use Optim instead of Roots because it solves more robustly

Author: alecloudenback

Project.toml

@@ -5,12 +5,11 @@ version = "0.2.0"
 
 [deps]
 Dates = "ade2ca70-3891-5945-98fb-dc099432e06a"
-Roots = "f2b01f46-fcfa-551c-844a-d8ac1e96c665"
+Optim = "429524aa-4258-5aef-a3af-852621145aeb"
 
 [compat]
 julia = "^1.1"
-Roots = "1"
-
+Optim = "≥ 0.20.0"
 [extras]
 Test = "8dfed614-e22c-5e08-85e1-65c5234f0b40"
 

src/ActuaryUtilities.jl

@@ -2,6 +2,7 @@ module ActuaryUtilities
 
 using Dates
 using Roots
+using Optim
 
 """
     Years_Between(d1::Date, d2::Date)
@@ -82,36 +83,48 @@ function duration(issue_date::Date, proj_date::Date)
 end
 
 """
-    internal_rate_of_return(cashflows::vector; search_interval::Tuple{Real,Real})
+    internal_rate_of_return(cashflows::vector)
     
 Calculate the internal_rate_of_return of a series of equally spaced cashflows, assuming the first 
-element occurs at time zero. By default searches the `search_interval`in the numeric range `[-1,1]`.
+element occurs at time zero. First tries to find a positive rate in the interval `[0.0,1.0]`. If none is found,
+will extend search to [-1.0,1.0]. If still not found, will return `nothing`.
 
 """
-function internal_rate_of_return(cashflows;search_interval::Tuple{Real,Real}=(-1.0,1.0))
-
-    f(i) = pv(i,cashflows[2:end]) + cashflows[1]
+function internal_rate_of_return(cashflows)
+    
 
-    return find_zero(f,search_interval)
+    return internal_rate_of_return(cashflows,[t for t in 0:(length(cashflows)-1)])
+    
 end
 
 """
-    internal_rate_of_return(cashflows::Vector, timepoints::Vector; search_interval::Tuple{Real,Real})
+    internal_rate_of_return(cashflows::Vector, timepoints::Vector)
 
 Calculate the internal_rate_of_return with given timepoints. 
-By default searches the `search_interval`in the numeric range `[-1,1]`.
+First tries to find a positive rate in the interval `[0.0,1.0]`. If none is found,
+will extend search to [-1.0,1.0]. If still not found, will return `nothing`.
 
 ```jldoctest
 julia> internal_rate_of_return([-100,110],[0,1]) # e.g. cashflows at time 0 and 1
-0.10000000000000005
+0.10000000001652906
 ```
 """
-function internal_rate_of_return(cashflows,times;search_interval::Tuple{Real,Real}=(-1.0,1.0))
-
-    f(i) = sum(cashflows[1:end] .* [1/(1+i)^t for t in times])
-
-    return find_zero(f,search_interval)
-
+function internal_rate_of_return(cashflows,times)
+    # Optim requires the optimizing variable to be an array, thus the i[1]
+    f(i) = sum(cashflows .* [1/(1+i[1])^t for t in times])
+    result = optimize(x -> f(x)^2, 0.0,1.0)
+    if abs(f(result.minimizer)) < 1.0e-3 # arbitrary that seems to work
+        return result.minimizer
+    else
+        # try finding a negative irr
+        result = optimize(x -> f(x)^2, -1.0,1.0)
+        if abs(f(result.minimizer)) < 1.0e-3 # arbitrary that seems to work
+            return result.minimizer
+        else
+            return nothing
+        end
+    end
+    
 end
 
 """

test/run_doctests.jl

@@ -0,0 +1,2 @@
+# using Documenter
+# doctest(ActuaryUtilities)

test/runtests.jl

@@ -47,16 +47,16 @@ end
         v = [-70000,12000,15000,18000,21000,26000]
 
         # per Excel (example comes from Excel help text)
-        @test irr(v[1:2]) ≈ -0.8285714285714
-        @test irr(v[1:3]) ≈ -0.4435069413346
-        @test irr(v[1:4]) ≈ -0.1821374641455
-        @test irr(v[1:5]) ≈ -0.0212448482734
-        @test irr(v[1:6]) ≈  0.0866309480365
+        @test isapprox(irr(v[1:2]), -0.8285714285714,atol = 0.001)
+        @test isapprox(irr(v[1:3]), -0.4435069413346,atol = 0.001)
+        @test isapprox(irr(v[1:4]), -0.1821374641455,atol = 0.001)
+        @test isapprox(irr(v[1:5]), -0.0212448482734,atol = 0.001)
+        @test isapprox(irr(v[1:6]),  0.0866309480365,atol = 0.001)
 
         # 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 irr(cfs) ≈ 0.06463163963925866
+        @test isapprox(irr(cfs), 0.06463163963925866,atol=0.001)
 
         # test the unsolvable
 
@@ -67,8 +67,8 @@ end
     @testset "xirr with float times" begin
 
 
-        @test irr([-100,100],[0,1]) ≈ 0.0
-        @test irr([-100,110],[0,1]) ≈ 0.1
+        @test isapprox(irr([-100,100],[0,1]), 0.0, atol =0.001)
+        @test isapprox(irr([-100,110],[0,1]), 0.1, atol =0.001)
 
     end
 
@@ -78,11 +78,11 @@ end
     dates = Date(2019,12,31):Year(1):Date(2024,12,31)
     times = yearfrac.(dates[1],dates,Thirty360)
     # per Excel (example comes from Excel help text)
-    @test irr(v[1:2], times[1:2]) ≈ -0.8285714285714
-    @test irr(v[1:3], times[1:3]) ≈ -0.4435069413346
-    @test irr(v[1:4], times[1:4]) ≈ -0.1821374641455
-    @test irr(v[1:5], times[1:5]) ≈ -0.0212448482734
-    @test irr(v[1:6], times[1:6]) ≈  0.0866309480365
+    @test isapprox(irr(v[1:2], times[1:2]), -0.8285714285714, atol = 0.001)
+    @test isapprox(irr(v[1:3], times[1:3]), -0.4435069413346, atol = 0.001)
+    @test isapprox(irr(v[1:4], times[1:4]), -0.1821374641455, atol = 0.001)
+    @test isapprox(irr(v[1:5], times[1:5]), -0.0212448482734, atol = 0.001)
+    @test isapprox(irr(v[1:6], times[1:6]),  0.0866309480365, atol = 0.001)
 
     end
 end
@@ -104,3 +104,6 @@ end
         @test isnothing(breakeven([-10,-15,2,3,4,8],times,0.10))
     end
 end
+
+
+include("run_doctests.jl")