# Group 7: Emily Dinan, Meg Donohue, Ashley Vega
# Problem 1A
# r(n) will be the interest rate for each month n
def r(n):
if n==1:
return (0.10)
else:
return (0.10 + .01*(n-1))
# d(n) will be the deposit made at the end of each month n
def d(n):
return RDF(100)
# P(n) will refer to the total balance at the end of each month n
def P(n):
if n==0:
return RDF(2000)
else:
return RDF((1+(r(n)/12))*(P(n-1))+ d(n))
print " n P";
for n in range(0,36): print "%2s "%(n),; print "%16s "%(P(n))
# It is clear that the balance surpasses $10,000 when n=35.
Q = [[n, P(n)] for n in range(36)]
pplot= scatter_plot(Q);
pplot.show()
n P 0 2000.0 1 2116.66666667 2 2236.06944444 3 2358.43013889 4 2483.97979873 5 2612.95956305 6 2745.62155758 7 2882.22984502 8 3023.06143449 9 3168.40735601 10 3318.57380581 11 3473.88336924 12 3634.6763282 13 3801.31206089 14 3974.17054205 15 4153.65395289 16 4340.18841025 17 4534.2258258 18 4736.24590688 19 4946.75831137 20 5166.30497057 21 5395.46259483 22 5634.84537853 23 5885.10792196 24 6146.94838981 25 6421.11192752 26 6708.39435874 27 7009.6461895 28 7325.77694701 29 7657.75988367 30 8006.63707989 31 8373.52498255 32 8759.62041946 33 9166.20713414 34 9594.66288978 35 10046.4671957 n P 0 2000.0 1 2116.66666667 2 2236.06944444 3 2358.43013889 4 2483.97979873 5 2612.95956305 6 2745.62155758 7 2882.22984502 8 3023.06143449 9 3168.40735601 10 3318.57380581 11 3473.88336924 12 3634.6763282 13 3801.31206089 14 3974.17054205 15 4153.65395289 16 4340.18841025 17 4534.2258258 18 4736.24590688 19 4946.75831137 20 5166.30497057 21 5395.46259483 22 5634.84537853 23 5885.10792196 24 6146.94838981 25 6421.11192752 26 6708.39435874 27 7009.6461895 28 7325.77694701 29 7657.75988367 30 8006.63707989 31 8373.52498255 32 8759.62041946 33 9166.20713414 34 9594.66288978 35 10046.4671957 |
# Problem 1B
def r(n):
if n==1:
return (0.10)
else:
return (0.10 + .01*(n-1))
def d(n):
if n==1:
return (100)
else:
return RDF(100*((n-1)/2))
def P(n):
if n==0:
return RDF(2000)
else:
return RDF(((1+(r(n)/12))*(P(n-1)))+ d(n))
print " n P";
for n in range(0,18): print "%2s "%(n),; print "%16s "%(P(n))
# It is clear that the balance surpasses $10,000 when n=17.
Q = [[n, P(n)] for n in range(18)]
pplot= scatter_plot(Q);
pplot.show()
n P 0 2000.0 1 2116.66666667 2 2186.06944444 3 2307.93013889 4 2482.93271539 5 2711.90026374 6 2995.79901704 7 3335.74300393 8 3732.99936315 9 4188.9943536 10 4705.32009753 11 5283.74209916 12 5926.20758589 13 6634.85472497 14 7412.02277386 15 8260.26322934 16 9182.35204662 17 10181.3030076 n P 0 2000.0 1 2116.66666667 2 2186.06944444 3 2307.93013889 4 2482.93271539 5 2711.90026374 6 2995.79901704 7 3335.74300393 8 3732.99936315 9 4188.9943536 10 4705.32009753 11 5283.74209916 12 5926.20758589 13 6634.85472497 14 7412.02277386 15 8260.26322934 16 9182.35204662 17 10181.3030076 |
# Problem 1C
def r(n):
if n==1:
return (0.10)
else:
return (0.10 + .01*(n-1))
def d(n):
if n==1:
return (100)
else:
return RDF(d(n-1)+(100*((n-1)/2)))
def P(n):
if n==0:
return RDF(2000)
else:
return RDF((1+(r(n)/12))*(P(n-1))+ d(n))
print " n P";
for n in range(0,11): print "%2s "%(n),; print "%16s "%(P(n))
# It is clear that the balance surpasses $10,000 when n=10.
Q = [[n, P(n)] for n in range(11)]
pplot= scatter_plot(Q);
pplot.show()
n P 0 2000.0 1 2116.66666667 2 2286.06944444 3 2558.93013889 4 2986.65188206 5 3621.49615402 6 4516.76485594 7 5726.98838736 8 7308.12072284 9 9317.74253369 10 11815.2734571 n P 0 2000.0 1 2116.66666667 2 2286.06944444 3 2558.93013889 4 2986.65188206 5 3621.49615402 6 4516.76485594 7 5726.98838736 8 7308.12072284 9 9317.74253369 10 11815.2734571 |
# Problem 2
pp = vector(RDF, 11);
pp[0]= 25.7;
pp[1]= 18.5;
pp[2]= 14.3;
pp[3]= 12.2;
pp[4]= 11.1;
pp[5]= 9.5;
pp[6]= 9.1;
pp[7]= 8.5;
pp[8]= 8.1;
pp[9]= 7.5;
pp[10]= 7.4;
Q = [[pp[i], pp[i+1]] for i in range(10)]
Q;
pplot= scatter_plot(Q);
pplot.show()
testline= plot((3/5)*x + 3, (x,0,27),rgbcolor=hue(0.4))
show(pplot+testline)
# We found r=(3/5) and k=3 to give the best estimation.
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