Patrolling a Pipeline

Abstract

A pipeline network can potentially be attacked at any point and at any time, but such an attack takes a known length of time. To counter this, a Patroller moves around the network at unit speed, hoping to intercept the attack while it is being carried out. This is a zero-sum game between the mobile Patroller and the Attacker, which we analyze and solve in certain cases.

Appendix: Omitted Proofs

Proof of Lemma 1 . The attack must be taking place during the time interval [0, r]. Let A be the set of points that the Patroller intercepts in this time interval. Then clearly A must have length no greater than r and so the probability the attack takes place at a point in A is $r/\mu$. It follows that $P \le r/\mu$.    $\square$

Proof of Lemma 2 . Suppose the attack starts at time $t_0$ at some point $x \in Q$. Then the attack is certainly intercepted if $\bar{w}$ is at x at time $t_0$. Let $t_x \in [0,\bar{\mu }]$ be such that $w(t_0+t_x) = x$, so that the attack is intercepted by $\bar{w}$ if $T=t_x$. Let A be the set of times $t \in [0,\bar{\mu }]$ such that $t_x-r \le t \le t_x$ or $t \ge t_x+\bar{\mu }-r$, so if $T \in A$, then the attack is intercepted by $\bar{w}$. But the measure of A is r, so the probability that T is in A is $r/\bar{\mu }$ and hence $P \ge r/\bar{\mu }$.    $\square$

Proof of Lemma 3 . Take a Patroller path w. We can assume that w starts at an endpoint, otherwise it is weakly dominated by a strategy that does. To see this, suppose the Patroller starts at an interior point before traveling directly to an endpoint, arriving there at time $t<1$. Now consider the Patroller strategy that is the same but in the time interval [0, t] the Patroller remains at the endpoint. Then clearly the second strategy intercepts the same set of attacks as the first one. Without loss of generalization we assume w starts at $x=0$.

We only need to consider the path in the time interval $[0,1+r]$, after which time the attack has been completed with probability 1. Since $r<2$ the walk cannot go between the two ends more than twice, so there are three possibilities.

The first is that w stays at $x=0$ for the whole time, in which case the probability the attack is intercepted is $P=1/2 \le r/2$.

The second possibility is that w stays at $x=0$ for time $t_1$, then goes to $x=1$ and stays there for time $t_2$. We can assume it takes the Patroller time 1 to go between the endpoints since any path taking longer than that would be dominated, so $t_1+t_2=r$. The attack is intercepted at $x=0$ if it starts sometime during $[0,t_1]$, which has probability $(1/2)t_1$. It is intercepted at $x=1$ if it ends sometimes during $[1+r-t_2,1-r]$, which has probability $(1/2)t_2$. Hence $P=(1/2)(t_1 + t_2)=r/2$.

The final possibility is that w stays at $x=0$ for time $t_1$, then goes directly to $x=1$ for time $t_2$, then goes directly back to $x=0$ for time $t_3$, in which case we must have $t_1+t_2+t_3=r-1$. This time the attack is intercepted at $x=0$ in the case of either of the two mutually exclusive events that it starts in $[0,t_1]$ or ends in $[1+r-t_3,1-r]$, which have total probability $(1/2)(t_1+t_3)$. If the attack takes place at $x=1$, it must be taking place during the whole of the time interval [1, r]. But w must reach $x=1$ sometime during this time interval, since it must have time to travel from $x=0$ to $x=1$ and back again, and hence intercepts the attack with probability 1. So the overall probability the attack is intercepted is $(1/2)(t_1+t_3) + 1/2 \le (1/2)(t_1+t_2+t_3)+1/2=r/2$.    $\square$

Proof of Theorem 3 . We already showed that $r/(1+r)$ is a lower bound for the value and now we show that it is also an upper bound. Now, suppose that the Attacker plays the r-attack strategy. The Patroller could:

• 1. Stay at any attack point but will not win with probability greater than $\frac{r}{1+r}$.

• 2. Travel between consecutive external attacks and if possible try to reach the internal attack: Suppose the Patroller is at point 0 up to time t: If $t \in [0,r]$ and then leaves for point r, she will reach point r at times in the range [r, 2r]. This gives total interception probability $f(t) + f(t+r) = \frac{t}{1+r} + \frac{2r-(t+r)}{1+r} = \frac{r}{1+r}$. Note that if the Patroller continues to the next attack along the unit interval, if it is the internal attack she will reach it at times greater than $r+\frac{r+\rho }{2} = 2r - \frac{r-\rho }{2}$, when the internal attack has been completed, and if it is an external attack she will reach it at time greater than 2r, where all external attacks have been completed. If $t \in [r,2r]$ then all attacks at point 0 have been intercepted but the Patroller arrives at point r after all attacks have been completed, which gives interception probability of $\frac{r}{1+r}$.

• 3. Travel between last two external attacks, crossing internal attack in the middle (this is the same as doing a roundtrip from one of the last external attacks to the internal attack and back): Suppose the Patroller leaves point $(q-1)r$ at time t, toward the internal attack point and the last external attack point 1: If $t \in [0,r - \rho ]$, she will reach the internal attack point at times $\left[ \frac{r+\rho }{2},r - \rho + \frac{r+\rho }{2}\right] = \left[ \frac{r+\rho }{2}, 2r - \frac{r+\rho }{s}\right]$, and she will reach the external attack at point 1 at times $[r+\rho , 2r]$. This sums to a probability of $f(t)\,+\,g\left( t\,+\,\frac{r\,+\,\rho }{2}\right) \,+\,f(t\,+\,r\,+\,\rho )\,=\,\frac{t}{1\,+\,r}\,+\,\frac{\rho }{1+r}+\frac{2r-(t+r+\rho )}{1+r} = \frac{r}{1+r}.$ If $t \in [r-\rho ,r]$, she will reach the internal attack point at times $\left[ 2r - \frac{r+\rho }{2}, 2r - \frac{r-\rho }{2}\right]$, and the external attack point 1 at times greater than 2r. This sums to a probability of $f(t) + g\left( t+\frac{r+\rho }{2} \right) = \frac{t}{1+r} + \frac{2r - \frac{r-\rho }{2} - \left( t+\frac{r+\rho }{2}\right) }{1+r} = \frac{r}{1+r}$. Finally, if $t \in [r,2r]$, the Patroller will intercept all attacks at point $(q-1)r$ and will not make it in time for the internal attack nor the attack at point 1, this gives the desired probability.    $\square$