# A comparison between the best and the worst strategies

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## Q range values in the case

In this section, we shall find numerically the limit value of Q that characterizes the range values in this third form. This case corresponds to a value of quota that allows to have a singular arc on [t1; T ]:
We can observe the behavior of the control with values below and above Qlim2 to validate our results:

### Problem with rain

At first, we’ll study the problem of maximization of the biomass without rain with the following parameters:
The maximum biomass in this case is : 0:39. Here are the figures of the variation of humidity S, the biomass B, V and the control u found by BocopHJB :
We will calculate the quantity of water Qmax that will allow us to have the maximum biomass using the code of the previous sections. We find that this value is equal to Qmax = 0:2, it will allow us to obtain a biomass B = 0:501.
We will introduce now the rain that is pre-defined with a quantity of water Qrain. At first, we will study the case where the sum of the quantity of water Q and the quantity of water brought by rain Qrain is equal to Qmax.

Near Qma

We will use a quantity of water Q = 0:1 and we will change the quantity of water brought by rain. The values Qrainthat we’ll use are:
We will define the rain as a piecewise constant function on the interval [0; T ] with an amplitude of:
p = kT2 Qrain zz In the first example, one could see that the variation of humidity is similar to the case where we have an initial quantity of water Q = 0:1 + Qrain because the rain intensity is weak during the totality of [0; T ]. The precipitations brought by rain in the second example allow us to always maintain the humidity above S and to have an almost max-imum biomass. In the last example, the rain brings a quantity of water Qrain that allows the system to have a maximum biomass because Qmax = Q + Qrain and the humidity is maintained at S = 1.
The goal of the next section is to find out how we shall distribute the quantity of water brought by rain Qrain that doesn’t allow us to get to Qmax and to maximaze the biomass B .

Distribution of rain

At first, we will see that the change of the distribution of Qrain (= 0:01 in this case) on the interval [0; T ] allows us to have diﬀerent values of biomass.
Furthermore, because it seems impossible to study all the possible scenarios of rain, we will proceed by studying 4 scenarios. The first one is to bring constant precipitations on the interval [0; T ]. The second one is to bring constant precipitations on the interval [t1; t2] where the humidity S is equal to S . The third one is to bring strong precipitations on the interval [0; T =10] and weak ones on the interval [9T =10; T ]. The last one is to bring weak precipitations on the interval [0; T =10] and strong ones on the interval [9T =10; T ]. The graph of rain on each scenario is:
We found out that the best scenario is the second one where the precipitations are present on the interval [t1; t2], if we compare this graph with the one without rain, we can see that the humidity is longer maintained at S and that explains why we have a better final biomass. Same for the first scenario but with less biomass produced as the humidity is also maintained at S but for a shorter duration than on the second scenario. For the third and fourth scenarios, we can spot that the duration of the singular arc on S is shorter than on the previous scenarios, the rain brought isn’t well distributed to help maintaining S on S.
From these four scenarios, we can conclude that having precipitations on the interval [t1; t2] is the most favourable thing to do to have a maximum biomass.

In reality, it’s diﬃcult to predict the distribution of rainfall on the whole season, that’s why we will use adaptive control to solve the problem with rain, this type of scenario is closer to the real scenario than solving knowing at first the distribution of rain.
The program will adapt its dynamic on each step taking on consideration if it rains at that specific moment or not. Let’s take an example where the real distribution of rain is as follows (here not known in advance):
The final biomass is:
Problem without rain Problem with rainfall known at first Problem with adaptive solving As we can see, solving the problem with rainfall known at first gives us the maximum biomass because the solving mechanism can decide even before starting on how to dis-tribute the quantity of water brought to get a maximum biomass, whereas in the adaptive solving, the solution of the problem is to be adapted to the real time rainfall.

#### References

[1] Kenza Boumaza, Nesrine Kalboussi, Alain Rapaport, Sébastien Roux, and Carol Sin-fort. Optimal control of a crop irrigation model under water scarcity. May 2021.
[2] Bocop HJB the optimal control solver.

1 Introduction
2 A comparison between the best and the worst strategies
3 The control of the minimisation problem
3.1 Multiple possible forms of control
3.2 The first form
3.2.1 Range of Q values
3.2.2 Singular arcs and corner points
3.2.3 Water consuming comparison
3.2.4 Biomass production comparison
3.3 The second form
3.3.1 The value of  in this form
3.4 The third form
3.4.1 Q range values in the case
4 Problem with rain
4.1 Near Qmax
4.2 Distribution of rain
4.2.1 First scenario
4.2.2 Second scenario
4.2.3 Third scenario
4.2.4 Fourth scenario