format version = 1.0 begin solution of problem simpleLin2 date = 9-Feb-09 10:53:56 url = http://www.cs.chalmers.se/~dag/identification // Settings for the Gennemark & Wedelin (2007) algorithm randomStarts_ = 60 1 smoothingParameter = 2.6 randomSeed = 4 // VARIABLE 3 reaction_1 of variable_3 has type = uniMolecularMassAction has variable X1 = variable_3 has parameter k1 = -0.975728061525E+00 reaction_2 of variable_3 has type = biMolecularMassAction has variable X1 = variable_4 has variable X2 = variable_1 has parameter k1 = 0.989014896643E+00 // VARIABLE 4 reaction_1 of variable_4 has type = uniMolecularMassAction has variable X1 = variable_3 has parameter k1 = 0.975728061525E+00 reaction_2 of variable_4 has type = biMolecularMassAction has variable X1 = variable_4 has variable X2 = variable_1 has parameter k1 = -0.989014896643E+00 reaction_3 of variable_4 has type = biMolecularMassAction has variable X1 = variable_4 has variable X2 = variable_2 has parameter k1 = -0.944645190160E+00 reaction_4 of variable_4 has type = uniMolecularMassAction has variable X1 = variable_5 has parameter k1 = 0.909149010310E+00 // VARIABLE 5 reaction_1 of variable_5 has type = biMolecularMassAction has variable X1 = variable_4 has variable X2 = variable_2 has parameter k1 = 0.944645190160E+00 reaction_2 of variable_5 has type = uniMolecularMassAction has variable X1 = variable_5 has parameter k1 = -0.909149010310E+00 // INITIAL VALUES variable_3 of experiment_1 has initialValue = 0.9794E+00 variable_4 of experiment_1 has initialValue = 0.1001E-01 variable_5 of experiment_1 has initialValue = 0.9913E-02 variable_3 of experiment_2 has initialValue = 0.9497E+00 variable_4 of experiment_2 has initialValue = 0.1020E-01 variable_5 of experiment_2 has initialValue = 0.1053E-01 variable_3 of experiment_3 has initialValue = 0.1025E+01 variable_4 of experiment_3 has initialValue = 0.1124E-01 variable_5 of experiment_3 has initialValue = 0.1026E-01 variable_3 of experiment_4 has initialValue = 0.9570E+00 variable_4 of experiment_4 has initialValue = 0.1054E-01 variable_5 of experiment_4 has initialValue = 0.9949E-02 variable_3 of experiment_5 has initialValue = 0.9573E+00 variable_4 of experiment_5 has initialValue = 0.1025E-01 variable_5 of experiment_5 has initialValue = 0.1032E-01 variable_3 of experiment_6 has initialValue = 0.9727E+00 variable_4 of experiment_6 has initialValue = 0.9640E-02 variable_5 of experiment_6 has initialValue = 0.1044E-01 variable_3 of experiment_7 has initialValue = 0.9580E+00 variable_4 of experiment_7 has initialValue = 0.9887E-02 variable_5 of experiment_7 has initialValue = 0.9636E-02 variable_3 of experiment_8 has initialValue = 0.9771E+00 variable_4 of experiment_8 has initialValue = 0.9799E-02 variable_5 of experiment_8 has initialValue = 0.9588E-02 // ERROR // error = -L + lambda * K // residual = -L error = 140.177161 residual = 132.177161 end of solution of problem simpleLin2