1. EML4550 2007 1 EML4550 - Engineering Design Methods Engineering-Economics Hyman: Chapter 8

2. EML4550 -- 2007 Economic Decision Rules (from the simplest to the comprehensive) nSimple criteria lLowest initial cost (capital cost) lLowest life-cycle cost (no time value of money) lAverage annual rate of return lPayback period nTime value of money? lPresent value lAnnualized cost lUnequal lifetimes analysis §Common multiple of lifetimes §Annualized cost of ownership and operation

3. EML4550 -- 2007 Unequal lifetimes nWhat happens if A and B had different lifetimes? nSuppose now that Ajax lasts 6 years (with salvage value of $4k), and Blaylock only lasts 4 years (with no salvage value). Also, neglect the productivity increase with Ajax. nRecalculate the Net Present Value (NPV) for each option

4. EML4550 -- 2007 Summary of NPV F or AP P Initial cost3020 Salvage-4-1.34 =-4*P/F(20%,6) Maintenance13.33 =1*P/U(20%,6) 25.18 =2*P/U(20%,4) Electricity3(+5%)11.02 =3*P/G(20%,5%,6) 3.5 (+5%) 9.86 =3.5*P/G(20%,5%,4) TOTAL (NPV)43.0134.84 Ajax (n=6)Blaylock (n=4)i = 20% P/F(20%,6)=(1+0.2) -6 =0.335; P/U(20%,6)=3.326; P/U(20%,4)=2.589 P/G(20%,5%,6)=3.675; P/G(20%,5%,4)=2.759

5. EML4550 -- 2007 Conclusions of NPV for unequal lifetimes nB is better (lowest cost) nBut this is unfair! lPenalize A for lasting longer nHow to solve? lBase analysis over 4 years §No, “A” would have 2 years left lBase analysis on 6 years §No, need to replace B after 4 years and then B would have 2 years left lCommon multiple of lifetime

6. EML4550 -- 2007 Common multiple of lifetimes = 12 years F or AP P Initial cost (n=0)3020 Initial cost (n=4)20 (2 nd B)9.64 Initial cost (n=6)30 (2 nd A)10.05 Initial cost (n=8)20 (3 rd B)4.65 Salvage (n=6)-4-1.34 Salvage (n=12)-4-0.45 Maintenance14.4428.88 Electricity3(+5%)15.973.5(+5%)18.63 TOTAL (NPV)58.6761.80 Ajax (n=12)Blaylock (n=12)i = 20% P/F(20%,4)=0.482 (for 2 nd B), P/F(20%,6)=0.335 (for 2 nd A and 1 st A salvage), P/F(20%,8)=0.233 (for 3 rd B), P/F(20%,12)=0.112 (for 2 nd A salvage) P/U(20%,12)=4.44 for maintenance, P/G(20%,5%,12)=5.324 for electricity

7. EML4550 -- 2007 Conclusions nNo unexpended life at the end of the period nOption A is better

8. EML4550 -- 2007 Annualized cost of ownership and operation F or PU U Initial cost309.02207.73 Salvage-4-0.40 Maintenance12 Electricity3(+5%)3.323.5(+5%)3.73 TOTAL (U)12.9413.46 Ajax (n=6)Blaylock (n=4)i = 20% U/P(20%,6)=0.301, U/P(20%,4)=0.386, U/F(20%,6)=0.101 U/G(20%,5%,6)=1.105, U/G(20%,5%,4)=1.066

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