Max-Flow Min-Cut Applications: Open-Pit Mining
Mining for Gold • Divide a cross-section of earth into a grid of equally-sized blocks. • Given the cost of excavating a full block and the market value of a full block of gold, determine an optimal shape for the mine. • Constraints on the shape of your mine require that a block of earth cannot be excavated unless the three blocks above it (directly, and diagonally left and right) have also been excavated.
Solution • To get the gold in block (2,4), we need to excavate blocks (1,3), (1,4), (1,5) and (2,4). Revenue = $500, cost = $400, profit = $100. Take block (2,4). • To get the gold in block (3,1), we need to excavate blocks (3,1), (2,1), (2,2), (1,1), (1,2) and (1,3). Revenue = $500, cost = $500*, profit = $0. Don’t take block (3,1). * Assuming (1,3) already excavated.
Maximum Flow Formulation • Each block is a node. • Add infinite capacity arc from node (i,j) to node (i-1,j-1), (i-1,j),(i-1,j+1). • Add source node s and sink node t.
t Maximum Flow Network 1,1 1,2 1,3 1,4 1,5 2,1 2,2 2,3 2,4 2,5 3,1 3,2 3,3 3,4 3,5 4,1 4,2 4,3 4,4 4,5 s
Maximum Flow Formulation • Wij = Vij - C = Vij = value of g old in block (i,j) C = cost of excavating a block • If Wij < 0 add an arc from node (i,j) to t with capacity –Wij block (i,j) has value 0 and cost Wij • If Wij > 0 add an arc from s to node (i,j) with capacity Wij block (i,j) has value Wij and cost 0
100 100 100 100 100 100 100 t 400 400 Maximum Flow Network 1,1 1,2 1,3 1,4 1,5 2,1 2,2 2,3 2,4 2,5 3,1 3,2 3,3 3,4 3,5 4,1 4,2 4,3 4,4 4,5 s
t 2,1 2,4 3,1 s Optimal Solution: v = 700 1,1 1,2 1,3 1,4 1,5 (100, 100) 2,2 (100,) (100,) (300, ) (300,400) (400,400)
1,3 1,4 1,5 t 2,1 2,4 3,1 s Minimum Cut 100 1,1 1,2 100 100 2,2 400
Meaning of the Minimum Cut • Excavate all blocks in S • Only finite-capacity arcs in minimum cut • Blocks in S can be excavated without excavating blocks in T • Arcs in [S,T] • Arcs from s to nodes in T • Capacity = value of gold blocks in T • Arcs from nodes in S to t • Capacity = cost of excavating S • u[s,t] = value of T + cost of S
t (i,j) Minimum Cut Example Arcs with capacity c from each block needed to mine (i,j) will be in the minimum cut s capacity arcs can’t be in the minimum cut
Meaning of the Minimum Cut • u[S,T] = 400 + 300 = 700 • Value of Gold in T: VG(T) = 400 • Cost of Mining S: CM(S) = 300 • u[S,T] = VG(T) + CM(S) • Min u[S,T] = Min (VG(T) + CM(S)) • Min (VG(T) + CM(S)) = Max (-VG(T) – CM(S))
Maximizing Profit Solving the max flow problem gives us the Max –VG(T) – CM(S) Let G*= total value of available gold G* is a constant Max –VG(T) – CM(S) = Max G*-VG(T)-CM(S) = Max VG(S) +VG(T) - VG(T) – CM(S) = Max VG(S) – CM(S) = profit from mining blocks in S
