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Networks, STD2 EQ-Bank 30

A weighted and directed network diagram is shown.
 

  1. What is the outflow from vertex \(C\) ?   (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

  2. Calculate the maximum flow through the network.   (2 marks)

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  3. The capacity of ONE saturated edge is to be increased in order to produce a new network with the maximum possible flow.
  4. State an edge which could have an increased capacity AND find the new maximum flow through this new network.   (2 marks)

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Show Answers Only

a.    \(\text{Inflow of \(C\) = Outflow of \(C\) = 14}\)

b.    \(\text{Method 1:}\)

\(\text{Minimum cut through}\ \ sC-Bt-At:\)

\(\text{Maximum flow} = 14+20+11=45\)
 

\(\text{Method 2:}\)
 

     

\(sAt\ \ 11\ \text{(leaving an excess flow capacity of } s A=11 \text { )}\)

\(sABt\ \ 11\ \text{(leaving an excess flow capacity of } AB=7 \text { and } B t=9 \text { )}\)

\(sBt\ \ 9\ \text{(leaving an excess flow capacity of } s B=7 \text { )}\)

\(sCt\ \ 14\ \text{(leaving an excess flow capacity of } C t=3)\)

\(\text{Maximum Flow} = 11+11+9+14=45\)
 

c.    \(\text{Answers could include one of the following:}\)

\(\text{B} t \ \text{could be increased (by 7) leading to a maximum flow of 52.}\)

\(\text{A} t \ \text{could be increased (by 11) leading to a maximum flow of 52.}\)

Show Worked Solution

a.    \(\text{Inflow of \(C\) = Outflow of \(C\) = 14}\)

b.    \(\text{Method 1:}\)

\(\text{Minimum cut through}\ \ sC-Bt-At:\)

\(\text{Maximum flow} = 14+20+11=45\)
 

\(\text{Method 2:}\)
 

     

\(sAt\ \ 11\ \text{(leaving an excess flow capacity of } s A=11 \text { )}\)

\(sABt\ \ 11\ \text{(leaving an excess flow capacity of } AB=7 \text { and } B t=9 \text { )}\)

\(sBt\ \ 9\ \text{(leaving an excess flow capacity of } s B=7 \text { )}\)

\(sCt\ \ 14\ \text{(leaving an excess flow capacity of } C t=3)\)

\(\text{Maximum Flow} = 11+11+9+14=45\)
 

c.    \(\text{Answers could include one of the following:}\)

\(\text{B} t \ \text{could be increased (by 7) leading to a maximum flow of 52.}\)

\(\text{A} t \ \text{could be increased (by 11) leading to a maximum flow of 52.}\)

Filed Under: Network Flow (Y12) Tagged With: Band 4, Band 5, smc-6915-10-Min Cut/Max Flow, smc-6915-35-Saturated Edges Method, syllabus-2027

Networks, STD2 N3 EQ-Bank 5 MC

The network diagram below flows from the source \((S)\) to sink \((T)\).

Which of the edges is not at maximum capacity?
 

   

  1. \(AC\)
  2. \(BC\)
  3. \(CT\)
  4. \(DT\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Consider the minimum cut line: all edges are saturated.}\)
 


 

\(\text{Edge \(CT\) is not at maximum capacity.}\)

\(\Rightarrow C\)

Filed Under: Flow Networks and Minimum Cuts, Network Flow (Y12) Tagged With: Band 4, smc-6915-30-Flow Capacity, smc-6915-35-Saturated Edges Method, smc-915-30-Flow Capacity

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