2.1 Problem statement
The key to solving the problem of optimal layout of feeder automation equipment is to describe the relationship between the power outage time caused by a specific fault with specific equipment layout through reasonable mathematical language. Figure 1 illustrates the cooperation of fault indicator, Itype switch and Vtype switch in the FLSR process.
The feeder in Fig. 1 has poles No. 1 to No. 4 which connect to load area \(j_{1}\) to \(j_{4}\), respectively. There is a normally closed breaker at the outlet of the line, and a normally open tie switch for load transfer at the end of line. For feeder automation equipment, its configuration position is generally on both sides of a pole. Under normal conditions, the entire line load is powered by the substation breaker.
Assuming that a permanent fault occurs in the Tconnected line where load area \(j_{3}\) is located, in order to reduce the impact of transient faults, the FLSR process is mainly divided into four stages as follows.
Stage 1: Overcurrent protection of the nearest Itype switch upstream of the fault will trip, and the downstream load area will lose power at the same time. All the corresponding Vtype switches will be opened because of voltage disappearance, while the tie switch will detect voltage disappearance on one side and start timing for closing. Fault current flows through the Vtype switch on the No. 2 pole, but does not flow through the fault indicator. Therefore, the dispatch center judges that the fault occurs in the line between No. 2 pole and No. 3 pole, and the Tconnected line in No. 3 pole according to the signal.
Stage 2: The tripped Itype switch is reclosed for the first time and the downstream Vtype switch is reclosed after a delay, in an attempt to deliver power segment by segment, so as to eliminate power outage caused by the transient fault. At the same time, the two nearest Vtype switches upstream and downstream of the fault both detect voltage.
Stage 3: Since the fault still exists, the Itype switch is tripped for the second time because of overcurrent protection action, and the downstream Vtype switch opens again. The two nearest Vtype switches upstream and downstream of the fault have openlockout to isolate the fault because the time for detecting voltage has not reached the fixed value.
Stage 4: The tripped Itype switch is reclosed for the second time to restore power supply to the nonfault area. The tie switch automatically closes after the oneside voltage loss reaches a fixed value and the downstream nonfault area restores power supply segment by segment. For the load in the fault area, the power supply can only be restored after the line patrol has found the fault and carried out the repair.
In summary, the three types of automation equipment jointly determine power outage time \(T_{i,j}^{{{\text{outage}}}}\) of the load point \(j\) caused by fault \(i\), and can be divided into the following three categories:
$$T_{i,j}^{{{\text{outage}}}} = \left\{ {\begin{array}{*{20}l} 0 \hfill & {j_{1} } \hfill \\ {t^{{{\text{iso}}}} } \hfill & {j_{2} ,j_{4} } \hfill \\ {t^{{{\text{loc}}}} + T_{i}^{{{\text{patrol}}}} + t^{{{\text{rep}}}} } \hfill & {j_{3} } \hfill \\ \end{array} } \right.$$
(1)
where \(t^{loc}\) is the judgment time for fault location, \(t^{{{\text{iso}}}}\) is the operating time for the Vtype switch, \(t^{{{\text{rep}}}}\) is the expected repair time for the fault, and \(T_{i}^{{{\text{patrol}}}}\) is the patrol time for fault \(i\) where this time is a variable related to the layout of the fault indicator and Vtype switch.
This paper will build an explicit expression for the relationship between the layout of the feeder automation equipment and the power outage time around Eq. (1), and linearize the nonlinear terms involved, to obtain the optimal layout model of that can obtain a global optimal solution.
2.2 Patrol time
It can be seen from Fig. 1 that for any segment \(s\) of the feeder, as long as there is at least one fault indicator or Vtype switch between itself and fault \(i\), it is not necessary to patrol. From [25], we define the Boolean variable \(b_{i,s}\). When its value is 0 this means the segment needs to be patrolled, and 1 means the patrol not needed. The formulae are:
$$X_{i,s} = \sum\limits_{{n \in \Omega_{i,s} }} {x_{n} }$$
(2)
$$Z_{i,s} = \sum\limits_{{n \in \Omega_{i,s} }} {z_{n} }$$
(3)
$$\frac{{X_{i,s} + Z_{i,s} }}{M} \le b_{i,s} \le X_{i,s} + Z_{i,s}$$
(4)
where \(x_{n}\) and \(z_{n}\) represent the Boolean configuration variable of the fault indicator and Vtype switch corresponding to the nth configuration position on the feeder. When its value is 1 this indicates that the corresponding equipment is configured at the corresponding position, and when 0 it indicates that it has not been configured. \(\Omega_{i,s}\) is the set of configuration positions between fault \(i\) and segment \(s\). \(X_{i,s}\) and \(Z_{i,s}\) are the quantity of fault indicator and Vtype switch corresponding to \(\Omega_{i,s}\). M is a maximum value, which is taken as 1000 in this paper. From Eq. (4), it can be seen that as long as there is at least one fault indicator or Vtype switch in \(\Omega_{i,s}\), \(b_{i,s}\) will equal 1, which means that the line segment \(s\) does not need to be patrolled. Because the fault indicator and Vtype switch generally use GRPS to communicate, there is a functional failure probability of communication. Thus, the patrol time for fault \(i\) is further established as:
$$t_{i,s}^{{{\text{search}}}} = \frac{{l_{s} }}{v} \times \left( {1  \left( {1  \rho_{i,s} } \right) \times b_{i,s} } \right)$$
(5)
$$T_{i}^{{{\text{patrol}}}} = \sum\limits_{{s \in \Omega_{s} }} {t_{i,s}^{{{\text{patrol}}}} }$$
(6)
where \(t_{i,s}^{{{\text{patrol}}}}\) is the expected patrol time of segment \(s\) for fault \(i\). \(l_{s}\) is the length of segment \(s\), \(v\) is the patrol speed, and \(\rho_{i,s}\) is the probability that all communication functions of the fault indicators and Vswitches between fault \(i\) and line \(s\) fail at the same time. \(\Omega_{s}\) is the segment set of the feeder.
\(\rho_{i,s}\) should be an exponential function with the functional failure probability of single equipment as the base, and the quantity of equipment between fault \(i\) and line \(s\) as the index, representing the probability of simultaneous functional failure of all equipment between fault \(i\) and line \(s\). Although it can be linearized by Taylor expansion, the error of loworder expansion is significant, whereas highorder expansion will bring too many additional terms. Therefore, the \(3\delta\) principle in statistics is introduced. The specific method is to consider that the functional failure probability decreases exponentially with the increase of quantity of equipment, but after it becomes less than 0.0027, it no longer changes. Thus, it transforms the exponential function into a piecewise function, as shown in Fig. 2, in which the functional failure probability of a single piece of equipment is 0.2.
For the convenience of expression, the piecewise function is defined as:
$$\rho _{{i,s}} = f(\rho ,X_{{i,s}} + Z_{{i,s}} ) = \left\{ {\begin{array}{*{20}l} {\rho ^{{X_{{i,s}} + Z_{{i,s}} }} } \hfill & {X_{{i,s}} + Z_{{i,s}} < G} \hfill \\ {\rho ^{G} } \hfill & {X_{{i,s}} + Z_{{i,s}} \ge G} \hfill \\ \end{array} } \right.$$
(7)
where \(\rho\) is the functional failure probability of single equipment, \(G\) is the number of segments, and has the following value constraints:
$$\left\{ \begin{gathered} \rho^{G  1} \ge 0.0027 \hfill \\ \rho^{G} < 0.0027 \hfill \\ \end{gathered} \right.$$
(8)
The piecewise function is implemented using a SpecialOrdered Set of type 2, namely SOS2 constraints [14], as:
$$f\left( {\rho ,X_{i,s} + Z_{i,s} } \right) = \sum\limits_{g = 1}^{{G{ + }1}} {\gamma \left( g \right) \times \rho \left( g \right)}$$
(9)
$$\rho (g) = \left\{ {\begin{array}{*{20}l} {p^{g} } \hfill & {g \le G} \hfill \\ {\rho ^{G} } \hfill & {g = G + 1} \hfill \\ \end{array} } \right.$$
(10)
$$\sum\limits_{g = 1}^{G + 1} {\gamma \left( g \right) \times \tau \left( g \right) = X_{i,s} + Z_{i,s} }$$
(11)
$$\tau (g) = \left\{ {\begin{array}{*{20}l} g \hfill & {g \le G} \hfill \\ {sum\left( {\Omega _{{i,s}} } \right)} \hfill & {g = G + 1} \hfill \\ \end{array} } \right.$$
(12)
$$0 \le \gamma (g) \le 1\quad 1 \le \forall g \le G + 1$$
(13)
$$\sum\limits_{g = 1}^{G + 1} {\gamma \left( g \right) = 1}$$
(14)
$$\gamma \left( g \right) \le b\left( g \right)\quad 1 \le \forall g \le G + 1$$
(15)
$$\sum\limits_{g = 1}^{G + 1} {b\left( g \right) \le 2}$$
(16)
$$\left\{ {\begin{array}{*{20}l} {b(g) + b(g + \delta ) \le 1} \hfill \\ {1 \le \forall g \le G + 1} \hfill \\ {1 < \forall \delta \le G + 1  g} \hfill \\ \end{array} } \right.$$
(17)
where \(\gamma \left( g \right)\) is the SOS2 coefficient corresponding to segment \(g\), \(\rho \left( g \right)\) is the failure probability corresponding to segment \(g\), and \(\tau \left( g \right)\) is the quantity of equipment corresponding to segment \(g\). \(b\left( g \right)\) is the auxiliary Boolean variable corresponding to segment \(g\), \(sum\left( {\Omega_{i,s} } \right)\) is the quantity of elements in \(\Omega_{i,s}\), and \(\delta\) is the auxiliary integer variable.
Equations (15)–(17) show that in SOS2 constraint, \(\gamma \left( g \right)\) can only take nonzero values at 1 or 2 segments, and the two segments must be continuous. The following is further illustrated by Fig. 3.
In Fig. 3, the functional failure probability of a single piece of equipment is 0.2, and the maximum number of pieces of equipment is 50. From the above assumption, when the number of pieces is fewer than or equal to 4, the probability decreases exponentially. Thus, G equals 4, and after there are more than four pieces the probability does not change. The continuous curve is divided into 5 segments, and the values of \(\rho \left( g \right)\) and \(\tau \left( g \right)\) are listed in Table 2.
According to the amount of equipment, it is divided into the following two cases:

(1)
If the number of pieces is 5, between segment G and G + 1, in order to satisfy Eq. (11), \(b\left( 4 \right)\) and \(b\left( 5 \right)\) are both equal to 1, \(\tau \left( 4 \right)\) and \(\tau \left( 5 \right)\) are 45/46 and 1/46, respectively. Substituting them into Eq. (9) obtains \(f\left( {0.2,5} \right)\) as 0.0016

(2)
If the number of pieces is 3 and fewer than G, Eq. (11) is satisfied only if \(b\left( 3 \right)\) is 1. As \(\tau \left( 3 \right)\) is 1, \(f\left( {0.2,3} \right)\) is 0.008.
The piecewise functions involved in the following are all processed in the above manner.
In Eq. (5), there is a nonlinear term multiplied by a Boolean variable and a continuous variable. These can be linearized by the bigM method [18], as:
$$\frac{v}{{l_{s} }} \times t_{i,s}^{{{\text{patrol}}}} + \left( {M  1} \right) \times \left( {b_{i,s}  1} \right) \le \rho_{i,s}$$
(18)
$$\frac{v}{{l_{s} }} \times t_{i,s}^{{{\text{patrol}}}} + \left( {1  M} \right) \times \left( {b_{i,s}  1} \right) \ge \rho_{i,s}$$
(19)
$$\frac{v}{{l_{s} }} \times t_{i,s}^{{{\text{patrol}}}} + \left( {1  M} \right) \times b_{i,s} \le 1$$
(20)
$$\frac{v}{{l_{s} }} \times t_{i,s}^{{{\text{patrol}}}} + \left( {1 + M} \right) \times b_{i,s} \ge 1$$
(21)
Similar nonlinear terms involved in the followings are also processed in the above manner.
2.3 Outage time
The FLSR process described in Fig. 2 is based on a linear distribution network, while load and fault can only be related to each other upstream and downstream. However, the actual distribution network structure is complex, as load and fault may be located in different branches under a public node. For the public node, if it cannot be isolated from the fault, it will be involved in a power outage, and this will indirectly lead to load power outage. In contrast, if there is at least one Itype switch on the path from the public node to the fault, the fault will be isolated directly from the path, without causing load power outage. Therefore, in order to establish the explicit expression of power outage time of automation equipment, \(\Omega_{i,j}^{i}\) is defined as the set of configuration positions between the last public node of fault \(i\) and load \(j\) to fault \(i\), whose schematic diagram is shown in Fig. 4.
There are three cases according to the location relationship between load \(j\) and fault \(i\), as follows:

Case 1: Load is downstream of the fault. For example, for load \(j_{4}\), the fault between poles No. 1 and 2 is upstream and the fault location itself is the last public node, so \(\Omega_{i,j}^{i}\) will be an empty set.

Case 2: Load is upstream of the fault. For example, for load \(j_{1}\), the fault between poles No. 2 and 4 is downstream. No. 1 pole is the public node, and the 2nd, 3rd and 6th configuration positions are in \(\Omega_{i,j}^{i}\).

Case 3: Load and fault are in different branches. For example, load \(j_{4}\) and the fault between poles No. 2 and 3 are in different branches. No. 2 pole is the public node, and the 4th configuration position is in \(\Omega_{i,j}^{i}\).
The Itype switch may not operate correctly because of improper parameter setting. However, as long as there is at least one Vtype switch between the load and the fault, the outage time will not be longer than the fault isolation time. According to [25, 27], the power outage time \(T_{i,j}^{{{\text{outage}}}}\) can be established as:
$$Y_{i,j}^{i} = \sum\limits_{{n \in \Omega_{i,j}^{i} }} {y_{n} }$$
(22)
$$Z_{i,j} = \sum\limits_{{n \in \Omega_{i,j} }} {z_{n} }$$
(23)
$$T_{i,j}^{{{\text{outage}}}} \ge f\left( {\rho ,Y_{i,j}^{i} } \right) \times t^{{{\text{iso}}}}$$
(24)
$$T_{i,j}^{{{\text{outage}}}} \ge \left( {1  Y_{i,j}^{i} } \right) \times t^{iso}$$
(25)
$$T_{i,j}^{{{\text{outage}}}} \ge \left( {1  Y_{i,j}^{i}  Z_{i,j} } \right) \times \left( {t^{{{\text{loc}}}} { + }T_{i}^{{{\text{search}}}} + t^{{{\text{repair}}}} } \right)$$
(26)
where \(y_{n}\) represents the Boolean configuration variable of the Itype switch corresponding to the nth configuration position on the feeder. \(Y_{i,j}^{i}\) is the number of Itype switches corresponding to \(\Omega_{i,j}^{i}\), and \(\Omega_{i,j}\) is the set of configuration positions between fault \(i\) and load \(j\). \(Z_{i,j}\) is the number of Vtype switches corresponding to \(\Omega_{i,j}\). \(f\left( {\rho ,Y_{i,j}^{i} } \right)\) is the functional failure probability of Itype switches corresponding to \(\Omega_{i,j}^{i}\), and \(f\left( {\rho ,Y_{i,j}^{i} } \right)\) times \(t^{{{\text{iso}}}}\) represents the expected isolation time when all Itype switches in \(\Omega_{i,j}^{i}\) fail.
The relationship between the power outage time and the automation equipment is described by the three inequalities from Eqs. (24)–(26). These are now explained in detail.

(1)
If there is no Itype switch in \(\Omega_{i,j}^{i}\) and no Vtype switch in \(\Omega_{i,j}\), in the mathematical sense, both \(Y_{i,j}^{i}\) and \(Z_{i,j}\) are equal to 0. Among the right terms of the three equations, the right term of Eq. (26) is the largest, which is the minimum value of \(T_{i,j}^{{{\text{outage}}}}\). In a physical sense, it means that if the load cannot be isolated from the fault by the Itype and Vtype switches, at least it needs to wait for the mainstation to roughly locate the fault, and then restore the power supply after searching for the actual fault location and carrying out repair.

(2)
If there is no Itype switch in \(\Omega_{i,j}^{i}\), but at least one Vtype switch in \(\Omega_{i,j}\), then similar to above, the minimum value of \(T_{i,j}^{{{\text{outage}}}}\) is the right term of Eq. (25). It means that although the load cannot be isolated from the fault by the Itype switch without causing a power outage, it can still be isolated by the Vtype switch and power supply can be quickly restored.

(3)
As long as there is at least one Itype switch in \(\Omega_{i,j}\), the minimum value of \(T_{i,j}^{{{\text{outage}}}}\) is \(f\left( {\rho ,Y_{i,j}^{i} } \right)\) times \(t^{{{\text{iso}}}}\).
In particular, if the optimization goal involves minimizing the derivative of the power outage time, such as a goal of minimizing the customer interrupt cost, it is desirable that the power outage time of any load under any fault should be as small as possible, around the minimum value that can be obtained. The power outage time in such a condition can be precisely expressed by the following explicit expression:
$$T_{{i,j}}^{{{\text{outage}}}} = \left\{ {\begin{array}{*{20}l} {f\left( {\rho ,Y_{{i,j}}^{i} } \right) \times t^{{{\text{iso}}}} } \hfill & {Y_{{i,j}}^{i} \ge 1} \hfill \\ {t^{{{\text{iso}}}} } \hfill & {Y_{{i,j}}^{i} = 0,Z_{{i,j}} \ge 1} \hfill \\ {t^{{{\text{loc}}}} {\text{ + }}T_{i}^{{{\text{patrol}}}} + t^{{{\text{repair}}}} } \hfill & {Y_{{i,j}}^{i} + Z_{{i,j}} = 0} \hfill \\ \end{array} } \right.$$
(27)
2.4 Reliability index
To effectively help power companies to carry out distribution automation transformation, we improve the two main assessment indicators of the State Grid for a distribution network, namely the normal power supply time and the number of trips of a substation breaker.
Based on the above requirements, the average service availability index (ASAI) is used as one of the reliability indicators. This reflects the annual proportion of normal power supply through a percentage value, as:
$${\text{ASAI = }}\left( {1{  }\frac{{\sum\limits_{{i \in \Omega_{i} }} {\sum\limits_{{j \in \Omega_{j} }} {\lambda_{i} N_{j} T_{i,j}^{{{\text{outage}}}} } } }}{{8760 \times \sum\limits_{{j \in \Omega_{j} }} {N_{j} } }}} \right) \times 100\%$$
(28)
where \(\Omega_{i}\) is the set of faults, \(\Omega_{j}\) is the set of loads. \(\lambda_{i}\) is the probability of fault \(i\), and \(N_{j}\) is the quantity of users at load \(j\). The number of hours in one year is 8760.
The system average interruption frequency index (SAIFI) for a breaker is used as another reliability indicator. This counts the annual expected number of trips of a breaker in a substation, as:
$$Y_{i} = \sum\limits_{{n \in \Omega_{i,0} }} {y_{n} }$$
(29)
$$\frac{{Y_{i} }}{M} \le b_{i} \le Y_{i}$$
(30)
$${\text{SAIFI}} = \sum\limits_{{i \in \Omega_{i} }} {\lambda_{i} \times \left( {1  b_{i} \times \left( {1  f\left( {\rho ,Y_{i} } \right)} \right)} \right)}$$
(31)
where \(\Omega_{i,0}\) is the set of configuration positions between fault \(i\) and the substation breaker, and \(Y_{i}\) is the number of Itype switches corresponding to \(\Omega_{i,0}\). \(b_{i}\) is the auxiliary Boolean variable used to judge whether there is an Itype switch in \(\Omega_{i,0}\), and \(f\left( {\rho ,Y_{i} } \right)\) is the piecewise function of failure probability with respect to \(Y_{i}\).
2.5 Economic indicators
In this paper, the equipment life cycle cost (ELCC) is used as one of the economic indicators. This includes the equipment installation and maintenance costs during the entire equipment life cycle, as:
$$ELCC = \left( {\sum\limits_{{n \in \Omega_{n} }} {\left( {x_{n} inv^{{\text{F}}} + y_{n} inv^{{\text{I}}} + z_{n} inv^{{\text{V}}} } \right)} } \right)\left( {1 + \sum\limits_{{t \in \Omega_{t} }} {\frac{1}{{\left( {1 + DR} \right)^{t} }}} } \right)$$
(32)
where \(\Omega_{n}\) is the set of all configuration positions on feeder. \(inv^{{\text{F}}}\), \(inv^{{\text{I}}}\) and \(inv^{{\text{V}}}\) are the purchase and configuration costs of fault indicator, Itype switch and Vtype switch, respectively. \(\Omega_{t}\) is the service life of distribution automation equipment, and \(DR\) is the maintenance rate.
The customer interrupt cost (CIC) is used as the second economic index. This counts the electricity bills that cannot be received because of power outages during the entire equipment life cycle, and is given as:
$$CIC{ = }\sum\limits_{{t \in \Omega_{t} }} {\sum\limits_{{i \in \Omega_{i} }} {\sum\limits_{{j \in \Omega_{j} }} {\sum\limits_{{k \in \Omega_{j,k} }} {\lambda_{i} T_{i,j}^{{{\text{outage}}}} P_{j,k} R_{k} \left( {1 + \mu } \right)^{t} } } } }$$
(33)
where \(\Omega_{j,k}\) is the set of all load types of load \(j\), \(P_{j,k}\) is the load power of the kth type load at load point \(j\), \(R_{k}\) is electricity price of the kth type load, and \(\mu\) is the average load growth rate.
2.6 Optimization layout model
In order to obtain the most reasonable equipment layout to meet the requirements of power supply reliability with the minimal economic investment, in this paper, two types of reliability indicators are used as constraints, and the sum of equipment life cycle cost and customer interrupt cost is minimized as the goal to build an optimal layout model of feeder automation equipment. Since both the fault indicator and Vtype switch have communication function, and the Itype switch and Vtype switch both have isolation function, repeated installation in the same position will cause a waste of money. Thus, the corresponding constraints, i.e., the same position cannot have two equipment with overlapping functions, are added. The model is given as:
$$\begin{gathered} minimize\;ELCC{\text{ + }}CIC \hfill \\ s.t.\left\{ {\begin{array}{*{20}l} {{\text{ASAI}} \ge {\text{ASAI}}^{{\lim }} } \hfill \\ {{\text{SAIFI}} \le {\text{SAIFI}}^{{\lim }} } \hfill \\ {x_{n} + z_{n} \le 1} \hfill \\ {y_{n} + z_{n} \le 1} \hfill \\ {\forall n \in \Omega _{n} } \hfill \\ \end{array} } \right. \hfill \\ \end{gathered}$$
(34)
where \({\text{ASAI}}^{\lim }\) is the lower limit of the reliability index ASAI. So the ASAI reliability of the model needs to be higher than this value. \({\text{SAIFI}}^{\lim }\) is the upper limit of the reliability index SAIFI, so the SAIFI reliability of the model needs to be lower than this value.