The challenge: What is cheap and what is expensive? - page 7
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Yuriy, why exactly 347? Is it mathematically derived or just an example? And what should be the profit of these pending orders? It follows from the problem that even if the price is a couple of pips higher, it is the best price among all previous ones, but it may be the best even with a couple of points difference compared to the spread and then it will have practically no profit compared to a possible loss.
See ready solution of the problem: http://www.mccme.ru/mmmf-lectures/books/books/book.25.pdf
Not only "better" notions play a role here, but also the number of grooms to be compared, and in the trading context - bars to choose from. That's why the expiration of orders is exactly the same as in the optimal solution of the problem.
Better means better. I apologise for the tautology, but just one pip better already corresponds to an optimal solution.
The quality of potatoes is the same everywhere in the problem.
Pure mathematics, physics, etc.: problems for brain training, not related to trade in any way.
In general, when an order is placed to execute a large volume, for example a purchase, the quality criterion is the average purchase price in relation to the weighted average price for the day's volume. If the manager bought below it, then good
Of course, the fact that the further away from the entry, the lower the price, is a known fact and we do not take this fact into account.
In this case, there is no time. How do you propose to calculate this criterion? The relationship of averages with different periods? Do the previous 3-5 broadsides have more adequate prices? Or do all prices have equal weights?
Of course, the fact that the further away from the entry, the lower the price, is a well-known fact and we do not take this fact into account.
In this setting, the day is about getting around all the dough. To buy well is to buy below the average price of the broads. I.e. if we bought a bag for 102 on average and the average price of all the broads is 105, then we have bought cheap
Then it's better over there _
See ready solution of the problem: http://www.mccme.ru/mmmf-lectures/books/books/book.25.pdf
Not only "better" notions play a role here, but also the number of grooms to be compared, and in the trading context - bars to choose from. That's why the expiration of orders is exactly the same as in the optimal solution of the problem.
Better means better. I apologize for the tautology, but just one pip better already corresponds to an optimal solution.
Yuri, it's not clear why you have such a great amount of bars.
The flow of bars, as opposed to the flow of grooms, is infinite. It seems that you can select any number of bars but in this case it reduces the solution to an undefined one.
Explain, please.
The princess has princes on a rank scale (worse/better than the previous prince). The broads have sack prices on an absolute scale.
There is a suspicion that the princess optimal solution would not be optimal for the baboks, although the opposite is true (by analogy like Spearman and Pearson correlation coefficients).
Yuri, it's not clear why you have so many bars.
Add up the numbers and you get that the princess' choice is limited to 1000 grooms. Denoted by the symbol n in the problem statement.
The flow of bars, as opposed to the flow of grooms, is infinite. It seems you can choose any number of bars, but then it reduces the solution to indefinite.
The flow of princesses is also infinite. So with each new bar we can start a new casting. At the same time all previous ones will be valid within their expiry.
By the terms of the problem, for each princess the choice is limited to the number of n potential princes. Therefore any number will not work and everything is strictly defined according to the optimal solution.
The flow of princesses is not limited either. That is, with each new bar we can start a new casting. At the same time all the previous ones will be valid within their expiration date.
By the terms of the problem, for each princess the choice is limited to the number of n potential princes. Therefore any number will not work and everything is strictly defined according to the optimal solution.