In: Finance
6. Calculate the value of the cash flows at time 6. The interest rate is 9.9%.
0 |
1 |
2 |
$1,120 |
$2,163 |
$3,290 |
7. In 1998, the average price of a gallon of gas was $1.01. Today, the average price of a gallon of gas is $2.87. At what annual rate has a gallon of gas increased over the last 20 years? (Answer as a percent. Enter only numbers and decimals in your response. Round to 2 decimal places.)
8. You are buying an investment property for $393,580 today. Through research, you have determined that the value of the property is likely to grow at a rate of 7.16% per year, on average. How long until the value of the property grows to $546,287?
9. Your goal is to have $76,386. If you can earn 10.78% per year and you invest $19,812 today, how many years until you reach your goal?
10. You are looking at an investment that will pay you $22,339 in year 2, $43,709 in year 4 and $45,031 in year 6. If your required return is 8.16%, what is the most you should pay for the investment? (In other words, how much is the project worth today?)
Solution 6 | ||||
Year | Cashflow | FV factor, (1+9.90%)^(6-t) | Cashflow*FV factor | |
0 | $ 1,120.00 | 1.761919875 | $ 1,973.35 | |
1 | $ 2,163.00 | 1.603202798 | $ 3,467.73 | |
2 | $ 3,290.00 | 1.458783256 | $ 4,799.40 | |
Total | $10,240.47 | |||
Solution 7 | ||||
1998 price | 1.01 | |||
Price after 20 years | 2.87 | |||
Time period | 20 | |||
Price after 20 years= | Price in 1998*(1+r)^time | |||
2.87= | 1.1*(1+r)^20 | |||
2.87/1.1= | (1+r)^20 | |||
2.609090909 | =(1+r)^20 | |||
(1+r)= | (2.60909090909091)^(1/20) | |||
(1+r)= | 1.0491183 | |||
R= | 0.0491183 | |||
R= | 4.91% | |||
Solution 8 | ||||
Investment | $ 393,580 | |||
Amount | $ 546,287 | |||
Period in years | T | |||
Interest | 7.16% | |||
Amount= | Investment* (1+Interest)^Time | |||
546287= | 393580*(1+7.16%)^T | |||
546287/393580= | (1.0716)^T | |||
1.387994817 | (1.0716)^T | |||
Log 1.3879948168098= | T Log 1.0716 | |||
0.142387844335619= | T*0.0300327049361713 | |||
T= | 0.142387844335619/0.0300327049361713 | |||
T= | 4.74 | Years | ||
Solution 9 | ||||
Investment | $ 19,812 | |||
Amount | $ 76,386 | |||
Period in years | T | |||
Interest | 10.78% | |||
Amount= | Investment* (1+Interest)^Time | |||
76386= | 19812*(1+10.78%)^T | |||
76386/19812= | (1.1078)^T | |||
3.85554209569958= | (1.0716)^T | |||
Log 3.85554209569958= | T Log 1.1078 | |||
0.586085449 | T*0.0444613608106648 | |||
T= | 0.586085449214303/0.0444613608106648 | |||
T= | 13.18 | Years | ||
Solution 10 | ||||
Year | Cashflow | PV factor, 1/(1+8.16%)^(t) | Cashflow*PV factor | |
2 | $22,339.00 | 0.854804191 | $19,095.47 | |
4 | $43,709.00 | 0.730690205 | $31,937.74 | |
6 | $45,031.00 | 0.62459705 | $28,126.23 | |
Present worth | $79,159.44 | |||