You can't, at least up to VA 8.5 (as far as I know). Every time you need to change the custom graph, reimport it in your report, and replace your object. There is something that can be done with BIRD to change the embedded code of the custom graph (it's just a string) but is way too complicate. ... View more

I'm looking at "layers" and "layersOut" but it's not working, the output is weird with columns like "_layeract_0_img_0_" which contain characters. It seems something related to images but my model has nothing to do with images. What's wrong? Any help?   EDIT: Even with the example I couldn't find a way to see all the layers values row by row: https://documentation.sas.com/?docsetId=casdlpg&docsetTarget=p024gf3bj0rmwcn1tfkyy68dzs4g.htm&docsetVersion=8.5&locale=en   Thank you very much Regards ... View more

Thanks a lot!   Just as an exercise I'm trying to recreate the standard MLP Autoencoder. Here's what I did:   cas session sessopts=(caslib="CASUSER"); libname CASUSER cas caslib="CASUSER"; data CASUSER.TRAIN; do id=1 to 10000; x=id/10000; /* Fake relation */ y=x+2; z=x*2-sqrt(x); output; end; run; data CASUSER.SCORE; set CASUSER.TRAIN; if id = 1 then z=z*10; /* anomaly */ if id > 10 then delete; run; proc cas; deepLearn.buildModel / modelTable={name="MODELTABLE" replace=true} type="DNN"; run; deepLearn.addLayer / layer={type="INPUT" std="STD"} modelTable={name="MODELTABLE"} name="LInput"; run; deepLearn.addLayer / layer={type="FC" n=1} modelTable={name="MODELTABLE"} name="LHidden" srcLayers="LInput"; run; deepLearn.addLayer / layer={type="OUTPUT"} modelTable={name="MODELTABLE"} name="LOutput" srcLayers="LHidden"; run; deepLearn.modelInfo / modelTable={name="MODELTABLE"}; run; deepLearn.dlTrain / inputs={"x" "y" "z"} modelTable={name="MODELTABLE"} modelWeights={name="MODELWEIGHTS" replace=true} table={name="TRAIN"} optimizer={loglevel=1}; run; quit; proc cas; deepLearn.dlScore / initWeights={name="MODELWEIGHTS"} modelTable={name="MODELTABLE"} table={name="SCORE"} casout={name="SCORED" replace=true} copyvars={"id"}; run; quit; However I can't see the scored layers values to calculate the mse row by row. What am I doing wrong? Is this simple code correct? Thanks! ... View more

Thanks, however nnet seems to support only the standard Autoencoder and not the LSTM form. I believe the model should be created "by hand" layer per layer with the deepLearn action set, but I don't know how to implement it. ... View more

AFAIK, there is no way to massively update a CAS table "in place" (i.e. without dropping it and recreating it), as well as DELETE rows.   I understand CAS is for analytical, but I believe it lacks A LOT of database features. I really hope that in the next releases the CAS will be more robust from this point of view, we had to rewrite tons of processes that worked correctly in LASR environment (which supports the UPDATE for instance) to implement workarounds for the things that cannot be made in CAS. ... View more

Yes the constraints are nonlinear, but they are "nice" (in the sense that they are almost always polynomials and in general continuous and derivable, that is why I expect this sense of continuity of the solutions over the grid).   I picked up these two instances:   1) P=6, H=85. With nlp: Q1=9.58, Q2=0.00, Q=9.58 (status=OPTIMAL). With lso: Q1=5.30, Q2=7.60, Q=12.90 (status=MAXGEN).   2) P=10, H=91.5. With nlp: Q1=13.63, Q2=0.00, Q=13.63 (status=OPTIMAL). With lso: Q1=7.26, Q2=9.52, Q=16.78 (status=ABSFCON).   I'll try to rerun it with the double solve command. In the mean time I think I'll start to implement approach no.2, let's see if it's reliable.   Thanks again for all you suggestions! Regards ... View more

Unfortunately the solution got worse... Here's the result, and most of the points have status MAXGEN instead of ABSFCON.   ... View more

  Thanks, we are getting closer    About the first approach with lso. I tried to run the model without changing the constraints (so my original problem) to see what are the results comparing nlp with lso. Let me explain the result: the variables the model finds are Q1 and Q2, for every P and H in a grid (as you can see in the code); then I calculate P1 and P2, and since P1+P2=P I calculate the ratio P1/P to get one number as result; in this way, when the ratio is 100% it means that P1=P and P2=0, when it is 0% it means that P1=0 and P2=P, otherwise 0 < P1,P2 < P. In the following graph I display in the P,H grid, this ratio with color: yellow (ratio=100%) blue (ratio=0%) brownish (0 < ratio < 100).   Here's the solution with nlp:   Here's the solution with lso: The solution with nlp makes a lot of sense (from the physical point of view of the thing I'm optimizing): when P is low, P1=P and P2=0; when P is medium, P1=0 and P2=P; when P is high, a combination of P1 and P2 is used. And you can see from the picture a sense of "continuity" in the grid. I know it may sound obscure but trust me this result is good!   Now the solution with lso seems a lot more noisy: the sense of "continuity" seems not to apply, and also the P low/medium/high is unclear. So before modify the constraint to fit the new problem of this thread, I'd like to get a more stable solution with lso, as "good" as the nlp one. Do you know if the lso can be configured to be more accurate? I read the doc and I saw a lot of parameters, I'm not sure about what to do to improve the results.     As for the second approach: yes I taught about this, and I believe it works fine. However for now I wrote the model with two "units" (you can see the reference to indexes "1" and "2" all over the code), but I'm sure I will be asked later to generalize it to work with N "units". So with N "units", I should have to solve 2^N-1 problems with all the combinations of P_i (zero or non zero) and then compare them, and it sounds hard...   As for the thirds approach, I haven't tried yet, but I think it will not be accurate at all since in this way the error on constraints won't have a metric to see how wrong it is, it would only get a flag "correct" "incorrect".   Thanks again, Regards ... View more

Here is the whole model:   proc fcmp outlib=work.funcs.f; function f1(Q, H); return ( (7.18895803108255)+(Q)*(0.90898166129053)+(H)*(-0.52518338771685)+(Q**2)*(-0.07243939757267)+(Q)*(H)*(-0.0101157174044)+(H**2)*(0.01140027697845)+(Q**3)*(0.00271247731414)+(Q**2)*(H)*(0.00078575662965)+(Q)*(H**2)*(-8.0331130545237E-7)+(H**3)*(-0.00010011945215)+(Q**4)*(-0.00005707623607)+(Q**3)*(H)*(-0.0000145464761)+(Q**2)*(H**2)*(-4.0740624504878E-6)+(Q)*(H**3)*(5.6715539778245E-7)+(H**4)*(3.0399728809992E-7)+(Q**5)*(6.2706019750555E-7)+(Q**4)*(H)*(-4.9275589970883E-8)+(Q**3)*(H**2)*(1.1402170133176E-7)+(Q**2)*(H**3)*(-1.1450221400824E-8) ); endsub; function f2(Q, H); return ( (9.56532591440209)+(Q)*(1.56477959347719)+(H)*(-0.58834848839521)+(Q**2)*(-0.14923333670627)+(Q)*(H)*(-0.02042582402532)+(H**2)*(0.01042030607175)+(Q**3)*(0.00551139580932)+(Q**2)*(H)*(0.00200710134061)+(Q)*(H**2)*(0.00006229314138)+(H**3)*(-0.00006493899805)+(Q**4)*(-0.00009774971067)+(Q**3)*(H)*(-0.00004756024059)+(Q**2)*(H**2)*(-0.00001062838926)+(Q)*(H**3)*(3.7950591182214E-7)+(Q**5)*(8.5938390272548E-7)+(Q**4)*(H)*(1.7658335207072E-7)+(Q**3)*(H**2)*(2.2050828738805E-7)+(H**5)*(8.7276517928309E-10) ); endsub; function g1(Q1,Q2); return ( 0.003134 * ((Q1+Q2)**2) + 0.003089 * (Q1**2) ); endsub; function g2(Q1,Q2); return ( 0.003134 * ((Q1+Q2)**2) + 0.003089 * (Q2**2) ); endsub; run; %let Q1max = 30.4; %let Q2max = 29.5; %let P1min = 2.5; %let P2min = 3.3; %let P1max = 21.7; %let P2max = 21.6; %let PSETL = 1; %let PSETR = 43; %let HSETL = 70; %let HSETR = 94; %let STEP = 0.5; options cmplib=work.funcs; proc optmodel nthreads=64; number P; number H; var Q1 >= 0 <= &Q1max; var Q2 >= 0 <= &Q2max; impvar HL1 = H - g1(Q1,Q2); impvar HL2 = H - g2(Q1,Q2); impvar P1 = Q1 * HL1 * f1(Q1, HL1) * 9.81 / 1000; impvar P2 = Q2 * HL2 * f2(Q2, HL2) * 9.81 / 1000; minimize Q = Q1 + Q2; constraint P = P1 + P2; constraint 0 <= P1 <= &P1max; /* (*) */ constraint 0 <= P2 <= &P2max; /* (*) */ set PSET = &PSETL .. &PSETR by &STEP; set HSET = &HSETL .. &HSETR by &STEP; num Qopt {PSET, HSET}; num Q1opt {PSET, HSET}; num Q2opt {PSET, HSET}; num P1opt {PSET, HSET}; num P2opt {PSET, HSET}; str solstatus {PSET, HSET}; num opterror {PSET, HSET}; do P = PSET; do H = HSET; solve with nlp / maxiter=10000 multistart=(maxstarts=10) seed=54321; put P= H= HL1= HL2= P1= P2= Q1= Q2= _solution_status_=; Qopt[P,H] = Q.sol; Q1opt[P,H] = Q1.sol; Q2opt[P,H] = Q2.sol; P1opt[P,H] = P1; P2opt[P,H] = P2; solstatus[P,H] = _solution_status_; opterror[P,H] = _OROPTMODEL_NUM_['OPTIMALITY_ERROR']; end; end; create data Qdata from [P H] P1opt P2opt Q1opt Q2opt Qopt solstatus opterror; quit;   Look at the rows marked with (*) constraint 0 <= P1 <= &P1max; /* (*) */ constraint 0 <= P2 <= &P2max; /* (*) */ As you can see, P1 and P2 have 0 as lower bound, and it worked very well. Now I'm asked to restrict P1 and P2 interval from [0,PMAX] to [PMIN, PMAX], but zero must remain allowed, so the constraints should become (P1 = 0) OR (&P1min <= P1 <= &P1max) (P2 = 0) OR (&P2min <= P2 <= &P2max) Hence this question. I hope I made myself clear now!   Thanks again, Regards ... View more

Thanks! Unfortunately my problem is non linear, so the solver is nlp. And using binary variables, it says: ERROR: The NLP solver does not allow integer variables. ... View more