About RobPratt

RobPratt · ‎06-09-2012

Here is how to discretize a bounded variable y: proc optmodel; var y >= 0 <= 3; num num_steps = 10; set STEPS = 0..num_steps-1; num a {s in STEPS} = (s - y.lb) / (y.ub - y.lb); var yb {STEPS} binary; con assign: sum {s in STEPS} yb = 1; con link: sum {s in STEPS} a * yb = y; quit;

RobPratt · ‎06-09-2012

And here's how you can do the sums if you declare wt and sp as in my earlier reply: impvar wt_sum {i in w} = sum {j in 1..n} wt[i,j]; impvar sp_sum {i in w} = sum {j in 1..n} sp[i,j];

RobPratt · ‎06-09-2012

Here's an alternative approach that uses fewer PROC OPTMODEL statements: %let n = 3; proc optmodel; num n = &n; set w; num weeks{w}; num dep{w}; num m{w,1..n}; read data M into w = [_n_] dep {j in 1..n} <m[_N_,j]=col('m'||j)>; var wt {w,1..n} >= 0 <= 1 init 1; impvar sp {i in w, j in 1..n} = m[i,j] * wt[i,j]; print sp; quit;

RobPratt · ‎05-28-2012

If you multiply both sides by y, you get x + y z = c y. If y and z are binary variables, you can replace the product y z with a new binary variable w and add some constraints. If only one of y and z is binary and the other is bounded, you can linearize the product in a similar manner. If neither of y and z is binary, you can discretize one of them and apply the same techniques. For details, please see Chapters 10 and 24 in the new book of mathematical programming examples available as part of the SAS/OR documentation: https://support.sas.com/pubscat/bookdetails.jsp?pc=64980 (click on View This Title Online for a free pdf version).

RobPratt · ‎05-25-2012

We have a new book of mathematical programming examples available as part of the SAS/OR documentation: https://support.sas.com/pubscat/bookdetails.jsp?pc=64980 (click on View This Title Online for a free pdf version). READ DATA and CREATE DATA are very powerful and expressive statements that are probably best learned via examples, and the book includes several examples of reading and creating dense and sparse data sets. For your particular example, the code below does what I think you have in mind. I have also eliminated set S and populated set N during the READ DATA. You could also populate wee by using READ DATA if you have a new data set with one observation per week. data move2; input week prod base; datalines; 1 1 200 1 2 105 1 3 40 1 4 70 2 1 180 2 2 135 2 3 60 2 4 71 ; proc optmodel; set N; /* Number of Products */ set wee = 1..2; /* Weeks */ number base{wee, N}; number cost{N}; number elast{N}; read data Prod into N=[prods] cost elast; read data move2 into [week prod] base;

RobPratt · ‎05-25-2012

No, (x1+x2) / (y1+y2) = 2 is not a linear equation and in fact it is not exactly equivalent to the linear equation x1+x2 = 2*(y1+y2) since the latter allows the solution (0,0,0,0). When you linearize constraints of this form (and even more so for inequalities), you have to be careful about the sign of the denominator. For example, (x1+x2) / (y1+y2) >= 2 is equivalent to x1+x2 >= 2*(y1+y2) if and only if y1+y2 > 0.

RobPratt · ‎05-10-2012

Even in the quadratic case, the code I sent (tom022012_heur.sas) to solve the continuous problem and then find the closest integer solution is only a heuristic. The resulting solution is integer feasible, but not necessarily optimal, to the original problem. But as long as the optimal MILP objective value is small, the nonlinear objective value of the resulting integer solution will not be much different than the optimal continuous nonlinear objective value, so you should get very tight bounds on the optimal objective (lower bound from the continuous solve, upper bound from the integer feasible solution).

RobPratt · ‎05-09-2012

You cannot have this circular dependency with IMPVAR. Everything on the right-hand side has to be declared before you can declare anything that depends on it. It sounds like you instead need to declare explicit variables (VAR) instead of implicit variables (IMPVAR), and then you can use constraints (CON) to enforce the equalities you want to hold. Something like this: var updatedRisk {2..n}; var calculatedRisk {2..n}; con mycon1 {j in 2..n}: updatedRisk = sum {i in 1..j-1} calculatedRisk * newRisk[n-j+1+i]; con mycon1 {j in 2..n}: calculatedRisk = updatedRisk + actualRisk ;

RobPratt · ‎05-09-2012

That is correct. Declarations (IMPVAR, VAR, NUM, etc.) cannot be nested in other statements (DO, FOR, etc.). See http://support.sas.com/documentation/cdl/en/ormpug/63975/HTML/default/viewer.htm#ormpug_optmodel_sect019.htm.

RobPratt · ‎05-09-2012

1. Yes, you can have multiple statements within a DO loop. 2. No, you cannot update the formula for an IMPVAR after the declaration. I gave correct syntax for the updatedRisk IMPVAR declaration in the previous reply. I think you just need a second IMPVAR statement for actualRisk. I'm not sure, but it looks to me like you might be trying to calculate your objective function by writing a DO loop. In OPTMODEL, objectives are instead declared with a single algebraic expression. In your case, you might need to use the SUM operator with multiple-dimensional indices, something like: sum {i in 1..n, j in 1..i, k in 1..j} mynum[i,j,k] * myvar[i,j,k]. If you write out your objective function on paper as a single algebraic expression, the conversion to the corresponding OPTMODEL code will be immediate.

RobPratt · ‎05-09-2012

Regarding linear interpolation, PROC OPTMODEL does not have anything built in. But you can try some of the ideas suggested in this blog posting: http://blogs.sas.com/content/iml/2012/03/16/linear-interpolation-in-sas/ and this Usage Note: http://support.sas.com/kb/24/560.html You can do the interpolation outside of OPTMODEL and then read the results into an OPTMODEL array.

RobPratt · ‎05-09-2012

I will reply to #1 first and #2 in a separate post. If you want to use IMPVAR, you have to include the formula in the declaration. You cannot update it later. The other problem is that the SUM operator has different syntax than what you tried. Here's what I think you wanted (a dot product of two vectors): impvar updatedRisk {i in 2..n} = sum {j in 1..i-1} calculatedRisk * newRisk[n-i+1+j];

RobPratt · ‎05-08-2012

Yes, you can do all of these things, but it will likely be inefficient to introduce new variables whose values are equal to existing variables. Instead, you can express everything in terms of calculatedRisk for different values of i. For example, to achieve #1, you can do the following: num n = card(OBS); impvar calculatedRiskReverse {i in OBS} = calculatedRisk[n-i+1]; But you can avoid introducing a new variable by just using calculatedRisk[n-i+1] everywhere you need it: for {i in OBS} put calculatedRisk[n-i+1]=; For #2: impvar newRisk {i in OBS} = (if i = 1 then calculatedRisk[1] else 0); For #3: calculatedRisk * newRisk[n-i+1] Again, I think you will be better off not introducing additional variables unless you need to. If you write out the full expression for your objective, I suspect you will find that you can eliminate the additional variables by using logical conditions in the SUM operator.

RobPratt · ‎05-08-2012

You cannot currently call PROC IML from within PROC OPTMODEL. In release 12.1 due out later this year, you can use a SUBMIT block to call generic SAS code from within PROC OPTMODEL. But even that functionality will not enable you to implement a black-box objective function. Instead, you can indicate matrix multiplication by using the SUM operator: c[i,j] = sum {k in 1..n} a[i,k] * b[k,j]

RobPratt · ‎05-08-2012

Thanks for the clarification. If I understand your example correctly, the following code solves the problem in one PROC OPTMODEL call, without using any macros. Maybe your more complicated objective can be handled similarly. data bounds; input lower upper; datalines; 0.4 8 0.05 1000 ; proc optmodel; set OBS; num hours {OBS}; num actualRisk {OBS}; read data test into OBS=[_N_] hours actualRisk; set VARS; var x {VARS}; read data bounds into VARS=[_N_] x.lb=lower x.ub=upper; impvar calculatedRisk {i in OBS} = CDF('WEIBULL',hours,x[1],x[2]); min SSE = sum {i in OBS} (actualRisk - calculatedRisk)^2; solve; print x; print actualRisk calculatedRisk; create data x_sol from [var] x; quit;

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