Formulation of mathematical functions
Yield and revenue function:
Let us consider a crop in which x cm water is applied during the crop period. Let the grain (primary yield) and straw yield (secondary yield) of the crop be expressed in functional form as:
Yg = a1 + b1x + c1 x2 ………………..(1)
Ys = a2 + b2x + c2 x2 ………………..(2)
where Yg and Ys are the grain and straw yield (t/ha), respectively; x is the depth of irrigation water applied (cm); a1 is the intercept, and b1, c1 are the coefficients for the grain yield function; and a2 is the intercept, and b2, c2 are the coefficients for the straw yield function.
Next, let us consider that:
p1 = unit price of grain (or primary yield), $/t
p2 = unit price of straw (or secondary yield), $/t
Thus, the gross income from one hectare of land (GI, in $), is:
GI (x) = (Yg× p1) + (Ys× p2)
= (a1 p1+ b1 p1x + c1 p1 x2) + (a2 p2+ b2 p2x + c2 p2 x2)
In other words,
GI (x) = x2(c1 p1+ c2 p2) + x (b1 p1+ b2 p2) + (a1 p1+ a2 p2) ……………(3)
- Cost function
Next, let the cost function be expressed as (in $/ha):
C(x) = d1 + d2 (x) + d3 (x) …………………(4)
where …….
- The case of a water-limiting condition
Let us assume that an x cm irrigation depth is applied to achieve a targeted yield.
For the x cm irrigation depth, the total volume of water per hectare = 1 ha × x cm = x ha-cm Let p3 = cost of the unit water (cost of 1 ha-cm water) (including the application cost, e.g., the incurred labour, if any), $, p4 = cost of the unit harvest, transportation, threshing, cleaning/processing, and other steps, $/t
Then,
d2 (x) = p3x
d3 (x) = p4 × (Yg + Ys)
Thus, equation. 4 can be written as:
C(x) = d1 + d2 (x) + d3 (x)
= d1 + (p3 × x) + p4 (Yg + Ys)
= d1 + (p3 × x) + p4 (a1 + b1x + c1 x2) + p4 (a2 + b2x + c2 x2)
Irrigation depth for maximising the net income
Here, the net income from one hectare land,
NI (x) = GI(x) - C(x)
= x2(c1 p1+ c2 p2) + x (b1 p1+ b2 p2) + (a1 p1+ a2 p2) – [d1 + (p3 × x) + p4 (a1 + b1x + c1 x2) + p4 (a2 + b2x + c2 x2) ]
In other words,
NI (x)= x2 (c1 p1+ c2 p2 - c1 p4- c2 p4) + x(b1 p1+ b2 p2 – p3 - b1 p4 – b2 p4) + (a1 p1+ a2 p2 – d1 - a1 p4 - a2 p4)
………………….(5)
Let the total available water in the area = V ha-m
For the x cm irrigation depth, the irrigable area,
