Annualized Costs(Ownership and Operation) • Instead of converting all amounts to a present worth (P), one can convert all amounts to an annual cost (U) • From lump sums, to series of cash flows • Capital Recovery Factor: uniform series present worth factor TPU,i,n converts U into P. The reverse of the factor is the capital recovery factor (the ratio of a constant annuity to the present value of receiving that annuity for a given length of time. Distribute the present cost over the given time.) [U/P]

  15. Annual cost calculations • Initial cost • U/P(20%, 5yrs) = 0.334 • 30x0.334 = 10.02, 20x0.334 = 6.68 • Rebuilding • P = P/F(20%, 3 yrs) = 1/(1+0.2)3 = 0.58 • PV-Rebuild = $3kx0.58 = $1.73k That is: one needs to invest $1.73k now in order to obtain $1.73k(1+0.2)3=$3k to rebuild the motor • U/P(20%, 5yrs) = 0.334 (the same as before for the initial cost) • A-Rebuild = $1.73kx0.334 = $0.58k (rebuilding cost distributed annually)

  16. Summary of Annual Cost Ajax i = 20% Blaylock

  17. Annual cost calculations • Sinking Fund Factor • Convert the future value to the present value: • Use capital recovery factor to convert P into U: • Salvage • U/F(20%, 5 yrs) = 0.134 • -$4k x 0.134 = -$0.54k

  18. Summary of Annual Cost Ajax i = 20% Blaylock

  19. Annual cost calculations (Cont’d) • Maintenance • Already an annual amount • Benefits • Already an annual amount

  20. Annual cost calculations (Cont’d) • Electricity • Escalating amount, how to transform to a uniform series of amounts? • U/G ? (P/G) then (U/P)

  21. Annual cost calculations (Cont’d) • U/G(20%, 5%, 5yrs) = 1.087 • 3x1.086 = 3.26, 3.5x1.086 = 3.8

  22. Summary of Annual Cost i = 20% Ajax Blaylock

  23. Conclusions of Annualized Cost Analysis • Blaylock is still better than Ajax • Same result as in NPV, as it should be, we did the same comparison but reducing all amounts to different basis

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