today, an engineer deposited $28,000 into an account that pays interest at 8% per year compounded…

today, an engineer deposited $28,000 into an account that pays interest at 8% per year compounded semiannually. if there is no interperiod compounding and withdrawals of $1000 in months 2, 11, and 23 are already planned, what will be the future value at the end of 3 years? the future value would be $
Answer
Explanation:
Step1: Calculate the interest rate per period and number of periods
The annual interest rate $r = 8%=0.08$. Since it is compounded semi - annually, the interest rate per period $i=\frac{0.08}{2}=0.04$, and the number of periods $n = 3\times2=6$ for the initial deposit.
Step2: Calculate the future value of the initial deposit
The future - value formula for compound interest is $F = P(1 + i)^n$, where $P=$28000$. So, $F_1=28000\times(1 + 0.04)^6$. [F_1=28000\times1.04^6=28000\times1.265319018=35428.9325]
Step3: Calculate the future value of the first withdrawal
The first withdrawal of $P_1 = 1000$ is made at month 2. The number of months from month 2 to the end of 3 years (36 months) is $36 - 2=34$ months. In terms of semi - annual periods, $k_1=\frac{34}{6}\approx5.67$ periods. The future value of this withdrawal is $F_{w1}=1000\times(1 + 0.04)^{5.67}$. First, calculate the exponent value: $1.04^{5.67}\approx1.2519$. So, $F_{w1}=1000\times1.2519 = 1251.9$.
Step4: Calculate the future value of the second withdrawal
The second withdrawal of $P_2 = 1000$ is made at month 11. The number of months from month 11 to the end of 3 years is $36-11 = 25$ months. In terms of semi - annual periods, $k_2=\frac{25}{6}\approx4.17$ periods. The future value of this withdrawal is $F_{w2}=1000\times(1 + 0.04)^{4.17}$. Calculate $1.04^{4.17}\approx1.1827$. So, $F_{w2}=1000\times1.1827 = 1182.7$.
Step5: Calculate the future value of the third withdrawal
The third withdrawal of $P_3 = 1000$ is made at month 23. The number of months from month 23 to the end of 3 years is $36 - 23=13$ months. In terms of semi - annual periods, $k_3=\frac{13}{6}\approx2.17$ periods. The future value of this withdrawal is $F_{w3}=1000\times(1 + 0.04)^{2.17}$. Calculate $1.04^{2.17}\approx1.0907$. So, $F_{w3}=1000\times1.0907 = 1090.7$.
Step6: Calculate the net future value
The net future value $F_{net}=F_1-(F_{w1}+F_{w2}+F_{w3})$. [F_{net}=35428.9325-(1251.9 + 1182.7+1090.7)] [F_{net}=35428.9325 - 3525.3] [F_{net}=31903.6325\approx31903.63]
Answer:
$31903.63$