Get minimum of the sum of 12 consecutive numbers in an array of 48
in [Matlab]
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From: Barry on 11 Aug 2010 11:24 I need to determine the minimum of each 12 consecutive in an array of 48 in simulink. So what i have is a feed in of an array (48x365) which is half hourly electricty cost data for each day. I need to get the time to start storing, over 6 hours which will be the minimum of 12 consecutive numbers in the array. Any ideas woud be great? so (48x365) need to find lowest sum of 12 consecutive numbers.
From: ImageAnalyst on 11 Aug 2010 13:02 On Aug 11, 11:24 am, "Barry " <barryoconne...(a)gmail.com> wrote: > I need to determine the minimum of each 12 consecutive in an array of 48 in simulink. So what i have is a feed in of an array (48x365) which is half hourly electricty cost data for each day. I need to get the time to start storing, over 6 hours which will be the minimum of 12 consecutive numbers in the array. Any ideas woud be great? > > so (48x365) need to find lowest sum of 12 consecutive numbers. ------------------------------------------------------------------------------------------------------- Did you look at all input and output arguments of the min() function?
From: Roger Stafford on 11 Aug 2010 21:07 "Barry " <barryoconnell2(a)gmail.com> wrote in message <i3ufal$gbv$1(a)fred.mathworks.com>... > I need to determine the minimum of each 12 consecutive in an array of 48 in simulink. So what i have is a feed in of an array (48x365) which is half hourly electricty cost data for each day. I need to get the time to start storing, over 6 hours which will be the minimum of 12 consecutive numbers in the array. Any ideas woud be great? > > so (48x365) need to find lowest sum of 12 consecutive numbers. - - - - - - - - - - Call the 48 by 365 data array D. h = 12; C = [zeros(1,size(D,2));cumsum(D,1)]; [S,K] = min(C(h+1:end,:)-C(1:end-h,:)); S is a 1 by 365 row vector of the minimum sum interval for each column and K contains the start index of that interval for each column. Roger Stafford
From: Barry on 12 Aug 2010 13:32 Thanks very much that was excellent really appreciate it.
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