Explain successive quadratic programming (SQP)in a very simple way?
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Think of a near-sighted mountain climber in a terrain with multiple peaks (represented by a function), and you'll see the difficulty posed for an algorithm that tries to move from point to point only by climbing up hill.   If, as well, the objective function is quadratic at most, this problem is called Quadratic Programming (QP). A quadratic aX^2+bX+C. Successive or sequential quadratic programming is through successive approximation a solution to a quadratic is shown.
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To do optimization problems, set up an equation in terms of one variable, take its derivative, and solve for 0. In this case, write an equation describing the area of the garden. Take the derivative and set it equal to 0, then solve.\n\nI'd be more help, but I don't quite understand the question...
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P1V1 = nRT = P2V2\nP1/P2 = V2/V1
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P1 x V1 / T1 = P2 x V2 / T2\n\nYou have all you need.  Calculate it.
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Hi!\nI'm suprised nobody used the easiest way that i will demonstrate now.\n\nLets assume further that x is positive real.\n\nWe start with a simple fact that ANY squares are nonnegative. The idea is we "choose" a square intelligently to suit our purpose. Also not that x does NOT equal zero.\n\nBy experience, \nI choose the square ( x^(1/2) - 1 / [ x^(1/2)] )^2.\nIt must be nonnegative for whatever positive real values of x.\nThus,\n\n    ( x^(1/2) - 1 / [ x^(1/2)] )^2 >= 0\n\nExpanding and regrouping, we get,\n\n    x - 2 + 1/x >= 0 \n      \nor equivalently,\n\n      x + 1/x >= 2\n\nThus, minimum is 2. \nWith equality when x^(1/2) = 1 / [ x^(1/2)], or\n x = 1.\n\nDone.\n\nAlternative Method.\n\nOne can use two variable AM - GM,\nFor any two positive reals m and n, we have\n\n ( m + n )/ 2 >= (mn)^(1/2)\n\nWith equality if and only if m = n.\n\nJust set m = x and n = 1/x, to get,\n\n( x + 1/x )/ 2 >= 1\n\nor  x + 1/x >= 2\n\nAgain, the minimum values is 2 with equality if and only if x = 1 .\n\nRegards
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This normally means that you are going to be using Gaussian (or Gauss-Jordan) reduction, where ther eare three "moves" that are allowed.\n\n1) You can switch any two rows in the matrix.\n2) You can multiply any row in the matrix by a non-zero scalar.\n3) You can add a multiple of one row to another.\n\nUsing these three "moves," the goal is to reduce the matrix into a usable form (the echelon form, for example).
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Below is a link to a simplified tutorial.
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the quadratic formula is a useful tool in solving an equation involving a variable and its square without breaking it down into smaller pieces. It can only be used when it is in the\nax^2+bx+c=0  format. (a, b, and c can be negative.)\n\nIn this problem, a=2, b=-2, and c=-1. The equation for the quadratic is:\n\nx= (-b+/- sq.rt (b^2-4ac))/2a\n\nSolving for the equation given, you get:\n\nx= (2 +/- sq.rt.(4-4*2*-1))/4\n   = (2+/- sq.rt 12)/4\n   = 1.366, 0.366
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The greatest mystery here is "WHAT IS THE PROBLEM TO BE SOLVED?"\n\nIf a knowledge of the 'knapsack' problem is required, please copy it here for all to ponder............
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Your equation would be:\n\n= ((1/p) - (1/q))/(1/pq)\n\n= ((q - p)/pq) * pq\n\n= q - p\n\nHope I am clear.
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playing a simple game of catch invovles the quadratic equation. Im positive nasa uses the quadratic equation for determining flight angles and all that stuff.
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Derivation of the quadratic formula by completing the square.\n\nax^2 + bx + c = 0\nx^2 + (b/a)x + (c/a) = 0\nx^2 + (b/a)x = -c/a\nx^2 + (b/a)x + b^2/(4a^2) =  b^2/(4a^2) - c/a \n(x + b/2a)^2 =  b^2/(4a^2) - 4ac/(4a^2) \nx + b/2a = ±√[ b^2/(4a^2) - 4ac/(4a^2) ]\n\nx = -b/2a ± √[ b^2/(4a^2) - 4ac/(4a^2) ]\n= -b/2a ± √[ b^2- 4ac] / 2a\n\nx = {-b ± √[ b^2- 4ac] } / 2a
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You have to solve the equation so that you can eliminate the sqrt by raising it to the second power .for instance,\n\nsqrt(2x+10)=5     you square it and get\n\n2x+10=25   easy, isnt it?\n\nbe careful if you have something like this\n\nsqrt(x+3)+sqrt(x-2)= 5\n\nhere you cannot do that you have to distribute one sqrt on each side of the equation.\n\nsqrt(x+3)=5-sqrt(x-2)\n\nnow you can square them
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This progression is created by squaring the previous number and adding 1.\n0 squared = 0, add 1 and you get 1.\n1 squared = 1, add 1 and you set 2\n2 squared = 4, add 1 and you get 5,\nand so on.
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Optimisation problems are normally done using partial derivatives, however you're problem is discreet so it will be handled differently. First make a map of the network lets say with coloured lines to represent the individual paths from the beginning to the end of the manufacturing process and nodes where they intersect, i.e. nodes that represent each machine. Then there are multiple ways to run the algorithm. For the case of minimum number search through the nodes to see if there is one that bundles all the lines through it if not start looking for pairs that would have all the lines going through one or the other of the nodes and so on until you find a solution. Unless visually it is a simple graph you will need a computer to do the processing.\nThe problem with a cost assigned is more difficult. In this case each node has a value associated with it. In this case generate an ordering from cheapest to dearest of the possible costs of selecting any number of nodes. The test each of these in succession until you find one that corresponds to having all the lines pass through at least one of the selected nodes associated with that particular cost.\nIn knowing details of the way in which the graph would look or the cost for certain locations it may be possible to modify the algorithm so it runs quicker, but this should do it for the information given.
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Diamond-Square is a great height-map generation algorithm. It's kind of cool that something that simple could create realistic landscapes.
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to know about the SPANNING TREE OPTIMAL PATH METHOD SPECIALLY THE WARSHALL METHOD WITH SUITABLE EXAMPLE visit the below mentioned site u will get all the info. \nhttp://www.me.utexas.edu/~jensen/ORMM/models/unit/combinatorics/tree.html\nyou can also see in wikipedia.com