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Q3) Answer from Question 3:
d1 = [ln(Pa/Pe) + (r + 0·5s2)t]/(st1/2) (from formulae sheet given in examination)
d1 =[ln(340/350) + ((0·04 + 0·5 x 0·42) x 1/6)]/(0·4 x 1/60·5) = –0·055
–d1 = 0·055
N(–d1) = 0·5 + (0·0199 + (0·0239 – 0·0199)/2) = 0·5219[/u]
Put contracts to be bought for a delta hedge = 200,000/(0·5219 x 1,000) = 383·2 rounding to 383 contracts.
Can someone assist how to get the N(-d1)? What formula is this, I am self studying so there is something i am figuring out. Thanks for your help.- Topics: 0
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Figure for N(d) need to take reference with the standard normal distribution table.
If d1 > 0, then we need to add 0.5 to the data find out in the table
If d1 < 0, then we need to minus 0.5 to the data find out in the table now d1 is 0.055, so we need to find out the figure of 0.055 from the standard normal duistribution table
0.05 is 0.0199
0.06 is 0.0239as 0.055 is between 0.0199 & 0.0239, that is 0.0199 + [(0.0239 + 0.0199)/2] = 0.0219
so n(d1) = 0.5 + 0.0219 = 0.5219
Without the help of the study group in Open Tuition, I can’t passed P4, I will support the group forever.
If d1<0, either (0.5 - data) or (data - 0.5) is correct ???? Please clarify for me.Thanks.
I learn somewhere that (0.5 – data) is correct. But I am not sure.- Topics: 0
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I am afraid I gave a wrong explanation. Let me have some time to find the notes out first!
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I checked my notes,
if d1 and d2 have a positive value, (data + 0.5)
if d1 and d2 is negative, then 0.5 – dataI am sorry for any inconvenience causes.
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what is the value option of option at the moment for organization. of course it for value of project but if project is not being started now, will it affect value of entity now ?
and when ever this option is excercised, will it affect organisation’s value ?
because at the moment project is instigated, the option is no more !
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