1. Examples Workshop

2. Workshop Supplement January 30, 2001 Inventory #001450 EX-2 Problem: Minimize the function y = 10 - (5 + x/4)sin(x) with x ranging from 0.1 rad to 10 rad. This problem is of interest because of the local minimum and the difficulty of approximating the entire range with a least squares fit x y pi/25pi/2 10 True minimum Local minimum x min =.1 radx max =10 rad Example A Minimization of a Sine Function

3. Workshop Supplement January 30, 2001 Inventory #001450 EX-3 Example A Minimization of a Sine Function Solution: Start with 5 design sets obtained from a sweep, then solve using both the subproblem and first- order methods. ! 5-pt sweep run /title, ( 10 - expanding sine wave ) a=1.5 b=10-(5+a/4)*sin(a) finish /opt opvar,a,dv,.1,10 opvar,b,obj optype,sweep opsweep,,5 opexe opdel,1! remove initial solution opsave,,opt0 ! Subproblem method opresu,,opt0 optype,subp opexe oplist,all ! First-order method opresu,,opt0 optype,first opexe oplist,all finish

4. Workshop Supplement January 30, 2001 Inventory #001450 EX-4 Example A...Minimization of a Sine Function Subproblem Results SET 2 SET 3 SET 4 *SET 5* (FEASIBLE) (FEASIBLE) (FEASIBLE) (FEASIBLE) A (DV) 0.10000 2.5750 5.0500 7.5250 B (OBJ) 9.4983 6.9707 15.909 3.4878 SET 6 SET 7 SET 8 SET 9 (FEASIBLE) (FEASIBLE) (FEASIBLE) (FEASIBLE) A (DV) 10.000 5.8992 6.2375 7.5026 B (OBJ) 14.080 12.426 10.299 3.5446 First Order Results SET 2 SET 3 SET 4 SET 5 (FEASIBLE) (FEASIBLE) (FEASIBLE) (FEASIBLE) A (DV) 0.10000 2.5750 5.0500 7.5250 B (OBJ) 9.4983 6.9707 15.909 3.4878 SET 6 *SET 7* SET 8 (FEASIBLE) (FEASIBLE) (FEASIBLE) A (DV) 10.000 7.8861 7.8470 B (OBJ) 14.080 3.0321 3.0384

5. Workshop Supplement January 30, 2001 Inventory #001450 EX-5 Problem: Find the value of 'a' to minimize the length of the two cables. Exact solution is a=3.0. Solution: We will use the subproblem method. a=5 /title, Minimize cable length /prep7 k,1,,3 k,2,a k,3,10,7 ltot=distkp(1,2)+distkp(2,3) finish /opt opvar,a,dv,0,10 opvar,ltot,obj optype,subp opexe oplist,all finish /exit Example B Cable Length Minimization SET 1 SET 2 SET 3 SET 4 (FEASIBLE) (FEASIBLE) (FEASIBLE) (FEASIBLE) A (DV) 5.0000 8.4733 4.5773 3.2235 LTOT (OBJ) 14.433 16.153 14.328 14.146 *SET 5* (FEASIBLE) A (DV) 2.9583 LTOT (OBJ) 14.142 a10-a 3 7 Cables

7. Workshop Supplement January 30, 2001 Inventory #001450 EX-7 /title,Lowest point on a sphere ! Spherical surface, concave up a=9 b=9 c=10-sqrt(5**2-(a-6)**2-(b-6)**2) /opt opvar,a,dv,3,9 opvar,b,dv,3,9 opvar,c,obj optype,first opexe oplist,all finish /exit Example C Minimum Point on a Spherical Surface Problem: Find the minimum point on a 10-inch hemispherical surface. Solution: A spherical surface is the easiest minimization problem for the First Order method because the gradient at any point on the surface points directly toward the minimum location. The input shown here represents a 10-inch diameter sphere centered at (6,6,10). Results from the subproblem method are also shown for comparison. Note the rapid (and exact) convergence of the first-order method. SET 1 SET 2 SET 3 SET 4 *SET 5* (FEASIBLE) (FEASIBLE) (FEASIBLE) (FEASIBLE) (FEASIBLE) A (DV) 9.0000 8.0830 5.3280 6.2254 5.8146 B (DV) 9.0000 5.7431 7.1542 6.7075 6.6649 C (OBJ) 7.3542 5.4618 5.1817 5.0554 5.0479 SET 1 SET 2 *SET 3* SET 4 (FEASIBLE) (FEASIBLE) (FEASIBLE) (FEASIBLE) A (DV) 9.0000 5.2918 6.0000 5.7295 B (DV) 9.0000 5.2918 6.0000 5.7295 C (OBJ) 7.3542 5.1013 5.0000 5.0147 Subproblem Results First Order Results

9. Workshop Supplement January 30, 2001 Inventory #001450 EX-9 Example D Shape Optimization of a Block Problem: Maximize the volume of a block subject to a surface area constraint of 100 units. Solution: This is another trivial problem which solves instantly with the First Order method. The problem is constrained using the surface area as a state variable. /title,block shape optimization a=1 b=1 c=1 v=100-a*b*c ! Convert to minimization form d=2*a*b+2*b*c+2*a*c ! Surface area /opt opvar,a,dv,0,100 opvar,b,dv,0,100 opvar,c,dv,0,100 opvar,d,sv,99,101 ! 99 100 opvar,v,obj optype,first opexe oplist,all finish /exit SET 1 SET 2 SET 3 *SET 4* (INFEASIBLE) (FEASIBLE) (FEASIBLE) (FEASIBLE) D (SV) > 6.0000 100.54 100.70 100.73 A (DV) 1.0000 4.0934 4.2121 4.2126 B (DV) 1.0000 4.0934 4.2121 4.2126 C (DV) 1.0000 4.0934 3.8709 3.8715 V (OBJ) 99.000 31.411 31.324 31.296

10. Workshop Supplement January 30, 2001 Inventory #001450 EX-10 SET 1 SET 2 SET 3 SET 4 (INFEASIBLE) (INFEASIBLE) (INFEASIBLE) (INFEASIBLE) D (SV) > 6.0000 17900. 22672. 8078.9 E (SV) 6.0000 > 17900. > 22672. > 8078.9 A (DV) 1.0000 84.733 69.268 84.800 B (DV) 1.0000 45.773 53.802 0.43256 C (DV) 1.0000 38.861 61.830 46.963 V (OBJ) 99.000 -0.15062E+06 -0.23033E+06 -1622.6 SET 5 SET 6 SET 7 SET 8 (INFEASIBLE) (INFEASIBLE) (INFEASIBLE) (FEASIBLE) D (SV) 18068. 576.84 > 55.559 99.320 E (SV) > 18068. > 576.84 55.559 99.320 A (DV) 32.097 6.0313 13.373 24.146 B (DV) 79.294 43.956 1.0024 1.0146 C (DV) 58.255 0.46619 1.0000 1.0000 V (OBJ) -0.14816E+06 -23.594 86.596 75.501 SET 9 SET 10 SET 11 SET 12 (INFEASIBLE) (INFEASIBLE) (INFEASIBLE) (INFEASIBLE) D (SV) > 82.613 20040. 1957.6 4811.9 E (SV) 82.613 > 20040. > 1957.6 > 4811.9 A (DV) 20.271 0.10000 15.588 57.994 B (DV) 0.98893 100.00 14.246 12.084 C (DV) 1.0000 100.00 25.364 24.331 V (OBJ) 79.953 -900.00 -5532.6 -16952. Example D...Shape Optimization of a Block Note that the Subproblem method is not very effective for this case despite converting the surface area constraint from an equality constraint to two one-sided constraints.