Ratio of gaussian CDFs in Black-scholes option pricing formula

What is meant by $\frac {\Phi (d_2)}{\Phi (d_1)}$ in the Black Scholes call option price?

I found it in a solution as $\frac{\text{short position in cash}}{(\text{number of shares})(\text{strike price discounted to time zero})}$

Reference can be found here

Q. Number 4 on page number 19, and Its solution on page number 26

• We need more background to understand your question. Jan 27 '16 at 19:40
• Hi @Gordon, please see in the Q, I edited it. Jan 28 '16 at 5:01
• This term is not part of the BS formula for a call price as you can see here: en.wikipedia.org/wiki/Black%E2%80%93Scholes_model in the section "Black-Scholes formula".
– Ric
Jan 28 '16 at 6:57
• @Rchard Yes but N (d_1) and N (d_2) are parts of the BS formula... and in solution given in link used N (d_2) divide by N (d_1) which I am confised about. Jan 28 '16 at 7:57
• Your question is really hard to understand, it does not become clear to me, what you are asking for. Consider editing it to improve the quality (and quantity) of answers. Jan 28 '16 at 8:08

For a call option with price given by \begin{align*} c = S_0 \Phi(d_1) - K e^{-rT}\Phi(d_2), \end{align*} the delta hedge ratio $\Phi(d_1)$ is the number of shares to hold. That is, $S_0 \Phi(d_1)$ is the total holding share value for hedging, while $K e^{-rT}\Phi(d_2)$ is the total cash amount in short.
In the question, it says that, for $N$ options, 250,000 shares of the stock are hold, and the amount of $£413,057$ is in short. The strike price is $K=2.0$. Therefore, \begin{align*} N \Phi(d_1) = 250000, \mbox{ and } N K e^{-rT}\Phi(d_2) = £413057. \end{align*} Consequently, \begin{align*} \frac{\Phi(d_2)}{\Phi(d_1)} &= \frac{N K e^{-rT}\Phi(d_2)}{N\Phi(d_1)}\frac{N}{N K e^{-rT}}\\ &=\frac{N K e^{-rT}\Phi(d_2)}{N\Phi(d_1) K e^{-rT}} \\ &=\frac{413057}{250000 \times 2.0 \times e^{-0.03 \times 0.5}}\\ &= 0.8386. \end{align*}
• $\Phi (d_1)$ is the numbers of shares hold PER OPTION, no? Jan 28 '16 at 16:55