Avalanche - page 409
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80 pips - between orders 1 and 2, as well as between the upper level of Breakeven and order 1, and between the lower level of Breakeven and order 2.
40 pips between order 1 and 3, and between order 3 and 2.
20 points - between 3 and 4, 4 and 2 orders.
I see. Why is the space between orders 1 and 3 not used? There is an empty space in the middle of this channel like a channel between orders 3 and 2.
Dear John, there is a "cheat" in your variant, after the second reversal we have a channel of 20pp. ( between 3 and 4 orders) the b/e level of which, at the top, is the width of 6 channels, at the bottom the b/e level is the width of 5 channels.
Thus, there will be a lot of reversals in the channel with such remoteness of the level b/u, it will be fatal many which more than compensates for the small coefficient of increase of the next order.
P.S. Why do you give examples with lock? Or you don't know how to calculate netting as I do?
Take the working corridor k............................
Do the following spreads yourself.
I made a test version of netting avalanche,
Here you can manually adjust values of lots and TP SL for each iteration (9 pieces maximum),
you can use two independent avalanches at the same time, each with its own settings...
It's possible to adjust the shift in pips between the avalanches, which would spread the load on larger iterations...
Zero iteration is defined by an indicator - random or trend
I haven't optimized the values, as avalanche is made as a procedure, it can be easily packed into Expert Advisor where you can use any principle of parameter calculation.
I'm asking, if anyone gets anything interesting on this basis, let me know...
it's into indicators
Я ?! Excuse me, but don't I have to run to get you a beer or shine your shoes?
This is for you! A bonus! We've got plenty of that stuff. (win-win avalanche).
Dear John, there is a "cheat" in your variant, after the second reversal we have a channel of 20pp. ( between 3 and 4 orders) the b/e level of which, at the top, is the width of 6 channels, at the bottom the b/e level is the width of 5 channels.
So, there will be a lot of reversals in the channel at such a distance from the b/u level, fatal many, which more than compensates for the small coefficient of increase of the next order.
Second, even if the number of reversals in a small corridor will be larger than in the initial working corridor, at the same deposit size "Avalanche 2", compared to a classic "Avalanche", allows you to handle a much larger number of reversals. For example, total order volumes on the channel borders in a classic "Avalanche" grow as follows: 0.10 - 0.20 - 0.40 - 0.80 - 1.60 - 3.20 - 6.40 - 12.80 - 25.60, and so on. This is a total of 7 reversals. For the same deposit in "Avalanche 2" with an excessive arithmetic progression (more favourable conditions than in the classic "Avalanche" - the distance to the break-even level decreases with every new order, allowing to make a profit quicker), the total volumes of the oppositely directed orders increase as follows: 0.10 - 0.20 - 0.30 - 0.50 - 0.70 - 1.00 - 1.30 - 1.70 - 2.10 - 2.60 - 3.10 - 3.70 - 4.30 - 5.00 - 5.70 - 6.50 - 7.30 - 8.20 - 9.10 - 10.10 - 11.10 - 12.20 - 13.30 - 14.50 - 15.70 - 17.00 - 18.30 - 19.70 - 21.10 - 22.60 - 24.10 - 25.70 and so on. This is 30 spreads! At the same deposit size. If to use fractional coefficients of increase (at which the level of breakeven remains on place), then at the same conditions "the Avalanche 2" allows to sustain about 50 reversals!
1. First, despite the fact that orders are placed at a small distance from each other, the working corridor is 80 pips and is located between 1 and 2 orders. This is equivalent to opening an 80-point corridor in a classic "Avalanche". This is the only way they can be compared. Price is more likely to stay within the 80 point corridor than the 20 point corridor. In the latter case, it just has to shift 20-30 pips to the side and it can hang there as long as it wants - it will not be able to make any more reversals in the small corridor.
2. Second, even if the number of reversals in a small corridor will be larger than in the initial working corridor, at the same deposit size "Avalanche 2", as compared to a classic "Avalanche", allows you to handle a much larger number of reversals. For example, total order volumes on the channel borders in a classic "Avalanche" grow as follows: 0.10 - 0.20 - 0.40 - 0.80 - 1.60 - 3.20 - 6.40 - 12.80 - 25.60, and so on. This is a total of 7 reversals. For the same deposit in "Avalanche 2" with an excessive arithmetic progression (more favourable conditions than in the classic "Avalanche" - the distance to the break-even level decreases with every new order, allowing to make a profit more quickly), the total volumes of the oppositely directed orders increase as follows: 0.10 - 0.20 - 0.30 - 0.50 - 0.70 - 1.00 - 1.30 - 1.70 - 2.10 - 2.60 - 3.10 - 3.70 - 4.30 - 5.00 - 5.70 - 6.50 - 7.30 - 8.20 - 9.10 - 10.10 - 11.10 - 12.20 - 13.30 - 14.50 - 15.70 - 17.00 - 18.30 - 19.70 - 21.10 - 22.60 - 24.10 - 25.70 and so on. This is 30 spreads! At the same deposit size. If we use fractional coefficients of increase (at which the level of breakeven remains in place), then under the same conditions, "Avalanche 2" allows you to make about 50 reversals!
1. No, in your example, we are talking exclusively about a 20 pt corridor as a working corridor and a flat trajectory within that 20pp. and will give an increase in volumes. You need to analyse this channel and the corresponding b/o level from its borders, which is 5/6 times wider than the channel itself. That's where you should look at the trajectories instead of hoping that "price is more likely to stay within the 80-point corridor than the 20-point one".
Why would it? Yes. It may or may not shift.
2. 30 reversals, from your calculation, gave an increase in volume compared to the initial one (0.1) by 257 times and amounted to 25.70... Not enough...? It's all over the place... Here are 44 reversals, just recently, b/w levels and borders of the channel itself are marked:
2. 30 reversals, from your calculation, gives 257 times the volume increase from the initial one (0.1) to 25.70. Not enough...? It's all over the place... Here's 44 reversals, just recently, b/w levels and borders of the channel itself marked:
Do you distinguish between an example and real trading? The 20 pips are taken from the ceiling for this example. For trading, the minimum band is adapted to each currency pair, just as I recommend for the classic "Avalanche" - using a standard indicator Average True Range.
Besides, the channel you have drawn is not correct - with arithmetic progression the breakeven level is continuously approaching, continuously reducing the distance the price has to cover to start taking profit.
You yourself cited statistics on the number of reversals over the last ten years - there are no more than 16 reversals. Take for full confidence 20 reversals - with the initial lot 0.10, the maximum total volume in "Avalanche 2" will be 12.20. Every profitable point will be more or equal to 30 rubles, the deposit in the DC without compensation, the leverage of 1:500 and EURUSD after 20 reversals will be approximately 100 000 rubles. According to your own statistics, there is no such number of reversals. You deposit 100 000 to your trading account, set the profit target, for example, 10 pips and get +300 roubles for each exit from "Avalanche 2". Since price moves in any direction are equally likely, in 50% of cases you will get +300 rubles immediately, even without using "Avalanche 2". For example, on an average day you will have 5 exits with a profit. This +1500 rubles per day or +7500 rubles per week or +30 000 rubles per month. And the deposit will be intact! It is impossible to lose it due to lack of funds - you have plenty of them.
1. Do you distinguish between an example and real trading? 20 pips is taken from the ceiling for this example. For trading, the minimum corridor is selected for each currency pair, you can do it as I recommended for the classic "Avalanche" - using a standard Average True Range indicator.
2. Besides, the channel you have drawn is not correct - with arithmetic progression the breakeven level is continuously approaching, continuously reducing the distance which the price has to cover to start taking profit.
1.:)
How about you?
Whatever you take from the ceiling, according to your scheme of placing orders, keeping proportions and increasing volumes (0.10 - 0.20 - 0.20 - 0.24 - 0.28 - 0.30 - 0.35... - your example), the channel with a smaller width and very remote breakeven level, which stays in place, works. I have shown in the example what to look out for.
2. Is the example wrong? It's for b/w that stays in place..:
JonKatana:
Volume chain for the first seven orders: 0.10 - 0.20 - 0.20 - 0.24 - 0.28 - 0.30 - 0.35...
The breakeven level remains in place - the distance to it does not change regardless of the number of reversals. And the volumes grow very slowly. With the same deposit size, "Avalanche 2" allows you to make several times as many reversals as the classic "Avalanche", while maintaining all its advantages.JonKatana:
I didn't calculate the formula, it is not needed. I was counting from breakeven level so that it would remain in place.