Minmax Regret 1-Facility Location on Uncertain Path Networks Haitao Wang Utah State University ISAAC 2013

Motivations A highway P connecting multiple cities Goal: choose a location on P to build an evacuation center such that when an emergency happens evacuate people in all cities to the center minimize the evacuation time a highway P a city an evacuation center

Motivations The highway P has a capacity, e.g., four driving lanes Cities have different populations 20k 20k 12k 31k

An uncertainty model The population wi of each city xi is uncertain days, nights, holidays, weekdays, weekends but it is known that wi is in an interval [li, ri] 15k ~ 23k 20k 20k 18k ~ 27k 12k 10k ~ 13k 31k 28k ~ 33k

A scenario At any moment, called a scenario, each city has a fixed value of population, i.e., wi is a fixed value 15k ~ 23k 20k 20k 18k ~ 27k 12k 10k ~ 13k 31k 28k ~ 33k

Defining the regret Suppose we build an evacuation center at location x For any scenario s T(x,s): the evacuation time xs : the optimal location for s The regret: R(x,s) = T(x,s) – T(xs,s) worst-case loss, opportunity loss x xs 12k 20k 20k 31k the highway an evacuation center

The problem objective Find a location x for the center, such that Rmax(x), the maximum R(x, s) over all scenarios s, is minimized Difficulty: There are infinitely many scenarios A crucial observation given by Cheng et al. 13’ only need to consider at most 2n scenarios, one of which is a worst-case scenario Previous work: O(nlog2n) time and O(nlog n) space, Cheng et al. 13’ Our result: O(nlogn) time and O(n) space

The definition of T(x, s) T(x, s) = max{TL(x, s), TR(x, s)} TL(x,s): the time for evacuating the cities on the left side TR(x,s): the time for the right side Suppose x is in (xk, xk+1] TL(x,s) = max1≤i≤k { (x-xi) * a + ∑it=1 wt } fiL(x,s) = (x-xi)*a + ∑it=1 wt a is a positive constant, the time for traversing a unit distance x1 x2 x3 xi xk x xk+1

T(x,s) : a unimodal function of x T(x,s) = max{TL(x,s), TR(x,s)} TL(x,s) = max1≤i≤k { (x-xi)*a + ∑it=1 wt } fiL(x,s) = (x-xi)*a + ∑it=1 wt T(x,s) TL(x,s) TR(x,s) f iL(x,s) x xs xi

Computing Rmax(x) Rmax(x) = maxall scenarios s{R(x,s) = T(x,s) – T(xs,s)} Difficulty: There are infinitely many scenarios A crucial observation by Cheng et al. 13’: Only need to consider a set E of 2n scenarios Rmax(x) = max s Є E R(x,s) Define E: E = EL ∪ ER EL = {sL(i) | 1 ≤ i ≤ n} sL(i) : r1 r2 … ri li+1 li+2 …. ln sL(i+1) : r1 r2 … ri ri+1 ri+2 …. ln a unimodal function each city population wi is in [li, ri]

The algorithm Goal: find a location x = x* that minimizes Rmax(x) Rmax(x) = max s Є E R(x,s) R(x,s) = T(x,s) – T(xs,s) The algorithm: Do binary search on x1 x2 … xn to determine the interval [xi, xi+1] that contains x* The main difficulty: For any x, compute R(x,s) in O(n) time for all 2n scenarios in E Rmax(x) x xi x* xi+1

Computing R(x,s) for all scenarios s in E in O(n) time R(x,s) = T(x,s) – T(xs,s) Two major procedures: As preprocessing, compute xs and T(xs,s) for all s of E O(nlog n) time in total Given any x, compute T(x,s) for all s of E O(n) time

Computing xs and T(xs,s) for all s of E Only discuss EL = {sL(i) | 1 ≤ i ≤ n} si = sL(i) : r1 r2 … ri li+1 li+2 …. ln TL(x,si) TR(x,si) x xs

The function TL(x, si+1) TL(x,s) = max1≤i≤k { (x-xi)*a + ∑it=1 wt } fiL(x,s) = (x-xi)*a + ∑it=1 wt si = sL(i) : r1 r2 … ri li+1 li+2 …. ln si+1 = sL(i+1) : r1 r2 … ri ri+1 ri+2 …. ln wi+1 increases from li+1 to ri+1 shift upwards by ri+1 – li+1 x xi xi+1

A monotonicity property of xs From si to si+1 xs moves to the left if xi+1 ≤ xs TR(x,si+1) TL(x,si+1) TL(x,si) TR(x,si) x xs xi xi+1

A data structure For any x and any scenario si in EL used in our algorithm compute TL(x, si) and TR(x, si) in O(log n) time Preprocessing: O(n) time and space

Computing T(x,s) for all s of E T(x,s) = max{ TL(x,s), TR(x,s)} Only discuss how to compute TL(x,s) for s in EL = {sL(i) | 1 ≤ i ≤ n} Compute TL(x, s) for the first scenario s = s1 = sL(1)

Observations Suppose x is in (xj-1, xj] and TL(x,s1) = fk(x,s1) Three cases for computing TL(x,si) for si : i is in [2, k] i is in [k+1, j-1] i is in [j, n] TL(x, s1) fk(x,s1) x xk xj-1 xj x

Case 1: i is in [2, k] s1: r1 l2 l3 …. ln s2: r1 r2 l3 …. ln TL(x, si) = TL(x,s1) + ∑it=2 (rt - lt) TL(x, s2) = TL(x,s1) + (r2 – l2) TL(x, s1) fk(x,s1) x xk xj-1 xj x

Case 3: i is in [j, n] TL(x, si) = TL(x, sj-1) TL(x, s1) fk(x,s1) x xk xj-1 xj x

Case 2: i is in [k+1, j-1] The difficult case After changing from sk from sk+1 The function ft shifts upwards for t > k, but fk does not TL(x, sk+1) = TL(x, sk+2) = max{fk(x,sk), fk+3(x,sk+2)} TL(x, sk+3) = max{fk(x,sk), fk+3(x,sk+3)} TL(x, sk+4) = ? max{fk(x,sk), fk+3(x,sk+1)} the second largest function at x x xk xk+1 xk+2 xk+3 xj-1 xj x

Thank You