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Permalink Reply by Phil B Brubaker on November 7, 2008 at 10:28am

Calculus Programming Example #1

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A.C. Motor Design
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Figure 1 is a case study of re-engineering -- a simulation program for A.C. motor design that was reapplied as an optimization program and fully tested in about an hour. The original program had been used for trial and error design calculations by a west coast manufacturer. Restructuring this mature engineering model into a design optimization program, only required introducing a FIND statement that defined a constrained optimization problem with 12 unknown parameters and 7 constraints. A FIND statement invokes a solution technique or solver that consists of automatic differentiation with a numerical algorithm. The solver involved in this case is a nonlinear programming algorithm called THOR. Figure 2 (part of output) is a report automatically generated by the solver, THOR, summarizing the iteration steps. This partial summary shows initial guesses of the parameters and the final results.

This case study illustrates the economic benefits of AD based software for reapplying existing application software in a higher-productivity mode. The existing engineering simulation model was used "as is". It was automatically elevated to optimization by the hidden differential arithmetic. There was no mathematical analysis or algorithm design required. Because this was a re-engineering of an existing program, no debugging was necessary, and since the optimization solvers are interchangeable, different algorithms could be substituted to verify solution correctness.

Improved Productivity Example #1 Source Code:

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Figure 1. Calculus Code ... A.C. Motor (Optimal) Design

global allproblem acmotor   dimension cns(7) 
  dynamic botm, bnd 

  call input 

  call design 

  print *, '---------------initial design---------------' 

  call output



  find coiltrns, ! number of turns per coil 



&     EPDIAM, ! Separating diameter 

&     STASLOTW, ! Stator slot opening width 

&     ROTSLOTW, ! Rotor slot opening width 

&     AIRGAP, ! Air gap 

&     STATOOTW, ! Stator tooth width 

&     STABAKIR, ! Stator back iron 

&     ROTTOOTW, ! Rotor tooth width 

&     ROTBAKIR, ! Rotor back iron 

&     STASLOTO, ! Stator slot opening depth 

&     ROTSLOTO, ! Rotor slot opening depth 

&     SLIP ! Slip 

&   IN DESIGN; BY THOR(TCON); 

&   WITH BOUNDS BND; AND LOWERS BOTM; 

&   HOLDING CNS; TO MAXIMIZE EFF



  print *, '----------------optimized design-----------------' 

  call output 

end 

model design 

  f1 = 231258.0438 / coiltrns 

  q1 = sepdiam * pi / 56 : q2=sepdiam * pi / 69 : q3=sepdiam * pi / 28 

  c1 =q1/((q1-staslotw)+(staslotw *(.7 -(.036 * staslotw/ airgap)))) 

  c2 =q2/((q2-rotslotw)+(rotslotw *(.7-(.036 * rotslotw/airgap))))

  crnol = f1 * airgap * c1 * c2 / (6.96 * coiltrns * q3) 

  q4 = (statorod - stabakir - stabakir - sepdiam - .1) / 2 - .01 

  q5= (sepdiam+q4+.1) * pi / 56 - statootw - .016 : a3=q4 * q5 

  z1 = 7000 * a3 

  q6 = coiltrns * 2 / z1 : z3= - 2 : z4=32+z3 : q7=1.26**z3 * 162 

  q8=64 / 1.26**z3 : z5=stacklen+stacklen+pi * statorod / 14 

  rdc =z5 * coiltrns * 28 * q7 / 12000 

  v1 = (sepdiam - .06 - rotorid - rotbakir - rotbakir) / 2 

  v2 = (sepdiam - .06 - v1) * pi / 69 - rottootw 

  a4 = v1 * v2 : r2 = (.0133 / a4) * (stacklen / 12000) * u1 

  rrot= (56 * coiltrns)**2 * r2 / 138 : r4 = rrot 

  g3 = 3.2 * (q4 / (3 * q5)+.03 / (staslotw+q5)+stasloto / staslotw) 

  g4 = 3.2 * (v1 / (3 * v2)+.03 / (rotslotw+v2)+rotsloto / rotslotw) 

  g5 = (1 / c1+1 / c2 - 1)**2 * .26 / airgap 

  g6= (g3+g5 * q1) / 56+(g4+g5 * q2) / 69 

  xleak = xc * pi * coiltrns**2 * (.05+g6 * stacklen) ! leakage reactance 

  crstal = 115 / (sqrt(((rdc+r4)**2)+xleak**2))       ! stall current 

  crfl= 115 / (sqrt(((rdc+r4 / slip)**2)+xleak**2))   ! full load current 

  storq = 1.5792 * (crstal**2) * r4                   ! stall torque 

  rtorq = 1.5792 * (crfl**2) * r4 / slip              ! running torque 

  p8=rtorq * (1 - slip) * 1.264944649 : p6= (crstal**2) * (rdc+r4) 

  p7 = (crfl**2) * r4 / slip 

  flux = (1.4 * f1 / (statootw * stacklen) / 6450     ! flux density 

  b7 = (sqrt(crfl**2+crnol**2) * flux / crnol) / 1.4 : b2 =.13 * b7**1.9

  b3 = (((statorod**2 - sepdiam**2) * pi / 4) - 56 * a3) * 0.28 * stacklen 

  feloss = b3 * b2                                    ! iron losses 

  culsta = (b7 * crnol) / flux)**2 * rdc        ! stator copper losses 

  eff = .5 * p8 / (p7+feloss / 2+culsta)        ! efficiency - maximize this 

  rac =13225 / feloss                           ! a.c. resistance 

  xmag =115 / crnol                             ! magnetizing reactance 

  culrot = crfl * crfl * rrot / staslotw        ! rotor copper losses 

  cns(1) = statorod - .1 - sepdiam     ! stator o.d. > = separation diam+.1 

  cns(2) = sepdiam - rotorid+.1        ! separation diam > = rotor i.d. - .1 

  cns(3) = ((sepdiam * pi / 69) - .035) - rottootw   ! rotor tooth geometry 

  cns(4)=.5*(sepdiam - rotorid) -.025 -rotbakir ! rotor back iron geometry 

  cns(5) = storq - 60                           ! stall torque > = 60 

  cns(6) = 19 - flux                            ! flux density < = 19 

  cns(7) =.05 - slip                            ! slip < = 5 percent 

end

As this example illustrates, the benefits of multiple - parameter optimization in practical engineering calculation can often be dramatic. Because of the large number of parameters, something on the order of a billion to a trillion computer cases would have been necessary to achieve this result if using the original program. The bottom line benefit is dramatically reduced cost, higher engineering productivity, and immediate results to meet tight schedules.

Improved Productivity Example #1 Solver Output:

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Figure 2. (Partial) Output Report ... A.C. Motor (Optimal) Design

---- Thor Summary, Invoked At Acmotor[29] For Model Design---- Convergence Condition After 35 Iterations     o
     o

     o

Loop Number......     [Initial]         35

Unknowns

coiltrns        2.900000e+01    3.214302e+01sepdiam         4.000000e+00    5.010000e+00staslotw        6.000000e-02    6.570807e-02
rotslotw        3.500000e-02    3.500000e-02

airgap          8.000000e-03    8.000000e-03

statootw        1.350000e-01    1.687119e-01

stabakir        1.500000e-01    3.955060e-01

rottootw        1.100000e-01    8.885508e-02

rotbakir        4.700000e-01    9.700004e-01

stasloto        3.500000e-02    2.500000e-02

rotsloto        1.500000e-02    0.000000e+00

slip            3.000000e-02    5.000000e-02 

Objective 

Eff             4.020854e-01    7.312523e-01 

Inequality Constraints 

cns ( 1)        1.010000e+00    0.000000e+00cns ( 2)        1.225000e+00    2.235000e+00cns ( 3)        3.712132e-02    1.042519e-01
cns ( 4)        6.750000e-02    7.249963e-02

cns ( 5)       -7.555936e+00    6.415480e-04

cns ( 6)        1.258933e+01    1.437190e+01

cns ( 7)        2.000000e-02    0.000000e+00 

--- End Of Loop Summary

Another improved productivity example do to using Calculus (level) programming.

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Permalink Reply by Masood Humayun on November 12, 2008 at 7:31pm

This is very good.! Especially for Engineers and people in Engineering field. Can we also have Linear Programming models for solving sales or marketing problems leading to generation of profits. Can someone help. - Masood

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Permalink Reply by Phil B Brubaker on November 13, 2008 at 8:25am

Yes, do you have such a math problem? If so, I'll gladly solve it if you'll add a short description similar to the above example. - Phil

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