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Networks, GEN2 2024 VCAA 14

A manufacturer \((M)\) makes deliveries to the supermarket \((S)\) via a number of storage warehouses, \(L, N, O, P, Q\) and \(R\). These eight locations are represented as vertices in the network below.

The numbers on the edges represent the maximum number of deliveries that can be made between these locations each day.
 

  1. When considering the possible flow of deliveries through this network, many different cuts can be made.   
  2. Determine the capacity of Cut 1, shown above.   (1 mark)

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  3. Determine the maximum number of deliveries that can be made each day from the manufacturer to the supermarket.   (1 mark)

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  4. The manufacturer wants to increase the number of deliveries to the supermarket.
  5. This can be achieved by increasing the number of deliveries between one pair of locations.
  6. Complete the following sentence by writing the locations on the lines provided:
  7. To maximise this increase, the number of deliveries should be increased between
    locations ____ and  ____.
       (1 mark)

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Show Answers Only

a.    \(46\)

b.    \(37\)

c.    \(\text{R and S}\)

Show Worked Solution

a.    \(13+18+6+9=46\)

\(\text{(Reverse flow}\ Q → O\ \text{is not counted.)}\)
 

b.  

\(\text{Max deliveries (min cut)}\ =13+5+11+8=37\)

♦ Mean mark (b) 29%.

 
c.   
\(\text{The number of deliveries should be increased between}\)

\(\text{locations R and S.}\)

♦ Mean mark (c) 22%.

Filed Under: Flow Problems Tagged With: Band 3, Band 5, Band 6, smc-625-10-Cut Capacity, smc-625-20-Max Flow/Min Cut, smc-625-25-Network adjustments

NETWORKS, FUR2 2020 VCAA 4

Training program 1 has the cricket team starting from exercise station `S` and running to exercise station `O`.

For safety reasons, the cricket coach has placed a restriction on the maximum number of people who can use the tracks in the fitness park.

The directed graph below shows the capacity of the tracks, in number of people per minute.
 


 

  1. How many different routes from `S` to `O` are possible?   (1 mark)

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When considering the possible flow of people through this network, many different cuts can be made.

  1. Determine the capacity of Cut 1, shown above.   (1 mark)

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  2. What is the maximum flow from `S` to `O`, in number of people per minute?   (1 mark)

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Show Answers Only
  1. `10`
  2. `52`
  3. `50`
Show Worked Solution
a.   `text(Routes: )` `SMO, STUNO, STUVO, STUVPO, SRQUNO,SRQUVO,`
    `SRQUVPO, SRQVO, SRQVPO, SRQPO`

♦♦ Mean mark part (a) 30%.

 
`:. 10\ text(routes)`

 

b.   `text{Capacity (Cut 1)}` `= 20 + 12 + 20`
    `= 52`

 

c.   `text(Max flow/minimum cut)`

♦♦ Mean mark part (c) 32%.

`= 20 + 10 + 20`

`= 50`
 

Filed Under: Flow Problems Tagged With: Band 4, Band 5, smc-625-10-Cut Capacity, smc-625-20-Max Flow/Min Cut

NETWORKS, FUR1-NHT 2019 VCAA 4 MC

Which one of the following flow diagrams shows a cut that has a capacity of 19?

A.  
B.  
C.  
D.  
E.  
Show Answers Only

`E`

Show Worked Solution

`text(Consider each option):`

`text(Capacity of:)`

`text(Cut)\ A` `= 5 + 2 + 7 + 10 + 7 = 31`
`text(Cut)\ B` `= 5 + 2 + 7 + 7 = 21`
`text(Cut)\ C` `= 5 + 2 + 10 = 17`
`text(Cut)\ D` `= 5 + 2 + 7 = 14`
`text(Cut)\ E` `= 5 + 7 + 7 = 19`

 
`=>  E`

Filed Under: Flow Problems Tagged With: Band 4, smc-625-10-Cut Capacity

NETWORKS, FUR1 2019 STD2 40

A museum is planning an exhibition using five rooms.

The museum manager draws a network to help plan the exhibition. The vertices `A`, `B`, `C`, `D` and `E` represent the five rooms. The number on the edges represent the maximum number of people per hour who can pass through the security checkpoints between the rooms.
 


 

  1. What is the capacity of the cut shown?   (1 mark)

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  2. The museum manager is planning for a maximum of 240 visitors to pass through the exhibition each hour. By using the 'minimum cut-maximum flow' theorem, the manager determines that the plan does not provide sufficient flow capacity.

     

    Draw the minimum cut onto the network below and recommend a change that the manager could make to one or more security checkpoints to increase the flow capacity to 240 visitors per hour.   (2 marks)

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Show Answers Only
  1. `290`
  2.   

Show Worked Solution
a.    `text(Capacity)` `= 130 + 90 + 70`
    `= 290`

♦♦ Mean mark 32%.
COMMENT: In part (a), edge BC flows from the exit to the entry and is therefore not counted.

b.   `text(Maximum flow capacity:)`
   

`text(Minimum cut = 80 + 40 + 65 + 45 = 230)`

♦♦♦ Mean mark 19%.
COMMENT: In part (b), edge BC now flows from entry to exit in the new “minimum” cut and is counted.

`text(If security is improved to increase the flow)`

`text(between Room C and Room B by 10 visitors)`

`text(per hour, the network’s flow capacity increases)`

`text(to 240.)`

Filed Under: Flow Problems Tagged With: Band 5, Band 6, smc-625-10-Cut Capacity, smc-625-20-Max Flow/Min Cut

NETWORKS, FUR1 2019 VCAA 3 MC

The flow of water through a series of pipes is shown in the network below

The numbers on the edges show the maximum flow through each pipe in litres per minute.
 

 
The capacity of Cut `Q`, in litres per minute, is

  1. 11
  2. 13
  3. 14
  4. 16
  5. 17
Show Answers Only

`C`

Show Worked Solution

`text(Capacity of cut)\ Q`

`= 5 + 6 + 3`

`= 14`

`=>  C`

Filed Under: Flow Problems Tagged With: Band 4, smc-625-10-Cut Capacity

NETWORKS, FUR2 2018 VCAA 1

The graph below shows the possible number of postal deliveries each day between the Central Mail Depot and the Zenith Post Office.

The unmarked vertices represent other depots in the region.

The weighting of each edge represents the maximum number of deliveries that can be made each day.
 


 

  1.  Cut A, shown on the graph, has a capacity of 10.

     

     Two other cuts are labelled as Cut B and Cut C.
     i.  Write down the capacity of Cut B.   (1 mark)

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    ii.  Write down the capacity of Cut C.   (1 mark)

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  1. Determine the maximum number of deliveries that can be made each day from the Central Mail   Depot to the Zenith Post Office.   (1 mark)

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Show Answers Only
  1.  i.  `9`
    ii.  `13`
  2.  `7`
Show Worked Solution
a.i.    `text{Capacity (Cut B)}` `= 3 + 2 + 4`
    `= 9`

 

a.ii.    `text{Capacity (Cut C)}` `= 3 + 6 + 4`
    `= 13`

♦ Mean mark part (b) 32%.
COMMENT: Review carefully! Most common incorrect answer was 9.

 

b.  `text(Minimum cut) = 2 + 2 + 3 = 7`

`:.\ text(Maximum deliveries) = 7`

Filed Under: Flow Problems Tagged With: Band 3, Band 5, smc-625-10-Cut Capacity, smc-625-20-Max Flow/Min Cut

NETWORKS, FUR1 2008 VCAA 6 MC

networks-fur1-2008-vcaa-6-mc

 
For the graph above, the capacity of the cut shown is

A.   `33`

B.   `36`

C.   `40`

D.   `42`

E.   `46`

Show Answers Only

`=> D`

Show Worked Solution

`text(Capacity of the cut)`

♦♦ Mean mark 35%.
MARKER’S COMMENT: For an individual flow to contribute to the “cut”, it must flow from the source to the sink.

`= 4 + 2 + 7 + 9 + 8 + 6 + 6`

`= 42`

`=> D`

 

Filed Under: Flow Problems Tagged With: Band 5, smc-625-10-Cut Capacity

NETWORKS, FUR1 2010 VCAA 6-7 MC

In the network below, the values on the edges give the maximum flow possible between each pair of vertices. The arrows show the direction of flow. A cut that separates the source from the sink in the network is also shown.
 

vcaa-networks-fur1-2010-6-7

 
Part 1

The capacity of this cut is

A.   `14`

B.   `18`

C.   `23`

D.   `31`

E.   `40`

 

Part 2

The maximum flow between source and sink through the network is

A.    `7`

B.   `10`

C.   `11`

D.   `12`

E.   `20`

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ B`

Show Worked Solution

Part 1

`text(Capacity of the cut)`

♦ Mean mark 50%.
COMMENT: A quarter of students incorrectly included the “8” which is flowing in the opposite direction.

`= 11 + 5 + 7`

`= 23`

`=>  C`

 

Part 2

vcaa-networks-fur1-2010-6-7i

`text(The maximum flow)`

♦♦♦ Mean mark 24%.

`=\ text{minimum cut (see above)}`

`= 4 + 2 + 3 + 1`

`= 10`

`=>  B`

Filed Under: Flow Problems Tagged With: Band 4, Band 5, smc-625-10-Cut Capacity, smc-625-20-Max Flow/Min Cut

NETWORKS, FUR1 2006 VCAA 6 MC

networks-fur1-2006-vcaa-6-mc

 
In the directed graph above the weight of each edge is non-zero.

The capacity of the cut shown is

A.   a + b + c + d + e

B.   a + c + d + e

C.   a + b + c + e

D.   a + b + c – d + e

E.   a – b + c – d + e 

Show Answers Only

`C`

Show Worked Solution

`text(Capacity of the cut)`

`= a + b + c + e`

`rArr C`

Filed Under: Flow Problems Tagged With: Band 4, smc-625-10-Cut Capacity

NETWORKS, FUR2 2007 VCAA 3

As an attraction for young children, a miniature railway runs throughout the new housing estate.

The trains travel through stations that are represented by nodes on the directed network diagram below.

The number of seats available for children, between each pair of stations, is indicated beside the corresponding edge.
 

NETWORKS, FUR2 2007 VCAA 3

 
Cut 1, through the network, is shown in the diagram above.

  1. Determine the capacity of Cut 1.   (1 mark)

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  2. Determine the maximum number of seats available for children for a journey that begins at the West Terminal and ends at the East Terminal.   (1 mark)

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On one particular train, 10 children set out from the West Terminal.

No new passengers board the train on the journey to the East Terminal.

  1. Determine the maximum number of children who can arrive at the East Terminal on this train.   (1 mark)

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Show Answers Only
  1. `43`
  2. `22`
  3. `7`
Show Worked Solution

a.   `text(The capacity of Cut 1)`

♦♦ Mean mark for all parts (combined) was 33%.
MARKER’S COMMENT: A common error was counting the edge with “10” in the reverse direction (it should be ignored).

`=14 + 8 + 13 + 8`

`= 43`

 

b.    networks-fur2-2007-vcaa-3-answer
`text(Maximum seats)` `=\ text(minimum cut)`
  `= 6 + 7 + 9`
  `= 22`

 

c.  `text{The path (edge weights) of the train setting out with}`

`text(10 children starts with: 11 → 13.)`

`text(At the next station, a maximum of 7 seats are available)`

`text(which remain until the East Terminal.)`
  

`:.\ text(Maximum number of children arriving is 7.)`

Filed Under: Flow Problems Tagged With: Band 5, Band 6, smc-625-10-Cut Capacity, smc-625-20-Max Flow/Min Cut

NETWORKS, FUR2 2013 VCAA 3

The rangers at the wildlife park restrict access to the walking tracks through areas where the animals breed.

The edges on the directed network diagram below represent one-way tracks through the breeding areas. The direction of travel on each track is shown by an arrow. The numbers on the edges indicate the maximum number of people who are permitted to walk along each track each day.
 

NETWORKS, FUR2 2013 VCAA 31
 

  1. Starting at `A`, how many people, in total, are permitted to walk to `D` each day?   (1 mark)

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One day, all the available walking tracks will be used by students on a school excursion.

The students will start at `A` and walk in four separate groups to `D`.

Students must remain in the same groups throughout the walk.

  1. i. Group 1 will have 17 students. This is the maximum group size that can walk together from `A` to `D`.
    Write down the path that group 1 will take.  (1 mark)

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  2. ii. Groups 2, 3 and 4 will each take different paths from `A` to `D`.
    Complete the six missing entries shaded in the table below.   (2 marks)

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NETWORKS, FUR2 2013 VCAA 32

Show Answers Only
  1. `37`
    1. `A-B-E-C-D`
    2.  `text{One possible solution is:}`
       
      Networks, FUR2 2013 VCAA 3_2 Answer1
Show Worked Solution
a.    `text(Maximum flow)` `=\ text(minimum cut through)\ CD and ED`
    `= 24 + 13`
    `= 37`
♦ Mean mark of all parts (combined) was 41%.

 

`:.\ text(A maximum of 37 people can walk)`

`text(to)\ D\ text(from)\ A.`
  

b.i.   `A-B-E-C-D`
  

b.ii.   `text(One solution using the second possible largest)`

  `text(group of 11 students and two groups from the)`

  `text(remaining 9 students is:)`

Networks, FUR2 2013 VCAA 3_2 Answer1

Filed Under: Flow Problems Tagged With: Band 4, Band 5, Band 6, smc-625-10-Cut Capacity, smc-625-20-Max Flow/Min Cut

NETWORKS, FUR1 2015 VCAA 4 MC

The arrows on the diagram below show the direction of the flow of waste through a series of pipelines from a factory to a waste dump.

The numbers along the edges show the number of megalitres of waste per week that can flow through each section of pipeline.
  

NETWORKS, FUR1 2015 VCAA 4 MC

 
The minimum cut is shown as a dotted line.

The capacity of this cut, in megalitres of waste per week, is

A.     `6`

B.   `18`

C.   `26`

D.   `32`

E.   `34`

Show Answers Only

`C`

Show Worked Solution

`text{Minimum Cut (ML/week)}`

`= 5 + 2 + 12 + 7`

`= 26`

`=> C`

Filed Under: Flow Problems Tagged With: Band 4, smc-625-10-Cut Capacity

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