7 September 2023 8:48:26.950 AM uncmin_test(): FORTRAN77 version. Test uncmin(). TEST01 Test OPTIF0, the simple interface to UNCMIN. RESULT - Iterate 0 Step 0.00000 0.00000 X 1.00000 1.00000 Function value = 783.98151627258801 Gradient vector 759.398 2395.63 Hessian matrix 1.00000 0.00000 0.00000 1.00000 RESULT - Iterate 2 X 20.0000 -20.1959 Function value = 90.999999390008171 Gradient vector 0.629141E-06 -0.600787E-06 Output from UNCMIN: Error code= 1 EXPLAIN: UNCMIN returned a termination code of 1 This code has the following meaning: The gradient is relatively close to zero. The current iterate is probably a good solution. F(X*) = 90.999999390008171 X* = 20.000000015249810 -20.195924304682116 (Partial reference results for comparison:) 19.9145 -20.6011 -5.26250 19.9900 -20.6230 19.9145 20.0100 -20.6230 19.9145 19.9900 -20.6023 19.9145 Error code = 1 F(X*) = 91.0001 X* = 19.9900 -20.6230 uncprb function of 3 variables. Solve problem using line search. RESULT - Iterate 0 Step 0.00000 0.00000 0.00000 X -1.00000 0.00000 0.00000 Function value = 95579.451357303798 Gradient vector -0.00000 -61831.9 -6183.19 Hessian matrix 200.000 -61831.9 0.00000 -61831.9 20000.0 2000.00 0.00000 2000.00 202.000 RESULT - Iterate 15 X 1.00000 -0.855374E-13 -0.830642E-12 Function value = 8.4434898490436188E-025 Gradient vector -0.610623E-11 -0.494644E-10 0.328516E-11 termination code itrmcd= 1 EXPLAIN: UNCMIN returned a termination code of 1 This code has the following meaning: The gradient is relatively close to zero. The current iterate is probably a good solution. return code msg= 0 x* = 0.99999999999996947 -8.5537399812474531E-014 -8.3064179595742699E-013 f(x*) = 8.4434898490436188E-025 gradient= -6.1062266354425922E-012 -4.9464404334689922E-011 3.2851568415540391E-012 uncprb function of 3 variables. Solve problem using double dogleg method. RESULT - Iterate 0 Step 0.00000 0.00000 0.00000 X -1.00000 0.00000 0.00000 Function value = 95579.451357303798 Gradient vector -0.00000 -61831.9 -6183.19 Hessian matrix 200.000 -61831.9 0.00000 -61831.9 20000.0 2000.00 0.00000 2000.00 202.000 RESULT - Iterate 9 X 1.00000 0.839933E-08 0.659617E-07 Function value = 2.1160045229905760E-012 Gradient vector -0.288384E-04 0.360632E-04 -0.347439E-05 termination code itrmcd= 1 EXPLAIN: UNCMIN returned a termination code of 1 This code has the following meaning: The gradient is relatively close to zero. The current iterate is probably a good solution. return code msg= 0 x* = 0.99999985580777384 8.3993303859146700E-009 6.5961742827351511E-008 f(x*) = 2.1160045229905760E-012 gradient= -2.8838445534733249E-005 3.6063151243759206E-005 -3.4743911429409752E-006 uncprb function of 3 variables. Solve problem using more-hebdon method. RESULT - Iterate 0 Step 0.00000 0.00000 0.00000 X -1.00000 0.00000 0.00000 Function value = 95579.451357303798 Gradient vector -0.00000 -61831.9 -6183.19 Hessian matrix 200.000 -61831.9 0.00000 -61831.9 20000.0 2000.00 0.00000 2000.00 202.000 RESULT - Iterate 10 X 1.00000 -0.568346E-12 -0.524357E-11 Function value = 4.7738109688649686E-023 Gradient vector -0.188960E-10 -0.879783E-09 0.774912E-10 termination code itrmcd= 1 EXPLAIN: UNCMIN returned a termination code of 1 This code has the following meaning: The gradient is relatively close to zero. The current iterate is probably a good solution. return code msg= 0 x* = 0.99999999999990552 -5.6834601901630654E-013 -5.2435687256218238E-012 f(x*) = 4.7738109688649686E-023 gradient= -1.8895995879620185E-011 -8.7978292908362984E-010 7.7491155457112088E-011 uncmin_test(): Normal end of execution. 7 September 2023 8:48:26.950 AM