time - f# break function evaluation -

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i have minimize quite complicate function. minimization use nonlinearprogram extreme optimization library. since there´s no way find global minimum, use different startpoints , choose, "best minimum". problem there can startpoints, evaluatin can take long time. there general way in f# or special method in extreme optimization, stop evaluation let´s after 10 min , give list [nan; nan; nan; nan; nan; nan] back?

let funcfindpara (startpoint:float list) func =       let nlp = new nonlinearprogram(6)      // add function     nlp.objectivefunction <- (fun x -> func x.[0] x.[1] x.[2] x.[3] x.[4] x.[5])      // add lineare constraints     nlp.addlinearconstraint("a + d > 0", vector.create(1.0, 0.0, 0.0, 1.0, 0.0, 0.0), 1.0e-5, infinity) |> ignore     nlp.addlinearconstraint("c > 0", vector.create(0.0, 0.0, 1.0, 0.0, 0.0, 0.0), 1.0e-5, infinity) |> ignore     nlp.addlinearconstraint("d > 0", vector.create(0.0, 0.0, 0.0, 1.0, 0.0, 0.0), 1.0e-5, infinity) |> ignore     nlp.addlinearconstraint("gamma > 0", vector.create(0.0, 0.0, 0.0, 0.0, 1.0, 0.0), 1.0e-5, infinity) |> ignore     nlp.addlinearconstraint("0 < rho_infty <= 1", vector.create(0.0, 0.0, 0.0, 0.0, 0.0, 1.0), 1.0e-5, 1.0) |> ignore      // add nonlinear constrains      // gamma <= -ln(rho_infty)     nlp.addnonlinearconstraint((fun (x : vector) -> x.[4] + log(x.[5])), constrainttype.lessthanorequal, 0.0, (fun (x : vector) -> fun (y : vector) ->            y.[0] <- 0.0            y.[1] <- 0.0            y.[2] <- 0.0           y.[3] <- 0.0           y.[4] <- 1.0           y.[5] <- 1.0 / x.[5]           y         )     ) |> ignore      // add starting point     nlp.initialguess <- vector.create(startpoint.[0], startpoint.[1], startpoint.[2], startpoint.[3], startpoint.[4], startpoint.[5])      // solve     let solution = nlp.solve()      // return list parameters     list.init 6 (fun index -> solution.[index])

you wrap function async { } , pass runsynchronously along timeout:

let withtimeout f timeout defaultvalue =     try async.runsynchronously((async { return f() }), timeout)     :? system.timeoutexception -> defaultvalue  let longfn() =      system.threading.thread.sleep(5000)     [1.0; 2.0; 3.0]  //usage withtimeout longfn 2000 [nan; nan; nan]

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