1. EML4550 2007 1 EML4550 - Engineering Design Methods Engineering-Economics Introduction Economic Decision Rules Hyman: Chapter 8

2. EML4550 -- 2007 What are we missing from previous economical analysis? Time value of money

3. EML4550 -- 2007 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. EML4550 -- 2007 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*[T P 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 =[T FP i,n ] Convert future transaction value (F) into the present value (P).

5. EML4550 -- 2007 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, 2 nd period U+U(1+i) and so on..  F n : future worth after n periods

6. EML4550 -- 2007 Transformation Table [P/F] [P/U] [P/A] [P/G] [U/F] [U/G]

7. EML4550 -- 2007 Summary of NPV FPFP Initial cost3020 Rebuilding (3 rd year)31.73 Salvage (5 th year)-4-1.61 Maintenance (every year for 5 years)1?2? Production benefits (5 years)-0.5? Electricity (5 years)3 (+5%) 3.5 (+5%) TOTAL (NPV) Ajax Blaylocki = 20%

8. EML4550 -- 2007 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 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 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 P/U (20%, 5 yrs) = 2.99  1x2.99 and 2x2.99=5.98  Benefits  Same P/U as maintenance

9. EML4550 -- 2007 Summary of NPV FPFP Initial cost3020 Rebuilding31.73 Salvage-4-1.61 Maintenance12.9925.98 Production benefits-0.5-1.5 Electricity3 (+5%) ?3.5 (+5%) ? TOTAL (NPV) Ajax Blaylocki = 20%

10. EML4550 -- 2007 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. EML4550 -- 2007 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. EML4550 -- 2007 Summary of NPV FPFP Initial cost3020 Rebuilding31.73 Salvage-4-1.61 Maintenance12.9925.98 Production benefits-0.5-1.5 Electricity3 (+5%) 9.753.5 (+5%) 11.38 TOTAL (NPV)39.6339.09 Ajax Blaylocki = 20%

13. EML4550 -- 2007 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. EML4550 -- 2007 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 T PU,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. EML4550 -- 2007 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. EML4550 -- 2007 Summary of Annual Cost PUPU Initial cost3010.02206.68 Rebuilding30.58 Salvage-4? Maintenance12 Production benefits-0.5 Electricity3 (+5%) 3.5 (+5%) TOTAL (Annual Cost) Ajax Blaylocki = 20%

17. EML4550 -- 2007 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. EML4550 -- 2007 Summary of Annual Cost PUPU Initial cost3010.02206.68 Rebuilding30.58 Salvage-4-0.54 Maintenance12 Production benefits-0.5 Electricity3 (+5%) ?3.5 (+5%) ? TOTAL (Annual Cost) Ajax Blaylocki = 20%

19. EML4550 -- 2007 Annual cost calculations (Cont’d)  Maintenance  Already an annual amount  Benefits  Already an annual amount

20. EML4550 -- 2007 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. EML4550 -- 2007 Annual cost calculations (Cont’d)  U/G(20%, 5%, 5yrs) = 1.087  3x1.086 = 3.26, 3.5x1.086 = 3.8

22. EML4550 -- 2007 Summary of Annual Cost PUPU Initial cost3010.02206.68 Rebuilding30.58 Salvage-4-0.54 Maintenance12 Production benefits-0.5 Electricity3 (+5%) 3.263.5 (+5%) 3.80 TOTAL (Annual Cost)13.2413.06 AjaxBlaylocki = 20%

23. EML4550 -- 2007 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