Thanks Rick: Data sim; Input A B C Beta P; cards; 1 2  3  0.5  0 2 3  4  0.5  1 3 4.5 3  0.5 1 3 5 6  0.5   0 ; Run; proc iml; use sim; read all var _NUM_ into DM; close; A = DM[,1]; B = DM[,2]; C = DM[,3];  Beta = DM[,4];  P = DM[,5]; n = nrow(DM); start Func(x) global( Beta_i, C_i , A_i, P_i); if P_i=0 then do;    return(  exp(-x)*exp(2-x +Beta_i*C_i)#(cdf("Normal", x-2+Beta_i*C_i + A_i)-cdf("Normal", x-1+C_i + A_i)) ); end; else do; return(  exp(-x)*exp(4-x +Beta_i*C_i)#(cdf("Normal", x-4 +Beta_i*C_i + A_i)-cdf("Normal", x-1+C_i + A_i)) ); end; finish; answer = j(nrow(DM),1); do i = 1 to nrow(DM);    Beta_i = Beta; C_i = C; A_i=A; P_i=P; /* set global variables */    call quad(result, "Func", A || B,);    answer = result; end; create kaplan1n var{A  B C Beta P Answer }; append; quit; ... View more

Thanks a lot Rick, I will read the blog as well ... View more

Hello Rick, Thank you, I am just trying to integrate this function exp(x-1)*exp(2-x +Beta*C)#(cdf("Normal", x-10+Beta*C)-cdf("Normal", x-4+C))) numerical and have an answer for each row. The only contant is "BETA" which is 0.5.  The limit is A and B that changes for each row. Also C changes for each row. I want to return a scalar. However, once I have additional variable on each row like in the case above, it doesnt give me any answer or runs with error. My question is how to incorporate the variable C into the integral?? I want to to something similar like the code below, just that in this case I have a additional variable C in the exponential funtion and CDF: Data sim; Input A B; cards; 1 3 2 7 3 10 4 12 ; Run; proc iml; use sim; read all  var{A B} into DM; close; A = DM[,1]; B = DM[,2]; n = nrow(DM); start Func(x);    return(  exp(2-x)#cdf("Normal", x) ); finish; answer = j(nrow(DM),1); do i = 1 to nrow(DM); call quad(result, "Func", A || B,); answer = result; end; create kaplan1n var{A  B Answer }; append; quit; ... View more

Hello Rick, Thanks for the help. I tried to fix it , however it runs without error but there is no answer because it never converges. Data sim; Input A B C Beta; cards; 1 2  7  0.5 1 2.5  9  0.5 1 2.5 8  0.5 1 3  9  0.5 ; Run; proc iml; use sim; read all var _NUM_ into DM; close; A = DM[,1]; B = DM[,2]; C = DM[,3];  Beta = DM[,4]; n = nrow(DM); start Func(x); do i = 1 to nrow(DM);    return(  exp(-x)*exp(2-x +Beta*C)#(cdf("Normal", x-2+Beta*C)-cdf("Normal", x-1+C))); end; finish; answer = j(nrow(DM),1); do i = 1 to nrow(DM); call quad(result, "Func", A || B,); answer = result; end; create kaplan1n var{A  B C Beta Answer }; append; quit; LOG FILE 417  Data sim; 418  Input A B C Beta; 419  cards; NOTE: The data set WORK.SIM has 4 observations and 4 variables. NOTE: DATA statement used (Total process time):       real time           0.06 seconds       cpu time            0.00 seconds 424  ; 425  Run; 426  proc iml; NOTE: IML Ready 427  use sim; 428  read all var _NUM_ into DM; 429  close; NOTE: Closing WORK.SIM 430  A = DM[,1]; 430!             B = DM[,2]; 430!                         C = DM[,3]; 430!                                      Beta = DM[,4]; 431  n = nrow(DM); 432 433  start Func(x); 434  do i = 1 to nrow(DM); 435     return(  exp(-x)*exp(2-x +Beta*C)#(cdf("Normal", x-2+Beta*C)-cdf("Normal", 435! x-1+C))); 436  end; 437  finish; NOTE: Module FUNC defined. 438  answer = j(nrow(DM),1); 439  do i = 1 to nrow(DM); 440  call quad(result, "Func", A || B,); 441  answer = result; 442  end; Convergence could not be attained over the subinterval              (1 , 2 ) The function might have one of the following:      1) A slowly convergent or a divergent integral.      2) Strong oscillations.      3) A small scale in the integrand: in this case         you can change the independent variable         in the integrand, or,         provide the optional vector describing roughly         the mean and the standard deviation of the         integrand      4) Points of discontinuities in the interior         in this case, you can provide a vector containing         the points of discontinuity and try again. Convergence could not be attained over the subinterval              (1 , 2.5 ) The function might have one of the following:      1) A slowly convergent or a divergent integral.      2) Strong oscillations.      3) A small scale in the integrand: in this case         you can change the independent variable         in the integrand, or,         provide the optional vector describing roughly         the mean and the standard deviation of the         integrand      4) Points of discontinuities in the interior         in this case, you can provide a vector containing         the points of discontinuity and try again. Convergence could not be attained over the subinterval              (1 , 2.5 ) The function might have one of the following:      1) A slowly convergent or a divergent integral.      2) Strong oscillations.      3) A small scale in the integrand: in this case         you can change the independent variable         in the integrand, or,         provide the optional vector describing roughly         the mean and the standard deviation of the         integrand      4) Points of discontinuities in the interior         in this case, you can provide a vector containing         the points of discontinuity and try again. Convergence could not be attained over the subinterval              (1 , 3 ) The function might have one of the following:      1) A slowly convergent or a divergent integral.      2) Strong oscillations.      3) A small scale in the integrand: in this case         you can change the independent variable         in the integrand, or,         provide the optional vector describing roughly         the mean and the standard deviation of the         integrand      4) Points of discontinuities in the interior         in this case, you can provide a vector containing         the points of discontinuity and try again. 443  create kaplan1n var{A  B C Beta Answer }; 444  append; 445  quit; NOTE: Exiting IML. NOTE: The data set WORK.KAPLAN1N has 4 observations and 5 variables. NOTE: PROCEDURE IML used (Total process time):       real time           0.15 seconds       cpu time            0.11 seconds ... View more

Hello Rick, Thanks for the help. I did that but still having issue: I think the problem is with this line :  return(  exp(x-1)*exp(2-x +Beta*C)#(cdf("Normal", x-10+Beta*C)-cdf("Normal", x-4+C))); Data sim; Input A B C Beta; cards; 1 3  3  0.5 2 7  4  0.5 3 10 3  0.5 4 12 6  0.5 ; Run; proc iml; use sim; read all var _NUM_ into DM; close; A = DM[,1]; B = DM[,2]; C = DM[,3];  Beta = DM[,4]; n = nrow(DM); start Func(x);    return(  exp(x-1)*exp(2-x +Beta*C)#(cdf("Normal", x-10+Beta*C)-cdf("Normal", x-4+C))); finish; answer = j(nrow(DM),1); do i = 1 to nrow(DM); call quad(result, "Func", A || B,); answer = result; end; create kaplan1n var{A  B C Beta Answer }; append; quit; ... View more

Thanks Tom. ... View more

Data sim; Input A B; cards; 1 3 2 7 3 10 4 12 ; Run; proc iml; use sim; read all  var{A B} into DM; close; A = DM[,1]; B = DM[,2]; n = nrow(DM); start Func(x);    return(  exp(2-x)#cdf("Normal", x) ); finish; answer = j(nrow(DM),1); do i = 1 to nrow(DM); call quad(result, "Func", A || B,); answer = result; end; create kaplan1n var{A  B Answer }; append; quit; ... View more

Thanks a lot. Yes you are right, This just give me one answer and I need a result for each row. This code below gives me and answer. How ever, just as you said , it is not giving me the answer for each row.  How do I loop it from 1 to 4 so that each row has an answer. ? Data sim; Input A B; cards; 1 3 2 7 3 10 4 12 ; Run; proc iml; use sim; read all  var{A B} into DM; close; A = DM[,1]; B = DM[,2]; n = nrow(DM); start Func(x);    return(  exp(2-x)#cdf("Normal", x) ); finish; do i = 1 to nrow(DM); call quad(answer, "Func", A || B,); end; create kaplan1n var{A  B Answer }; append; quit; ... View more