John wants to save for his retirement, which will occur 30 years from now, by depositing monthly into an accou

John wants to save for his retirement, which will occur 30 years from now, by depositing monthly into an account earning 8% interest compounded monthly. His goal is to be able to draw $30,000 each year during a 40 year retirement. How much should John save each month? a.$200.00 b.$220.03 c.$240.03 d.$260.03

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1. first detirmine the value of the account needed at 30 years to fund retirement for 40 years P at 70 = 0.00 P at 30 = $359,600.00 P at 0 = 0, and monthly deposits of 240.03 i wish i could give you the equations for these answers, but they are on my calculator. they are P given A, and A given F using the website below, you can solve both equations, and it gives you the equations.
2. The formula that represents the nest egg developed over 30 years by making a payment n at an 8% interest compounded monthly: nest egg = n * integral of e^(k*t) dt from t1=1 month to t2 = 12*30 = 360 months with k = 8%/12 = .08/12 the integral of this is n *(1/k)*(e^(k*t2) - e^(k*t1)) nest egg = n * (12/.08) * ( e^(.08*360/12) - e^(.08*1/12)) = n * 1502.473 If the $30000/yr is withdrawn at a monthly rate of $2500 over 40 years with nest egg earning 8% interest, then: 1st month nest egg * (1+.08/12) - 2500*(1+.08/12) 2nd month nest egg * (1 + .08/12)^2 - 2500 * ((1+.08/12)+ (1+.08/12)^2) 3rd month nest egg * (1 + .08/12)^3 - 2500 * ((1+.08/12)+(1+.08/12)^2+(1+.08/12)^3) .... formula that represents this is amount remaining at the end of 40 years = (nest egg)*(1+.08/12)^(12*40) - $2500 * integral of (1+.08/12)^t dt from t1 = 0 to t2 = 12*40 = 480 = (nest egg)*(1+.08/12)^480 - 2500*((1+.08/12)^(480+1) - (1+.08/12))/(1+.08/12 - 1) =(nest egg)*((1+.08/12)^480) - $8785703.04 So if the amount remaining is zero then you can solve for the minimum amount required for nest egg nest egg = ($8785703.04)/(1+.08/12)^480 = $361948 = n * 1502.473 So assuming monthly withdrawal of $2500 over 40 years the monthly savings over the first 30 years should be: n = $361948 / 1502.473 = $240.90 In this case c would be close but it would be under. I would suggest d. If the withdrawal of the $30000 was once yearly at the beginning of the year then: 1st year nest egg * (1+.08/12)^12 - 30000*(1+.08/12)^12 2nd year nest egg * (1+.08/12)^24 - 30000 * ((1+.08/12)^12 + (1+.08/12)^24) 3rd year nest egg * (1+.08/12)^36 - 30000 * ((1+.08/12)^12+(1+.08/12)^24+(1+.08/12)^36) formula that represents this is amount remaining at the end of 40 years = (nest egg)*(1+.08/12)^(40*12) - $30000 * integral of (1+.08/12)^12t dt from t1 = 1 to t2 = 40 I need to go now. I will finish this later. I expect this answer to require more money saved than the first assumption since more money is being withdrawn at an earlier time than in the previous assumption. If I had to choose between the four answers I would choose d in this case. One thing to try is to use excel and crunch the numbers without a formula. I think you will find that depending on the assumption of withdrawal you may find that $240.03 is not sufficient and you will have a loss. Using excel, with a monthly investment of $240.03 for 360 months at 8% compounded monthly, I obtained a nest egg of $357748.36. With a withdrawal rate of $2500/month starting at month 361 and nest egg earning 8%, the funds would be depleted in month 805. At month 840 at the end of John's desired retirement he would have a loss of $101939.13. With a monthly investment of $260.03 for 360 months at 8% compounded monthly, John would have a nest egg of $387557 and at month 840, at the end of retirement, John would still have $621617.42. If the withdrawal rate is $30000 at the beginning of each year in retirement, then with a savings rate of $240.03 for the first 30 years at 8% compounded monthly, then the nest egg would be depleted by month 721 and by month 840 at the end of desired retirement the loss would be $426554.7. If the savings rate is $260.03, John would still have $297001.85 at the end of retirement.