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Goldstein-Price test case #11
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f1(x1, x2) = x1 + x2 + 1
f2(x1, x2) = 19 - 14*x1 + 3*x1^2 - 14*x2 + 6*x1*x2 + 3*x2^2
f3(x1, x2) = 2*x1 - 3*x2
f4(x1, x2) = 18 -32*x1 + 12*x1^2 + 48*x2 - 36*x1*x2 + 27*x2^2
f(x1, x2) = (1 + f1(x1,x2)^2 * f2(x1,x2)) * (30 + f3(x1,x2)^2 * f4(x1,x2))
f(X) = f(X[1], X[2])
julia> minimise(f, IntervalBox(-2..2, 2)) |
julia> @time globalmin, minimisers = minimise(f, IntervalBox(-2..2, 2));
globalmin
6.653330 seconds (3.40 k allocations: 7.525 MiB)
julia> globalmin
[2.21929, 3.00003] |
With the
But using smaller tolerances takes a very long time. |
Using the mean value form instead is very fast, so closing. |
This should be added as a test and benchmark. |
This seems to have got slow again. |
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minimise struggles with this.
https://github.com/JuliaOpt/SumOfSquares.jl/blob/master/examples/goldsteinprice.jl
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