[@Hel83] and the method involves solving the following equation for N$_{UL}$, $$\label{eq1}
 \beta I\left(\frac{ -N_{exc}}{\sigma_{exc} }\right)=  I\left(\frac{ N_{UL} -N_{exc}}{\sigma_{exc} }\right)$$ where (1-$\beta$)$\times$ 100$\%$ is the confidence level, N$_{exc}$ is number of excess events with $\sigma_{exc}$ as its standard deviation. The function I(x) is given by $$\label{eq2}
I(x)=  \frac{1}{\sqrt{2\pi}} \int^{\infty}_{x} \; e^{-t^2/2} \; dt=  \frac{1}{2} \;erfc( \frac{x}{\sqrt{2}})$$ Where erfc(x) is the complementary error function. On solving equation 1 with N$_{exc}$= 2, $\sigma_{exc}$= 56 (refer $\alpha$–plot shown in Figure 1(b)) and $\beta$=0.01, we get N$_{UL}$ $\approx$ 146 as 99$\%$ confidence level upper limit on the excess events from the source. Knowing that the $\gamma$–ray rate of (14.91$\pm$2.82)h$^{-1}$ corresponds to 1 CU, the resulting 99$\%$ limit on the rate of excess events from 1ES1218+304 translates to 3.68h$^{-1}$ (i.e 146/39.62h or 0.25 CU). Alternatively, on dividing the upper limit on the excess events by the product of effective collection area $\sim 3.0\times 10^8$ cm$^{2}$ (obtained after applying the Dynamic Supercuts) and observation time $\sim$ 39.62 h, we find the 99$\%$ upper limit on the integral flux to be 3.41$\times10^{-12}$ photons $cm^{-2} s^{-1}$ above threshold energy of 1.1 TeV. The values of effective collection area and threshold energy used above correspond to zenith angle of 