Combining STOP orders with STOP LOSS orders

• 11/06/2019 at 12:16 PM #112264
  Wilko

  I have a question that I hope someone on the forum can help me answer.

  This is best explained by an example:

  Let’s suppose I have a system that issues a buy-order and at the same time sets a stop loss:

  IF NOT LongOnMarket AND Close > Close[1] THEN BUY 1 CONTRACT AT MARKET SET STOP %LOSS 1 ENDIF

  1 2 3 4

  IF NOT LongOnMarket AND Close > Close[1] THEN BUY 1 CONTRACT AT MARKET SET STOP %LOSS 1 ENDIF

  Let’s say I then at a later bar want to issue a different STOP order

  IF BarIndex - TradeIndex >= 2 AND Close > PositionPrice THEN SELL AT close STOP ENDIF

  1 2 3

  IF BarIndex - TradeIndex >= 2 AND Close > PositionPrice THEN SELL AT close STOP ENDIF

  Will the system accept the new order without first cancelling the first STOP LOSS order? Ie will it reject the second STOP order because there is already a STOP order in the system? Or can a system have one STOP LOSS order at the same time as a different SELL STOP order?

  In other words do I have to run the below instruction before I can issue the second STOP order?

  SET STOP %LOSS 0

  1	SET STOP %LOSS 0

  Many thanks for any clarification!

   

  /F

  

  Wilko

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  11/06/2019 at 1:01 PM #112269
  robertogozzi

  You can write the second line 2 as:

  SELL AT MARKET

  1	SELL AT MARKET

  since that code will be always executed at closing time, thus CLOSE is the current price.

  ProOrder does’n care about previous settings, whenever there’s a LONG trade open and you order it to SELL… it.

  Your broker may have some objections if you don’t abide to its distance requirements.

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  robertogozzi

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  11/06/2019 at 1:13 PM #112270
  Wilko

  Thanks! I didn’t realize it was a bad example!  I will change above code to the below to hopefully make it clear what I am trying to do:

  SELL AT close * 0.995 STOP

  1	SELL AT close * 0.995 STOP

  I am aware that I need to respect minimum distance requirements for any STOP or LIMIT orders

  

  Wilko

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  11/06/2019 at 1:46 PM #112271
  robertogozzi

  It will place the SELL order at your requested price level. As all pending orders it will expire after 1 bar and you will have to place it again, if needed.

  But I’m not sure if this will affect your previous SL. in case your SELL order is not triggered and expires and you don’t place it again will your previous SL be remembered or you may end up without a SL.

  When I trail stop loss, any time a new pending stop order is executed, IG updates the status of my trade to set a new SL, and it remains there because I set it continuously candle after candle, but I never tried to not place it again to see what happens.

  It would be worth trying.

  

  robertogozzi

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  11/06/2019 at 2:11 PM #112272
  robertogozzi

  I just tested this strategy, it sets a SL at 200 pips from entry price, then, when onmarket, sets a new SELL pending order at 100 pips from entry price without placing it again next bar.

  The original SL is kept, while for just 1 bar there’s a further pending order at -100 pips:

  ONCE Flag = 0 // place a SL to 100 pips for just this time to let it expire IF OnMarket AND Flag = 0 THEN SELL AT Tradeprice - 100 * pipsize STOP Flag = 1 ENDIF IF Not OnMarket THEN BUY AT MARKET SET TARGET pPROFIT 200 SET STOP pLOSS 200 ENDIF

  1 2 3 4 5 6 7 8 9 10 11

  ONCE Flag = 0 // place a SL to 100 pips for just this time to let it expire IF OnMarket AND Flag = 0 THEN    SELL AT Tradeprice - 100 * pipsize STOP    Flag = 1 ENDIF IF Not OnMarket THEN    BUY AT MARKET    SET TARGET pPROFIT 200    SET STOP   pLOSS   200 ENDIF

   

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  robertogozzi

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  11/06/2019 at 2:21 PM #112273
  Wilko

  Thanks!

  If that is correct, that’s good news for me. I remember seeing an order rejected with error message “Stop order already active”, so therefore my question. But I guess one of my systems managed to issue two stop orders during same instance in that case.

  

  Wilko

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  11/06/2019 at 5:28 PM #112282
  Nicolas

  Wilko wrote:

  But I guess one of my systems managed to issue two stop orders during same instance in that case.

  In that case, the first stop order read in the code is the one that will be set.

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  Nicolas

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