Political problem

Political problem Suppose a politician trying to win an election 3 types of areas --- urban, suburban, rural. Certain issues --- road, gun control, farm subsidies, gasoline tax. Try to find out the minimum amount of money you need to win 50,000 urban, 100,000 suburban, 25,000 rural votes. Policy urban suburban rural Build roads -2 5 3 Gun control 8 2 -5 Farm subsidies 0 0 10 Gasoline tax 10 0 -2 The effects of policies on votes.

4 variables : x1 is the # of thousands of dollars on advertising on building roads. x2 is the # of thousands of dollars on advertising on gun control. x3 is the # of thousands of dollars on advertising on farm subsidies. x4 is the # of thousands of dollars on advertising on gasoline tax. We format this problem as: Minimize: x1 + x2 + x3 + x4 Subject to: -2x1 + 8x2 + 0x3 + 10x4  50 5x1 + 2x2 + 0x3 + 0x4  50 3x1 + 5x2 + 10x3 + 2x4  50 x1, x2, x3, x4  0 The solution of this liner program will yield an optimal strategy for the politician.

An overview of liner programming: Two forms: standard and slack. Linear program with two variables: Maximize x1 + x2 Subject to: 4x1 - x2  8 2x1 + x2  10 5x1 - 2x2  -2 x1, x2  0 Feasible solution : if x1 and x2 satisfies all the constraints.

Formulating problems as linear programs • Shortest paths: minimize d[t] subject to: d[v]  d[u] + w( u, v ) for each (u, v) E. d[s] = 0. • Maximum flow: maximize f[s,v] subject to: f[u,v]  c[u,v] for each u, vV, f[u,v] = -f[v,u] for each u, vV, f[u,v] = 0 for each uV – {s,t}.

Minimum-cost-flow problem x x c=5 a=2 c=2 a=7 1/2 a=7 2/5 a=2 c=1 a=3 1/1 a=3 s t s t c=2 a=5 2/2 a=5 c=4 a=1 3/4 a=1 y y An example of minimum cost flow problem. A solution to minimum cost flow problem.

Vector Form Maximize: cx Subject to : Ax  b c = (c1, c2, …, cn) x = b = A = Summation Form Maximize: cixi Subject to: a1ixi  b1 a2ixi  b2 . . amixi  bm Linear Programming (LP) x1 . . xn b1 . . bn a11 … a1n …… …… an1 … amn

n = 2; m = 4 x1 + x2 max x1 0  (-1)x1 + 0x2 0 x2 0  0x1 + (-1)x2 0 x1 5  (-1)x1 + 0x2 5 x2 6  0x1 + 1x2 6 c = (1, 1) b = A = Optimal solution is the unique point of intersection of the objective with the hyperplane feasible polytope. x2 optimal solution x1 = 5 ; x2 = 6 objective: x1 + x2 = 11 x1 Example LP 0 0 5 6 Feasible solution region -1 0 0 -1 1 0 0 1

Vector Form Maximize: cx Subject to : Ax  b and x  {0,1} c = (c1, c2, …, cn) x = b = A = Summation Form Maximize: cixi Subject to: a1ixi  b1 a2ixi  b2 . . amixi  bm and x  {0,1} Integer Linear Programming (ILP) x1 . . xn b1 . . bn a11 … a1n …… …… an1 … amn

ILP for MIS • Maximum Independent Set (MIS) - Find the maximum subset of nodes in graph G = (V, E) which are pairwise non-adjacent • ILP - For any v  V make a variable xv  {0, 1} xv = 0  v  MIS which means 0 is not chosen xv = 1  v  MIS which means 1 is chosen - Maximize vV xv Subject to:  e = (u, v)  V, xu + xv  1

Max: x1 + x2 + x3 + x4 + x5 + x6 subject to: x1 + x6 1 x1 + x2 1x2 + x3 1x3 + x6 1x3 + x5 1x6 + x5 1x3 + x4 1x4 + x5 1and x1, x2,…, x6 {0,1} Graph Example ILP of MIS 2 1 3 6 4 5

ILP - xi  max - Subject to: xi + xj  1  (i, j)  E ILP for MaxClique • MaxClique • - Given G = (V, E) • - Find • X  V  x, x’  X • (x, x’)  E • |X|  max

Matching - Given G = (V, E) - Find X  E  e, e’  X e and e’ don’t share endpoint. |X|  max ILP - for any e  E xe {0, 1} + 0 is not in matching + 1 is in matching - eE xe  max - Subject to: e incident to V xe  1 v  V e2 xe1 +xe2 +xe3  1 only one edge from e1 e3 matching can be incident to v ILP for Matching v

MISP xi  max, i  V xi + xj  1, for each edge (xi,xj) E xi {0, 1} LPR  xi  max, i  V xi + xj  1, for each edge (xi,xj) E 0  xi  1 LP relaxation (LPR) vs. ILPLP relaxation (LPR) for MAX independent set problem (MISP) gives larger value than the maximum size of independent set.

xi  max x1 +x6  1 x1 +x2  1 x5 +x6  1 x5 +x2  1 x5 +x4  1 x4 +x3  1 x4 +x2  1 x2 +x3  1 ILP x1 = x3 = x5 = 1 xi = 3 LPR xi = ½ xi = 3 Example 1 of ILP vs. LPR 3 4 2 1 5 6

x1 +x2 + x3 max x1 +x2  1 x1 +x3  1 x2 +x3  1 0  x1  1 0  x2  1 0  x3  1 LPR ()  3/2 Implies LPR () = 3/2 So x1 = x2 = x3 = ½ LPR ()  3/2 ILP () = 1 Integrality Gap (IG) = LPR / ILP = 3/2 What is the integrality gap for (MISP) For a complete graph ILP (Kn) = 1 LPR (Kn) = n/2 Integrality Gap (IG) = LPR / ILP Integrality gap may be as large as n/2 MISP Integrality Gap 2 3 1

MVCP xi  min, i  V xi + xj  1, for each e= (xi,xj) E xi {0, 1} LPR  xi  min , i  V xi + xj  1, for each e= (xi,xj) E 0  xi  1 LPR vs. ILP LP relaxation (LPR) for Minimum Vertex Cover problem (MVCP) gives smaller value than the minimum size of vertex cover

x1 +x2 + x3 min x1 +x2 1 x1 +x3 1 x2 +x3 1 0  x1  1 0  x2  1 0  x3  1 LPR ()  3/2 Implies LPR () = 3/2 So x1 = x2 = x3 = ½ LPR ()  3/2 ILP () = 2 Integrality Gap (IG) = ILP / LPR = 4/3 What is the integrality gap for (MVCP) For a complete graph ILP (Kn) = n - 1 LPR (Kn) = n/2 Integrality Gap (IG) = ILP / LPR Integrality gap may be as large as 2 – (2 / n) For more information: http://www-unix.mcs.anl.gov/otc/Guide/faq/linear-programming-faq.html MVCP Integrality Gap 2 1 3

Political problem