# 3.10.jl
# create 2023-01-02.
# $Id: 3.10.jl 1.1 2023/01/02 00:00:00 s Exp $

using Printf

# k : period, 期
# n : the number of years remaining, 満期までの年数
# m : the number of coupon payments per year, 1年当り利払回数
# λ : the yield to maturity, 満期利回り
# f : face value, 額面価値
# r : coupon rate, クーポン率
# c : payment, 支払

# input
λ = 0.1
n = 10
r = 0.08
m = 2
f = 100

# calculating
c = f * r / m
j = Array(1: n*m)
A = Array(1/m: 1/m: n)
B = fill(c, n*m)
B[n*m] += f
C = 1 ./ (1 .+ λ ./ m) .^ j
D = B .* C
price = sum(D)
E = D ./ price
F = A .* E
duration = sum(F)

# output
println("3.10.jl")
println(repeat("-", 65))
@printf("%10s%10s%10s%15s%10s%10s\n","A", "B", "C", "D", "E", "F")
@printf("%10s%10s%10s%15s%10s%10s\n","", "", "Discount", "Present Value", "", "")
@printf("%10s%10s%10s%15s%10s%10s\n","", "", "factor", "of payment", "Weight", "")
@printf("%10s%10s%10s%15s%10s%10s\n","Year","Payment","(@10%)","(BxC)","(D/Price)","AxE")
println(repeat("-", 65))
for k = 1:length(A)
    @printf("%10.1f%10.0f%10.3f%15.3f%10.3f%10.3f\n",A[k], B[k], C[k], D[k], E[k], F[k])
end
println(repeat("-", 65))
@printf("%10s%20s%15.3f%10.3f%10.3f\n", "Sum", "", price, sum(E), duration)
@printf("%45s%20s\n", "Price", "Duration")
println(repeat("-", 65))
# eof