2.1 Reducing energy during specified time intervals

This section reviews work directly related to the problem of managing energy use for a fleet of trains during specified time intervals.

A feasibility study by Pam et al. [Reference Pam, Letrouv, Zhou, Yee, Pudney and Grellier8] simulated a fleet of $105$ TGV trains to show that it is possible to reduce power demand that initially varies between 220 MW and 290 MW during a 1-hour peak demand period to a constant average power of 200 MW for each 10-minute interval. Experiments conducted on individual trains running on the SNCF South Europe Atlantic high-speed line confirmed that significant power reductions could be achieved in practice. Their algorithm for adjusting train schedules to reduce energy use in 10-minute intervals worked by adjusting the locations of trains at the beginning and end of each interval during a 1-hour peak demand period, but did not attempt to find an optimal solution to the problem.

Scheepmaker et al. [Reference Scheepmaker, Goverde and Kroon12] gave a comprehensive review of the background literature on optimal train control. Our previous work using Pontryagin’s principle is summarized by Pudney [Reference Pudney, Su, Tian and Goverde9], with details given in a pair of papers by Albrecht et al. [Reference Albrecht, Howlett, Pudney, Vu and Zhou1, Reference Albrecht, Howlett, Pudney, Vu and Zhou2]. Pontryagin’s principle states necessary conditions for an optimal control $u^*$ that can be found by maximizing the control Hamiltonian $\mathcal {H}$ [Reference Pudney, Su, Tian and Goverde9]. There are five possible optimal control modes for a train equipped with continuous control travelling along a track with known gradient and fixed journey time: maximum acceleration, speedhold using partial acceleration, coast, speedhold using partial regenerative braking (if available) and maximum brake [Reference Albrecht, Howlett, Pudney, Vu and Zhou2]. Rao et al. [Reference Rao, Feng, Wang and Sun10, Reference Rao, Feng, Wang and Sun11] extend this analysis for scenarios where regeneration energy can be shared between trains but not exported back to the grid, and identify two additional control modes: driving to absorb power generated by other trains and braking to supply power to other trains.

Howlett et al. [Reference Howlett, Pudney and Albrecht6] proposed a driving strategy for a single train on a level track to minimize energy use when there are energy constraints during specified time intervals. They used classical methods of constrained optimization to develop an optimal strategy that uses a period of coasting followed by speedholding at an optimal driving speed during intervals where it is necessary to decrease speed, and a period of maximum acceleration followed by speedholding at an optimal driving speed during intervals where it is necessary to increase speed.

In this paper, we extend the work of Howlett et al. [Reference Howlett, Pudney and Albrecht6] by considering multiple trains. Our aim is to find general characteristics for an optimal solution, so we start with simple train models with piecewise constant speed profiles (Section 3). This is equivalent to assuming that the energy use for a time interval depends on the average speed of the train during that interval. Section 4 formulates the problem of minimising total energy use of a fleet of trains with one peak demand interval. Section 5 extends the problem with multiple consecutive peak demand intervals. In Section 6, we discuss the implications for more realistic train models.

2.2 Methods for reducing peak power demand for train operations

There are several papers that consider reducing peak power demand for train operations, but not the specific problem of managing power demand during specified time intervals.

Su et al. [Reference Su, Tang and Wang14] evaluated general strategies for reducing energy use on railways. They considered timetable optimization, reducing train mass, improving the design of gradients along the track, increasing traction and braking forces, using regenerative braking, and improving running resistance. Although these strategies can provide energy savings, some require drastic system changes such as changing the train schedules, operating trains made of lighter materials, and with larger traction and braking forces, adjusting the slope of the track, installing regenerative braking systems, reducing aerodynamic drag, and reducing the rolling resistance of the train. Even if these strategies are implemented and trains are operating as efficiently as possible, a method for managing energy use for a fleet of trains during specified time intervals is still required.

Other literature has focused on energy-efficient scheduling to prevent high peaks in power demand. Albrecht [Reference Albrecht3] examined the possibility of using train running time control for synchronizing acceleration and braking phases to reduce power peaks and energy use, and showed that optimized timetables for multiple trains gave smaller power peaks and less energy use. Albrecht [Reference Albrecht3] also looked at the effects of stochastic station dwell times, which made it harder for the trains to run on time and preserve running time reserve, and still found that the modification of train running times can contribute to significantly reducing power peaks and energy use, and thereby reducing energy cost.

Li and Lo [Reference Li and Lo7] used a genetic algorithm to jointly optimize the scheduling and speed control of metro rail systems such that the net energy use is minimized by coordinating arrival and departure times so that regenerative energy from braking trains is used by nearby accelerating trains. Their results showed that a larger headway leads to a smaller energy saving rate, and the maximum energy saving of roughly 25% is achieved when the minimum allowable headway is used.

There is much recent literature related to energy efficient train operations, but none address the problem of managing a fleet of trains during specified time intervals to reduce peak power demand.

4.1 Minimizing total energy

We can now define an optimization problem. We wish to find constant speeds $w_i$ and $z_i$ for each train $i \in \{1,\ldots ,n\}$ to minimize the total energy use E defined by (4.4), such that the energy use $E_1$ during the peak demand period, defined by the summation in (4.1), is constrained to the pre-defined level $(1-\alpha )E_0$ . Speed $w_i$ is the optimal speed of train i during the peak demand interval, and speed $z_i$ is the optimal speed of train i before and after the peak demand period. Equation (4.2) must be satisfied for each train i to ensure that the journey covers the correct distance $X_i$ . To solve this problem, we form a Lagrangian function $\mathcal {L} = \mathcal {L}(z_i,w_i)$ given by

$$ \begin{align*} \mathcal{L} = & \sum_{i = 1}^n k_i \varphi_i(z_i) +E_1 \\ &+ \lambda \bigg(\sum_{i=1}^n h_i \varphi_i(w_i) - E_1\bigg) \\ &+ \sum_{i=1}^n \delta_i (X_i - k_i z_i - h_i w_i) , \end{align*} $$

where $\lambda $ and $\delta _i$ are Lagrange multipliers. The optimal solution must satisfy the Karush–Kuhn–Tucker conditions [Reference Beck4]:

$$ \begin{align*} \frac{\partial \mathcal{L}}{\partial w_i} = \frac{\partial \mathcal{L}}{\partial z_i} = 0 \quad \text{for all } i \in \{1,\ldots,n\}, \end{align*} $$

and the complementary slackness conditions

$$ \begin{align*} \lambda \bigg(\sum_{i=1}^n h_i \varphi_i(w_i) - E_1\bigg) = 0 \end{align*} $$

and

$$ \begin{align*} \delta_i (X_i - k_i z_i - h_i w_i) = 0 \quad \text{for all } i = \{1,\ldots,n\}. \end{align*} $$

The partial derivatives with respect to the new speeds $w_i$ and $z_i$ are

(4.5) $$ \begin{align} \frac{\partial \mathcal{L}}{\partial w_i} &= h_i (\lambda \varphi^{\prime}_i(w_i) - \delta_i) , \end{align} $$

(4.6) $$ \begin{align} \ \kern1.5pt\frac{\partial \mathcal{L}}{\partial z_i} &= k_i (\varphi^{\prime}_i(z_i) - \delta_i).\quad \end{align} $$

Equation (4.6) can equal zero when either $k_i=0$ or when $\delta _i=\varphi ^{\prime }_i(z_i)$ . If a train journey starts and finishes within the peak demand period, then $k_i=0$ , and then (4.2) gives $w_i = v_i$ . In other words, journeys entirely within the peak demand interval remain unchanged. Otherwise, if journey i starts before or finishes after the peak demand period where $k_i> 0$ , then setting the partial derivatives from (4.5) and (4.6) to zero and rearranging gives

(4.7) $$ \begin{align} \delta_i = \lambda \varphi^{\prime}_i(w_i) = \varphi^{\prime}_i(z_i). \end{align} $$

Substituting $z_i$ from (4.3) gives

(4.8) $$ \begin{align} \lambda \varphi^{\prime}_i(w_i) = \varphi^{\prime}_i\bigg( \frac{X_i - h_i w_i}{k_i} \bigg). \end{align} $$

Equation (4.8) defines the relationship between the Lagrange multiplier $\lambda $ and the speed $w_i$ for each train i during the peak demand interval. Notice that the optimal speed $w_i$ for each train depends only on $\lambda $ and not on the speeds of the other trains. Furthermore, we have the following.

• • If $\lambda = 1$ , then from (4.7), we have $\varphi ^{\prime }_i(w_i) = \varphi ^{\prime }_i(z_i) \implies w_i = z_i$ , which implies that the speed during the peak demand interval $w_i$ is the same as the speed before and after the peak demand interval $z_i$ for train i. That is, if $\lambda = 1$ , then the optimal journey for each train is unchanged.

• • Now, suppose $\lambda> 1$ . For trains that have some part of their journey before or after the peak demand period (that is, $k_i> 0$ ), (4.7) implies $w_i < z_i$ , since $\varphi ^{\prime }_i$ is an increasing function (Figure 2). That is, each train that has some part of its journey before or after the peak demand period speeds up before the peak demand period, slows down during the peak demand period and speeds up after the peak demand period. Increasing $\lambda $ will decrease the speed of each train during the peak demand period and, hence, decrease the energy use during the peak demand period.

  

  Figure 2 Relationship between $w_i$ and $z_i$ for a given $\lambda> 1$ .

• • If $\lambda < 1$ , then trains will speed up during the peak demand period.

If we pick a value for $\lambda $ , then we can calculate $w_i$ by numerically solving (4.8) for each train $i \in \{1,\ldots ,n\}$ . If (4.1) is not satisfied, then we simply adjust $\lambda $ and try again; if the energy use during the peak demand interval is too high, then $\lambda $ must increase, otherwise $\lambda $ must decrease.

4.2 Single peak demand interval example

Consider four trains, with:

• • journey distances $\mathbf {X} = [600\,000, 382\,500, 114\,000, 200\,000];$

• • start times $\mathbf {s} = [0, 500, 2500, 3000]$ ;

• • finish times $\mathbf {f} =[8000, 5000, 4400, 7000]$ .

The peak demand period starts at time $p = 1800$ and finishes at time $q = 5400$ . All units are SI. The initial train speeds are $\mathbf {v} = [75, 85, 60, 50]$ . If $\varphi _i(v) = v^3$ , then the energy use during the peak demand interval is $E_0 \approx 4.19 \times 10^9$ . If we want to reduce this by $10\%$ , then $\alpha = 0.1$ and the target energy use during the peak demand interval is $E_1 \approx 3.77 \times 10^9$ . Numerical solution of (4.8) gives the Lagrange multipliers $\lambda \approx 1.21$ and ${\delta } \approx [0, 0, 13\,070, 0]$ . We note that $\delta _i> 0$ only when a train journey starts and finishes within the peak demand period ( $k_i = 0$ ). The new speeds within the peak demand period are $\mathbf {w} \approx [71.09, 82.61, 60.00, 48.08]$ , and the new speeds before and after the peak demand period are $\mathbf {z} \approx [78.20, 90.87, 0, 52.88]$ . This solution is shown in Figure 3 and was calculated in approximately $0.005$ seconds. Total energy increases from $7.049 \times 10^9$ to $7.091 \times 10^9$ . This is a $0.60\%$ increase.

Figure 3 Initial (pale, dashed) and adjusted (bold) speed profiles for four trains. Shaded region is the peak demand period.

5.1 Minimizing total energy

For the problem with multiple peak demand intervals, the optimization problem is to find constant speeds $w_{ij}$ and $z_i$ for each peak demand interval $j \in \{1,\ldots ,q\}$ and for each train $i \in \{1, \ldots , n\}$ to minimize total energy use E defined by (5.5) such that the energy use $E_{1j}$ during peak demand interval j, defined by the summation in (5.1), is constrained to the pre-defined level $(1-\alpha _j)E_{0j}$ for each peak demand interval $j \in \{1,\ldots ,q\}$ . Speed $w_{ij}$ is the optimal speed of train i during peak demand interval j, and speed $z_i$ is the optimal speed of train i before and after the peak demand period. Equation (5.2) must be satisfied for each train i to ensure that the journey covers the correct distance $X_i$ . To solve this problem, we once again form a Lagrangian function $\mathcal {L} = \mathcal {L}(z_i,w_{ij})$ , this time given by

$$ \begin{align*} \mathcal{L} =&\sum_{i=1}^n k_i \varphi_i(z_i) + \sum_{j=1}^q E_{1j}\\ &+ \sum_{j=1}^q \lambda_j \bigg(\sum_{i=1}^n h_{ij} \varphi_i(w_{ij}) - E_{1j} \bigg) \\ &+ \sum_{i=1}^n \delta_i \bigg( X_i - k_i z_i - \sum_{j=1}^q h_{ij} w_{ij} \bigg), \end{align*} $$

where $\lambda _j$ and $\delta _i$ are Lagrange multipliers. The optimal solution must satisfy the Karush–Kuhn–Tucker conditions

$$ \begin{align*} \frac{\partial \mathcal{L}}{\partial w_{ij}} = \frac{\partial \mathcal{L}}{\partial z_i} = 0 \quad \text{for all } i \in \{1,\ldots,n\}, j \in \{1,\ldots,q\}, \end{align*} $$

and the complementary slackness conditions

$$ \begin{align*} \lambda_j \bigg(\sum_{i=1}^n h_{ij} \varphi_i(w_{ij}) - E_{1j} \bigg) = 0 \quad \text{for all } j \in \{1,\ldots,q\} \end{align*} $$

and

$$ \begin{align*} \delta_i \bigg( X_i - k_i z_i - \sum_{j=1}^q h_{ij} w_{ij} \bigg) = 0 \quad \text{for all } i \in \{1,\ldots,n\}. \end{align*} $$

The partial derivatives with respect to the new speeds $w_{ij}$ and $z_{i}$ are

$$ \begin{align*} \frac{\partial \mathcal{L}}{\partial w_{ij}} &= h_{ij} (\lambda_j \varphi_i'(w_{ij}) - \delta_i) \quad \text{for all } i \in \{1,\ldots,n\}, \text{for all } j \in \{1,\ldots,q\},\\ \frac{\partial \mathcal{L}}{\partial z_i} &= k_i (\varphi_i'(z_i) - \delta_i) \quad \text{for all } i \in \{1,\ldots,n\}. \end{align*} $$

For the problem with multiple peak demand intervals, we now have a Lagrange multiplier $\lambda _j$ for each peak demand interval j, and the speed for each train in peak demand interval j is determined by the choice of $\lambda _j$ and not by the speeds of the other trains. The optimization problem can be solved as follows.

1. (1) Pick an initial value for each $\lambda _j$ .

2. (2) Find values $w_{ij}$ for each train i as follows:

   1. (a) if $k_i> 0$ , then calculate $w_{ij}$ numerically by solving

      (5.6) $$ \begin{align} \lambda_j \varphi_i'(w_{ij}) = \varphi_i'(z_i), \end{align} $$
      where $z_i$ is given by (5.3);

   2. (b) otherwise, if $k_i = 0$ , then pick a value for $\delta _i$ and calculate $w_{ij}$ numerically by solving

      (5.7) $$ \begin{align} \lambda_j \varphi_i'(w_{ij}) = \delta_i. \end{align} $$
      If (5.4) is not satisfied, then adjust $\delta _i$ and try again.

3. (3) If (5.1) is not satisfied, then adjust the $\lambda _j$ values and try again from Step (2).

The outer loop, which finds values for $\lambda _j$ , can be solved using a multi-dimensional root-finding method. We used the SciPy fsolve function [13].

5.2 Multiple peak demand interval example

Consider the same four trains from our single peak demand interval example (Section 4.2). As before, the peak demand period starts at time $p_0 = 1800$ and finishes at time $p_q = 5400$ , and the initial train speeds are $\mathbf {v} = [75, 85, 60, 50]$ . The peak demand period is divided into 12 equal intervals of 5-minutes, so we have 12 initial energy values $E_{0j}$ and 12 target energy values $E_{1j} = 0.9\,E_{0j}$ . Numerical solution of (5.6) and (5.7) give 12 Lagrange multipliers $\lambda _j$ each in the range $[1.173, 1.221]$ and $\delta \approx [0, 0, 13\,160, 0]$ . Similarly to the case with a single peak demand interval, $\delta _i> 0$ only when a train journey starts and finishes within the peak demand period ( $k_i = 0$ ). We calculate new speeds $w_{ij}$ within each peak demand interval j. The average speeds across the entire peak demand period are $\overline {\mathbf {w}} \approx [71.11, 82.60, 60.01, 48.10]$ , and the new speeds before and after the peak demand period are $\mathbf {z} = [78.18, 90.94, 0, 52.86]$ . This solution is shown in Figure 4 and was calculated in approximately $0.08$ seconds. Total energy increases from $7.049 \times 10^9$ to $7.0913 \times 10^9$ . This is a $0.60\%$ increase.

Figure 4 Initial (pale, dashed) and adjusted (bold) speed profiles for four trains. The shaded region is the peak demand period divided into $5$ -minute peak demand intervals.

Notice that the average speeds $\overline {\mathbf {w}}$ across the entire peak demand period are different to the holding speeds $\mathbf {w}$ shown in the example with a single peak demand interval in Section 4.2. Within the peak demand period, whenever a journey starts or finishes, the speeds $w_{ij}$ of the other trains change, including journeys completely within the peak demand period. However, this change is usually small.

5.3 One hundred trains

Ultimately, we want to calculate optimal journey profiles for a fleet of 100 trains or more that reduce their energy use in each 10-minute interval for a 1-hour peak demand period. As a final example with simple trains, we generated 100 trains with random journey distances $X_i \in [39\,100, 736\,500]$ , start times $s_i \in [0,5000]$ , finish times $f_i \in [2300, 10800]$ , and initial train speeds $v_i \in [60,85]$ . The peak demand period was divided into six equal intervals of 10 minutes:

$$ \begin{align*} \mathbf{p} = [1800,2400,3000,3600,4200,4800,5400] \end{align*} $$

so we have six initial energy values $E_{0j}$ and six target energy values $E_{1j} = 0.9\,E_{0j}$ . Numerical solution of (5.6) and (5.7) give the Lagrange multipliers

$$ \begin{align*} {\lambda} \approx [1.235, 1.142, 1.135, 1.133, 1.133, 1.133] \end{align*} $$

and $\delta _i \in [0, 1\,596\,000]$ , from which we can calculate new speeds within each peak demand interval $w_{ij} \in [0, 83.99]$ and new speeds outside the peak demand period ${z_i \in [0, 89.38]}$ . This solution is shown in Figure 5 and was calculated in approximately $0.97$ seconds. Total energy increases from $168.86 \times 10^9$ to $169.22 \times 10^9$ . This is a $0.21\%$ increase.

Figure 5 Adjusted (bold) speed profiles for 100 trains. The shaded region is the peak demand period containing six consecutive 10-minute peak demand intervals.