CODEWITHSUNDEEP
exceljulia
BL_7
BL_7

Reputation: 63

Does Julia have a way to solve for unknown variables

Is there a function in Julia that is similar to the solver function in Excel where I can provide and equation, and it will solve for the unknown variable? If not, does anybody know the math behind Excel's solver function? I am not expecting anybody to solve the equation, but if it helps:

Price = (Earnings_1/(1+r)^1)+(Earnings_2/(1+r)^2)++(Earnings_3/(1+r)^3)+(Earnings_4/(1+r)^4)+(Earnings_5/(1+r)^5)+(((Earnings_5)(RiskFreeRate))/((1+r)^5)(1-RiskFreeRate))

The known variables are: Price, All Earnings, and RiskFreeRate. I am just trying to figure out how to solve for r.

Upvotes: 1

Views: 663

Answers (2)

Chris Rackauckas
Chris Rackauckas

Reputation: 19132

Write this instead as an expression f(r) = 0 by subtracting Price over to the other side. Now it's a rootfinding problem. If you only have one variable you're solving for (looks to be the case), then Roots.jl is a good choice. 

fzero(f, a::Real, b::Real)

will search for a solution between a and b for example, and the docs have more choices for algorithms when you don't know a range to start with and only give an initial condition for example.

In addition, KINSOL in Sundials.jl is good when you know you're starting close to a multidimensional root. For multidimensional and needing some robustness to the initial condition, I'd recommend using NLsolve.jl.

Upvotes: 2

Bathsheba
Bathsheba

Reputation: 234715

There's nothing out of the box no. Root finding is a science in itself.

Luckily for you, your function has an analytic first derivative with respect to r. That means that you can use Newton Raphson, which will be extremely stable for your function.

I'm sure you're aware your function behaves badly around r = -1.

Upvotes: 1

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