Implied Volatility of the S&P Today

The Black-Scholes-Merton Formula is: $C(S_0,t) = S_0N(d_1) - Ke^{-r(T-t)}N(d_2)$ With $N(d_1)$ and $N(d_2)$ being: $d_1 = \frac{\ln{\frac{S_0}{K}} + \big(r + \frac{\sigma^2}{2}\big)(T-t)}{\sigma\sqrt{T-t}}$ $d_2 = d_1 - \sigma\sqrt{(T-t)} = \frac{\ln{\frac{S_0}{K}} + \big(r - \frac{\sigma^2}{2}\big)(T-t)}{\sigma\sqrt{T-t}}$

So plugging what we know in here: $\$7.94 = \$216N(d_1) - \$210 e^{-0.018*\frac{30}{365}}N(d_2)$ $d_2 = d_1 - \sigma\sqrt{(T-t)} = \frac{\ln{\frac{\$216}{\$210}} + \big(0.018 + \frac{\sigma^2}{2}\big)(\frac{30}{365})}{\sigma\sqrt{\frac{30}{365}}} - \sigma\sqrt{\frac{30}{365}}$

This would be a great time to use Microsoft Excel's goal seek. If you're actually an investor you'll definitely have and use this tool for efficiency's sake!

Alternatively, since the option value is an increasing function of volatility, we may use a root approximation method (like [[wiki-https://brilliant.org/wiki/root-approximation-bisection/|Bisection) to test multiple values by hand.

We sthart by guessing what a volatility could be - say 20%, plug that in and see price the Black-Scholes-Merton formula spits out: $\sigma = 0.20 \Rightarrow C(\$216,30 \text{days})=\$8.66 > \$7.93$ Let's try another volatility that gives a value below $7.93: $\sigma = 0.10 \Rightarrow C(\$216,30 \text{days})=\$6.79 < \$7.93$ Then, let's test the midpoint $\frac{1}{2}(0.20 + 0.10) = 0.15$ : $\sigma = 0.15 \Rightarrow C(\$216,30 \text{days})=\$7.64 < \$7.93$ Again, let's test the midpoint $\frac{1}{2} ( 0.20 + 0.15) = 0.175$ .
$\sigma = 0.175 \Rightarrow C(\$216,30 \text{days})=\$8.14 > \$7.93$ Again, let's test the midpoint $\frac{1}{2} ( 0.175 + 0.15) = 0.1625$ .
$\sigma = 0.1625 \Rightarrow C(\$216,30 \text{days})=\$7.88 < \$7.93$ Again, let's test another point in the range, which we will use $0.1675$ :
$\sigma = 0.1675 \Rightarrow C(\$216,30 \text{days})=\$7.99 > \$7.93$ Again, let's test another point in the range, which we will use $0.165$ :
$\sigma = 0.165 \Rightarrow C(\$216,30 \text{days})=\$7.935 \text{ truncate d} = \$7.93 _\square$