In: Economics
Please assist if familiar with economics. thanks.
An equipment has an initial cost of $20,000 and a useful life of 7 years. Each year the equipment decreases in value by 15%. The operating and maintenance cost is $1200 the first year and increases thereafter by $1000. Additionally, it has a breakdown cost of $2500 the first year and increases by $1500 henceforth. Create a table to display all necessary costs and show the annual marginal cost. Interest rate is 10%.
Marginal cost = Loss in market value + Forgone interest + O&M cost + Breakdown cost
Year | Salvage | Loss in market value | Forgone interest | O&M cost | Breakdown | Total MC |
0 | 20000.00 | |||||
1 | 17000.00 | 3000.00 | 2000.00 | 1200.00 | 2500.00 | 8700.00 |
2 | 14450.00 | 2550.00 | 1700.00 | 2200.00 | 4000.00 | 10450.00 |
3 | 12282.50 | 2167.50 | 1445.00 | 3200.00 | 5500.00 | 12312.50 |
4 | 10440.13 | 1842.38 | 1228.25 | 4200.00 | 7000.00 | 14270.63 |
5 | 8874.11 | 1566.02 | 1044.01 | 5200.00 | 8500.00 | 16310.03 |
6 | 7542.99 | 1331.12 | 887.41 | 6200.00 | 10000.00 | 18418.53 |
7 | 6411.54 | 1131.45 | 754.30 | 7200.00 | 11500.00 | 20585.75 |
Showing work in Excel
Year | Salvage | Loss in market value | Forgone interest | O&M cost | Breakdown | Total MC |
0 | 20000 | |||||
1 | =20000*0.85 | =O13-O14 | =O13*0.1 | =1200 | 2500 | =P14+Q14+R14+S14 |
2 | =O14*0.85 | =O14-O15 | =O14*0.1 | =R14+1000 | =S14+1500 | =P15+Q15+R15+S15 |
3 | =O15*0.85 | =O15-O16 | =O15*0.1 | =R15+1000 | =S15+1500 | =P16+Q16+R16+S16 |
4 | =O16*0.85 | =O16-O17 | =O16*0.1 | =R16+1000 | =S16+1500 | =P17+Q17+R17+S17 |
5 | =O17*0.85 | =O17-O18 | =O17*0.1 | =R17+1000 | =S17+1500 | =P18+Q18+R18+S18 |
6 | =O18*0.85 | =O18-O19 | =O18*0.1 | =R18+1000 | =S18+1500 | =P19+Q19+R19+S19 |
7 | =O19*0.85 | =O19-O20 | =O19*0.1 | =R19+1000 | =S19+1500 | =P20+Q20+R20+S20 |