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.
Maximum Flow Network
t
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
Maximum Flow Network
100
100
t
100 100 100
1,2
1,3
1,4
1,1
100
1,5
100
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
400
400
s
Optimal Solution: v = 700
t
1,1
2,1
1,2
2,2
1,3
1,4
1,5
2,4
(100, 100)
(100,)
(100,)
(300, )
(300,400)
3,1
(400,400)
s
Minimum Cut
t
1,1
1,2
100
100
1,3
2,1
100
1,4
1,5
2,2
2,4
3,1
400
s
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
Minimum Cut Example
Arcs with capacity c from each block needed to mine (i,j)
will be in the minimum cut
t



 

s






(i,j)
 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