7- Math for Finance

Mathema Math ematics tics of Fina Finance nce

Simple interest Notations  p

–

inte in tere rest st rat ratee per co conv nvers ersion ion per period iod (i (in n per perce cent nt))

t

–

time ti me in un unit itss th that at co corre rrespon spond d to th thee rat rate, e, mo mome men nt

K 0

–

openi ope ning ng ca capi pita tal, l, ori origin ginal al pri princ ncip ipal, al, pr pres esen entt valu aluee

K t

–

amou am oun nt, ca capi pita tall at th thee mo mome men nt t

Z t

–

simp si mple le in inte tere rest st ea earn rned ed in th thee te term rm t

i

–

interest rate: i =

T 

–

number of days

p

100

The most freque frequentl ntly y period conside considered red is the year, but a half-y half-year, ear, quarter, month etc. can be used either. The number of days per year or month differs 30 act act from country to country. Usual usances are , , , where act means 360 360 act the actual number of days. In what follows, in all formulas containing the quantity T  the underlying period is the year with 360 days, each month havin ha vingg 30 days (i. e., the first usance usance is use used). d).

•

Basic interest formulas T  = 30 (m2

·

− m1 ) + d2 − d1

p · 100 · t = K 0 · i · t K 0 · i · T  K 0 · p · T  Z  = = 360 100 · 360 100 · Z  Z  K 0 = =  p · t i·t 100 · Z   p = K 0 · t Z t = K 0 T  T 

t

t

t

i=

Z t K 0 t

· 100 · Z  t= K 0 · p

t

– numbe numberr of days for for which which inter interest est is paid; m1 , m2 – months; d1 , d2 – days – sim simpl plee int inter eres estt – sim simple ple int intere erest st on a day day basis basis – ca capi pita tall (a (att t = 0) – in inter terest est rate (in perce percent nt)) – in inte tere rest st rat ratee

=

Z t K 0 i

·

– te term rm,, ti time me

B. Luderer et al., Mathematical Formulas for Economists, Economists, 4th ed., DOI 10.1007/978-3-642-04079-5_6, 10.1007/978-3-642-04079-5_6, © Springer-Verlag Berlin Heidelberg 2010

32

Mathematics of Finance

Capital at the moment K t = K 0 (1 + i t) = K 0

·

K 0 =

t



T  1+i 360

K t K t = T  1+i t 1 + i 360

·

– amount, capital at the moment t – present value

·

K n K 0 K n K 0 = 360 K 0 t K 0 T 

− · · K  − K 0 t= K 0 · i K  − K 0 T  = 360 · K 0 · i i=

·



− ·

n

– interest rate – term, time

n

– number of days for which interest is paid

Periodical payments 1 Dividing the original period into m parts of length m and assuming periodical payments of size r at the beginning (annuity due) and at the end (ordinary annuity), respectively, of each payment interval, the following amount results:

•

 

 ·  ·

m+1 R = r m+ i 2 R = r m+

m

−1

– annuity due

i – ordinary annuity 2 Especially m = 12 (monthly payments and annual interest): R = r(12 + 6, 5i) – annuity due;

R = r(12 + 5, 5i) – ordinary annuity

Different methods for computing interest Let ti = Di M i Y i be day, month and year of the i-th date (i = 1: begin, i = 2: end); let t = t2 t1 describe the actual number of days between begin and end; let T i denote the number of days in the ’broken’ year i; basis i = 365 oder 366.

−

method 30/360 act/360 act/act ∗

formula t = [360 (Y 2

− Y 1) + 30 · (M 2 − M 1) + D2 − D1] /360 t = (t2 − t1 )/360 T 1 T 2 t= + Y 2 − Y 1 − 1 + basis1 basis 2

·

∗

If the term considered lies within one year, only the first summand remains.

Compound interest

33

Compound interest When considering several conversion periods, one speaks about compound  interest  if interest earned is added to the capital (usually at the end of every conversion period) and yields again interest. Notations  p

– rate of interest (in percent) per period

n

– number of conversion periods

K 0

– opening capital, present value

K n

– capital after n periods, amount, final value

i

– (nominal) interest rate per conversion period: i =

q

– accumulation factor (for 1 period): q = 1 + i

qn

– accumulation factor (for n periods)

m

– number of parts of the period

d

– discount factor

im , ˆim

– interest rates belonging to each of the m parts of the period

δ

– intensity of interest

p 100

Conversion table of basic quantities p

i

q

 p

p

100i

100(q

i

p 100

i

q

q

1+

p 100

1+i

d

p 100 + p

i 1+i

δ

p ln 1 + 100





ln(1 + i)

− 1)

d 100

1

−d

−d

100(eδ eδ

1 1

−1 ln

1

  1

−d

− 1)

−1 eδ

−d d

q

ln q

1 d

−1 q

q

d

δ

1

−e

δ

−

δ

34

Mathematics of Finance

Basic formulas K n = K 0 (1 + i)n = K 0 qn

·

K 0 =

K n K n = (1 + i)n qn

n

≈

– present value at compound interest, amount at time t = 0

   − 

 p = 100 n=

– compound amount formula

·

n

K n K 0

1

– rate of interest (in percent)

log K n log K 0 log q 69  p

−

– term, time – approximate formula for the time within which a capital doubles

K n = K 0 q1 q2 . . . qn

– final value under changing rates of inpj terest pj , j = 1, . . . , n (with qj = 100 )

· · · ·

 pr =100



1+i 1+r

 − ≈ 1

100(i r) – real rate of interest considering a rate of inflation r

−

Mixed, partly simple and partly compound interest K t = K 0 (1 + it1 )(1 + i)N (1 + it2 ) – capital after time t Here N  denotes the integral number of conversion periods, and t1 , t2 are the lengths of parts of the conversion period where simple interest is paid.

•

To simplify calculations, in mathematics of finance usually the compound amount formula with noninteger exponent is used instead of the formula of  mixed interest, i. e. K t = K 0 (1 + i)t , where t = t1 + N  + t2 .

•

Anticipative interest (discount) In this case the rate of interest is determined as part of the final value (◮ discount factor p. 33). d=

K 1

− K 0 = K  − K 0 K 1 K  · t t

t

K n =

K 0 (1 d)n

– final value

−

K 0 = K n (1

– (discount) rate of interest under anticipatice compounding

n

− d)

– present value

Compound interest

35

More frequently compounding K n,m = K 0 im =

i m

ˆim =

m

i n·m m

·  1+

– amount after n years when interest rate is converted m times per year – relative rate per period

√1 + i − 1

ieff  = (1 + im )m

– equivalent rate per period

−1

– effective rate of interest per year

p m  peff  = 100 (1 + ) 100m



 − 1

– effective rate of interest per year (in percent)

Instead of one year one can take any other period as the original one. Compounding interest m times per year with the equivalent rate per period ˆim leads to the same final value as compounding once per year with the nominal rate i; compounding interest m times per year with the relative rate per period im leads to that (greater) final value, which results when compounding once per year with the effective rate ieff .

• •

Continuous compounding K n,

= K 0 ei n

–

amount at continuous compounding

δ = ln(1 + i)

–

intensity of interest (equivalent to the interest rate i)

i = eδ

–

nominal rate (equivalent to the intensity δ)

∞

·

−1

·

Average date of payment Problem: At which date of payment tm equivalently the total debt K 1 + K 2 + . . . + K k has to be payed? simple interest: K 1 + K 2 + . . . + K k tm = i compound interest:

− K 0 ,

K 1 0

K 2

t1

t2

K k -

...

tk

liabilities to pay

where K 0 =

K 1 K k + ... + 1 + it1 1 + itk

ln(K 1 + . . . + K k ) ln K 0 K 1 K k , where K 0 = t + . . . + t ln q q qk continuous compounding:

−

tm =

tm =

ln(K 1 + . . . + K k ) δ

− ln K 0 ,

1

where K 0 = K 1 e

δt 1

−

δt k

−

+. . .+ K k e

36

Mathematics of Finance

Annuities Notations  p n R q

– – – –

rate of interest term, number of payment intervals or rent periods periodic rent, size of each payment p accumulation factor: q = 1 + 100

Basic formulas Basic assumption: Conversion and rent periods are equal to each other. qn due F n = R q

· · q −−11 R q −1 P due = · q 1 q−1 q −1 F ord = R · q−1 R q −1 P ord = · q−1 q

– amount of an annuity due, final value

n

n

n−

– present value of an annuity due

n

n

– amount of an ordinary annuity, final value

n

n

n

P due =

Rq q 1

P ord =

R

∞

∞

– present value of a perpetuity, payments at the beginning of each period

−

q

– present value of an ordinary annuity

– present value of a perpetuity, interest used as earned

−1





1 q 1 1 n= log F nord +1 = log log q R log q R

·

· −

·

R

− P ord(q − 1)

– term

n

Factors describing the amount of 1 per period due qn 1 s¨n | = q q 1

amount of 1 per period present worth of 1 per period

· −− q −1 1 (q − 1)

ordinary qn 1 sn | = q 1

n

a ¨n | =

qn

−

− − q −1 = | q (q − 1) n

an

n

Conversion period > rent period If  m periodic payments (rents) are made per conversion period, then in the above formulas the payments R are to be understood as R = r m + m2+1 i (annuities due) and R = r m + m2 1 i (ordinary annuities), resp. These amounts R arise at the end of the conversion period, so with respect to R one always has to use formulas belonging to ordinary annuities.



−

·





·

38

Mathematics of Finance

Dynamic annuities Arithmetically increasing dynamic annuity Cash flows (increases proportional to the rent R with factor δ): R R(1 + δ )

R(1+(n − 1)δ )

R(1+ δ )

R

R(1+(n − 1)δ )

-

0

...

1

F ndue = P ndue = F nord =

Rq q

−1

qn

−



R 1 (q

−

R q

n 1

−1



qn

qn



qn





Rq

P due = ∞

q

−1

1+



n

q q



n

q q

−1+δ

δ q

0

−1



−1 −1

−1+δ

−1+δ

R P nord = n qn q (q 1)

−

n

−1+δ

− 1)

-



−1 −1



1

...

2

n

 − n

qn 1 q 1

− −n −



 − n

qn 1 q 1



− −n −

P ord =

,

∞

R q

−1



1+

δ q

−1



Geometrically increasing dynamic annuity R 0

Rb 1

Rb2 2

Rbn

1

−

R -

... n 1

−

n

P due n

F ord n

qn = Rq q

0

n

= = =

P due

=

n

∞

n

Rq q

−b

,

...

2

n

b = q;

F ndue

=

Rnq n ,

b=q

, b = q;



P ndue

=

Rn,

b=q

b = q;



F nord

=

Rnq n

b = q;



P nord

=

Rn , q

b < q;

P ord

=



n

n

P ord

1

of succeeding terms is characterized by

n

n−

n

s

100

n

· −− bb , R q −b · q 1 q−b q −b R· , q−b R q −b · q−b , q

1

−

-

n

The constant quotient b = 1 + the rate of increase s. F due

Rbn

Rb

n

∞

R q

1

−

− b,

, b=q b=q b 0 its maturity yield is higher. If several possibilities of  investments are for selection, then that with the highest net present value is preferred.

•

Method of internal rate of return The internal rate of return (yield-to-maturity) is that quantity for which the net present value of the investment is equal to zero. If several investments are possible, then that with the highest internal rate of return is chosen. Annuity method qn (q 1) F A = qn 1 AI  = K I  F A A = K E F A AP  = AI  A

· − − · · −

–

annuity (or capital recovery) factor

– – –

income annuity expenses annuity net income (profit) annuity

For AI  = A the maturity yield of the investment is equal to p, for AI  > A the maturity yield is higher than the conventional rate of interest p.

•

Depreciations

43

Depreciations Depreciations describe the reduction in value of capital goods or items of  equipment. The difference between original value (cost price, production costs) and depreciation yields the book-value. n A wk Rk

– – – –

term of utilization (in years) original value depreciation (write-down) in the k-th year book-value after k years (Rn – remainder, final value)

Linear (straight-line) depreciation wk = w = Rk = A

A

−R

n

n k w

− ·

–

annual depreciation

–

book-value after k years

Arithmetically degressive depreciation (reduction by d each year) wk = w1 (k 1) d nw1 (A Rn ) d=2 n(n 1)

− − · · − −−

–

depreciation in the k-th year

–

amount of reduction

wk = (n k + 1) d 2 (A Rn ) d= n(n + 1)

–

depreciation in the k-th year

–

amount of reduction

Sum-of-the-years digits method (as a special case): wn = d

−

·

· −

Geometrically degressive (double-declining balance) depreciation (reduction by s percent of the last year’s book value in each year) Rk = A s = 100

s 100

k

· −      · −   · · − 1

1

s wk = A 100

n

1

Rn A

s 100

k−1

–

book-value after k years

–

rate of depreciation

–

depreciation in the k-th year

Transition from degressive to linear depreciation Under the assumption Rn = 0 it makes sense to write down geometrically degressive until the year k with k = n + 1 100 and after that to write s down linearly.

⌈⌉

−

44

Mathematics of Finance

Numerical methods for the determination of zeros Task: Find a zero x of the continuous function f (x); let ε be the accuracy bound for stopping the iteration process. ∗

Table of values For chosen values x find the corresponding function values f (x). Then one obtains a rough survey on the graph of the function and the location of zeros. Interval bisection Given xL with f (xL ) < 0 and xR with f (xR ) > 0. 1. Calculate xM  = 12 (xL + xR ) and f (xM ). 2. If  f (xM ) < ε, then stop and take xM  as an approximation of  x . 3. If  f (xM ) < 0, then set xL := xM  (xR unchanged), if  f (xM ) > 0, then set xR := xM  (xL unchanged), go to 1.

|

|

∗

Method of false position (linear interpolation, regula falsi) Given xL with f (xL ) < 0 and xR with f (xR ) > 0. 1. Calculate xS = xL

− f (xx ) −− xf (x R

L

L)

R

f (xL ) and f (xS ).

2. If  f (xS ) < ε, then stop and take xS as an approximation of  x . 3. If  f (xS ) < 0, then set xL := xS (xR unchanged), if  f (xM ) > 0, then set xR := xS (xL unchanged), go to 1.

|

|

• For f (x

L)

∗

> 0, f (xR ) < 0 the methods can be adapted in an obvious way.

Newton’s method Given x0 U (x ); let the function f  be differentiable.

∈

∗

1. Calculate xk+1 = xk

− f f (x(x )) . k

′

k

2. If  f (xk+1 ) < ε, then stop and take xk+1 as an approximation of  x . 3. Set k := k + 1, go to 1.

|

|

∗

• If  f  (x ) = 0 for some k, then restart the iteration process with another starting point x0 . • Other stopping rule: |x − x | < ε or |x +1 − x | < ε. ′

k

L

R

k

k

Descartes’ rule of signs. The number of positive zeros of the polynomial  n



k=0

ak xk is equal to w or w

− 2, w − 4, . . . , where w is the number of changes

in sign of the coefficients ak (not considering zeros).