Please help me find a solution in the form of a formula/algorithm.
Given percentage levels(Fibonacci levels)
Let zero be the current price i.e. 1.5 and 100% be 1.7, number 100 (e.g. the price may be 1.55). If the price grows, we open orders at the levels that are going to sell.
We have to find the volume of the lot to be opened, so that when the price moves from any level to the lower one (ideally we set for how many levels lower) all accumulated losses would be covered and the initial lot would be used for profit.
To do this you need to calculate the prices of all levels, and start from them, there is a formula for averaging, to derive a lot from it and it will be the required volume...
Dear Alex, I have long wanted to tell you that what matters for success is not the intensity of your efforts, but their right direction.
Although it is not relevant to your topic, the point I was trying to make, in case you didn't get it, is this: Look in the wrong place.
Dear Alex, what matters for success is not the intensity of your efforts, but their right direction.
Although this is not relevant to your topic, my point, in case you didn't get it, is: Look in the wrong place.
This is the picture that emerged
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But if you make the closure in two levels instead of one, you get a more acceptable result
Levl | Levl % | Price | LossPoint | Lot's 1lvl | Lot's 2lvl | |
7 | 123,6 | 1,7472 | 0,0472 | 50,99 | 10,30 | |
6 | 100 | 1,7 | 0,0472 | 25,50 | 7,72 | |
5 | 76,4 | 1,6528 | 0,0292 | 15,52 | 5,67 | |
4 | 61,8 | 1,6236 | 0,0236 | 8,47 | 3,84 | |
3 | 50 | 1,6 | 0,0236 | 4,24 | 2,45 | |
2 | 38,2 | 1,5764 | 0,0292 | 2,00 | 2,00 | |
1 | 23,6 | 1,5472 | 0 | 1,00 | 1,00 | |
0 | 1,5 | 0 | 1,5 | -0,0472 | 0 | 0 |
Total: | 107,71 | 32,98 |
But if you make the closure in two levels instead of one, you get a more acceptable result
Levl | Levl % | Price | LossPoint | Lot's 1lvl | Lot's 2lvl | |
7 | 123,6 | 1,7472 | 0,0472 | 50,99 | 10,30 | |
6 | 100 | 1,7 | 0,0472 | 25,50 | 7,72 | |
5 | 76,4 | 1,6528 | 0,0292 | 15,52 | 5,67 | |
4 | 61,8 | 1,6236 | 0,0236 | 8,47 | 3,84 | |
3 | 50 | 1,6 | 0,0236 | 4,24 | 2,45 | |
2 | 38,2 | 1,5764 | 0,0292 | 2,00 | 2,00 | |
1 | 23,6 | 1,5472 | 0 | 1,00 | 1,00 | |
0 | 1,5 | 0 | 1,5 | -0,0472 | 0 | 0 |
Total: | 107,71 | 32,98 |
-Aleks-:
I can tell you that I only did it more deliberately, with a good rebound that would overlap, the calculated lot would be enormous...
-Aleks-:
I can tell you that I only did it more thoughtfully, with a good no-backlash to overlap, the calculated lot would be enormous...
What do you mean by "more thoughtful"?
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Please help me find a solution in the form of a formula/algorithm.
Given percentage levels(Fibonacci levels)
Let zero be the current price i.e. 1.5 and 100% be 1.7, number 100 (e.g. the price may be 1.55). If the price grows, we open orders at the levels that are going to sell.
We have to find the volume of the lot to be opened, so that when the price moves from any level to the lower one (ideally we set for how many levels lower) all accumulated losses would be covered and the initial lot would be used for profit.