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allow IRR to work on Cashflows #13

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Feb 9, 2025
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allow IRR to work on Cashflows #13

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1 change: 1 addition & 0 deletions src/Contracts.jl
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Original file line number Diff line number Diff line change
Expand Up @@ -58,6 +58,7 @@ end

maturity(c::C) where {C<:Cashflow} = c.time
Base.:-(c::C) where {C<:Cashflow} = Cashflow(-c.amount, c.time)
Base.zero(c::C) where {C<:Cashflow} = Cashflow(zero(c.amount), c.time)

"""
amount(x)
Expand Down
5 changes: 3 additions & 2 deletions src/FinanceCore.jl
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Expand Up @@ -6,12 +6,13 @@ using Dates
include("Rates.jl")
export Rate, rate, discount, accumulation, Periodic, Continuous, forward

include("irr.jl")
export irr, internal_rate_of_return

include("Contracts.jl")
export Cashflow, Quote, maturity, timepoint, amount, Composite

include("irr.jl")
export irr, internal_rate_of_return

include("pv.jl")
export pv, present_value

Expand Down
69 changes: 67 additions & 2 deletions src/irr.jl
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Expand Up @@ -23,6 +23,21 @@ function internal_rate_of_return(cashflows)
return internal_rate_of_return(cashflows, 0:length(cashflows)-1)
end

function internal_rate_of_return(cashflows::Vector{C}) where {C<:Cashflow}
# first try to quickly solve with newton's method, otherwise
# revert to a more robust method

v = irr_newton(cashflows)

lower, upper = -2.0, 2.0
r = rate(v)
if isnan(r) || (r >= upper && r <= lower)
return irr_robust(cashflows)
else
return v
end
end

function internal_rate_of_return(cashflows, times)
# first try to quickly solve with newton's method, otherwise
# revert to a more robust method
Expand Down Expand Up @@ -54,6 +69,20 @@ function irr_robust(cashflows, times)

end

function irr_robust(cashflows::Vector{C}) where {C<:Cashflow}
f(i) = sum(amount(cf) / (1 + i)^timepoint(t) for (cf, t) in cashflows)
# lower bound at -.99 because otherwise we can start taking the root of a negative number
# when a time is fractional.
roots = Roots.find_zeros(f, -0.99, 2)

# short circuit and return nothing if no roots found
isempty(roots) && return nothing
# find and return the one nearest zero
min_i = argmin(roots)
return Periodic(roots[min_i], 1)

end


function irr_newton(cashflows, times)
@assert length(cashflows) >= length(times)
Expand All @@ -68,12 +97,23 @@ function irr_newton(cashflows, times)

end

function irr_newton(cashflows::Vector{C}) where {C<:Cashflow}
# use newton's method with hand-coded derivative
r = __newtons_method1D_irr(
cashflows,
0.001,
1e-9,
100)
return Periodic(exp(r) - 1, 1)

end

# an internal function which calculates the
# present value and it's derivative in one pass
# for use in newton's method
function __pv_div_pv′(r, cashflows, times)
n = zero(typeof(first(cashflows) * 0.1))
d = zero(typeof(first(cashflows) * 0.1))
n = 0.0
d = 0.0
@turbo warn_check_args = false for i ∈ eachindex(cashflows)
cf = cashflows[i]
t = times[i]
Expand All @@ -84,6 +124,19 @@ function __pv_div_pv′(r, cashflows, times)
return n / d
end

function __pv_div_pv′(r, cashflows::Vector{C}) where {C<:Cashflow}
n = 0.0
d = 0.0
@turbo warn_check_args = false for i ∈ eachindex(cashflows)
cf = amount(cashflows[i])
t = timepoint(cashflows[i])
a = cf * exp(-r * t)
n += a
d += a * -t
end
return n / d
end

"""
irr(cashflows::vector)
irr(cashflows::Vector, timepoints::Vector)
Expand All @@ -104,4 +157,16 @@ function __newtons_method1D_irr(cashflows, times, x, ε, k_max)
k += 1
end
return x
end

function __newtons_method1D_irr(cashflows::Vector{C}, x, ε, k_max) where {C<:Cashflow}
k = 1
Δ = Inf
while abs(Δ) > ε && k ≤ k_max
# @show x,H(x), ∇f(x)
Δ = __pv_div_pv′(x, cashflows)
x -= Δ
k += 1
end
return x
end
81 changes: 45 additions & 36 deletions test/irr.jl
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Original file line number Diff line number Diff line change
@@ -1,77 +1,86 @@
#convenience function to wrap scalar into default Rate type
p(rate) = Periodic(rate,1)
p(rate) = Periodic(rate, 1)

@testset "irr" begin

v = [-70000,12000,15000,18000,21000,26000]
v = [-70000, 12000, 15000, 18000, 21000, 26000]

# per Excel (example comes from Excel help text)
@test isapprox(irr(v[1:2]), p(-0.8285714285714), atol = 0.001)
@test isapprox(irr(v[1:2]), p(-0.8285714285714), atol = 0.001)
@test isapprox(irr(v[1:3]), p(-0.4435069413346), atol = 0.001)
@test isapprox(irr(v[1:4]), p(-0.1821374641455), atol = 0.001)
@test isapprox(irr(v[1:5]), p(-0.0212448482734), atol = 0.001)
@test isapprox(irr(v[1:6]), p(0.0866309480365), atol = 0.001)
@test_throws MethodError irr("hello")
@test isapprox(irr(v[1:2]), p(-0.8285714285714), atol=0.001)
@test isapprox(irr(v[1:2]), p(-0.8285714285714), atol=0.001)
@test isapprox(irr(v[1:3]), p(-0.4435069413346), atol=0.001)
@test isapprox(irr(v[1:4]), p(-0.1821374641455), atol=0.001)
@test isapprox(irr(v[1:5]), p(-0.0212448482734), atol=0.001)
@test isapprox(irr(v[1:6]), p(0.0866309480365), atol=0.001)
@test_throws ArgumentError irr("hello")


# much more challenging to solve b/c of the overflow below zero
cfs = [t % 10 == 0 ? -10 : 1.5 for t in 0:99]

@test isapprox(irr(cfs), p(0.06463163963925866), atol = 0.001)
@test isapprox(irr(cfs), p(0.06463163963925866), atol=0.001)

# issue #28
cfs = [-8.728037307132952e7, 3.043754023830998e7, 2.963004184784189e7, 2.8803030748755097e7, 2.7956912111811966e7, 2.7092182051244527e7, 2.6209069543806538e7, 2.5307964329840004e7, 2.438961041057478e7, 2.3455084653011695e7, 2.2505925520018265e7, 2.154395414765592e7, 2.0571076113065004e7, 1.958930608135183e7, 1.8600627464895025e7, 1.7606980923262402e7, 1.661046149512893e7, 1.561312825963898e7, 1.461760481586352e7, 1.3626801207410209e7, 1.2644733969499402e7, 1.1675393687299855e7, 1.0722720151658386e7, 9.79075673433771e6, 8.883278741880089e6, 8.004445298876338e6, 7.1588010859461725e6, 6.351121678665243e6, 5.585860320479795e6, 4.8673895159943625e6, 4.19908059495347e6, 3.583538247530099e6, 3.022766488834396e6, 2.5181072324190177e6, 2.0701053881076649e6, 1.6782921224664208e6, 1.3410605489291362e6, 1.0556643097527474e6, 818348.5357315112, 624147.9373214925, 467849.788997191, 344241.752520618, 248285.65630649775, 175235.5475426321, 120677.87174498942, 80759.09804678289, 52186.83400936739, 32211.057718402008, 18589.51907385164, 9540.782278174447, 3688.4015341755294]
@test irr(cfs,0:50) ≈ p(0.3176680627111823)
@test irr(cfs, 0:50) ≈ p(0.3176680627111823)


@test irr([-100, 100]) ≈ p(0.0)
@test isnothing(irr([100, 100])) # answer is -1, but search range won't find it

@test irr([-100,100]) ≈ p(0.)
@test isnothing(irr([100,100])) # answer is -1, but search range won't find it

# test the unsolvable
@test isnothing(irr([-1e8,0.,0.,0.],0:3))
@test isnothing(irr([-1e8, 0.0, 0.0, 0.0], 0:3))

end

@testset "irr with fractional time" begin
irr1 = irr([-10,5,5,5],[0,1,2,3])
@test irr1 ≈ irr([-10,5,5,5])
irr2 = irr([-10,5,5,5],[0,1,2,3] ./ 2)
irr1 = irr([-10, 5, 5, 5], [0, 1, 2, 3])
@test irr1 ≈ irr([-10, 5, 5, 5])
irr2 = irr([-10, 5, 5, 5], [0, 1, 2, 3] ./ 2)

@test (1+rate(irr1))^2-1 ≈ rate(irr2)
@test (1 + rate(irr1))^2 - 1 ≈ rate(irr2)

end

@testset "numpy examples" begin

@test isapprox(irr([-150000, 15000, 25000, 35000, 45000, 60000]), p(0.0524), atol = 1e-4)
@test isapprox(irr([-100, 0, 0, 74]), p(-0.0955), atol = 1e-4)
@test isapprox(irr([-100, 39, 59, 55, 20]), p(0.28095), atol = 1e-4)
@test isapprox(irr([-100, 100, 0, -7]), p(-0.0833), atol = 1e-4)
@test isapprox(irr([-100, 100, 0, 7]), p(0.06206), atol = 1e-4)
@test isapprox(irr([-150000, 15000, 25000, 35000, 45000, 60000]), p(0.0524), atol=1e-4)
@test isapprox(irr([-100, 0, 0, 74]), p(-0.0955), atol=1e-4)
@test isapprox(irr([-100, 39, 59, 55, 20]), p(0.28095), atol=1e-4)
@test isapprox(irr([-100, 100, 0, -7]), p(-0.0833), atol=1e-4)
@test isapprox(irr([-100, 100, 0, 7]), p(0.06206), atol=1e-4)

# this has multiple roots, of which 0.709559 and 0.0886. Want to find the one closer to zero
@test isapprox(irr([-5, 10.5, 1, -8, 1]), p(0.0886), atol = 1e-4)
@test isapprox(irr([-5, 10.5, 1, -8, 1]), p(0.0886), atol=1e-4)
end

@testset "xirr with float times" begin


@test isapprox(irr([-100,100], [0,1]), p(0.0), atol = 0.001)
@test isapprox(irr([-100,110], [0,1]), p(0.1), atol = 0.001)
@test isapprox(irr([-100, 100], [0, 1]), p(0.0), atol=0.001)
@test isapprox(irr([-100, 110], [0, 1]), p(0.1), atol=0.001)

end

@testset "xirr with real dates" begin

v = [-70000,12000,15000,18000,21000,26000]
v = [-70000, 12000, 15000, 18000, 21000, 26000]
dates = Date(2019, 12, 31):Year(1):Date(2024, 12, 31)
times = map(d->DayCounts.yearfrac(dates[1], d, DayCounts.Thirty360()), dates)
# per Excel (example comes from Excel help text)
@test isapprox(irr(v[1:2], times[1:2]), p(-0.8285714285714), atol = 0.001)
@test isapprox(irr(v[1:3], times[1:3]), p(-0.4435069413346), atol = 0.001)
@test isapprox(irr(v[1:4], times[1:4]), p(-0.1821374641455), atol = 0.001)
@test isapprox(irr(v[1:5], times[1:5]), p(-0.0212448482734), atol = 0.001)
@test isapprox(irr(v[1:6], times[1:6]), p(0.0866309480365), atol = 0.001)
times = map(d -> DayCounts.yearfrac(dates[1], d, DayCounts.Thirty360()), dates)
# per Excel (example comes from Excel help text)
@test isapprox(irr(v[1:2], times[1:2]), p(-0.8285714285714), atol=0.001)
@test isapprox(irr(v[1:3], times[1:3]), p(-0.4435069413346), atol=0.001)
@test isapprox(irr(v[1:4], times[1:4]), p(-0.1821374641455), atol=0.001)
@test isapprox(irr(v[1:5], times[1:5]), p(-0.0212448482734), atol=0.001)
@test isapprox(irr(v[1:6], times[1:6]), p(0.0866309480365), atol=0.001)

end

@testset "irr with cashflows" begin
c = Cashflow.([-10, 0, 0, 15], [0, 1, 2, 3])
@test irr(c) ≈ Periodic((15 / 10)^(1 / 3) - 1, 1)

# issue #28
cfs = [-8.728037307132952e7, 3.043754023830998e7, 2.963004184784189e7, 2.8803030748755097e7, 2.7956912111811966e7, 2.7092182051244527e7, 2.6209069543806538e7, 2.5307964329840004e7, 2.438961041057478e7, 2.3455084653011695e7, 2.2505925520018265e7, 2.154395414765592e7, 2.0571076113065004e7, 1.958930608135183e7, 1.8600627464895025e7, 1.7606980923262402e7, 1.661046149512893e7, 1.561312825963898e7, 1.461760481586352e7, 1.3626801207410209e7, 1.2644733969499402e7, 1.1675393687299855e7, 1.0722720151658386e7, 9.79075673433771e6, 8.883278741880089e6, 8.004445298876338e6, 7.1588010859461725e6, 6.351121678665243e6, 5.585860320479795e6, 4.8673895159943625e6, 4.19908059495347e6, 3.583538247530099e6, 3.022766488834396e6, 2.5181072324190177e6, 2.0701053881076649e6, 1.6782921224664208e6, 1.3410605489291362e6, 1.0556643097527474e6, 818348.5357315112, 624147.9373214925, 467849.788997191, 344241.752520618, 248285.65630649775, 175235.5475426321, 120677.87174498942, 80759.09804678289, 52186.83400936739, 32211.057718402008, 18589.51907385164, 9540.782278174447, 3688.4015341755294]
@test irr(Cashflow.(cfs, 0:50)) ≈ p(0.3176680627111823)
end