I'm going to do this a bit at a time
Let it take y time periods to save 10,000,000
The idea is to minimize y
So i spend for x time periods then I save for y-x time periods
spending
end of first month after being paid T1= 25
end of second month after being paid T2= 25* 1.1^1
....
end of the xth month after being paid Tx= 25* 1.1^(x-1)
now I am going to start saving everything.
end of (x+1) month after being paid I'll have T(x+1)= 2*[25* 1.1^(x-1)]
end of (x+2) month after being paid I'll have T(x+2)= 3*[25* 1.1^(x-1)]
end of (x+(y-x)) month after being paid I'll have T(x+(y-x))= (y-x+1)*[25* 1.1^(x-1)]
we want
\((y-x+1)*[25* 1.1^{x-1}]=10^7\\ \text{I want y to be the subject}\\ (y-x+1)*[25* 1.1^{x-1}]=10^7\\ y[25* 1.1^{x-1}]+(-x+1)[25* 1.1^{x-1}]=10^7\\ y[25* 1.1^{x-1}]=10^7-(-x+1)[25* 1.1^{x-1}]\\ y[25* 1.1^{x-1}]=10^7+(x-1)[25* 1.1^{x-1}]\\ y=\frac{10^7+(x-1)[25* 1.1^{x-1}]}{[25* 1.1^{x-1}]}\\ \)
Now I want to find the stationary points for y dy/dx=0
\(y=\frac{10^7+(x-1)[25* 1.1^{x-1}]}{[25* 1.1^{x-1}]}\\ y=\frac{10^7}{[25* 1.1^{x-1}]}+\frac{(x-1)[25* 1.1^{x-1}]}{[25* 1.1^{x-1}]}\\ y=\frac{10^7}{25}*1.1^{1-x}+x-1\\\)
I'm too lazy to work this out with calculus so here is the graph.
\(y=\frac{10^7}{25}*1.1^{1-x}+x-1\\\)
So if I stop saving at around 111.7 months then it will take around 121.2 months to have $10,000,000
if x=111
y=10^7/25*1.1^(1-111)+111-1 = 121.1908899233864998 that is 122 months
if x=112
10^7/25*1.1^(1-112)+112-1 = 121.1735362939877271 that is 122 months too
if x=113
10^7/25*1.1^(1-113)+113-1 = 121.248669358170661 that is 122 months too
if x=114
10^7/25*1.1^(1-114)+114-1 = 121.4078812347006009 that is 122 months too
if x=115
10^7/25*1.1^(1-115)+115-1 = 121.6435283951823645 that is 122 months too
if x=116
10^7/25*1.1^(1-116)+116-1 = 121.9486621774385131 that is 122 months too
if x=117
10^7/25*1.1^(1-117)+117-1 = 122.3169656158531938 that is 123 months
if x=110
10^7/25*1.1^(1-110)+110-1 = 121.3099789157251498
if x=109
10^7/25*1.1^(1-109)+109-1 = 121.5409768072976648
if x=108
10^7/25*1.1^(1-108)+108-1 = 121.8950744880274312
if x=107
10^7/25*1.1^(1-107)+107-1 = 122.3845819368301744 that is 123 months
So if what I have done is correct
I can spend for 108 to 116 months and it will take still take 122 months to accumulate
$10,000,000
I cannot accumulate 10,000,000 in less than 122 months.
This needs to be carefully checked.
