Using Excel for Economic life analysis
Year | Discount factor | O&M cost | PV (O&M) | Cumulative (O&M) | Cumulative (O&M) + Initial Cost | Salvage value | PV (Salvage value) | NPV | (A/P,15%,n) | EUAC |
A | B | C | D=C*B | E | F=E+9000 | G | H=G*B | I=F-H | J | K = I*J |
1 | 0.86957 | 8600.00 | 7478.26 | 7478.26 | 16478.26 | 6000.00 | 5217.39 | 11260.87 | 1.15000 | 12950 |
2 | 0.75614 | 9600.00 | 7258.98 | 14737.24 | 23737.24 | 4500.00 | 3402.65 | 20334.59 | 0.61512 | 12508 |
3 | 0.65752 | 10600.00 | 6969.67 | 21706.91 | 30706.91 | 3375.00 | 2219.12 | 28487.79 | 0.43798 | 12477 |
4 | 0.57175 | 11600.00 | 6632.34 | 28339.25 | 37339.25 | 2531.25 | 1447.25 | 35892.00 | 0.35027 | 12572 |
5 | 0.49718 | 12600.00 | 6264.43 | 34603.68 | 43603.68 | 1898.44 | 943.86 | 42659.82 | 0.29832 | 12726 |
6 | 0.43233 | 13600.00 | 5879.66 | 40483.33 | 49483.33 | 1423.83 | 615.56 | 48867.77 | 0.26424 | 12913 |
7 | 0.37594 | 14600.00 | 5488.68 | 45972.01 | 54972.01 | 1067.87 | 401.45 | 54570.56 | 0.24036 | 13117 |
Discount factor | 1/(1+0.15)^n | |||||||||
(A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) |
As Minimum EUAC = 12477, Economic life is 3 yrs
it lies between 12458 - 12498
option B is the correct answer
Showing formula in Excel
Year | Discount factor | O&M cost | PV (O&M) | Cumulative (O&M) | Cumulative (O&M) + Initial Cost | Salvage value | PV (Salvage value) | NPV | (A/P,15%,n) | EUAC |
A | B | C | D=C*B | E | F=E+9000 | G | H=G*B | I=F-H | J | K = I*J |
1 | =1/(1.15)^A127 | 8600 | =C127*B127 | =D127 | =9000+E127 | =8000*0.75 | =G127*B127 | =F127-H127 | =0.15*((1 + 0.15)^A127)/((1 + 0.15)^A127-1) | =I127*J127 |
2 | =1/(1.15)^A128 | =C127+1000 | =C128*B128 | =E127+D128 | =9000+E128 | =G127*0.75 | =G128*B128 | =F128-H128 | =0.15*((1 + 0.15)^A128)/((1 + 0.15)^A128-1) | =I128*J128 |
3 | =1/(1.15)^A129 | =C128+1000 | =C129*B129 | =E128+D129 | =9000+E129 | =G128*0.75 | =G129*B129 | =F129-H129 | =0.15*((1 + 0.15)^A129)/((1 + 0.15)^A129-1) | =I129*J129 |
4 | =1/(1.15)^A130 | =C129+1000 | =C130*B130 | =E129+D130 | =9000+E130 | =G129*0.75 | =G130*B130 | =F130-H130 | =0.15*((1 + 0.15)^A130)/((1 + 0.15)^A130-1) | =I130*J130 |
5 | =1/(1.15)^A131 | =C130+1000 | =C131*B131 | =E130+D131 | =9000+E131 | =G130*0.75 | =G131*B131 | =F131-H131 | =0.15*((1 + 0.15)^A131)/((1 + 0.15)^A131-1) | =I131*J131 |
6 | =1/(1.15)^A132 | =C131+1000 | =C132*B132 | =E131+D132 | =9000+E132 | =G131*0.75 | =G132*B132 | =F132-H132 | =0.15*((1 + 0.15)^A132)/((1 + 0.15)^A132-1) | =I132*J132 |
7 | =1/(1.15)^A133 | =C132+1000 | =C133*B133 | =E132+D133 | =9000+E133 | =G132*0.75 | =G133*B133 | =F133-H133 | =0.15*((1 + 0.15)^A133)/((1 + 0.15)^A133-1) | =I133*J133 |
Discount factor | 1/(1+0.15)^n | |||||||||
(A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) |
Question 1 (1 point) A 5 year-old tooling kit that was purchased new for $50,000 and...
1. Assume that your private university's tuition is $28,000 a) If inflation rate for the tuition is 5% per year calculate what the tuition will cost 20 years from now? b) If general inflation rate for the economy is 3% per year, express that future tuition 20 years from now in real dollars, based on today's rate. 2. The local gas company has asked the planning department of your company to help them plan for future cash flows. They want...
CASE 1-5 Financial Statement Ratio Computation Refer to Campbell Soup Company's financial Campbell Soup statements in Appendix A. Required: Compute the following ratios for Year 11. Liquidity ratios: Asset utilization ratios:* a. Current ratio n. Cash turnover b. Acid-test ratio 0. Accounts receivable turnover c. Days to sell inventory p. Inventory turnover d. Collection period 4. Working capital turnover Capital structure and solvency ratios: 1. Fixed assets turnover e. Total debt to total equity s. Total assets turnover f. Long-term...