Question

You will need $100,000 per year for 25 years starting in year 45.    You are able to save $3000 per year from year 3 to year 15, inclusive. How much must you save from years 10 through 37, inclusive, if interest rates will be 4% from years 0 through the end of year 25 and 7% starting at the beginning of year 26 and on?    Solve for the unknown payments:

4% 7% 3 10 15 25 26 37 45 69 $3000/year PYMT=? $100,000/yr. for 25 years 2,400,000

Answer 1

No of withdrawals=25

r=7%

A=$100000

PW of withdrawal at emd of year 44 =A*(1-(1+r)^-n)/r

=100000*(1-(1+7%)^-25)/7%

=100000*(1-1.07^-25)/0.07

=100000*(1-0.1842)/0.07

=100000*0.8158/0.07

=$1165358.32

No of period between year 44 and 37=44-37=7

Hence PW of withdrawal at end of year 37=1165358.32/(1+7%)^7=1165358.32/1.07^7=1165358.32/1.6058=$725726.59

No of deposits in year 3 to 15=13

A=$3000

r=4%

Hence FW of deposit at end if year 15=A*((1+r)^n-1)/r

=3000*((1+4%)^13-1)/4%

=3000*(1.04^13-1)/0.04

=3000*(1.6651-1)/0.04

=3000*0.6651/0.04

=$49880.51

No of period between year 15 to 25=10

Hence FW of deposits at end of year 25=49880.51*(1+4%)^10=49880.51*1.04^10=49880.51*1.4802=$73835.34

No of period between year 25 to 37=12

FW of deposit at emd of year 37=73835.34*(1+7%)^12=73835.34*1.07^12=73835.34*2.2522=$166291.34

FW of unknown deposit at end of year 37 should be=725726.59-166291.34=$559435.25

No of deposit between year 10 to 25=16

FW=A*((1+4%)^16-1)/4%

=A*(1.04^16-1)/0.04

=A*(1.873-1)/0.04

=A*0.873/0.04

=$21.82A

No of period between year 25 to 37=12

FW of deposit at emd of year 37=21.82A*(1+7%)^12=21.82A*1.07^12=21.82A*2.2522=$49.15A

No of deposits between year 26to 37=12

FW at end of year 37=A*((1+7%)^12-1)/7%

=A*(1.07^12-1)/0.07

=A*(2.2522-1)/0.07

=A*1.2522/0.07

=$17.89A

Total FW of deposit=49.15A+17.89A=$67.04A

Hence 67.04A=559435.25

Or,A=559435.25/67.04=$8344.42