Abstract
The authors consider a continuous dynamic approximating model of a gas fields group and, on its basis, set maximum and minimum issues. The tasks proposed for research are optimal control problems with mixed constraints with free-final-time and moving right end. We analytically solve the rapid-action problem. The central mathematical apparatus is Pontryagin maximum principle in Arrow form, using Lagrange multipliers. The theoretically obtained results of the analysis are of particular interest.
1 Introduction
Natural gas refers to non-renewable minerals located under the surface of the earth at high pressure and temperature. It is located in the earth's layer in a gaseous state in the form of separate accumulations (gas deposits). The composition of natural gas includes many useful substances used in industry.
We extract natural gas from the earth using wells. We try to place the wells evenly throughout the field. Thus, we ensure the same pressure drop across an entire deposit [1].
Natural gas is a valuable product with an enormous demand. Russia is one of the wealthiest countries in the world in terms of natural gas reserves. Proven natural gas reserves in Russia for 2014 amount to 24.6% of world reserves [2].
However, the locations of gas production and its consumption are at a considerable distance. One of the most effective ways to deliver gas to consumers is by transporting it through a gas pipeline.
Gas pipeline capacity
We introduce the following notation:
by i we denote an ordinal number of the deposit in a group consisting of n elements;
by
by Ni(t) we denote an operating wells stock at time t;
by qi(t) we denote an average flow rate of producing wells at time t.
If at time t the following strict inequality holds
then there exists such an
is satisfied.
There are many such solutions. Therefore, on this set, it is possible to solve other both static and dynamic problems. As the flow rate of producing wells decreases, there comes a moment T at which the equality
is fulfilled.
Further, we can no longer fulfil the plan for gas production. Therefore, the value T is the shelf length of the gas fields group. The question arises: within what limits can be the shelf length of the gas deposits group be? Thus, it is necessary to solve two optimal control problems with a mixed restriction on the maximum and minimum. We can analytically solve these problems and obtain theoretically substantiated estimates of the gas fields shelf length. We devote this paper to this statement. As the central mathematical apparatus, we choose the Pontryagin maximum principle in the Arrow form.
2 Pontryagin maximum principle in Arrow form
About 50 years ago, K. Arrow, the 1972 Nobel Prize winner in economics, published an article [3], where he, taking Pontryagin maximum principle [4] as a basis, modifies it and formulates propositions that allow solving problems optimal control with mixed constraints. The proposals additionally include some elements of nonlinear programming, such as Lagrangian function, Lagrange multipliers, and the complementary slackness conditions. The author of this article used a modified principle in his papers [5, 6].
Among the Russian works on optimal control, we can distinguish monographs by A. M. Ter-Krikorov [7], S. M. Aseev, and A.V. Kryazhemsky [8]. An interesting international paper is an article by E. Balder [9], in which the author proves the existence theorem for optimal control problems with an infinite horizon.
We formulate the Arrow proposition.
Proposition 1
Let v*(t) be a choice of instruments (0 ≤ t ≤ T) which maximizes
(a)
(b) F[x(t), v(t), t] ≥ 0
If the Constraint Qualification (Regularity Conditions) holds, then there exist auxiliary variables p(t), such that, for each t,
(c) v*(t) maximizes H[x(t), v(t), p(t), t] subject to the constaints (b), where H(x, v, p, t) = U(x, v, t) + pS(x, v, t);
(d)
(e) L(x, v, p, t) = H(x, v, p, t) + qF(x, v, t), and the Lagrange multipliers q are such that
(f) ∂L/∂vk = 0, for x = x(t), v = v*(t), p = p(t), q(t) ≥ 0, q(t)F[x(t), v*(t), t] = 0, and
(g) p(T) ≥ 0, p(T)x(T) = 0.
Note that if a set of constraints (b) consists only of inequalities
(b) F[v(t)] ≥ 0 at t ∈ [0, T],
3 Model Description and Problems Posing
Let us consider a model for the functioning of a gas fields group with interacting wells [10, 11]. By Vi(t) we denote the recoverable gas reserve at time t. Between variables, we establish a differential relationship, which we describe as a system of differential equations:
under the initial conditions
and limt→⧜ Vi(t) = 0 when
There is a general limitation on the total shelf length of the gas deposits given by
We calculate cumulative gas production using the following formula:
The paper [6] considered two questions: the problem of maximizing the accumulated extraction, and the problem of maximizing the total shelf length for the gas fields group. We present their mathematical formulations.
On the maximization of accumulated production for the gas fields group
Problem 1
We maximize the functional
on the fixed interval [0, T] under the differential connections
with the initial conditions
where
We impose the following restrictions on the controls:
We also assume that
Here qi(t) for
The controls Ni(t) for
On maximizing the total shelf length for the gas fields group
Problem 2
We maximize the functional T under the differential connections (7), with initial conditions (8) and constraints (10) and
We assume that the initial conditions satisfy the strict inequality (12). As in Problem 1, the controls Ni(t) when
Proposition 2
For any admissible trajectory from sets (7), (8), and (10) the following strict inequalities hold: qi(T) > 0,
Proposition 2 follows from the differential equations in (7), the initial conditions (8) and restrictions on controls (10).
Proposition 3
For any admissible trajectory from the sets (7), (8), (10), and (13), and under the initial constraint (12), there exists T > 0 for which the following equality holds:
The point q(T) is the right end of an admissible trajectory q(t). Further development of the system (7) is impossible due to existing limitations.
Proof
Multiply both sides of each differential equation in (7) by the corresponding value
Taking into account (13), we continue the transformation to yield
Let
The right-hand side of the inequality (17) reaches
Proposition 4
Let the equality (14) holds on some trajectory at time T, then
at T < t.
The following proposition follows from Propositions 3 and 4.
Proposition 5
For any admissible trajectory from the sets (7), (8), (10), and (13), and under the initial constraint (12), the control on the right end is strictly definite, i.e.
The exact upper bound for the total length of the shelf can be obtained by solving the Problem 2. We have described the solution in [6]. The essence of the solution is as follows. We sort all deposits in ascending order of values
On minimizing a length of the total shelf for gas fields group
Problem 3
It is required to minimize the functional T under the differential connections in (7), with the initial conditions (8) and with the constraints (10), (13).
We assume that the initial conditions satisfy the strict inequality (12). As in Problems 1 and 2, controls Ni(t) when
In [6], the search for a solution to Problem 2 is carried out indirectly. First, we solve Problem 1. Then, by taking limits, we find the solution to Problem 2. However, the search for a solution to Problem 2 can be carried out directly using the Arrow proposition and the corresponding transversality conditions.
The transversality conditions applied below are not mentioned in the paper [3] along with the Arrow proposition. Nevertheless, they are used in solving similar problems of optimal control together with the Pontryagin maximum principle in the classical formulation [12].
We now reformulate Problems 2 and 3 as follows.
On maximizing a length of the total shelf for gas fields group
Problem 2a
It is required to find optimal controls Ñi when
On minimizing a length of the total shelf for gas fields group
Problem 3a
It is required to find optimal controls Ñi for
Problems 2a and 3a are optimal control problems with a moving right end and free-final-time. Moreover, Problem 3a refers to the optimal rapid action problem. The optimal solution for Problems 2a and 3a exists. This fact, for example, follows from the monograph [13]. In search of decisions to Problems 2a and 3a, there is a logical relationship. The existing differences are insignificant.
In [6] we have already found the optimal solution to Problem 2. Therefore, in this paper, we restrict ourselves only to searching for the optimal decision to Problem 3a.
4 Investigation of Problem 3a
We recall
Next, we order the phase variables in descending order of
where
Next, we describe a set of admissible controls G(q), on which we maximize Hamiltonian (20) as follows
The main objective of the study is to search a continuous vector function ψ(t) and control Ñ(t), which satisfy the adjoint system of differential equations (27) and transversality conditions:
where ν is some number. For each t ∈ [0, T] the control Ñ(t) maximizes the Hamiltonian
We represent the system of equations as
or
According to Proposition 1, the Lagrange multipliers β(t), γ(t), δ(t) must satisfy the equalities:
or
In nonlinear programming, the above relations (29), (30) and (31) are complementary slackness conditions. We introduce the notation
Let us consider the transversality conditions in more detail. It follows from (24) and (25) that the right end of the adjoint trajectory ψ(t) lies in the negative region. However, if we solved Problem 2a, then ψ(T) would belong to the positive domain. Using the transversality conditions (24) and (25), and taking into account (22), (35), and (36), we obtain
This implies the following proposition.
Proposition 6
The right end of the “adjoint” vector function φ(t) lies in the positive domain; that is, φi(T) > 0 when
We do not know the value of the vector
Theorem 1
Let phase variables qi(t) be in descending order
we determine the optimal trajectory by the formulas:
(42)the following rule uniquely determines the vector of optimal controls Ñ(t). For each t there exists an integer k ∈{1, . . . , n} such that the following double inequality holds
(43)(44)(45)we combine a set of values t related to one integer k into half-intervals (τk−1, τk]. Under the condition
Proof
Further, we will adhere to the condition formulated in paragraph (3) of Theorem 1. We install
Below, in Proposition 7, we show the same order of functions φi(t); that is, for any t ∈ [0, T] the strict inequalities hold: φ1(t) > φ2(t) > . . . > φn(t).
Maximizing the Hamiltonian (32) reduces to function maximization
Now we introduce the following notation:
under the conditions
At the same time, the coefficients φi decrease.
Taking (49) into account, we transform the linear function (48) and obtain the following linear programming problem at maximum
Note that the coefficients
The following procedure allows us to find such an optimal vector ũ, which gives the maximum of the linear function (51) and, accordingly, the maximum of the linear function (48).
At the first step, we choose
Then we go to the second component u2, and repeat the first step's procedure, but with the changed value of
If we stop at the k-th step, then
From the above proof, it follows that for any k-th component of the control vector N(t), the following statements hold
if
if ũk = 0, then Ñk(t) = 0;
if
Let's consider the optimal trajectory's dynamic behavior on the segment [0, T]. Suppose that at the initial time
the following relations hold
From the system of “adjoint” equations (33) we take into account (34), (30) and (31) at t ∈ (0, τ1) to get γ1(t) = 0; δ1(t) = 0; β(t) = φ1(t) = β1 = const; φi(t) = const for
Then there is a moment τ2 such that:
the following relations hold
From the system of “adjoint” equations (33), and taking into account (34), (30) and (31) at t ∈ (τ1, τ2), then we get γ2(t) = δ2(t) = 0; β(t) = φ2(t) = β2 = const; φi(t) = const for
On the segment [0, τ1], the function φ1(t) − φ2(t) is constant, and on [τ1, τ2] this function strictly increases. The order relations established among the components of the vector functions φ(t) at the time t = 0, do not change on the segment [0, τ2]. Thus, we divide the segment [0, T] into n parts [0, τ1], [τ1, τ2], ..., [τn−1, τn]. Here τn = T. At the end of each k − th segment the equality holds
Proposition 7
Let t ∈ (0, τ1), k = 1, then:
All functions φi(t) are constant on the segment [0, τ1]. On it, the strict inequalities hold φ1(t) > φ2(t) > . . . > φn(t).
Let t ∈ (τ1, τ2), k = 2, then:
The function φ1(t) strictly increases on the segment [τ1, τ2]. The remaining functions φi(t) for
Let t ∈ (τk−1, τk),
The functions φi(t), where
Proof
Using the relations (28)–(31), (33), (42) and taking into account the continuity of the vector function φ(t), we write out formulae for calculating the variables β(t), N(t), φ(t) on each of the n segments.
Taking into account the earlier established orders φ1(0) > φ2(0) > . . . > φn(0), we prove the strict inequalities φ1(t) > φ2(t) > . . . > φn(t) on each time segment.
For the dimension n = 3 in Figure 1, we schematically show the behaviour dynamics of optimal controls Ñ1(t), Ñ2(t), Ñ3(t), the functions φ1(t), φ2(t), φ3(t) and the Lagrange multiplier β(t) obtained by solving Problem 3a.
Behavior dynamics of optimal controls Ñ1(t), Ñ2(t), Ñ3(t), the functions φ1(t), φ2(t), φ3(t) and the Lagrange multiplier β(t) by solving Problem 3a of dimension n = 3.
Let us describe a scheme for determining the continuous vector function φ(t). First, we find τ1, τ2, . . . , τn = T and qi(T) for
Next, from the formula in (38), we calculate the exact values of the vector φ(T), whose components, according to Proposition 6, are in descending order.
The results in Proposition 7 are the basis for further description of the algorithm. We consider the segments sequentially starting with [τn−1, T] and ending with [0, τ1]. We know the Lagrange multiplier β(t) = βn on the n-th interval (τn−1, T), the function φn(t) = φn(0) = φn(T) = βn on the segment [0, T] and the values φi(T),
where
The following inequalities hold φi(t) > φn(t) for
We proceed to the sequential study of functions φk(t) on the k-th segment of [τk−1, τk] for
when
Thus, we find all φi(τk−1), where
Therefore, on the segment [0, T], we have constructed a single-valued vector function φ(t) satisfying the transversality conditions (38) and the maximum principle in the Arrow form. Moreover, this function is unique. Indeed, any other order of the components of the vector φ(0) does not allow us to hold the necessary optimality conditions.
Taking into account the existence of an optimal trajectory for the problem under study, we come to the following conclusion. We have found the only “adjoint” vector function φ(t) and the corresponding optimal trajectory. We have proved Theorem 1.
Next, we consider the case when all indicators
Proposition 8
For any admissible trajectories from sets (7)–(10) and (13) of the total shelf lengths under the condition
the lengths of their shelves are the same and they are calculated using the following formula
Proof
We determine the accumulated production by the formula:
Taking into account (5) and (54), we rewrite (56) as follows
According to Propositions 3, 4, and 5, at the end of any admissible trajectory, the equality
We now turn to the description of the algorithm for determining the minimum and maximum length of the total shelf.
Algorithm for calculating the minimum and maximum length of the total shelf for a group of gas fields
To minimize, we sort deposits in descending order of values
First step
We put the first field into development and, using Proposition 8 with n = 1 we calculate the temporary step length:
Denote
Note that
Second step
Denote by θ2 the temporary second step length. Denote by
We determine the temporary length of the second step from a solution of equation (61) with one unknown. Taking into account (4) and (5), τ2 = τ1 + θ2 and
We consider equality (61) as an equation with one unknown variable θ2 We denote the right-hand side of equality (62) by F(θ2). We differentiate the function F(θ2) and analyze it. As a result, we obtain the following inequality
Hence, the function F(θ2) is an increasing function for all positive values θ2.
The inequalities
k-th step,
k = 3 , n ¯
Denote by θk the temporary k-th step length. Denote by
We determine the temporary length of the k-th step from the solution of the equation with one unknown. Taking into account (4) and (5), τk = τk−1 + θk and
It has one solution. To prove this, we denote the right-hand side of the last equality by F(θk). It is easy to show the following properties of the function F(θk): F′(θk) > 0 for all θk > 0;
The last equation has no explicit solution. We can solve it numerically by dividing the segment in half. At the end of the k-th stage, we describe the producing wells flow rate dynamics of the k-th field
This completes the description of the algorithm.
5 Conclusion
When developing a group of gas fields, the main problem at any given time is the fulfilment of the current gas production plan. In the initial period of development of the gas fields group, we can achieve this effect in different ways, including by turning on or off wells. However, there comes a time when there is not enough capacity to fulfil the current gas production plan. All available wells in all fields are involved in the development. At this point, the process of developing the gas fields group is interrupted. Each development dynamics of the gas fields group has its shelf length. It is required to determine its exact estimates.
We solve the problems of optimal control for the minimum and maximum shelf lengths of gas fields group under strict restrictions on total gas production. To solve the set tasks, we sort all fields by the value of the parameter
We divide the entire optimal period into n parts equal to the number of deposits in the group. At the beginning of each part of the optimal period, we bring into development the next field in order. All previously commissioned fields continue to operate at their maximum capacities until the end of the optimal period and cannot ensure the implementation of the entire current production plan. Therefore, we include a new deposit in development.
At the end of the current part of the optimal period, all wells of the newly developing field are in development and are operating at full capacity.
At the end of the optimal period, all wells in the group of fields operate at full capacity with maximum efficiency. Any further development of the group only leads to non-fulfilment of the current plan for the volume of gas production. This fact interrupts the dynamic process. If we set all parameters
We have presented in this article an algorithm for calculating the minimum length of the total shelf for a group of gas fields.
Conflict of interest: Authors state no conflict of interest.
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