Reservoir capacity: rice, wheat, sugarcane

Given: canal loss = 18%, reservoir loss = 14%; area, base period and duty of each crop as in the table (the duty is at the field).

Method. For each crop, discharge needed at the field $Q = A/D$; volume for the base period $V = Q\times B\times 86400$ m$^3$. The losses are taken as a fraction of the water entering that part of the system, so field volume is divided by $(1-\text{loss})$.

Total volume at the field: $V_f = 336.370\times10^6$ m$^3$

Allow canal loss (18%):

$$V_{canal\ head} = \frac{336.370}{1-0.18} = 410.208\times10^6\ \text{m}^3$$

Allow reservoir loss (14%):

$$V_{reservoir} = \frac{410.208}{1-0.14} = 476.986\times10^6\ \text{m}^3$$

Answer: reservoir capacity $\approx$ 476.99 million m$^3$ (0.4770 Gm$^3$).