About ayaafarag

ayaafarag · ‎09-30-2015

Please, how to make the first two decision variables unrestricted in sign variables inside the attached code?

ayaafarag · ‎09-30-2015

Please, how to make the first two decision variables unrestricted in sign variables?

ayaafarag · ‎09-30-2015

Below is my code, I have 3 questions: 1- how to formulate it as a minimum problem? 2- how to formulate the constraints as equality constraints? 3- I want the first 6 decision variables to be greater than or equal zero and the rest 2 decision variables to be unrestricted in sign, so how to do that here? Proc iml; option nonotes; P=2 ; n=3 ; nruns=50; B= {1, 1}; X1= { 2.148910293 -1.756195507 -0.562194351 　　 }; x2={ 0.558425545 -0.0446071 -3.180084241 }; Xt1 = t(x1); xt2=t(x2); nx=xt1 || xt2; Do j=1 to nruns; E1= 1#normal(repeat(-1,n-1)); E2=3#NORMAL(REPEAT(-1,1)); E=E1//E2; nY=nx* B+E; 　　　/* the problem data */ obj = {1 1 1 1 1 1 0 0}; coef=I(3) || -I(3) || nx; bb = {0}//ny; /* embed the objective function in the coefficient matrix */ a = obj // coef; /* solve the problem */ call lp(rc, x, dual, a, bb); 　　 end; 　 Print devres ;

ayaafarag · ‎09-25-2015

if my problem is: x is 3*2 matrix y is 3*1 vector min dev(1)+dev(2)+dev(3)+dev(4)+dev(5)+dev(6) s.t. dev(7)*x11+dev(8)*x12+dev(1)-dev(4)=y1 dev(7)*x21+dev(8)*x22+dev(2)-dev(5)=y2 dev(7)*x31+dev(8)*x32+dev(3)-dev(6)=y3 dev(1),dev(2),dev(3),dev(4),dev(5),dev(6)>= 0 dev(7),dev(8) are unrestricted in sign so what is missing in the following code? Proc iml; option nonotes; P=2 ; n=3 ; nruns=50; B= {1, 1}; X1= { 2.148910293 -1.756195507 -0.562194351 　　 }; x2={ 0.558425545 -0.0446071 -3.180084241 }; nx=x1 || x2; Do j=1 to nruns; E1= 1#normal(repeat(-1,n-1)); E2=3#NORMAL(REPEAT(-1,1)); E=E1//E2; y=nx* B+E; 　 obj={1 1 1 1 1 1 0 0}; coef = I(3) || -I(3) || nx; rhs=y; call lp(rc,dev,dual,coef,rhs); 　end; 　　　　　　

ayaafarag · ‎09-25-2015

How define x matrix and y vector inside proc optmodel?

ayaafarag · ‎09-12-2015

An extra question please the output gives me negative values for dev[1] and dev[5] however I specify dev[1]-dev[6] are nonnegative decision variables, so how to solve for this problem?

ayaafarag · ‎09-10-2015

I tried the attached code, but I have two questions: 1-How to specify that the first three constraints and the last three are equality constraints and the rest are inequality constraints? 2-DO opt[10] and opt[11] are written correctly according to my problem?

ayaafarag · ‎09-10-2015

Here is my problem x is 3*2 matrix y is 3*1 vector min dev(1)+dev(2)+dev(3)+dev(4)+dev(5)+dev(6) s.t. dev(7)*x11+dev(8)*x12+dev(1)-dev(4)=y1 dev(7)*x21+dev(8)*x22+dev(2)-dev(5)=y2 dev(7)*x31+dev(8)*x32+dev(3)-dev(6)=y3 dev(1),dev(2),dev(3),dev(4),dev(5),dev(6)>= 0 dev(7),dev(8) are unrestricted in sign dev(1)*dev(4)=0 dev(2)*dev(5)=0 dev(3)*dev(6)=0

ayaafarag · ‎09-10-2015

Am using SAS 9.2

ayaafarag · ‎09-10-2015

I have defined nx and ny before writing the LP problem. I want to write this code inside proc iml command. So what is the write command to run " LP" model in this case ? And I need to put equality constraints but I do not know how to do this here? here is my detailed code: Proc iml; option nonotes; P=2 ; n=3 ; B= {1, 1}; X1= { 0.019831833 0.150408859 0.782582506 　　 }; x2={ 1.008931805 0.365347241 0.055856515 　　 }; Xt1 = t(x1); xt2=t(x2); xt=xt1 || xt2; E= 1#normal(repeat(-1,n)); Y=xt* B+E; my = sum(y)/n; m1 = sum(xt1)/n; m2 = sum(xt2)/n; ny =(y-my)/sqrt(t(y-my)*(y-my)); nX1 =(xt1-m1)/sqrt (t(xt1-m1)*(xt1-m1)); nX2 =(xt2-m2)/sqrt(t(xt2-m2)*(xt2-m2)); 　 nx= nx1||nx2; start fun(DEV) global (n,p); sumf=0; do i = 1 to 6; sumf = sumf + (DEV[I]); end; 　　 return (sumf); finish fun; 　 start con(DEV) global (lambda,lenght,thrd,estvar1,g,mseco,mspeco,i2,R,nx,n,ny,rxy,p); c=j(3,1,0); sumc1= DEV[1]-DEV[4]+(DEV[7]*NX[1,1])+(DEV[8]*NX[1,2]); sumc2=DEV[2]-DEV[5]+(DEV[7]*NX[2,1])+(DEV[8]*NX[2,2]); sumc3=DEV[3]-DEV[6]+(DEV[7]*NX[3,1])+(DEV[8]*NX[3,2]); 　　 c[1]= NY[1]-sumc1; c[2]=NY[2]-SUMC2; c[3]=NY[3]-SUMC3; return (c); finish con; call LP (rc, kres, "fun");

ayaafarag · ‎09-09-2015

I have two questions concerning this code: 1- how to solve this problem using linear programming? 2- how to make these constraints equality constraints and how to add another 2 inequality constraints? start fun(DEV) global (n,p); sumf=0; do i = 1 to 6; sumf = sumf + (DEV[I]); end; 　　 return (sumf); finish fun; 　 start con(DEV) global (lambda,lenght,thrd,estvar1,g,mseco,mspeco,i2,R,nx,n,ny,rxy,p); c=j(3,1,0); sumc1= DEV[1]-DEV[4]+(DEV[7]*NX[1,1])+(DEV[8]*NX[1,2]); sumc2=DEV[2]-DEV[5]+(DEV[7]*NX[2,1])+(DEV[8]*NX[2,2]); sumc3=DEV[3]-DEV[6]+(DEV[7]*NX[3,1])+(DEV[8]*NX[3,2]); 　　 c[1]= NY[1]-sumc1; c[2]=NY[2]-SUMC2; c[3]=NY[3]-SUMC3; return (c); finish con; call LP (rc, kres, "fun");

ayaafarag · ‎04-23-2015

After I remove all the do loops and specify the initial values, I got the following error message using NLPQN. How can I increase the number of iterations or number of function calls inside the optimization code? ERROR: QUANEW Optimization cannot be completed. ERROR: QUANEW needs more than 200 iterations or 500 function calls. WARNING: The point x is feasible only at the LCEPSILON= 0.1 range.

ayaafarag · ‎04-19-2015

I tried this code : return( sum((e*l+ (k##2 # g##2)) /(l+ k)##2) ); but I got this error message ERROR: (execution) Matrices do not conform to the operation. operation : # at line 4183 column 39 operands : _TEM1002, _TEM1003 _TEM1002 1 row 2 cols (numeric) 0 25 _TEM1003 2 rows 1 col (numeric) 0.1154035 0.3658517 statement : RETURN at line 4183 column 1 traceback : module FUN at line 4183 column 1 ERROR: (execution) Matrices do not conform to the operation. operation : NLPQN at line 4235 column 1 operands : *LIT1218, k, optn, , , , , , *LIT1219 *LIT1218 1 row 1 col (character, size 3) fun k 1 row 2 cols (numeric) 0 5 optn 1 row 11 cols (numeric) *LIT1219 1 row 1 col (character, size 3) con statement : CALL at line 4235 column 1

ayaafarag · ‎04-18-2015

I tried another initial values and this error message disappeared but I repeate this code 50000 times in simulation study. and now the runs do not stop even along the whole day. So can I use call nlpnms instead of call nlpqn for this optimization model?

ayaafarag · ‎04-17-2015

this my code and after I run it I got this error message: ERROR: Overflow error in NLPQN. operation : NLPQN at line 5279 column 1 operands : *LIT1222, k, optn, , , , , , *LIT1223 *LIT1222 1 row 1 col (character, size 3) fun k 1 row 2 cols (numeric) 0 0 optn 1 row 11 cols (numeric) *LIT1223 1 row 1 col (character, size 3) con statement : CALL at line 5279 column 1 ***************************************** where l and g are vectors the code is : proc iml; start fun(K) global (l,e,g,p); sumf=0; do i = 1to p; sumf = sumf + (((e*l)+ ( k**2 * g**2 )) /(l+ k)**2); end; return (sumf); finish fun; start con(k) global (l,lg,e1,n,ny,p,g2,z,g); c=j(5,1,0); sumc1=0; do i = 1to p; sumc1 = sumc1 + ( (g2*l**2) /(l+k)**2 ); end; sumc3=0; do i = 1to p; sumc3 = sumc3 + (((e1*l)+ ( k**2 * g**2 )) /(l+ k)**2); end; sumc41=0; do i = 1to n; sumc41 = sumc41 +( (ny-((z[i,1]*g[1]*l[1])/(la[1]+k[1]))-((z[i,2]*g[2]*l[2])/(l[2]+k[2])) )**2 ); end; sumc42=0; do i=1to n; sumc42=sumc42+(1-((z[i,1]**2)/(l[1]+k[1]))-((z[i,2]**2)/(l[2]+k[2])) ); end; sumc4=(n*sumc41)/(sumc42)**2; c[1]= lg-sumc1; c[2]=k[1]; c[3]=k[2]; c[4]=3-sumc3; c[5]=4-sumc4; return (c); finish con; k=j(1,2,0); optn=j(1,11,.); optn[11]=0; optn[10]=5; call nlpqn (rc, kres, "fun", k, optn) nlc="con";

Online Status	Offline
Date Last Visited	‎10-24-2015 09:30 AM

Unrestricted in sign decision variables

Re: LP

Re: LP

Re: LP

Re: LP

Re: LP

Re: LP

Re: LP

Re: LP

Re: LP

Unrestricted in sign decision variables

Re: LP

Re: LP

Re: LP

Re: LP

Re: LP

Re: LP

Re: LP

Re: LP

Re: LP

LP

Re: Problem in NLPQN

Re: Problem in NLPQN

Re: Problem in NLPQN

Problem in NLPQN