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

2. 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

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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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]

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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)

18. 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

19. 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

20. 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

21. 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

22. Thank You