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Unformatted text preview: tored in the reservoir at the end of any month t. For the six‐month design period, there will be 6 months of operation, say t 1, 2, , 6 . Let U designate the largest volume among the 6 months, and let L designate the smallest volume among the 6 months. Then (U L) is the range (Note: U L 0 ), and the function Z Min (U L) minimizes the range. Effectively, this minimization simultaneously pushes down the maximum value (U) while pushing up the minimum value (L). Note that the constraints St U , t 1, 2, , 6 in combination with Z Min (U L) will result in U taking on the maximum value of St , t 1, 2, , 6 (since Z Min (U L) will try to make U as small as possible, but at least as big as the largest value of St ). Similarly, the constraints St L , t 1, 2, , 6 in combination with Z Min (U L) will result in L taking on the minimum value of St , t 1, 2, , 6 (since Z Min (U L) will try to make L as large as possible, but not greater than the smallest value of St ). 2. Limiting the deviation of flow in the stream from the target interval, F X i E . Consider the deviation from the upper limit, i.e., X i E . Note that X i E may be either positive or negative; if it’s negative, then we don’t care; if it’s positive, then it would represent a deviation that is beyond acceptable. So, represent X i E by X i E U i U i , where U i ,U i 0 . Note that this encompasses all possible values of X i E (both negative and positive). Then, if we minimize U i we effectively limit the amount that X i E is positive, or the amount by which X i E . Similarly, consider the deviation from the lower limit, i.e., X i F . Note that X i F may be either positive or negativ...
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- Spring '14