Compute the remaining values and bounds after visiting node (4,1) in Example 5.6 (Section 5.7 .1

Compute the remaining values and bounds after visiting node...

Key Concepts

Branch and Bound

A method used in combinatorial optimization and search problems where the solution space is divided into smaller subproblems (branches) and bounds are computed for these subproblems to eliminate those that cannot produce a better solution than the current best. This technique effectively reduces the computational burden by pruning sections of the search tree.

Node Evaluation

The procedure of assessing the quality or potential of a partial solution at a specific node within a search tree. This involves calculating current solution values and estimating future outcomes which inform the subsequent decision-making process in the search algorithm.

Bound Computation

The calculation of lower and upper limits (bounds) for the value of potential solutions stemming from a node. These bounds provide insights into the best-case and worst-case outcomes from that point in the search, guiding the process of pruning suboptimal branches.

Search Tree Pruning

A crucial optimization strategy in search algorithms where branches of the tree that do not promise better outcomes based on the computed bounds are eliminated. This reduces unnecessary exploration, leading to increased efficiency in finding the optimal solution.

Transcript

00:01 Alright, so we have this function, very nice function, we have the function 1 plus 5 x squared, or all like this, 1 plus 5 x squared.

00:27 I want to approximate.

00:30 I want to compute the area in 0 and 2.

00:35 This is not up to scale, but it should be some point.

00:41 There so you want to approximate what is we want to find what is nearly this area here that region from zero up to two by a limit process and so also want to take n steps we're divided it into the area of the of rectangles between rectangles that go from zero to two and well there are aiming of those rectangles so and let's call it and you take n steps steps of t at the end so stepped and there is a t at you steps yeah um so the delta x the size of the width of each of these rectangles each of these rectangles is going to have a width of given by the difference between the points so 2 minus zero divided by n and so each rectangle you have a height given by the higher point x i will be just the function evaluated at that point with this being our function this is f of x and so let's take the right end point approximation so it's going to be the f at x i and we're excited being the right endpoint the i would be so 0 because that's our starting point 0 plus i times the delta x which is i times 2 over n i times 2 over n that is r xi so that our function and that x i will be equal to 1 plus 5 times our squared so 2 i over n squared that is our x i so well the area of this and in intervals and rectangles approximation would be the sound from i equals 1 up to n of delta x times the height at each xi and so replacing that with those values and also it's going to be the sum from i equals 1 up to n of delta x is 2 over n 2 over n times b h i which is that 1 plus 5 all that is a 2 square that is a 4 5 times 4 times i square times i squared over and n squared or n squared so is that and then well here for this sum we can we can bring outside the 2 over n and we can break this sum into the sum from i equals 1 up to n of the term plus the sum and bringing also the constants outside of 5 times 4 over n squared the sum the sum from i equals 1 up to n of i squared so you have uh so you have that this is going to be 2 over n times the sum of i equals 1 up to n of 1 plus 5 times 4 over n squared um the sum of i equals 1 up to n of 1 plus 5 times 4 over n squared squared to n from i equals 1 up to n of i squared and so well this sum here that is just n and this sum is equal to n n times n plus 1 and plus 1 times 2 n plus 1 times 2 n plus 1 x by 6 so that's well here you'll have 2 over n and that multiplying n plus 5 times 4 divided by n squared divided by n square times 6 and that times 6 on the bottom here coming from that sum that times n times n times n times n plus 1 times 2 and plus 1 and so here some very nice things happen is end cancels in that end and so we'll also multiply that we're gonna get a cancellation there so that this term will become 2 for that times up plus 2 and to the two multiplying that two times the two over and multiply that two times five it's four over n times n there was a remaining n there so n squared six is n there was gone that times n plus one times two n plus and so this expression will be so this six there that is three and this two and these two cancels with that two so this expression will be equal to two and then plus five times four over three and then that times n squared on the bottom right divided n plus 1 times 2 n plus 1 so what we have here well this is the area of our in rectangle approximation and rectangle approximation and right uproximation so the real area should be equal to it's going to be equal to the limit the same goes to infinity of the area of in rectangles area and rectangle approximation and rect approximation so well it's going to be equal to the limit the same goes to to infinity the real area will be equal to the limit assembles to infinity of this expression so it's going to be the limit of the same goes to infinity of 2 plus 5 times 4 over 3 times 4 over 3 times 10 and plus 1 times 2 and 2 times 2 and plus 1 and plus 1 times 2 and plus 1 divided by n squared divided by n and so well the limit of this expression will be equal to the sum of the limit so it's going to be equal to just 2 plus the limit of 2 is 2 the limit of plus 5 times 4 over 3 times the limit same goes infinity of this expression n plus 1 and plus 1 times 2 n plus 1 over n squared and so in this expression in this one while you take the limit it's gonna be infinity plus 1 times infinity or infinity so it's gonna be infinity or infinite so for this limit so for this limit we can apply lobito.

13:29 Logito.

13:33 So if we differentiate the top, we get the derivative of this times that.

13:40 So we can delimid a single infinity of 2 and plus 1 with the same goes infinity...

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