How is project value enhanced by uncertainty in market payoff? I was reading about the concept of real options valuation. It says that ROV has shown that uncertainty in market pay-off enhances the project value (V). How is that? Will it be possible to explain that with an example?

Here by 'uncertainty' we mean stochastic variability of parameter distributions. 

{V = f(performance,cost,time,market requirement,market payoff) }
In general, if you own an option, holding fixed the expected value of the underlying asset and the strike price of the option, the greater the variance in the distribution of the prices of the option the (weakly) more valuable the option is. Here is a graphical example of a call option at maturity. They grey line indicates the value of the option for each value of the underlying asset where the strike price is \$5. The blue diamonds indicate the payoffs for values of the underlying at 4 and 6. If each are equally likely (p=0.5) then the underlying has an expected value of 5 and the option of \$0.5 (the lower green square). This distribution as a standard deviation of 0.5. Increase the standard deviation to 1.0 but hold the expected value constant at 5 by  changing the two underlying outcomes to \$3 and \$7 (orange squares). The expected value of this option is 1 (higher green square), so the increase in variance of the distribution of the underlying asset has increased the value of the option even though the expected value of the underlying is unchanged.


The reason for this is essentially Jensen's inequality. We can model the value of an option as a convex function $f(x)$ where $x$ is the underlying. The value of the option is $g(x) = E[f(x)]$ This implies $g(x)$ is increasing in the variance of $x$. We can see this because when we do the Taylor expansion of the function f : $$g(x) \approx E[f(x_0) +  (x - x_0)\cdot f'(x_0) +  (x - x_0)^2 \cdot f''(x_0)]$$

Because of the convexity, the second order term $f''(x_0)$ will be positive. The additive part $E[(x - x_0)^2 \cdot f''(x_0)] = E[(x - x_0)^2] \cdot f''(x_0)$ is increasing in the variance $E[(x - \bar{x})^2]$ so the value of the option is increasing in the variance.