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From: raj74 on 2 Apr 2010 16:02 Hello!! I want to sumproduct for 2 vert. array (A1:A15 & B1:B15) but with a slight modification. While doing the sumproduct, i want the a const value say representing cell E1 will be subtracted from each value of array 2 and then multiply with the array 1 & finaly the addup of all multiplicated values. Can it be possible in excel. Regards
From: Paul on 2 Apr 2010 16:09 =sumproduct(a1:a15*(b1:b15-$e$1)) -- Paul - Paul ------------------------------------------------------------------------ Paul's Profile: 1697 View this thread: http://www.thecodecage.com/forumz/showthread.php?t=192599 http://www.thecodecage.com/forumz
From: Paul J. on 2 Apr 2010 16:12 Try: =SUMPRODUCT(A1:A15*(B1:B15-$E$1)) "raj74" wrote: > Hello!! > > I want to sumproduct for 2 vert. array (A1:A15 & B1:B15) but with a slight > modification. While doing the sumproduct, i want the a const value say > representing cell E1 will be subtracted from each value of array 2 and then > multiply with the array 1 & finaly the addup of all multiplicated values. > Can it be possible in excel. > Regards
From: David Billigmeier on 2 Apr 2010 16:14 Sure, you can just subtract the constant from the array, Excel is smart enough to know you want that subtracted from every cell in the range: =SUMPRODUCT(A1:A15,B1:B15-E1) -- Regards, Dave "raj74" wrote: > Hello!! > > I want to sumproduct for 2 vert. array (A1:A15 & B1:B15) but with a slight > modification. While doing the sumproduct, i want the a const value say > representing cell E1 will be subtracted from each value of array 2 and then > multiply with the array 1 & finaly the addup of all multiplicated values. > Can it be possible in excel. > Regards
From: raj74 on 2 Apr 2010 22:49
Paul, David Thanks a lot. Regards!!!!!!! Rajarshi "Paul" wrote: > > =sumproduct(a1:a15*(b1:b15-$e$1)) > > > -- > Paul > > - Paul > ------------------------------------------------------------------------ > Paul's Profile: 1697 > View this thread: http://www.thecodecage.com/forumz/showthread.php?t=192599 > > http://www.thecodecage.com/forumz > > . > |