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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 21 "Financial mathematics" }}
{PARA 18 "" 0 "" {TEXT 257 23 "Demonstrative exercises" }}}{SECT 1 
{PARA 3 "" 0 "" {TEXT -1 14 "Relative value" }}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 71 "i-th commodity, i=1,2,3 is being traded at time t=1,2,3,4
,5 and course " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "restart;
\nkappa:=(i,t)->abs(2*i^3+(5*i^2)*sin(i*t/6))+1;" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 111 "plot(\{kappa(1,t),kappa(2,t),kappa(3,t)\},t=0..6, co
lor=[GREEN,NAVY,RED],title=`Courses of commodities in time`);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 7 "Courses" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 115 "Digits:=3;\nj:='j';\nfor i from 0 to 5 do\nprint(i,e
valf(kappa('j',i)) $'j'=1..3);\n#print(i,(kappa(i,j)) $j=1..3);\nod;" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 71 "Course margin - increase (or decrease) in courses per u
nit of commodity" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 254 "for i from 0 t
o 5. do\nxxx:=evalf(kappa(j,i)):yyy:=evalf(kappa(j,i+1)):\nzzz:=([xxx,
yyy] $j=1..3);\nzzzz:=evalf(kappa(j,i+1)-kappa(j,i));\nprint(i,evalf(k
appa(j,i+1)-kappa(j,i)) \n$j=1..3);\nod:\nevalf([5, -3.70+5.*sin(1), -
19.9+20.*sin(2), -26.9+45.*sin(3)]);\n" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 359 "From point 0 to point
 2 there is the biggest nominal increase in the course of commodity #3
. Though, this information is not relevant for making investment decis
ions. Because it is mot important how much money do we make per unit o
f commodity, but how much do we make per unit of capital we have inves
ted in certain commodity. That's why we need to calculate:" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 87 "Relative course margin - increase (or dec
rease) in courses per unit of capital invested" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 108 "for i from 0.1 to 5.1 do\ni:=i-.1;\nprint(i,evalf(ka
ppa(j,i+1.)-kappa(j,i))/(kappa(j,i)) $j=1..3);\ni:=i+.1\nod:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 136 "for i from 1 to 3 do\nplot(
\{kappa(i,t),kappa(i,t+1)-kappa(i,t),(kappa(i,t+1)-kappa(i,t))/kappa(i
,t)\},t=0..6, color=[GREEN,NAVY,RED]);\nod;" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 349 "\n\nplot(\{kappa(1,t+1)-kappa(1,t),kappa(2,t+1)-k
appa(2,t),kappa(3,t+1)-kappa(3,t)\},t=0..6, color=[GREEN,NAVY,RED],tit
le=`Nominal course margin in time 1`);\n\nplot(\{(kappa(1,t+1)-kappa(1
,t))/kappa(1,t),\n(kappa(2,t+1)-kappa(2,t))/kappa(2,t),\n(kappa(3,t+1)
-kappa(3,t))/kappa(1,t)\},t=0..6, color=[GREEN,NAVY,RED],title=`Relati
ve course margin in time 1`);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
39 "Course derivation according to time is " }{XPPEDIT 18 0 "limit((ka
ppa(i,t)-kappa(i,t+delta(t)))/delta(t),delta(t)=0)" "6#-%&limitG6$*&,&
-%&kappaG6$%\"iG%\"tG\"\"\"-F)6$F+,&F,F--%&deltaG6#F,F-!\"\"F--F26#F,F
4/-F26#F,\"\"!" }{TEXT -1 50 " which shows instantaneous total change \+
in course." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 141 "\nplot(\{diff(kappa(
1,t),t),diff(kappa(2,t),t),diff(kappa(3,t),t)\},t=0..6, color=[GREEN,N
AVY,RED],title=`Course derivation according to time`);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "diff(Kappa(i,t)/Kappa(i,t0),t)=diff
(Kappa(i,t),t)/Kappa(i,t0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "Re
lative total change of course is " }{XPPEDIT 18 0 "limit((kappa(i,t)-k
appa(i,t+delta(t)))/(kappa(i,t)*delta(t)),delta(t)=0)=diff(kappa(i,t),
t)/kappa(i,t)" "6#/-%&limitG6$*&,&-%&kappaG6$%\"iG%\"tG\"\"\"-F*6$F,,&
F-F.-%&deltaG6#F-F.!\"\"F.*&-F*6$F,F-F.-F36#F-F.F5/-F36#F-\"\"!*&-%%di
ffG6$-F*6$F,F-F-F.-F*6$F,F-F5" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 186 "plot(\n\{diff(kappa(1,t),t)/kappa(1,t)\n,diff(k
appa(2,t),t)/kappa(2,t)\n,diff(kappa(3,t),t)/kappa(3,t)\n\},t=0..6, co
lor=[GREEN,NAVY,RED],title=`Relative derivation commodity values in ti
me`);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Digits:=12;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 120 "Supposing that we could trade com
modities continually, the most profitable is to buy and keep commodity
 #3 till time T1," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "T1:=fs
olve(diff(kappa(2,t),t)/kappa(2,t)=diff(kappa(3,t),t)/kappa(3,t));" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "then sell commodity #3 and buy co
mmodity #2. Keep commodity #2 till T2," }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 73 "T2:=fsolve(diff(kappa(1,t),t)/kappa(1,t)=diff(kappa(2
,t),t)/kappa(2,t));\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "and fina
lly sell commodity #2 and buy commodity #1. " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 55 "Compare profit realized in disc
rete and continual time:" }{MPLTEXT 1 0 1 " " }}}{SECT 1 {PARA 5 "" 0 
"" {TEXT -1 5 "Trade" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 281 "Supposing
 you preciously know courses of all the commodities in advance. How mu
ch you can make disposing of 100 CZK? Consider that the commodities ar
e ideally divisible; you are allowed to buy random amount of each comm
odity; and you start to trade at time t=0 ending at time t=4)?" }}}
{SECT 1 {PARA 5 "" 0 "" {TEXT -1 36 "Trade in discrete time (t=1,2,. .
 .)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 140 "NumberOfTitles:=3;\n
Capital[0]:=100;\nRateOfProfit:=(i,t)->kappa(i,t+1)/kappa(i,t);\nValue
:=t->Sum(kappa(i,t)*Number[i],i=1..NumberOfTitles);\n\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "Max:=t->max(RateOfProfit(i,t) $i=1.
.NumberOfTitles);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 131 "Vyber
:=if(t)\nglobal NumberOfTitle;\nfor i from 1 to NumberOfTitles do\nif \+
Max(t)=RateOfProfit(i,t)\nthen NumberOfTitle:=i;\nfi \nod\nend;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 401 "for t from  0 to 4 do \ni:=
'i';\nVyber(t); \n  for i from 1 to NumberOfTitles do\n   Number[i]:=0
;\n  od:\nNumber[NumberOfTitle]:=Capital[t]/kappa(NumberOfTitle,t);\nC
apital[t+1]:=Number[NumberOfTitle]*kappa(NumberOfTitle,t+1);\nprint(`a
t time `,t, ` I purchase `, evalf(Number[NumberOfTitle]),\n ` items of
 commodity `,   NumberOfTitle, `at time`, t+1, `will I have the capita
l of `, evalf(Capital[t+1]));\nod:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "In time t=3 and  t=4 is be
tter not to go to the market and keep commodity #1." }}}}{SECT 1 
{PARA 5 "" 0 "" {TEXT -1 24 "Trade in continual time:" }}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 137 "Until time t=5 you can make a tiny better prof
it if you realize the trade in time 0, T1 and T2 instead of time t0, t
1, t2. The amount is:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 163 "RateOfPro
fit := if (i, t1, t2) options operator, arrow; kappa(i,t2)/kappa(i,t1)
 end if; evalf(100*RateOfProfit(3,0,T1)*RateOfProfit(2,T1,T2)*RateOfPr
ofit(1,T2,5));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}}{SECT 0 
{PARA 3 "" 0 "" {TEXT -1 18 "Coumpound interest" }}{SECT 1 {PARA 4 "" 
0 "" {TEXT -1 14 "Common formula" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 147 "restart;\nPhi[1]:=(x0,xi,t)->x0*(1+xi)^(floor(t));\n\nPhi[2]:
=(x0,xi,t)->x0*(1+xi)^(floor(t))*(1+xi*(t-floor(t)));\n\nPhi[3]:=(x0,x
i,t)->x0*(1+xi)^(t);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Let's a
ssign values into variables: " }{XPPEDIT 18 0 "x0=10,xi=1.9" "6$/%#x0G
\"#5/%#xiG-%&FloatG6$\"#>!\"\"" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 125 "
Arg:=(10,1.9,t);\nplot(\{\nPhi[1](Arg),\nPhi[2](Arg),\nPhi[3](Arg)\},
\nt=0..5,\nthickness=3,\ncolor=[coral,gold,maroon],discont=true);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 165 "Distinction between the values of
 continual and compound interest; meaning the distinction between the \+
values of exponential function and the values on it's secants." }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "plot(Phi[2](Arg)-Phi[3](Arg),t=0..5
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 89 "We'll find the moment, when
 the distinction is equal to the half of the original deposit." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "T:=solve(Phi[2](Arg)-Phi[3](
Arg)=5,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "evalf(subs(t=
T,Phi[2](Arg)));\nevalf(subs(t=T,Phi[3](Arg)));" }}}}{SECT 0 {PARA 4 "
" 0 "" {TEXT -1 7 "Samples" }}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 200 "1.
  1. On my current account there are 10 golden coins(GC). On 1. March \+
I deposit another 10GC and on 1. September 20GC more. What will be my \+
portfolio on 1. 1. next year if the interest rate is 0.05?" }}{SECT 1 
{PARA 5 "" 0 "" {TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 407 "NumberOfDaysInM[1]:=31:\nNumberOfDaysInM[2]:=28:\nNu
mberOfDaysInM[3]:=31:\nNumberOfDaysInM[4]:=30:\nNumberOfDaysInM[5]:=31
:\nNumberOfDaysInM[6]:=30:\nNumberOfDaysInM[7]:=31:\nNumberOfDaysInM[8
]:=31:\nNumberOfDaysInM[9]:=30:\nNumberOfDaysInM[10]:=31:\nNumberOfDay
sInM[11]:=30:\nNumberOfDaysInM[12]:=31:\nNumberOfDaysTillEndOfY:=(d,m)
->\nNumberOfDaysInM[m]-d+sum(NumberOfDaysInM[i],i=m+1..12);\nNumberOfD
aysTillEndOfY(1,1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 159 "zet
a := (1+5/100)^(1/365)-1;  evalf(10*(1+zeta)^NumberOfDaysTillEndOfY(21
,1)+10*(1+zeta)^NumberOfDaysTillEndOfY(16,3)+20*(1+zeta)^NumberOfDaysT
illEndOfY(7,9));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "
NumberOfDaysTillEndOfY(21,1);\nNumberOfDaysTillEndOfY(16,3);\nNumberOf
DaysTillEndOfY(7,9);" }}}}}{PARA 4 "" 0 "" {TEXT -1 0 "" }}{SECT 1 
{PARA 0 "" 0 "" {TEXT -1 271 "When your current account (CA) reaches t
he nominal value of 120 giggles(G), if you deposit 90G at time t=0 and
 the interest rate is 0.05 during the time when the nominal value of t
he CA is below 100G and 0.07 during the time when the nominal value of
 the CA is above 100?" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 8 "Solution
" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "We have to calculate the time \+
before we reach 100G on our CA" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 39 "restart;  T[1]:=solve(90*1.05^t=100,t);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 72 "then we'll count the time when the value of the CA g
rows from 100 to 120" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "T[2
]:=solve(100*1.07^t=120,t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "fi
nally we add up T1 and T2. " }{MPLTEXT 1 0 14 "Tt:=T[1]+T[2];" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 146 "Considering that the time unit is
 the same in which the interest rate is being accredited. For example \+
if the interest is accredited only at time " }{XPPEDIT 18 0 "t=1/365,2
/365,3/365" "6%/%\"tG*&\"\"\"F&\"$l$!\"\"*&\"\"#F&F'F(*&\"\"$F&F'F(" }
{TEXT -1 130 ",... we can say, that the CA will never reach the nomina
l value of 120G. Yet the firs moment when we can expect 120G an our CA
 is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 171 "T[1]:=ceil(solve(T
[1]=x*1/365,x));\nT[2] := solve((90*1.05^(T[1]/365))*1.07^t = 120,t);
\n#(90*1.07^(T[1]/365))*1.07^T[2];\nT[2]:=ceil(solve(T[2]=x*1/365,x));
\nTt:=T[1]+T[2];\n\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "t-th mome
nt" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "evalf(Tt/365);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "the value of the CA at t-th moment
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "(90*1.05^(T[1]/365))*1
.07^(T[2]/365);" }}}}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 45 "Think of tw
o CAs. On the first one there are " }{XPPEDIT 18 0 "x1:=1234;" "6#>%#x
1G\"%M7" }{TEXT -1 42 "GC deposited. On the second one there are " }
{XPPEDIT 18 0 "x2 := 1230" "6#>%#x2G\"%I7" }{TEXT -1 81 "GK. What is t
he interest rate the CA must have been charged with for the period  " 
}{XPPEDIT 18 0 "T:=2;" "6#>%\"TG\"\"#" }{TEXT -1 12 " to achieve " }
{XPPEDIT 18 0 "x1" "6#%#x1G" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "x2" "6#
%#x2G" }{TEXT -1 9 ", if the " }{XPPEDIT 18 0 "x1" "6#%#x1G" }{TEXT 
-1 42 " is charged with simple interest rate and " }{XPPEDIT 18 0 "x2
" "6#%#x2G" }{TEXT -1 21 " with compound one?  " }}{SECT 1 {PARA 5 "" 
0 "" {TEXT -1 8 "solution" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 
"formula:=x[1]*(1+xi*T)=x[2]*(1+xi)^T;" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 56 "solution:=solve(\{formula,xi>0\},xi);\nallvalues(solution);" }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "zeta:=op(2,op(evalf(solution)));" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 189 "plot(x[2]*(1+xi)^T-x[1]*
(1+xi*T),xi=0..zeta+0.1,title=`The distinction between the current val
ue of simple and compond interrest in relation to the interest rate.`,
titlefont=[HELVETICA,7]);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 4 "" 0 "" {TEXT -1 0 "" }}}}
{SECT 0 {PARA 3 "" 0 "" {TEXT -1 33 "Depositing in equidistant moments
" }}{SECT 0 {PARA 0 "" 0 "" {TEXT -1 173 " Suppose a year of 360 days \+
with 12 months of 30 days. Your deposit is allways 100CZK. The interes
t rate is 0.05 p. a. How much is there going to be at the end of a yea
r if:" }}{PARA 15 "" 0 "" {TEXT -1 16 "You deposit once" }}{PARA 16 "
" 0 "" {TEXT -1 28 "at the beginning of the year" }}{PARA 16 "" 0 "" 
{TEXT -1 23 "at the half of the year" }}{PARA 15 "" 0 "" {TEXT -1 18 "
You deposit twice " }}{PARA 16 "" 0 "" {TEXT -1 45 "at the beginning a
nd at the half of the yearu" }}{PARA 16 "" 0 "" {TEXT -1 38 "at the ha
lf and at the end of the year" }}{PARA 15 "" 0 "" {TEXT -1 13 "Twelve \+
times " }}{PARA 16 "" 0 "" {TEXT -1 30 "at the beginning of each month
" }}{PARA 16 "" 0 "" {TEXT -1 25 "at the end of each month." }}{SECT 
1 {PARA 20 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 
"Trying to calculate Present value (PV) of the CA. Time unit is 1 day.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "restart;\nPV:=sum(Z(t)*
(1+xi)^((360-t)/360), t=1..360):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 
"PVdem:=Sum(Z(t)*(1+xi)^((360-t)/360), t=1..360):" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 8 "xi:=.05;" }{TEXT -1 24 ", t is a time in years, " }
{XPPEDIT 18 0 "z(t);" "6#-%\"zG6#%\"tG" }{TEXT -1 36 " is the value of
 a deposit at time t" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 201 "z[
1]:=t->piecewise(t=1,100,t<>0,0);\nz[1](t);\nz[2]:=t->piecewise(t=360/
2,100,t<>0,0);\nz[2](t);\nz[3]:=t->piecewise(t=1,100,t=360/2,100,0);\n
z[3](t);\nz[4]:=t->piecewise(t=360/2,100,t=360,100,0);\nz[4](t);\n\n" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "for i from 1 to 4 do\nZ:=
t->z[i](t);\nPresentValue[i]:=PV;\nod;" }}}{EXCHG {PARA 0 "" 0 "" 
{XPPEDIT 18 0 "PresentValue[3]=PresentValue[2]+PresentValue[1]" "6#/&%
-PresentValueG6#\"\"$,&&F%6#\"\"#\"\"\"&F%6#F,F," }{TEXT -1 2 ", " }
{XPPEDIT 18 0 "PresentValue[4]=Presentvalue[2]+100" "6#/&%-PresentValu
eG6#\"\"%,&&%-PresentvalueG6#\"\"#\"\"\"\"$+\"F-" }{TEXT -1 64 ", Beca
use the last deposit isn't being charged with an interest." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 102 "\nz[5]:=t->piecewise(type((t-1)/(30),integer)
,100,0);\nz[5](31);\npointplot(\{seq([n,z[5](n)],n=0..95)\});\n" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "# z[6]:=t->piecewise(type((t)/(30),
integer),100,0);\n# z[6](31);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 36 "Z:=t->z[5](t);\nPresentValue[5]:=PV;\n" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 44 "In fact we're summing the geometrical series" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "#restart;" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 119 "i:='i';Xi:=xi;Time:='Time':\nxi:='xi';\nxxx:=Sum(1
00*(1+xi)^(Time(i)/360),i=1..12)=sum(100*(1+xi)^(Time(i)/360),i=1..12)
;" }}{PARA 0 "" 0 "" {TEXT -1 5 "where" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 62 "Time := if (i) options operator, arrow; 360-(i-1)*30-
1 end if;" }{TEXT -1 61 " is time from depositing i-th deposit till th
e end of a year." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "xi:=Xi;
\nxxx;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "geometrical series with
 quotient" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "q:=(100*(1+xi)
^(Time(i+1)/360))/(100*(1+xi)^(Time(i)/360));" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "q:=simpl
ify(q);\na1:=subs(i=1,100*(1+xi)^(Time(i)/360));" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 56 "We could sum this series according to the common f
ormula" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "PresentValue[5]:=
a1*(1-q^12)/(1-q);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 206 "the case w
e deposit at the end of a month is different just in function Time and
 in present value of the first deposit. Quotient is the same like in t
he previous case and the number of series members also: " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "Time := i->360-(i-1)*30-30;\na1:=s
ubs(i=1,100*(1+xi)^(Time(i)/360));\nPresentValue[6]:=sum(100*(1+xi)^(T
ime(i)/360),i=1..12);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 "
\n" }{TEXT 256 7 "valid:\n" }{MPLTEXT 1 0 51 "(PresentValue[5]/Present
Value[6]=(1+xi)^(29/360));\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}}}{PARA 3 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" 
{TEXT -1 7 "Savings" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 56 "Saving on \+
compound interest through equidistant deposits" }}{SECT 1 {PARA 4 "" 
0 "" {TEXT -1 15 "General formula" }}{PARA 0 "" 0 "" {TEXT -1 74 "Let'
s assume we're depositing in the equidistant moments, with proportion \+
" }{XPPEDIT 18 0 "tau" "6#%$tauG" }{TEXT -1 2 ". " }{XPPEDIT 18 0 "eps
ilon" "6#%(epsilonG" }{TEXT -1 37 " is the moment of the first deposit
, " }{XPPEDIT 18 0 "epsilon+tau" "6#,&%(epsilonG\"\"\"%$tauGF%" }
{TEXT -1 16 " of the second, " }{XPPEDIT 18 0 "epsilon+2*tau" "6#,&%(e
psilonG\"\"\"*&\"\"#F%%$tauGF%F%" }{TEXT -1 18 " of the third and " }
{XPPEDIT 18 0 "epsilon+T" "6#,&%(epsilonG\"\"\"%\"TGF%" }{TEXT -1 23 "
 of the last one.      " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "#
restart;\nWholePart:=x->floor(x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
31 "We're depositing at the moment " }{XPPEDIT 18 0 "n*tau+epsilon,n =
 0 .. T;" "6$,&*&%\"nG\"\"\"%$tauGF&F&%(epsilonGF&/F%;\"\"!%\"TG" }
{TEXT -1 36 "\nHow much deposits do we do in time " }{XPPEDIT 18 0 "t \+
< T+epsilon;" "6#2%\"tG,&%\"TG\"\"\"%(epsilonGF'" }{TEXT -1 1 "?" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "numberA := WholePart((t-epsilon)/ta
u)+1:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "and in time  " }
{XPPEDIT 18 0 "t > T+epsilon;" "6#2,&%\"TG\"\"\"%(epsilonGF&%\"tG" }
{TEXT -1 2 "? " }{MPLTEXT 1 0 19 "numberB := T/tau+1;" }{TEXT -1 26 ",
 which is integer number." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Generally: If we're depositing at \+
time " }{XPPEDIT 18 0 "n*tau+epsilon,n = 0 .. T;" "6$,&*&%\"nG\"\"\"%$
tauGF&F&%(epsilonGF&/F%;\"\"!%\"TG" }{TEXT -1 20 "\nwe receive in time
 " }{XPPEDIT 18 0 "t" "6#%\"tG" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "n
umber := min(NumberA,NumberB);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 
"the number of deposits. At which moment is the i-th deposit being dep
osited" }{XPPEDIT 18 0 "%?;" "6#%#%?G" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 63 "moment := if (i) options operator, arrow; i*tau+epsilon end if;
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "How long is the interest paid
 on i-th deposit till the time " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT 
-1 8 " if the " }{XPPEDIT 19 1 "moment(i) <= t;" "6#1-%'momentG6#%\"iG
%\"tG" }{TEXT -1 1 "?" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "time := if
 (i) options operator, arrow; t-moment(i) end if; time(i);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 46 "What is the value of the i-th deposit at \+
time " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 26 " at(on) the interest
 rate " }{XPPEDIT 18 0 "xi" "6#%#xiG" }{TEXT -1 5 ", if " }{XPPEDIT 
19 1 "moment(i) <= t;" "6#1-%'momentG6#%\"iG%\"tG" }{TEXT -1 48 " and \+
if its value at the time of depositing was " }{XPPEDIT 18 0 "z" "6#%\"
zG" }{TEXT -1 1 "?" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "value := if (
i) options operator, arrow; z*(1+xi)^time(i) end if; value(i);" }
{TEXT -1 25 " is on the condition that" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 69 "If we intent to sum current values of all of deposits don
e at moment " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 33 " or before we
 became the series  " }{XPPEDIT 18 0 "Sum(value(i),i = 0 .. number);" 
"6#-%$SumG6$-%&valueG6#%\"iG/F);\"\"!%'numberG" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 112 "I-th + 1st divided by i-th member of the series is \+
independent on i. Afterwards we can call it quotient, because" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "quotient := value(i+1)/value(i); qu
otient := simplify(quotient);" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 74 "Thus we're daling  with a geometrical ser
ies determined by this quotient. " }}}{EXCHG {PARA 16 "" 0 "" {TEXT 
-1 24 "The number of members is" }}{PARA 16 "> " 0 "" {MPLTEXT 1 0 25 
"NumberOfMembers = number;" }}}{EXCHG {PARA 16 "" 0 "" {TEXT -1 13 "0t
h member is" }}{PARA 16 "> " 0 "" {MPLTEXT 1 0 25 "ZerothMember := val
ue(0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "Thus the sum equals to
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "Formula:=Sum(value(i),i=0..numb
er-1)=simplify(sum(value(i),i=0..number-1));" }}{PARA 0 "" 0 "" {TEXT 
-1 43 "Using this formula for defining functions:\n" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 92 "Formula:=simplify(op(2,Formula)):\n#Formula:=subs(
tau=1/k,Formula);\n#limit(subs(k=K,\"),K=k);\n" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 75 "SUMGeometricalSeries:=Sum(a[0]*q^i,i=0..n)=sim
plify(sum(a[0]*q^i,i=0..n));\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 74 "Formula2:=subs(a[0]=NultyClen,q=quotient,n=number-1,SUMGeometr
icalSeries):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Formula2:=op(2,Form
ula2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 180 "psi2:=(Tau,Epsil
on,Xi,Z,TT,time)->subs(tau=Tau,epsilon=Epsilon,xi=Xi,z=Z,T=TT,t=time,F
ormula2);\nprint(`----------`);\nsimplify(psi2(tau,epsilon,xi,z,T,t)-p
si(tau,epsilon,xi,z,T,t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "The
 first function determines the value of the saved up amount, from the \+
parameters given." }}{PARA 14 "" 0 "" {TEXT -1 61 "Cross distance of d
eposits in time (same for all neighboring)" }}{PARA 14 "" 0 "" {TEXT 
-1 50 "The moment at which the first deposit is deposited" }}{PARA 14 
"" 0 "" {TEXT -1 22 "Constant interest rate" }}{PARA 14 "" 0 "" {TEXT 
-1 34 "The constant value of each deposit" }}{PARA 14 "" 0 "" {TEXT 
-1 61 "The period of time for which the deposits are being deposited" 
}}{PARA 14 "" 0 "" {TEXT -1 0 "" }}{PARA 14 "" 0 "" {TEXT -1 71 "The m
oment at which we want to know the nominal balance on the account." }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "psi:=(Tau,Epsilon,Xi,Z,TT,time)->su
bs(tau=Tau,epsilon=Epsilon,xi=Xi,z=Z,T=TT,t=time,Formula);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 97 "If we intend to know the nominal residue \+
on the account while we're still depositing, it is also " }{XPPEDIT 
18 0 "trunc(t-epsilon/tau) <= T+epsilon+1;" "6#1-%&truncG6#,&%\"tG\"\"
\"*&%(epsilonGF)%$tauG!\"\"F-,(%\"TGF)F+F)F)F)" }{TEXT -1 10 " and the
n " }{XPPEDIT 18 0 "min(trunc((t-epsilon)/tau),T/tau)=trunc((t-epsilon
)/tau)" "6#/-%$minG6$-%&truncG6#*&,&%\"tG\"\"\"%(epsilonG!\"\"F-%$tauG
F/*&%\"TGF-F0F/-F(6#*&,&F,F-F.F/F-F0F/" }{TEXT -1 2 ". " }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 321 "Formula_:=subs(min(trunc((t-epsilon)/tau),T/t
au)=trunc((t-epsilon)/tau),Formula):\nFormula_:=subs(min((T/tau,trunc(
(t-epsilon+trunc(abs((t-epsilon)/tau))*tau+tau)/tau)-trunc(abs((t-epsi
lon)/tau))))=\ntrunc(T/tau,trunc((t-epsilon+trunc(abs((t-epsilon)/tau)
)*tau+tau)/tau)-trunc(abs((t-epsilon)/tau)))\n,Formula_):\nFormula_;\n
\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 17 "And the function " }{XPPEDIT 18 0 "psi" "6#%$psiG" }
{TEXT -1 73 " determines the value of the saved up amount from the par
ameters given:  " }}{PARA 14 "" 0 "" {TEXT -1 61 "Cross distance of de
posits in time (same for all neighboring)" }}{PARA 14 "" 0 "" {TEXT 
-1 50 "The moment at which the first deposit is deposited" }}{PARA 14 
"" 0 "" {TEXT -1 22 "Constant interest rate" }}{PARA 14 "" 0 "" {TEXT 
-1 34 "The constant value of each deposit" }}{PARA 14 "" 0 "" {TEXT 
-1 71 "The moment at which we want to know the nominal balance on the \+
account." }}{PARA 0 "" 0 "" {TEXT -1 3 "If " }{XPPEDIT 18 0 "trunc((t-
epsilon)/tau) <= T/tau;" "6#1-%&truncG6#*&,&%\"tG\"\"\"%(epsilonG!\"\"
F*%$tauGF,*&%\"TGF*F-F," }{TEXT -1 4 ", is" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 86 "psi_:=(Tau,Epsilon,Xi,Z,T)->subs(tau=Tau,epsilon=Epsi
lon,xi=Xi,z=Z,N=TT,t=T,Formula_);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
9 "If also  " }{XPPEDIT 18 0 "t-epsilon)/tau" "6#*&,&%\"tG\"\"\"%(epsi
lonG!\"\"F&%$tauGF(" }{TEXT -1 96 " is an integer number, meaning just
 at the moments of depositing, the previous function will be:" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "FormulaC:=subs(trunc((t-epsilon)/ta
u)=(t-epsilon)/tau,Formula_);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 85 "Phi:=(Tau,Epsilon,Xi,Z,T)->subs(tau=Tau,epsilon=Epsilon,xi=Xi,z=
Z,N=TT,t=T,FormulaC);" }}}{SECT 0 {PARA 20 "" 0 "" {TEXT -1 6 "Sample
" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "p2:=plot(psi2(1/3,1/6,3/4,5,2-1/3,t
),t=0..2.75,discont=false):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
165 "p3:=plot(psi2(1/3,1/6,3/4,5,2-1/3,1.4)*(1+3/4)^(t-1.4),t=1.4..2.7
5#,title=`the interest is paid from the balance on the acount, though \+
we don't deposit any more.`\n):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 155 "l := [[n/3+1/6,(n+1)*5] $n=0..5]:\np1:=plot(l, x=0..2.5, styl
e=point,symbol=circle#,title=`The points illustrate the sum of the nom
inal values deposited`\n):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
31 "with(plots):\ndisplay(p1,p2,p3);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 54 "`plotsetup/devices`[jpeg]:=[jpeg,`plot.jpg`,[],[],``]
:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 43 "#plotsetup(jpeg, plotoutput=`sporeni.jpg`);" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 49 "#plotsetup(plotoptions=`height=1200,width=1200`);
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "#eval(plotsetup):" }}}{EXCHG 
{PARA 12 "> " 1 "" {MPLTEXT 1 0 34 " evalf(psi2(1/3,1/6,1/3,5,2,0.1));
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 202 "ll := [trunc(100*(n/3+
1/6))/100 $n=0..5];\nplot(evalf(subs(tau=1/3,epsilon=1/6,T=2-1/3,t=t,n
umber)),t=0..4,y=0..7,xtickmarks=ll, title=`graph of the relation of t
he munber of amounts deposited on time`);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 49 "evalf(subs(tau=1/3,epsilon=1/6,T=2,t=.1,number));" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "evalf(subs(T=2,tau=1,t=10
/4,epsilon=1/2,number));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}}}{PARA 4 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT 
-1 55 " Saving on simple interest through equidistant deposits" }}
{SECT 1 {PARA 4 "" 0 "" {TEXT -1 15 "General formula" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 28 "#restart; \nWholePart:=floor;" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 74 "Let's assume we're depositing in the equi
distant moments, with proportion " }{XPPEDIT 18 0 "tau" "6#%$tauG" }
{TEXT -1 2 ". " }{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT -1 37 "
 is the moment of the first deposit, " }{XPPEDIT 18 0 "epsilon+tau" "6
#,&%(epsilonG\"\"\"%$tauGF%" }{TEXT -1 16 " of the second, " }
{XPPEDIT 18 0 "epsilon+2*tau" "6#,&%(epsilonG\"\"\"*&\"\"#F%%$tauGF%F%
" }{TEXT -1 18 " of the third and " }{XPPEDIT 18 0 "epsilon+T" "6#,&%(
epsilonG\"\"\"%\"TGF%" }{TEXT -1 19 " of the last one.  " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "at time " }
{XPPEDIT 18 0 "t < T+epsilon;" "6#2%\"tG,&%\"TG\"\"\"%(epsilonGF'" }
{TEXT -1 13 " we receive  " }{MPLTEXT 1 0 40 "numberA := WholePart((t-
epsilon)/tau)+1:" }{TEXT -1 14 " and at time  " }{XPPEDIT 18 0 "t > T+
epsilon;" "6#2,&%\"TG\"\"\"%(epsilonGF&%\"tG" }{TEXT -1 13 " we receiv
e  " }{MPLTEXT 1 0 17 "NumberB:=T/tau+1:" }{TEXT -1 38 " the number of
 deposits, thus at time " }{XPPEDIT 18 0 "t>0" "6#2\"\"!%\"tG" }{TEXT 
-1 13 " we receive  " }{MPLTEXT 1 0 31 "number := min(numberA,numberB)
:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
56 "deposits, i-th deposit is being deposited at the moment " }
{MPLTEXT 1 0 25 "moment:=i->i*tau+epsilon;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 60 "How long is the interest paid on i-th deposit till the ti
me " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 8 " if the " }{MPLTEXT 1 
0 31 "time:=i->t-okamzik(i): time(i);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 20 "its value at moment " }{XPPEDIT 18 0 "t" "6#%\"tG" }
{TEXT -1 4 " is " }{MPLTEXT 1 0 37 "value:=i->z*(1+xi*time(i)): value(
i);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "We must sum present values
 (original values + interest) of all of the deposits till the moment \+
" }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 59 ". The difference between \+
i-th and i-th + 1st deposit is    " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
41 "diference:=simplify(value(i+1)-value(i));" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 37 "So that the series is independent on " }{XPPEDIT 18 0 
"t" "6#%\"tG" }{TEXT -1 87 ", thus the series is arithmetic one. The v
alue of the 0th member of the series at time " }{XPPEDIT 18 0 "t" "6#%
\"tG" }{TEXT -1 4 " is " }{XPPEDIT 19 1 "value(0);" "6#-%&valueG6#\"\"
!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "and the value of the last de
posit at time " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 4 " is " }
{MPLTEXT 1 0 16 "value(number-1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
28 "Consequently the formula is:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 76 
"Formula:=Sum(value(i),i=0..number-1)\n=simplify(sum(value(i),i=0..num
ber-1));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Formula:=op(2,Formula);
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 58 "Formula2:=simplify(1/2*(value(0)+value(number-1))*num
ber);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "subs(min(T/tau,-tr
unc((-t+epsilon)/tau))=K,\n2*simplify(xxx/sum(value(i),i=0..number)));
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "simplify(subs(min(trunc
((t-epsilon)/tau),T/tau)=K, Formula-Formula2));" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 382 "W:=(Tau,Epsilon,Xi,Z,TT,time)->\n[evalf(subs(
tau=Tau,epsilon=Epsilon,xi=Xi,z=Z,T=TT,t=time,Formula)),evalf(subs(tau
=Tau,epsilon=Epsilon,xi=Xi,z=Z,T=TT,t=time,Formula2)),\nevalf(subs(tau
=Tau,epsilon=Epsilon,xi=Xi,z=Z,T=TT,t=time,value(0))),\nevalf(subs(tau
=Tau,epsilon=Epsilon,xi=Xi,z=Z,T=TT,t=time,value(number-1))),\nevalf(s
ubs(tau=Tau,epsilon=Epsilon,xi=Xi,z=Z,T=TT,t=time,number))\n];" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "W(1,0,1,5,10,1);\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 196 "psi:=(Tau,Epsilon,Xi,Z,TT,t
ime)->subs(tau=Tau,epsilon=Epsilon,xi=Xi,z=Z,T=TT,t=time,op(2,Formula)
);\npsi2:=(Tau,Epsilon,Xi,Z,TT,time)->subs(tau=Tau,epsilon=Epsilon,xi=
Xi,z=Z,T=TT,t=time,Formula2);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 50 "evalf(psi(1,0,1,5,2,1));\nevalf(psi2(1,0,1,5,2,1));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 268 "At the \+
beginning of the year we begin to deposit on an account. At the beginn
ing of each month we deposit 114,7 Oobols. At the end of the year we h
ave to pay the tax 1/15 of the interest rate. The interest rate itself
 is 1/50 p. m. How much do you save up in 11 years?" }}{SECT 1 {PARA 
5 "" 0 "" {TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 40 "restart;\nz:=114.7;xi:=1/50;delta:=1/15;\n" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 12 "In one year:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 58 
"xxx := 12.*z+(sum(z*(1+xi)^t,t = 1 .. 12)-12*z)*(1-delta);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "In 11 years:" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 42 "\nsum(xxx*(1+xi*(1-delta))^(12*t),t=0..10);" }}}}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 23 "
Internal rate of return" }}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 363 "Dan H
as bought a car via financial leasing. He should pay remaining 130 000
 CZK in 36 monthy part payments of the value of 4592.5 CZK. By the per
ipeties of life he was forced to repay the part payments from his savi
ngs. What interest rate must the savings be charged with to be able to
 repay the debt preferable to pay immediately the whole amount of 130 \+
000 CZK?" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 8 "Solution" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 40 "Resolving the equation 36 with variable " }{XPPEDIT 18 0 
"xi" "6#%#xiG" }{TEXT -1 2 ". " }{XPPEDIT 18 0 "xi" "6#%#xiG" }{TEXT 
-1 45 " is interest rate per month in this case.    " }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 51 "rce:=sum(4592.5*(1+xi)^i,i=0..35)=130000*
(1+xi)^36;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "This equation must \+
be solved numerically. Maple disposes of built-in function " }
{HYPERLNK 17 "fsolve:" 2 "fsolve" "" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "xxx:=fsolve(rce,xi)
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 174 "We're interested just in th
e non-negative solution out of two solutions of the equation. The inte
rest rate per month must be recalculated into the interest rate per a \+
year.  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "`It is worth re
paying the depth from savings if the interest rate per year is bigger \+
than:`, (1+xxx[2])^12-1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "This \+
equation can also be rewritten with help of the sries-summing formula \+
into:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "rce2:=4592.5*(1-(1
+xi)^36)/(1-(1+xi))=rhs(rce);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "
and be solved directly. Having to analyse the equation and find out eq
uation root:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "rce3:=facto
r((lhs(rce2)-rhs(rce2)))=0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "Th
e only one positive equation root is the interest rate per month:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "solve(\{rce3,xi>0\});" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "Surprisingly the same one as from \+
the previous procedure." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "The interest rate is the int
ernal rate of return of the cash flow, -130 000, 4592.5, 4592.5, . . .
 , 4592.5." }}}}{PARA 0 "" 0 "" {TEXT -1 127 "Let's assume that Dan wo
uld rather like to pay the last 13 part payments 'en block' altogether
. He would like to pay less than " }{XPPEDIT 18 0 "13*4592.5" "6#*&\"#
8\"\"\"-%&FloatG6$\"&Df%!\"\"F%" }{TEXT -1 451 " CZK indeed, 'cause he
 repays the money immediately en block. Consider that the demanded for
 the financial leasing services higher than supply, thus the leasing c
ompany can immediately re-invest the money on the same conditions (mea
ning the same interest rate), then the question can also be formulated
: what is the marginal value of the part payment (1 instead of 13) to \+
keep the profit of the leasing company? (while naglecting the transact
ion cost)" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 8 "Solution" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart; " }{TEXT -1 162 "Also in th
is case we have to compute the internal rate of return, which means th
e interest rate on which the company experiences the profit out of mon
ey invested." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "rce:=sum(45
925/10*(1+xi)^(-i),i=1..36)=130000:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 33 "Xi:=fsolve(rce,xi=0.013..0.014); " }{TEXT -1 76 "We h
ave to discont the walue of 13 part payments with this rate of interes
t." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "xi := Xi; `the reasonable pri
ce is`, sum(45925/10*(1+xi)^(-i),i = 0 .. 13);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}
{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Exactness" }}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 22 "Assuming the cash flow" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "CF:=[-1000,3600,-4310,1716];" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 34 "PV:=sum(CF[i]*(1+xi)^(-i),i=1..4);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "solve(PV);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{PARA 3 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 37 "
Speculation demand for money (sample)" }}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 539 "According to John Maynard Keynes one of the incentive for the \+
demand for money is the speculation motive. Keynes asks the question w
hy people demand higher amount of money than the amount demanded becau
se of transactional and cautiousness motive. He induced, that the mone
y is held because of uncertainty about the changes in the rate of inte
rest in the future and  in the consequence of the relations between th
e interest rate and the bond prices. The consequence of the grow of th
e interest rate is the loss experienced by holding bonds." }}}{SECT 1 
{PARA 0 "" 0 "" {TEXT -1 147 "Let's assume a bond which will never be \+
repaied. Meaning that it's principal will never be repaied whereas it \+
brings fixed coupon revenue per year." }}{PARA 0 "" 0 "" {TEXT -1 193 
"Supposing that the bond asset is bought for the price of 1000 CZK, wh
ile the annual fixed coupon revenue is 100 CZK. What is the real value
 of the bond if the interest rate falls to it's half? " }}{SECT 1 
{PARA 5 "" 0 "" {TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "restart; " }}{PARA 0 "" 0 "" {TEXT -1 243 "Holding thi
s kind of bond is similar to receiving a ethernal (everlasting) rent o
f 100 CZK p.a. out of 1000 CZK locked. The interest rate corresponding
 to the annual interest of 100 CZK out of 1000 CZK at the moment when \+
the bond was bought is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "x
i[0]:=solve(1000*xi=100);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "yet \+
the interest rate nowadays went half" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "xi[1]:=xi[0]/2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
38 "consequently the value of the bond is " }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 22 "Z:=solve(z*xi[1]=100);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 28 "Thus the profit is 1000 CZK." }}}}}}{SECT 0 {PARA 3 "" 0 
"" {TEXT -1 23 "Two examples - revision" }}{SECT 0 {PARA 0 "" 0 "" 
{TEXT -1 60 "what is the interest rate on the 3rd month in the period \+
if:" }}{PARA 15 "" 0 "" {TEXT -1 104 "the annual interest rate is 0.05
, and the interest rate per month is the same on all months in the yea
r?" }}{PARA 15 "" 0 "" {TEXT -1 134 "the interest rate during the firs
t month is 0.01, on the second month it's 0.02 and on the first quarte
r the rate of interest is 0.06?" }}{PARA 15 "" 0 "" {TEXT -1 307 "the \+
annual interest rate is 0.1 and  \non the 4th quarter the rate of inte
rest equals to 0.02 ,\non the 4th quarter the rate of interest is 0.03
,\non the 4th quarter the rate of interest is 0.02 again,\nand on the \+
1st month the interest rate is 0.003,\nfinally on the 2nd month the in
terest rate equals to 0.004?" }}}{SECT 0 {PARA 0 "" 0 "" {TEXT -1 138 
"How long will it take to repay the debt of 10 000 CZK by monthly afte
rdate (realized at the end of each month) part payments of 1000 CZK. \+
" }}{PARA 15 "" 0 "" {TEXT -1 32 "if the interest rate is 0 p. m.," }}
{PARA 15 "" 0 "" {TEXT -1 130 "if the interest rate is 0.05 p. m; of w
hat high the monthly part payments must have been to shorten the repay
 period by one month?" }}{PARA 15 "" 0 "" {TEXT -1 33 "if the interest
 rate is 0.1 p. m." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "pom:=S
um(1000*(1+xi)^(t-i),i=1..t)=simplify(sum(1000*(1+xi)^(t-i),i=1..t));
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "pom:=op(2,pom)=10000*(1
+xi)^t;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "solve(\{pom,xi=.
1\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "simplify(sum(a*q^i
,i=1..n));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "10*(1-1.1)^(2
0)/0.1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 3 "" 
0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 44 "Deposits de
nominated in different currenceis" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 
25 "The course-change problem" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 178 "
The course of  jen-min-piao (juan) is 36:37 to rupee. The expected cou
rse at following period is 34:39. Which of these currenceis is prefera
ble to keep? Quantify the destinction." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "restart;\nE[0]:=36/37;\nE[1]:=34/39;" }}}{EXCHG 
{PARA 0 "" 0 "" {XPPEDIT 18 0 "E[t]" "6#&%\"EG6#%\"tG" }{TEXT -1 112 "
 is the value of rupee denominated in juan at moment t. The value of j
uan denominated in rupee is reciprocal of " }{XPPEDIT 18 0 "E[t]" "6#&
%\"EG6#%\"tG" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "if E[1] < E[0
] then `The rupee course falls` else `The rupee course rises` end if;
" "6#@%2&%\"EG6#\"\"\"&F&6#\"\"!%7The~rupee~course~fallsG%7The~rupee~c
ourse~risesG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 155 "Profit measuring
: You will neither gain nor lose anything if we keep rupee. If we keep
 juan instead of rupee, your relative profit denominated in rupee is:
" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "x*E[0]/E[1]*`rupi\355`;" 
"6#**%\"xG\"\"\"&%\"EG6#\"\"!F%&F'6#F%!\"\"%&rupi|hyGF%" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 28 "x rupee invested gives back " }{XPPEDIT 
18 0 "x*E[0]/E[1]" "6#*(%\"xG\"\"\"&%\"EG6#\"\"!F%&F'6#F%!\"\"" }
{TEXT -1 30 " rupee. So that the profit is:" }}}{EXCHG {PARA 0 "> " 0 
"" {XPPEDIT 19 1 "E[0]/E[1]-1;" "6#,&*&&%\"EG6#\"\"!\"\"\"&F&6#F)!\"\"
F)F)F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "If we calculate the pr
ofit denominated in juan, we will neither gain nor lose anything, but \+
keeping rupee would mean loss. " }}{PARA 0 "" 0 "" {TEXT -1 74 "if we \+
invest x juan into rupee, we'll have (in the following period) just:" 
}}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "x*E[0]^(-1)*E[1]*juan;" "6#
**%\"xG\"\"\")&%\"EG6#\"\"!,$F%!\"\"F%&F(6#F%F%%%juanGF%" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 21 "and the lose would be" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 17 "E[0]^(-1)*E[1]-1;" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 280 "How is it possible that we didn't get a number with th
e same absolute value? That's because we have chosen different (grater
 in this case) unit. (Lose one juan is grater loss than lose one rupee
.) Therefore choosing one fixed currency for this kind of computation \+
seems suitable. " }}}{SECT 0 {PARA 0 "" 0 "" {TEXT -1 147 "The courses
 of currencies (1) jen-min-piao (2) ngultrum (Bhutan) a (3) kyata (Bar
ma) is to the Danish crown (DKK) at time, t=1,2,3,4,5 is equal to " }
{XPPEDIT 18 0 "kappa(i,t)=9/i+abs(sin(i*t*Pi/6))" "6#/-%&kappaG6$%\"iG
%\"tG,&*&\"\"*\"\"\"F'!\"\"F,-%$absG6#-%$sinG6#**F'F,F(F,%#PiGF,\"\"'F
-F," }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 196 "Supposing that we have 100 DKK disposable and we know \+
all the courses in advance. We can also change the money without trans
action cost in actual courses at time i=1,2,3,4,5. How much we can mak
e?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "restart;\nkappa:=(i,t)
->9/i+abs(sin(i*t*Pi/6));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "plot(
\{kappa(1,t),kappa(2,t),kappa(3,t)\},t=0..6, color=[navy,blue,aquamari
ne]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "kappa(1,1);" }}}
{SECT 0 {PARA 5 "" 0 "" {TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 137 "NumberOfTitlValueCapital[1]:=100;\nRateOfProfit:=
(i,t)->kappa(i,t+1)/kappa(i,t);\nNumber:=t->Sum(kappa(i)*Number[iCapit
alNumberOfTitles);\n\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "M
ax:=t->max(RateOfProfit(i,t) $i=1..NumberOfTitles);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 277 "Vyber:=if(t)\nglobal NumberOfTitle;\ni:=
'i';\n#print(evalf(RateOfProfit(i,t)) $i=1..NumberOfTitles);\nfor i fr
om 1 to NumberOfTitles do\nif Max(t)=RateOfProfit(i,t)\nthen NumberOfT
itle:=i;\n#print(`The biggest profit makes the currency`,i,`a to`,eval
f(RateOfProfit(i,t)));\nfi \nod\nend;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 658 "for t from  1 to 6 do \ni:='i';\nVyber(t); MAX:=eval
f(Max(t));#print(MAX);\nif MAX>1 then\n  for i from 1 to NumberOfTitle
s do\n   Number[i]:=0;\n  od:\nNumber[NumberOfTitle]:=Capital[t]/kappa
(NumberOfTitle,t);\nCapital[t+1]:=Number[NumberOfTitle]*kappa(NumberOf
Title,t+1);\nprint(`At time `,t, ` I buy `, evalf(Number[NumberOfTitle
]),\n ` units of `,   NumberOfTitle, `at time`, t+1, `I will have`, ev
alf(Capital[t+1]),`of the original currency`);\nelse\nCapital[t+1]:=Ca
pital[t];\nprint(MAX,evalf(RateOfProfit(i,t)) $i=1..NumberOfTitles,`At
 time`,t,`I will keep the original currency, at time`, t+1, `I will ha
ve`, evalf(Capital[t+1]),`of the original currency`)\nfi\nod:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 88 "The Cond
ition of uncovered interest parity condition, International interest a
rbitration" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "Assuming two curren
cies CZK and USD for instance. Their real courses at time 0 and expect
ed course at time 1. " }{XPPEDIT 18 0 "E(0)" "6#-%\"EG6#\"\"!" }{TEXT 
-1 5 " and " }{XPPEDIT 18 0 "E(1)" "6#-%\"EG6#\"\"\"" }{TEXT -1 84 " (
meaning E(i) as the price of USD denominated in CZK at time i) and int
erest rates " }{XPPEDIT 18 0 "xi[CZK]" "6#&%#xiG6#%$CZKG" }{TEXT -1 5 
" and " }{XPPEDIT 18 0 "xi[USD]" "6#&%#xiG6#%$USDG" }{TEXT -1 4 ".   \+
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "If we invest CZK at time 0 in
to USD deposits, the profit at time 1 will be equal to:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "
" {XPPEDIT 19 1 "Profit[US] := x*E(0)^(-1)*(1+xi[USD])*E(1)-x;" "6#>&%
'ProfitG6#%#USG,&**%\"xG\"\"\")-%\"EG6#\"\"!,$F+!\"\"F+,&F+F+&%#xiG6#%
$USDGF+F+-F.6#F+F+F+F*F2" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "What is the rate of interest (" }
{XPPEDIT 18 0 "xi[CZK]" "6#&%#xiG6#%$CZKG" }{TEXT -1 67 ") by which th
e profit will be the same as the profit of CZK chosens" }{XPPEDIT 18 
0 "%?;" "6#%#%?G" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Profit[
CZ]:=x*xi[CZK];" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "rce := Pro
fit[US] = Profit[CZ];" "6#>%$rceG/&%'ProfitG6#%#USG&F'6#%#CZG" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "xxx:=solve(rce,xi[CZK]);" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 75 "Modify. Let's define Q as expected rate of depreciation of CZK \+
against USD." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "Q:=(E(1)-E(
0))/E(0);\nxxx2:=sort(expand(xxx/Q),xi[USD]);\n" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 129 "EquilibriumCondition:=xi[CZK]=sort(\n(simplif
y((op(1,xxx2)-xi[USD])*Q)+xi[USD]*Q)+\nsimplify(op(2,xxx2)*Q+op(3,xxx2
)*Q)\n,xi[USD]);\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 14 "The expression" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
33 "op(2,op(2,EquilibriumCondition));" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 35 "is a neglectable one (Weather both " }{XPPEDIT 18 0 "xi[U
SD]" "6#&%#xiG6#%$USDG" }{TEXT -1 187 " and the rate of depreciation a
re low). Let's sum up: in the equilibrium the difference between the i
nterest rates is equal to the the interest payed on the expected rate \+
of depreciation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 143 "lhs(Eq
uilibriumCondition)=op(1,rhs(EquilibriumCondition))+\nfactor(\nsimplif
y(op(2,rhs(EquilibriumCondition))+op(3,rhs(EquilibriumCondition)))\n);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}}}}{MARK "2 2 1 0 0" 156 }{VIEWOPTS 1 1 0 
1 1 1803 0 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }