CO-351-1079-Final_solutions

# CO-351-1079-Final_solutions - CO 351 - Final Examination 1...

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Unformatted text preview: CO 351 - Final Examination 1 CO 351 Network Flow Theory Solutions to the Final Examination December 6, 2007 Instructor: Ashwin Nayak CO 351 - Final Examination 2 Question 1. [10 marks] Find an arborescence of shortest dipaths rooted at node 1 in the following digraph with arc-weights by running the Dual Algorithm. Start with the node labels on the top left copy of the digraph as the initial potential. Using as many copies of the digraph as you need, show the feasible potential by labeling the nodes, draw the equality arcs in bold, show the separator you pick, and indicate the increase in potential in each iteration of the algorithm. Answer: [Each iteration gets 2 marks; a numerical error with a domino effect is penalized only once.]-2 6 3 1-4 7 6 5 4 3 2 1 2 3 5-6-1 6 3 1-4 7 6 5 4 3 2 1 2 3 5-6 5 5 3 3 2 2 6-6 5 7 6 5 4 3 2 1 3-7 1-1-2 6 3 1-4 7 6 5 4 3 2 1-4-1 5-2-2-1-4 4 1 5-6-(-7)+1=2 6-(-3)+4=13 5-(-6)+5=16-7-3 4 1-6 5 increase in potential = 1 2-0+2=4-5 The bold arcs without arc 42 form an arborescence rooted at s-5 4 1 5-2-2-(-3)+0=1 5-(-3)+5=13-3-5 1 4 5 6-(-1)+4=11 5-(-1)-5=1 5-(-4)+5=14 3-4+2=1 3 1-4 7 6 5 4 3 2 1 2 3 5-6 5-1 6 increase in potential =1 increase in potential =1 increase in potential = 2 3 2 Iteration 0 2 2 2 Iteration 5 2 3 5-6 5-1-2-4 2 3 5-6 5-1-2 6 3 1-4 7 6 5 4 3 2 1-8 4 Iteration 4 Iteration 3 Iteration 2 Iteration 1 Initial potentials CO 351 - Final Examination 3 Question 2. [10 marks] Let D = ( N,A ) be a digraph with non-negative arc-capacities c ∈ R A , nodes s,t ∈ N , s negationslash = t . Show that the value of any feasible ( s,t )-flow in the digraph is bounded from above by the capacity of any ( s,t )-cut in D . Prove all the claims you make. Answer: Let x be any feasible ( s,t )-flow, and let Z be any ( s,t )-separator. We claim (as shown below) that the net flow out of Z equals the net flow out of s (i.e., the value of the flow x ) f out x ( Z ) = f out x ( s ) . [2 marks] So the value of the flow is bounded as f out x ( s ) = f out x ( Z ) = x ( δ out ( Z )) − x ( δ in ( Z )) ≤ x ( δ out ( Z )) ≤ c ( δ out ( Z )) , which is the capacity of the cut δ out ( Z ). [5 marks] We now prove the claim. By the flow conservation conditions ( f in x ( v ) = 0 = − f out x ( v ) for all v negationslash∈ { s,t } ) [1 mark], we have f out x ( s ) = summationdisplay v ∈ Z f out x ( v ) = summationdisplay v ∈ Z summationdisplay z : vz ∈ δ out ( v ) x vz − summationdisplay u : uv ∈ δ out ( v ) x uv = summationdisplay v ∈ Z summationdisplay z ∈ Z : vz ∈ δ out ( v ) x vz − summationdisplay u ∈ Z : uv ∈ δ out ( v ) x uv + summationdisplay v ∈ Z summationdisplay z negationslash∈ Z : vz ∈ δ out ( v ) x vz − summationdisplay u negationslash∈ Z : uv ∈ δ out ( v ) x uv = summationdisplay v,z ∈ Z : vz ∈ δ out ( v ) x vz − summationdisplay u,v ∈ Z : uv ∈ δ out ( v ) x uv + x ( δ out ( Z )) − x ( δ...
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## This note was uploaded on 11/18/2010 for the course CO 351 taught by Professor Various during the Fall '05 term at Waterloo.

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CO-351-1079-Final_solutions - CO 351 - Final Examination 1...

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