Discussion of article "The checks a trading robot must pass before publication in the Market" - page 6

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Andrey F. Zelinsky:
The article does not address the question of what to do if stoplevel =0. The question has been repeatedly raised on the forum. There are no recommendations in the article on this matter.
What is the question? There is no stoplevel - and good....
 
Oleksii Chepurnyi:
What's the question? There's no stoplight, and that's fine...

Zero doesn't mean absent. Zero means floating.

 
Andrey F. Zelinsky:

That's right. According to Roche's "methodology" -- zero stop levelling means zero and no need to work it out in any way. But what to do with questions and doubts on the forum about it. It is necessary to clarify and sanctify in the "methodology" the situation when the stop leveller is equal to zero.

We must.

Usually double spread is used. But that is "usually". And sometimes it is not "usually".

 
Andrey F. Zelinsky:

Artyom, what to ponder "usually" and "not usually". We are talking about the market requirements, which, as I understand, are oriented on the real correct working of zero stoplevel.

I.e. it is necessary to make a comment on zero stoplevel in the "methodology" and remove all questions about it. But it turns out that questions arise on the forum, the same Vladon has already rubbed his corns pressing the keys on this issue - and the official side is silent.

That's what I'm saying - it's necessary ;)
 
Oleksii Chepurnyi:

Somehow it's not clear about the sign.....

I wrote it like this.

It should be something like this

   for(int i=PositionsTotal()-1; i>=0; i--)
     {
      if(!pst.SelectByIndex(i)) ShowError;
      if(pst.Symbol()==CheckSymb)
        {
         if(oper==1 && pst.PositionType()==POSITION_TYPE_BUY)  sum_volume += pst.Volume();
         if(oper==2 && pst.PositionType()==POSITION_TYPE_SELL) sum_volume -= pst.Volume();
        }
     }
   for(int i=OrdersTotal()-1; i>=0; i--)
     {
      if(!ord.SelectByIndex(i)) ShowError;
      if(ord.Symbol()==CheckSymb)
        {
         if(oper==1 && (ord.OrderType()==ORDER_TYPE_BUY  || ord.OrderType()==ORDER_TYPE_BUY_LIMIT  || ord.OrderType()==ORDER_TYPE_BUY_STOP  || ord.OrderType()==ORDER_TYPE_BUY_STOP_LIMIT))  sum_volume += ord.VolumeCurrent();
         if(oper==2 && (ord.OrderType()==ORDER_TYPE_SELL || ord.OrderType()==ORDER_TYPE_SELL_LIMIT || ord.OrderType()==ORDER_TYPE_SELL_STOP || ord.OrderType()==ORDER_TYPE_SELL_STOP_LIMIT)) sum_volume -= ord.VolumeCurrent();
        }
     }
and then analyse not only the value, but also the sign of the sum. Look for yourself further.
 
Andrey F. Zelinsky:

That's right. According to Roche's "methodology" -- zero stop levelling means zero and no need to work it out in any way. But what to do with questions and doubts on the forum about it. It is necessary to clarify and sanctify in the "methodology" the situation when the stop levelling is equal to zero.

So write - what is the problem? To talk again

It has already been answered - zero stop levelling does not always mean that there is no such restriction.

 
Rashid Umarov:
and then analyse not only the value, but also the sign of the sum. Look further for yourself

Why? Because the output will be the sum of positions and orders in one direction only.

 
Oleksii Chepurnyi:

Why? Because the output will be the sum of positions and orders in one direction only.

Let me remind you of your question from

As far as I understand, the calculation should take into account the volume of all positions and orders only in the direction in which we are going to open a deal or place an order.

But here the Limitation on the number of lots for one symbol the direction is not taken into account at all....

Or have I misunderstood something?


What is wrong?
 
Rashid Umarov:

What's wrong?

Why do we need a sign? My example summarises the volumes of only one direction.....

 

OK

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