π
πππ§ππ€ππ
=
π
π‘
βπ
0
π
0
β 100%
π
2
=
1
1 + π
1
β1 β100%
premia/dyskonto =
π
0
βπ
0
π
0
β
12
π
ππππ πΔππ¦
β 100% (=
π
0
βπ
0
π
0
β
360
π
πππ
β 100%)
1 + π = 1 + π (1 + π)
π
ππππ
= π
πππ
β
1 + π
π
1 + π
β
π
β
β π
π
1 + π
π
=
π
β
β π
π
1 + π
π
π
π
π
0
=
1 + π
β
1 + π
π
π
π‘
π
0
=
(1 + π
β
)
π‘
(1 + π
π
)
π‘
π
πΉ
πΈ
= π
πΉ
1 β π (1 β π) β π
π
π»
πΈ
= π
π»
1 β T β dT
π
β
=
π
πΉ
β π
π»
1 + π
πΉ
π
π‘
π
0
=
1 + π
β
1 + π
π
1
2
3
4
5
6
7
8
9
10
11
12