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An international semiconductor company increases their productivity by an investment l in $. The purchased equipment has a salvage value of SV of the life time which is t. The increased productivity will create an annual Revenue of AW. Use a spreadsheet to evaluate the IRR. Is the investment a good one if you assume an MARR of 18% per year? Create a graph which shows the PW as a function of the interest rate. For your calculation use the following values: I = your student ID e.q. S00024356 means l = 24356 $. SV = I * 0.15 AW = I * 0.25 t = last digit of your ID e.q. 6. If the last digit is less than 5 use 5.
Expert Answer
| A | Initial Investment | 24356 | |||||||
| B=0.15*A | Salvage Value | 3653.4 | (24356*0.15) | ||||||
| C=0.25*A | Annual Revenue | 6089 | (24356*0.25) | ||||||
| D | number of years (t) | 6 | |||||||
| E | MARR | 0.18 | |||||||
| F=PV-A | PW of the project | ($1,705.76) | (Using excel PV function with rate=0.18, Nper=6, Pmt=6089,FV=3653.4) | ||||||
| VALUE OF PW AS A FUNCTION OF INTEREST RATE | |||||||||
| Interest Rate | PW | ||||||||
| 5% | $9,726 | ||||||||
| 10% | $4,225 | ||||||||
| 12% | $2,529 | ||||||||
| 15% | $267 | ||||||||
| 18% | ($1,706) | ||||||||
| 20% | ($2,883) | ||||||||
| 25% | ($5,976) | ||||||||
