Economic Epidemiology: An Operations Research Approach Fred Roberts, DIMACS

1. Economic Epidemiology: An Operations Research ApproachFred Roberts, DIMACS

2. Pearl Harbor, Hawaii, USADecember 1941

4. World War II Led to Creation of the Field of Operations Research Operations Research in WWII: logistics optimal scheduling inventory strategic planning Similar problems arise in industry and government in many fields and led to explosion of OR after WWII. Similar problems arise in Epidemiology

5. Operations Research Deals with Optimization Problems Find the best testing/treatment schedule. Find the cheapest medication. Find the least risky intervention. Find the ..

6. Optimization Problems in Economic Epidemiology Many problems in Economic Epi can be formulated as optimization problems: Find a solution that maximizes or minimizes some value. Find the optimal location for a hospital. Find the optimal assignment of health care workers to jobs. Optimize investment in health care supplies. Minimize the total cost of a series of medical tests. Control an outbreak with as small an investment in vaccines as possible.

7. Greedy Algorithms Often, the simplest approach to an optimization problem is a greedy algorithm: Choose the best (cheapest, highest-rated,…) available alternative at each step. In general, greedy algorithms will find locally optimal solutions, but not globally optimal ones. Global optimum Local optimum

8. Greedy Algorithms We give examples from Economic Epi: Some where a greedy solution achieves a global optimum Others where it doesn’t, but we can either make modifications or get a bound on how far from optimal we are.

9. Outline Classic Example I: Assigning Health Care Workers to Jobs Classic Example II: Investing in Health Care Options Classic Example III: Locating Health Care Facilities Classic Example IV: Rerouting Emergency Vehicles after Floods

10. Classic Example I: Assigning Health Care Workers to Jobs n workers W1, W2, …, Wn m jobs J1, J2, …, Jm We know which workers are qualified to do which jobs and the cost of using each worker. Goal: assign workers to jobs they are qualified for, each to at most one job, filling as many jobs as possible, and among all ways of filling as many jobs as possible, find the way to do it with minimum total cost. This is known as the Minimum Cost Assignment Problem

11. Assigning Health Care Workers to Jobs Greedy Algorithm: Create a set C of workers chosen. Choose the least expensive worker, assign the worker to any acceptable job. (Choose arbitrarily throughout all the steps if there are ties.) At each stage, add the next least expensive worker to C if there is an acceptable (feasible) assignment using all the workers in the new C obtained by adding this worker. If a next least expensive worker cannot be added to C, go to the next least expensive one in the list. Continue until you have gone through all the workers.

12. Assigning Health Care Workers to Jobs W1 qualified for J2, J4, J5; cost 90 W2 qualified for J2, J3; cost 50 W3 qualified for J2, J3; cost 80 W4 qualified for J2, J3; cost 60 W5 qualified for J1, J5; cost 60 W6 qualified for J3; cost 100. Comment: cost of Wi doing Jk could vary with k.

13. Assigning Health Care Workers to Jobs W1 qualified for J2, J3, J5; cost 90; W2 qualified for J2, J3; cost 50; W3 qualified for J2, J3; cost 80; W4 qualified for J2, J3; cost 60; W5 qualified for J1, J5; cost 60; W6 qualified for J3; cost 100. Step 1: pick W2 for C and assign it to, say, J2. Step 2: pick W4 and use the assignment W2-J2, W4-J3. C = {W2,W4} Step 3: pick W5 and use the assignment W2-J2, W4-J3, W5- J5,C = {W2,W4,W5} Step 4: pick W3. Note that W2, W3, W4 can only be assigned to J2 and J3, so there is no way to add W3 to the set C. Step 5: pick W1. This can only be assigned to J2, J3, or J5, all of which are already in the assignment. However, we could switch W5 to J1 and then assign W1 to J5, so we add W1 to C. Step 6: pick W6. Since one of W2, W4 has to be assigned to J3, there is no way to make an assignment for W6. C = {W2, W4, W5, W1} and only jobs J1, J2, J3, J5 are covered. This turns out to be optimal.

14. Assigning Health Care Workers to Jobs The greedy algorithm always gives an optimal job assignment.

15. Classic Example II: Investing in Health Care Options Suppose we are faced with a selection of health care options in which to invest. Option i has an estimated cost ci and an estimated value vi. Alternative health care facilities Alternative supplies for a clinic Alternative research programs Problem: Determine which ones to invest in so that the total cost is within budget and the total value is as large as possible.

16. Investing in Health Care Options This is an example of a Knapsack Problem. We have a fixed size knapsack, a number of items we can take, each of a certain size and value, and we want to choose items for the knapsack so as to maximize their value but not exceed the capacity of the knapsack. In general, knapsack problems are difficult to solve computationally – they are NP-complete.

17. Investing in Health Care Options Knapsack Problem Maximize i vixi Subject to i cixi≤ B where xi = number of items i chosen Variants xi = 0 or 1 xi  {0, 1, …, bi} Bounded Knapsack Problem xi is any integer Unbounded Knapsack Problem

18. Investing in Health Care Options Greedy Algorithm Due to George Dantzig 1957 Sort items in decreasing order of value per unit cost: vi/ci Pick as many copies of the first item as possible until no more are possible or until one more would violate i cixi≤ B. Continue in the same way with the second item, then the third, etc.

19. Investing in Health Care Options i vi ci Budget B = 205 AIDS Prevention Options Option 1: Condoms Option 2: Educational Posters Option 3: Clean Needles to Distribute Option 4: Testing Option 5: Funded Researchers

20. Investing in Health Care Options i vi ci B = 205 9 5 8 vi/ci 10 2.5 1. Pick two copies of item 1. Total cost = 70+70 = 140 2. Pick one copy of item 3. Total cost = 140+35 = 175 3. Pick three copies of item 5. Total cost = 175+24 = 199 4. Pick one copy of item 4. Total cost = 199+5 = 204.

21. Investing in Health Care Options i vi ci B = 99 4.7 vi/ci 5 1 Greedy: 9 copies of item 1 plus 9 copies of item 3, at total value of 459. Alternative: 9 copies of item 2, at total value of 468 So, the greedy algorithm does not give the maximum value

22. Investing in Health Care Options For the unbounded knapsack problem, if A is the maximum total value i vixi achievable, then the greedy algorithm always achieves a total value of at least A/2. Is this a satisfactory outcome? It depends upon the application and the tradeoff between the need to make a speedy decision and the ability to wait and do a complicated calculation. For the bounded knapsack problem, the algorithm might be quite a bit further from the optimal.

23. Classic Example III: Locating Health Care Facilities We have a number of users of a planned set of health care facilities. Where do we put the facilities and how do we assign a user to a facility?

24. Classic Example III: Locating Health Care Facilities Suppose we always associate a user to the facility she can get to with minimum cost. There are two costs associated with a choice of locations for the facilities. The cost fi of locating (building, opening) a facility at location i The cost cij of assigning a user at location j to a facility at location i, which is the distance between i and j (or in some applications the distance multiplied by the importance of j or the demand for facility use at j). We wish to minimize the sum of these two costs, summed over all facilities that are located and over all j corresponding to users.

25. Locating Health Care Facilities Given a set S of facilities,let F = the sum of costs fi over all facilities in S C = the sum of distances cij over all users j, where i is the closest facility to j We seek to find S so that F+C is as small as possible. This is the Uncapacitated Facility Location Problem Uncapacitated since we have no limit on the number of facilities.

26. 1 a f 1 1 e b 1 1 c d 1 Locating Health Care Facilities We consider a special case where the users and facilities are located on a network or graph and cij is the distance between i and j in the network. • 1’s represent distances along edges • d(x,y) = length of • shortest route from x to y • So, d(a,c) = 2.

27. 1 a f 1 1 e b 1 1 c d 1 Locating Health Care Facilities Cost 5 Cost 0.5 Cost 4 Cost 3 Cost 6 Cost 1 Cost of facility is in yellow.

28. 1 a f 1 1 e b 1 1 c d 1 Locating Health Care Facilities Cost 5 Cost 0.5 Cost 4 Cost 3 Cost 6 Cost 1 Given users at red circled locations, where do we locate facilities to minimize F+C?

29. Locating Health Care Facilities Greedy Algorithm Due to Charikar and Guha (2004) First find a preliminary solution S. Order the nodes of the network in order of increasing cost of locating a facility at the node. Choose p so that if S is the set of the first p facilities, then the cost F + C associated with S is as small as possible.

30. 1 a f 1 1 e b 1 1 c d 1 Locating Health Care Facilities Cost 5 Cost 0.5 Cost 4 Cost 3 Cost 6 Cost 1 Ordering nodes in order of increasing cost gives the order f, d, b, e, a, c.

31. 1 a f 1 1 e b 1 1 c d 1 Locating Health Care Facilities Cost 5 Cost 0.5 Cost 4 Cost 3 Cost 6 Cost 1 First, take S = {f}. F = cost of f = 0.5 C = d(f,f) + d(b,f) + d(c,f) = 0 + 2 + 3 = 5 F+C = 5.5

32. 1 a f 1 1 e b 1 1 c d 1 Locating Health Care Facilities Cost 5 Cost 0.5 Cost 4 Cost 3 Cost 6 Cost 1 Next, take S = {f,d}. F = cost of f + cost of d = 0.5 + 1 = 1.5 C = d(f,f) + d(b,f) + d(c,d) = 0 + 2 + 1 = 3 F+C = 4.5

33. 1 a f 1 1 e b 1 1 c d 1 Locating Health Care Facilities Cost 5 Cost 0.5 Cost 4 Cost 3 Cost 6 Cost 1 Next, take S = {f,d,b}. F = cost of f + cost of d + cost b = 0.5 + 1 + 3 = 4.5 C = d(f,f) + d(b,b) + d(c,d) = 0 + 0 + 1 = 1 F+C = 5.5

34. 1 a f 1 1 e b 1 1 c d 1 Locating Health Care Facilities Cost 5 Cost 0.5 Cost 4 Cost 3 Cost 6 Cost 1 All larger sets S have F > 4.5, so F+ C > 4.5 Thus, preliminary solution is S = {f,d}.

35. Locating Health Care Facilities At each step, we try to improve the cost of the current solution S, starting with the preliminary solution. Choose a node k at random. (It may be one already in S.) Choose a subset T of facilities in S to remove from S. If a node j is closer to k than to the facility i in S it is assigned to, change its assignment to k and add k to S. If node j was assigned to a facility that was removed from S, i.e., a facility in T, assign it to k. Call the resulting set of facilities S+ Over all possible sets T, find one to remove from S so that S+ minimizes the total cost F+C of the revised assignment. If the new set of facilities that results has a smaller total cost F+C than that associated with S, use it. Otherwise, stay with the previous S.

36. Locating Health Care Facilities Charikar and Guha show that, given , the algorithm is guaranteed to achieve a cost F+C that is at most 2F* + 3C* + (F* + C*) in at most O(nlog(n/ ) steps, where F* and C* are costs associated with an arbitrary optimal solution.

37. Locating Health Care Facilities Note that the way we define S+, it is possible that in some steps, node j is not assigned to its closest facility in S+. We could be tempted to change the definition of S+ to include a step that would reassign a user at node j to its closest facility i in S+ and then removing from S+ all nodes not used for facilities. However, this method does not attain any better bound on the outcome value of F+C and takes longer because we have to find closest facilities.

38. Classic Example IV: Rerouting Emergency Vehicles in Case of Floods New initiative in Climate and Health at DIMACS.

39. Climate and Health Concerns about global warming. Resulting impact on health Of people Of animals Of plants Of ecosystems

40. Climate and Health Some early warning signs: 1995 extreme heat event in Chicago 514 heat-related deaths 3300 excess emergency admissions 2003 heat wave in Europe 35,000 deaths Food spoilage on Antarctica expeditions Not cold enough to store food in the ice

41. Climate and Health Some early warning signs: Malaria in the African Highlands Dengue epidemics Floods, hurricanes

42. Extreme Events due to Global Warming We anticipate an increase in number and severity of extreme events due to global warming. More heat waves. More floods, hurricanes.

43. Extreme Heat Events Result in increased incidence of heat stroke, dehydration, cardiac stress, respiratory distress Hyperthermia in elderly patients can lead to cardiac arrest. Effects not independent: Individuals under stress due to climate may be more susceptible to infectious diseases

44. Problem 1: Evacuations during Extreme Heat Events One response to such events: evacuation of most vulnerable individuals to climate controlled environments. Modeling challenges: Where to locate the evacuation centers? Whom to send where? Goals include minimizing travel time, keeping facilities to their maximum capacity, etc. All involve tools of Operations Research: location theory, assignment problem, etc.

45. Problem 2: Rolling Blackouts during Extreme Heat Events A side effect of such events: Extremes in energy use lead to need for rolling blackouts. Modeling challenges (note many OR problems): Understanding health impacts of blackouts and bringing them into models Design efficient rolling blackouts while minimizing impact on health Lack of air conditioning Elevators no work: vulnerable people over-exertion Food spoilage Minimizing impact on the most vulnerable populations

46. Problem 3: Pesticide Applications after a Flood • Pesticide applications often called for after a flood • Chemicals used for pesticidal and larvicidal control of mosquito disease vectors are themselves harmful to humans • Maximize insect control while minimizing health effects

47. Problem 4: Emergency Rescue Vehicle Routing to Avoid Rising Flood Waters • Emergency rescue vehicle routing to avoid rising flood waters while still minimizing delay in provision of medical attention and still get afflicted people to available hospital facilities • OR problem called Minimum Spanning Tree arises here.

48. OR Problem: Minimum Spanning Tree Problem 2 26 10 14 15 22 20 8 28 16 • A spanning tree is a tree using the edges of the graph and containing all of the nodes. • It is minimum if the sum of the numbers on the edges used is as small as possible.

49. Minimum Spanning Tree Problem 2 26 10 14 15 22 20 8 28 16 • Red edges define a minimum spanning tree.

50. Minimum Spanning Tree Problem • Minimum spanning trees arise in many applications. • One example: Given a road network, find usable roads that allow you to go from any node to any other node, minimizing the lengths of the roads used. • This problem arises in the DIMACS Climate and Health project: Find a usable road network for emergency vehicles in case extreme events leave flooded roads.