Despite having sorted out my problem with the attached IML code, I am looking for an alternative way to get the following done:   I have a fixed-length (60 months evenly spaced) vector for Y that can take whatever shape when plotting it against months in a series plot. These values belong to the gap between the pending amount of debt on a typical loan and the residual value of the object being the collateral of the loan (i.e. a car). As the residual value can be subject to commercial decisions at discrete time points (12, 24 months, ...) and the down payment can differ, the gap vector can vary a lot in terms of shape.   Now I want to find such ever increasing function f(x) (f'(x) >= 0 and f''(x) >= 0 expressed mathematically I suppose...) that approximates the Y values from below and never exceeds them. Furthermore the function has to return 0 until the gap vector gets positive.   Before solving it in IML I thought about doing it with some kind of regression. But as the regressions I know cut the cloud of observed values by the middle thus violating the restriction I dropped this idea. Furthermore I'd need a regression whose solution is an exponential formula or something like a polynom of exp functions or line segments that get steeper and steeper...      Is there any PROC that can do what I am looking for? Is there an easy mathematical formulation for this?   Thanks a lot. Here's the valid solution for me that I obtained with the attached code in IML.     The graph belongs to a real world vector of gaps and represents the solution I'm fine with.   The code below uses a test vector but the logic applied is the same. I go through the months and the GAP vector gets updated after each assignment of available increments. The available increment is determined by the maximum value I can apply from the month in course till the last month without violating the restriction that the accumulated steps can never exceed the dynamic gaps. By doing so and going ahead with the months I ensure that the resulting "function" (totalling the increments for each month) is ever increasing.     proc iml; GAP={60 140 240 280 350 480}; print gap; /*I make use of arithmetic series and sums to force the obj function to be increasing for such cases when the obj function has equal increments due to the shape of the gap function. Here for simplicity desactivated*/ n=ncol(gap); Reihe=3; reparto=0.0; cake=gap[n]*reparto; tester=1:n; tester2=shape(tester,reihe); do i=1 to nrow(tester2); tester2[i,]=i/(reihe*(reihe+1)/2)/ncol(tester2); end; testcol=colvec(tester2); prime=testcol*cake; gap=gap-cusum(t(prime)); testsum=tester2[+,]*REPARTO; testsum=testsum[,+]; print tester tester2 testsum testcol prime gap; start Rel_dist(x, nn); idx0=loc(x<=0); pos=(ncol(idx0)=0); maxdiv=j(1,nn,1); print pos maxdiv; if pos then do; gap_dyn=x; maxdiv[1:nn]=1:nn; gap_dyn_dist=gap_dyn/maxdiv; end; if ^pos then do; be=max(idx0); gap_hembra=j(1,be,0); gap_dyn=gap_hembra||t(x[(be+1):nn]); maxdiv[(be+1):nn]=1:(nn-be); gap_dyn_dist=gap_dyn/maxdiv; end; return(gap_dyn_dist); finish; TEMP=J(n,n,0); gap_dyn=rel_dist(gap, n); I_DYN=1; do u=1 to n while(i_dyn=1); idx=loc(gap_dyn>0); gap_rel=gap_dyn[idx]; min_dyn=gap_rel[>:<,]; dim_gap=nrow(gap_rel); if min_dyn=dim_gap then do; temp[u,(n-dim_gap+1):n]=gap_rel[min_dyn]; i_dyn=0; end; if min_dyn ^= dim_gap then do; temp[u,(n-dim_gap+1):n]=gap_rel[min_dyn]; gap=gap-cusum(temp[u,]); gap_dyn=rel_dist(gap, n); end; end; testsum2=temp[+,]; testsum2=testsum2[,+]; print gap gap_dyn idx gap_rel min_dyn temp testsum2; ALL=TEMP//T(PRIME); ALL=ALL[+,]; PRINT ALL; CTR=CUSUM(ALL); PRINT CTR; ... View more

I run into an error while assigning a scalar with the loc function.  Everything works fine until I do not meet the condition in the loc.   Break_even = MIN(loc(result > 0));   This returns the month when the project breaks even. But if the project fails to do so, the code stops executing because the matrix Break_even isn't set.  I use this inside a Do Loop and I have tried to control this situation with If-then.  without success so far.  How do I handle situations like this?   Thank you for your help, bye, Arne     ... View more

In my autodidactic approach towards knowledge in SAS IML I read the interesting article:  Example 9.10 Quadratic Programming :: SAS/IML(R) 13.2 User's Guide   This prompted me to apply this technique to a  different problem exposed in the recommendable book "Schaum's Outline of Operations Research".   In doing so I cash in on the following advantages: to reassure myself of the correctness of the solution I get by solving it with IML and secondly being guided with te recipe to state the problem and define the objective function and the restrictions.   And it works! although I must confess that my higher mathematics university course lays far in the past and I don't succeed in understanding completely what's the reasoning with the eigenvalue... If someone wants to explain me that, I'm willing to put an effort in understanding it   And generally speaking about Operations Research in IML, can someone give me advice on how to define a Travelling Salesman Problem via IML (not OR) which is additionally subject to some restrictions? I manage to formulate most of the restrictions but I fail to force the solution to being a round trip (once arriving at node B continue travelling from there ...).      Here comes the case description as an excerpt from the mentioned book.         proc iml; start qp( names, c, H, G, rel, b, activity); if min(eigval(h))<0 then do; error={'The minimum eigenvalue of the H matrix is negative.', 'Thus it is not positive semidefinite.', 'QP is terminating.'}; print error; stop; end; nr=nrow(G); nc=ncol(G); /* Put in canonical form */ rev = (rel='<='); adj = (-1 * rev) + ^rev; g = adj# G; b = adj # b; eq = ( rel = '=' ); if max(eq)=1 then do; g = g // -(diag(eq)*G)[loc(eq),]; b = b // -(diag(eq)*b)[loc(eq)]; end; m = (h || -g`) // (g || j(nrow(g),nrow(g),0)); q = c // -b; /* Solve the problem */ call lcp(rc,w,z,M,q); /* Report the solution */ print ({'*************Solution is optimal***************', '*********No solution possible******************', ' ', ' ', ' ', '**********Solution is numerically unstable*****', '***********Not enough memory*******************', '**********Number of iterations exceeded********'}[rc+1]); activity = z[1:nc]; objval = c`*activity + activity`*H*activity/2; print objval[L='Objective Value'], activity[r=names L= 'Decision Variables']; finish qp; c = { -7.6, -5, 1}; h = { 0.08 0 0 , 0 0.04 0, 0 0 0}; g = { 0.2 0.2 0 , 0.8 0.3 0, 1 0 0, 0 1 0, 0 0 1}; /* constraints: */ b = { 20 , 60, 30, 30, 673.5 }; rel = { '<=', '<=', '>', '>', '=' }; names = 'cheese1':'cheese3'; names [3]= 'dummy'; run qp(names, c, h, g, rel, b, activity);     ... View more