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December 28th, 2018, 08:08 AM  #1 
Newbie Joined: Dec 2018 From: UK Posts: 3 Thanks: 0  Shares breakeven point question
I’ve posted this here as I’m sure it’s an algebra problem and it's driving me nuts. I want to calculate the breakeven point of 2 positions via a formula. For example if I have a long position of (+ 5) @ 1261 and a short position of ( 3) @ 1251 What is the easiest way of calculating breakeven point if the market keeps trending up? I know the answer is 1276 ,as I’ve plotted it manually, but just can figure it out via calculation. It has to do with the ratio of the size which is 1.66 Any insight would be much appreciated. 
December 28th, 2018, 02:15 PM  #2  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  Quote:
$5(p  1261) + 3(1251  p) = 0 \implies 5p  6305 + 3753  3p = 0 \implies$ $2p = 6305  3753 = 2552 \implies p = 1276.$  
December 29th, 2018, 02:04 AM  #3  
Newbie Joined: Dec 2018 From: UK Posts: 3 Thanks: 0  Quote:
This is great and does solve that problem, but there seems to be a variable missing. For example, if I have +8@1261 and 5@1251 using this calculation does not work out: 2p = 10088  6225 = p = 1931.5  
December 29th, 2018, 05:39 AM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  Quote:
$8(p  1261) = \text { realized gain (or loss) on } 8 \text { shares long.}$ $5(1251  p) = \text { realized gain (or loss) on } 5 \text { shares short.}$ $\therefore 8(p  1261) + 5(1251  p) = 0 \implies 8p  10088 + 6255  5p = 0 \implies$ $8p  5p = 10088  6255 \implies 3p = 3833 \implies p =1277 \text { and } \dfrac{2}{3}.$ Does that work out? $8 \left ( 1277 + \dfrac{2}{3}  1261 \right ) = 8 * \left ( 16 + \dfrac{2}{3} \right ) = 128 + \dfrac{16}{3} = 133 + \dfrac{1}{3}.$ $5 \left ( 1251  \left \{ 1277 + \dfrac{2}{3} \right \} \right ) = \ 5 * \left ( 26 + \dfrac{2}{3} \right ) = \ \left ( 130 + \dfrac{10}{3} \right ) = \ \left ( 133 + \dfrac{1}{3} \right ).$ Yes, that works exactly. In practice, of course, it may not be possible to achieve the exact breakeven price due to fractions. The general formula is as follows: $q = \text { purchase price on long transaction;}$ $r = \text { sales price on short transaction;}$ $m = \text { number of shares long;}$ $n = \text { number of shares short; and}$ $p = \text { breakeven price.}$ $m = n \text { and } q \ne r \implies p \text { does not exist.}$ $m = n \text { and } q = r \implies p \text { is anything.}$ $m \ne n \implies p = \dfrac{mq  nr}{m  n}.$ To derive the last: $m(p  q) + n(r  p) = 0 \implies mp  mq + nr  np = 0 \implies mp  np = mq  nr \implies$ $p =\dfrac{mq  nr}{m  n}.$  
December 30th, 2018, 08:48 AM  #5 
Newbie Joined: Dec 2018 From: UK Posts: 3 Thanks: 0 
Ah of course. I feel so stupid. Thank you so much for this. Sent you a PM. Last edited by skipjack; February 4th, 2019 at 12:02 AM. 

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