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Networks, GEN1 2024 NHT 40 MC

A project has 10 activities, labelled \(A\) to \(J\). The table below shows the immediate predecessor(s) for each activity. Each activity has a duration of at least one day.

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textbf{Activity} & \textbf{Immediate}\\
&\textbf{predecessor(s)} \rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}A \rule[-1ex]{0pt}{0pt}& - \\
\hline
\rule{0pt}{2.5ex}B \rule[-1ex]{0pt}{0pt}& - \\
\hline
\rule{0pt}{2.5ex}C \rule[-1ex]{0pt}{0pt}& A \\
\hline
\rule{0pt}{2.5ex}D \rule[-1ex]{0pt}{0pt}& B \\
\hline
\rule{0pt}{2.5ex}E \rule[-1ex]{0pt}{0pt}& B \\
\hline
\rule{0pt}{2.5ex}F \rule[-1ex]{0pt}{0pt}& D \\
\hline
\rule{0pt}{2.5ex}G \rule[-1ex]{0pt}{0pt}& D, E \\
\hline
\rule{0pt}{2.5ex}H \rule[-1ex]{0pt}{0pt}& C, F \\
\hline
\rule{0pt}{2.5ex}I \rule[-1ex]{0pt}{0pt}& E \\
\hline
\rule{0pt}{2.5ex}J \rule[-1ex]{0pt}{0pt}& G,H \\
\hline
\end{array}

Which one of the following statements about this project is not true?

  1. The earliest starting time of activity \(H\) could be two days.
  2. In the network for this project, there would be a dummy activity from the end of activity \(D\) to the start of activity \(G\).
  3. One of the paths through the network of this project is \(B D G J\).
  4. The latest starting time of activity \(E\) could be three days.
  5. The network for this project would require two dummy activities.
Show Answers Only

\(A\)

Show Worked Solution

\(\text{One strategy: draw a network diagram:}\)
 

\(\text{Earliest starting time of activity \(H\) = 3 days.}\)

\(\Rightarrow A\)

Filed Under: Critical Path Analysis Tagged With: Band 5, smc-621-10-Network table, smc-621-20-Critical Paths/EST

Networks, GEN2 2024 VCAA 15

An upgrade to the supermarket requires the completion of 11 activities, \(A\) to \(K\).

The directed network below shows these activities and their completion time, in weeks.

The minimum completion time for the project is 29 weeks.
 

 

  1. Write down the critical path.   (1 mark)

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

  2. Which activity can be delayed for the longest time without affecting the minimum completion time of the project?   (1 mark)

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

Use the following information to answer parts c-e.

A change is made to the order of activities.

The table below shows the activities and their new latest starting times in weeks.

\begin{array}{|c|c|}
\hline
\textbf{Activity} & \textbf{Latest Starting}\\
&\textbf{time} \text{(weeks)}\\
\hline A & 0 \\
\hline B & 2 \\
\hline C & 10 \\
\hline D & 9 \\
\hline E & 13 \\
\hline F & 14 \\
\hline G & 18 \\
\hline H & 17 \\
\hline I & 19 \\
\hline J & 25 \\
\hline K & 22 \\
\hline
\end{array}

A dummy activity is now required in the network.

  1. On the directed network below, draw a directed edge to represent the dummy activity. Include a label.  (1 mark)

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

     

  1. What is the new minimum completion time of the project?  (1 mark)

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

  2. The owners of the supermarket want the project completed earlier.
  3. They will pay to reduce the time of some of the activities.
  4. A reduction in completion time of an activity will incur an additional cost of $10 000 per week.
  5. Activities can be reduced by a maximum of two weeks.
  6. The minimum number of weeks an activity can be reduced to is seven weeks.
  7. What is the minimum amount the owners of the supermarket will have to pay to reduce the completion time of the project as much as possible?   (1 mark)

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

Show Answers Only

a.    \(A, C, H, J\)

b.    \(\text{Activity E}\)

c.    

d.    \(\text{30 weeks}\)

e.    \($50\,000\)

Show Worked Solution

a.    \(\text{Critical path:  }A, C, H, J\)
 

 
b.    
\(\text{Activity with the largest float time can be delayed the longest.}\)

\(\text{Consider Activity E:}\)

\(\text{EST = 11, LST}= 18-4=14\rightarrow\ \text{Float time = 3 weeks}\)

\(\therefore\ \text{Activity E can be delayed the longest.}\)
 

♦♦ Mean mark (b) 34%.

c.    

♦♦♦ Mean mark (c) 10%.

d.    \(\text{New minimum completion time is 30 weeks.}\)
 

♦♦♦ Mean mark (d) 27%.

e.    \(\text{Activities that can be reduced:}\)

\(-A\ \text{can be reduced by 2 weeks}\)

\(-B, D\ \text{can each be reduced by 1 week each}\)

\(-H\ \text{can be reduced by 1 week}\)

\(\text{Total reduction = 5 weeks}\)

\(\Rightarrow \ \text{Minimum payment}=$50\,000\)

♦♦♦ Mean mark (e) 7%.

Filed Under: Critical Path Analysis Tagged With: Band 4, Band 5, Band 6, smc-621-20-Critical Paths/EST, smc-621-40-Crashing/Reduce completion time, smc-621-45-Adding activities, smc-621-50-Dummy activities

Networks, GEN2 2023 VCAA 14

One of the landmarks in state \(A\) requires a renovation project.

This project involves 12 activities, \(A\) to \(L\). The directed network below shows these activities and their completion times, in days.
 

The table below shows the 12 activities that need to be completed for the renovation project.

It also shows the earliest start time (EST), the duration, and the immediate predecessors for the activities.

The immediate predecessor(s) for activity \(I\) and the EST for activity \(J\) are missing.

\begin{array} {|c|c|c|}
\hline
\quad \textbf{Activity} \quad & \quad\quad\textbf{EST} \quad\quad& \quad\textbf{Duration}\quad & \textbf{Immediate} \\
&  & & \textbf{predecessor(s)}  \\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & 0 & 6 & - \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 0 & 4 & - \\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 6 & 7 & A \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & 4 & 5 & B \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & 4 & 10 & B \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & 13 & 4 & C \\
\hline
\rule{0pt}{2.5ex} G \rule[-1ex]{0pt}{0pt} & 9 & 3 & D \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & 9 & 7 & D \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 13 & 6 & - \\
\hline
\rule{0pt}{2.5ex} J \rule[-1ex]{0pt}{0pt} & - & 6 & E, H \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & 19 & 4 & F, I \\
\hline
\rule{0pt}{2.5ex} L \rule[-1ex]{0pt}{0pt} & 23 & 1 & J, K \\
\hline
\end{array}

  1. Write down the immediate predecessor(s) for activity \(I\).  (1 mark)

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

  2. What is the earliest start time, in days, for activity \(J\) ?  (1 mark)

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

  3. How many activities have a float time of zero?  (1 mark)

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

The managers of the project are able to reduce the time, in days, of six activities.

These reductions will result in an increase in the cost of completing the activity.

The maximum decrease in time of any activity is two days.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Activity} \rule[-1ex]{0pt}{0pt} & \quad A \quad & \quad B  \quad&  \quad F \quad  &  \quad H \quad &  \quad I  \quad &  \quad K \quad \\
\hline
\rule{0pt}{2.5ex} \textbf{Daily cost (\$)} \rule[-1ex]{0pt}{0pt} &  1500 & 2000 & 2500 & 1000 & 1500 & 3000  \\
\hline
\end{array}

  1. If activities \(A\) and \(B\) have their completion time reduced by two days each, the overall completion time of the project will be reduced.
  2. What will be the maximum reduction time, in days?  (1 mark)

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

  3. The managers of the project have a maximum budget of $15 000 to reduce the time for several activities to produce the maximum reduction in the project's overall completion time.
  4. Complete the table below, showing the reductions in individual activity completion times that would achieve the earliest completion time within the $ 15 000 budget.  (1 mark)

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

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} &  \\
\hline
\end{array}

Show Answers Only

a.    \(\text{Immediate predecessors of}\ I:\ C, G\)

b.    \(\text{EST}(J) = 4+5+7 = 16\ \text{days}\)

c.    \(\text{5 activities have a float time of zero.}\)

d.    \(\text{Maximum reduction time = 2 days}\)

e.    

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  0\\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} &  1\\
\hline
\end{array}

Show Worked Solution

a.    \(\text{Immediate predecessors of}\ I:\ C, G\)

\(\text{Dummy activity before activity}\ C\ \text{does not effect this.}\)
 

b.    \(\text{Scan network:}\)

\(\text{EST}(J) = 4+5+7 = 16\ \text{days}\)
 

c.    \(\text{Critical Path:}\ A\ C\ I\ K\ L\)

\(\text{Activities on the critical path have a float time of zero.}\)

\(\Rightarrow \ \text{5 activities have a float time of zero.}\)
 

d.    \(\text{If activities}\ A\ \text{and}\ B\ \text{are reduced by 2 days,}\)

\(\text{the critical path remains:}\ A\ C\ I\ K\ L\ \text{(22 days)}\)

\(\text{Maximum reduction time = 2 days}\)

♦♦ Mean mark (c) 35%.
♦♦ Mean mark (d) 33%.

e.    

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  0\\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} &  1\\
\hline
\end{array}

♦♦♦ Mean mark (e) 9%.

Filed Under: Critical Path Analysis Tagged With: Band 4, Band 5, Band 6, smc-621-20-Critical Paths/EST, smc-621-30-Float time/LST, smc-621-40-Crashing/Reduce completion time

Networks, GEN1 2022 VCAA 6 MC

A landscaping project has 12 activities. The network below gives the time, in hours, that it takes to complete each activity.
 

The earliest start time, in hours, for activity \(G\) is

  1. 10
  2. 11
  3. 12
  4. 13
  5. 14
Show Answers Only

\(C\)

Show Worked Solution
\(\text{Critical Path}\) \(= CEGIL\ \ \text{(see diagram above)}\)  
\(\text{EST for}\ G\) \(= 5 + 7 = 12\ \text{hours}\)  

 
\(\Rightarrow C\)


♦ Mean mark 48%.

Filed Under: Critical Path Analysis Tagged With: Band 5, smc-621-20-Critical Paths/EST

NETWORKS, FUR2 2021 VCAA 4

Roadworks planned by the local council require 13 activities to be completed.

The network below shows these 13 activities and their completion times in weeks.
 

  1. What is the earliest start time, in weeks, of activity `K`?   (1 mark)

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

  2. How many of these activities have zero float time?   (1 mark)

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

  3. It is possible to reduce the completion time for activities `A, E, F, L` and `K`.
  4. The reduction in completion time for each of these five activities will incur an additional cost.
  5. The table below shows the five activities that can have their completion time reduced and the associated weekly cost, in dollars.
     
       
  6. The completion time for each these five activities can be reduced by a maximum of two weeks.
  7. The overall completion time for the roadworks can be reduced to 16 weeks.
  8. What is the minimum cost, in dollars, of this change in completion time?   (1 mark)

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

Show Answers Only
  1. `14 \ text{weeks}`
  2. `7`
  3. `$ 380 \ 000`
Show Worked Solution

a.    `text{Scan forwards:}`

`EST \ (text{activity} \ K)`

`= A \ E \ J`

`= 6 + 5 + 3`

`= 14 \ text{weeks}`
 

b.     `text{Scan backwards:}`

`text{Critical paths:}\ ADGLM\ text(and)\ AEHLM`

`:. \ text{7 activities have no float time:} \ ADEGHLM`
 

c.    `text{There are 9 possible paths}`

`ADGLM\ (19), AEHLM\ (19) , AEIM\ (14)`

`AEJK\ (17), BCEIM \ (13), BCEJK\ (16)`

`BCEHLM\ (18), BCDGLM \ (18), BFK (11)`

`text{Consider the 5 paths with completions over 16 weeks}`

` to \ text{all contain}\ A\ text{or}\ L \ text{or both.}`
  

`text{Consider} \ ADGLM, AEHLM \ (text{contains both} \ A\ text{and} \ L )`

`to \ text{reduce} \ L xx 2 \ ,  A xx 1 \ text{to reach 16 weeks}`

`to \ text{cheaper than} \ L xx 1 \ , \ A xx 2`
  

`text{Consider} \ BCEHLM , BCDGLM\ (text{both contain} \ L \ text{only})`

`to \ text{reduce} \ L xx 2 \ text{to reach 16 weeks}`
 

`text{Consider} \ AEJK \ ( text{contains} \ A \ text{only} )`

`to \ text{reduce} \ A xx 1 \ text{reach 16 weeks}.`
 

`:. \ text{Minimum cost to reduce time to 16 weeks}`

`= 2 xx 120\ 000 + 1 xx  140\ 000`

`= $ 380\ 000`

Filed Under: Critical Path Analysis Tagged With: Band 5, Band 6, smc-621-20-Critical Paths/EST, smc-621-40-Crashing/Reduce completion time

NETWORKS, FUR1 2021 VCAA 6 MC

The directed graph below shows the sequence of activities required to complete a project.

The time taken to complete each activity, in hours, is also shown.
 

The minimum completion time for this project is 18 hours.

The time taken to complete activity `E` is labelled  `x`.

The maximum value of  `x`  is

  1. 2
  2. 3
  3. 4
  4. 5
  5. 6
Show Answers Only

`B`

Show Worked Solution

`text{Consider all possible network paths:}`

♦ Mean mark 46%.

`ADHJ \ -> \ text{18 hours}`

`BEJ \ -> \ (12 + x) \ text{hours}`

`CGFEJ \ -> \  (15 + x) \ text{hours}`

`CGIJ \ -> \ text{18 hours}`
 

`text{Minimum completion time = 18 hours}`

`:. \ text{No path can be longer than 18 hours}`

`:. \ x_text{max} = 3 \ text{hours}`
 

`=> B`

Filed Under: Critical Path Analysis Tagged With: Band 5, smc-621-20-Critical Paths/EST

NETWORKS, FUR1 2020 VCAA 10 MC

The directed network below shows the sequence of activities, `A` to `I`, that is required to complete an office renovation.

The time taken to complete each activity, in weeks, is also shown.
 


 

The project manager would like to complete the office renovation in less time.

The project manager asks all the workers assigned to activity `H` to also work on activity `F`.

This will reduce the completion time of activity `F` to three weeks.

The workers assigned to activity `H` cannot work on both activity `H` and activity `F` at the same time.

No other activity times will be changed.

This change to the network will result in a change to the completion time of the office renovation.

Which one of the following is correct?

  1. The completion time will be reduced by one week if activity `F` is completed before activity `H` is started.
  2. The completion time will be reduced by three weeks if activity `F` is completed before activity `H` is started.
  3. The completion time will be reduced by one week if activity `H` is completed before activity `F` is started.
  4. The completion time will be reduced by three weeks if activity `H` is completed before activity `F` is started.
  5. The completion time will be increased by three weeks if activity `H` is completed before activity `F` is started.
Show Answers Only

`A`

Show Worked Solution

`text{Original forward scan (note}\ F\ text{is 6 origionally but}`

♦♦♦ Mean mark 24%.

`text{is reduced to 3 for the adjusted critical path):}`
 


 

`text(Original critical path is:)\ ACEFGI = 2+5+3+6+4+5=25`

`text(If activity)\ F\ text(is completed in 3 weeks, and then)`

`text(activity)\ H\ text(starts, new critical path is:)`

`ACEF\ text{(dummy)}\ HI = 24\ text(weeks)`

`=>  A`

Filed Under: Critical Path Analysis Tagged With: Band 6, smc-621-20-Critical Paths/EST, smc-621-40-Crashing/Reduce completion time

NETWORKS, FUR1 2020 VCAA 6 MC

The activity network below shows the sequence of activities required to complete a project.

The number next to each activity in the network is the time it takes to complete that activity, in days.

 

The minimum completion time for this project, in days, is

  1. 18
  2. 19
  3. 20
  4. 21
  5. 22
Show Answers Only

`E`

Show Worked Solution

`text(Scanning forward:)`
 

`text(Critical path)` `= BEIJM`
  `= 7 + 4 + 3 + 3 + 5`
  `= 22\ text(days)`

 
`=>  E`

Filed Under: Critical Path Analysis Tagged With: Band 4, smc-621-20-Critical Paths/EST

NETWORKS, FUR1 2018 VCAA 7 MC

A project requires nine activities (A–I) to be completed. The duration, in hours, and the immediate predecessor(s) of each activity are shown in the table below.

The minimum completion time for this project, in hours, is

  1. 14
  2. 19
  3. 20
  4. 24
  5. 35
Show Answers Only

`C`

Show Worked Solution

`text(Sketch the network:)`
 

 
`text(Completion time of paths:)`

♦ Mean mark 49%.

`ABEGI = 4 + 3 + 5 + 4 + 3 = 19\ text(hours)`

`ACFGI = 4 + 7 + 2 + 4 + 3 = 20\ text(hours)`

`ADHI = 4 + 2 + 5 + 3 = 14\ text(hours)`
 

`:.\ text{Minimum completion time (critical path) = 20 hours}`

`=> C`

Filed Under: Critical Path Analysis Tagged With: Band 5, smc-621-10-Network table, smc-621-20-Critical Paths/EST

NETWORKS, FUR1 2018 VCAA 5 MC

The directed network below shows the sequence of 11 activities that are needed to complete a project.

The time, in weeks, that it takes to complete each activity is also shown.
 


 

How many of these activities could be delayed without affecting the minimum completion time of the project?

  1. 3
  2. 4
  3. 5
  4. 6
  5. 7
Show Answers Only

`B`

Show Worked Solution

`text(Activities not on critical path can be delayed.)`

`text{Two critical paths exist (15 weeks)}`

♦ Mean mark 43%.

`ADHK\ text(and)\ BFJK`

`:.\ text(Activity)\ C, E, G\ text(and)\ I\ text(could be delayed)`

`=> B`

Filed Under: Critical Path Analysis Tagged With: Band 5, smc-621-20-Critical Paths/EST

NETWORKS, FUR2 2018 VCAA 3

At the Zenith Post Office all computer systems are to be upgraded.

This project involves 10 activities, `A` to `J`.

The directed network below shows these activities and their completion times, in hours.
 

  1. Determine the earliest starting time, in hours, for activity `I`.   (1 mark)

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

  2. The minimum completion time for the project is 15 hours.

     

    Write down the critical path.   (1 mark)

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

  3. Two of the activities have a float time of two hours.

     

    Write down these two activities.   (1 mark)

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

  4. For the next upgrade, the same project will be repeated but one extra activity will be added.
    This activity has a duration of one hour, an earliest starting time of five hours and a latest starting time of 12 hours.

     

    Complete the following sentence by filling in the boxes provided.   (1 mark)

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

     

    The extra activity could be represented on the network above by a directed edge from the

   end of activity   
 
  to the start of activity   
 
Show Answers Only
  1. `10\ text(hours)`
  2. `B-E-G-H-J`
  3.  `text(Activity)\ A\ text(and)\ C`
  4. `text(end of activity)\ E\ text(to the start of activity)\ J`
Show Worked Solution

a.  `text(Longest path to)\ I:`

`B -> E -> G`

`:.\ text(EST for)\ \ I` `= 2 + 3 + 5`
  `= 10\ text(hours)`

 
b.
  `B-E-G-H-J`
 

c.  `text(Scanning forwards and backwards:)`

♦ Mean mark 45%.

 


 

`:.\ text(Activity)\ A\ text(and)\ C\ text(have a 2 hour float time.)`
 

d.   `text(end of activity)\ E\ text(to the start of activity)\ J`

♦♦ Mean mark 25%.
 

`text(By inspection of forward and backward scanning:)`

`text(EST of 5 hours is possible after activity)\ E.`

`text(LST of 12 hours after activity)\ E -> text(edge has weight)`

`text(of 1 and connects to)\ J`

Filed Under: Critical Path Analysis Tagged With: Band 3, Band 4, Band 5, smc-621-20-Critical Paths/EST, smc-621-30-Float time/LST, smc-621-45-Adding activities

NETWORKS, FUR1 2017 VCAA 4-5 MC

The directed graph below shows the sequence of activities required to complete a project.

The time to complete each activity, in hours, is also shown.
 


 

Part 1

The earliest starting time, in hours, for activity `N` is

  1. 3
  2. 10
  3. 11
  4. 12
  5. 13

 

Part 2

To complete the project in minimum time, some activities cannot be delayed.

The number of activities that cannot be delayed is

  1. 2
  2. 3
  3. 4
  4. 5
  5. 6
Show Answers Only

`text(Part 1:)\ D`

`text(Part 1:)\ C`

Show Worked Solution

`text(Part 1)`

`text(EST of)\ N:\ CGJ`

`:.\ text(EST of)\ N` `= 4 + 3 +5`
  `= 12\ text(hours)`

`=> D`

 

`text(Part 2)`

`text(Critical path is:)\ CFHM`

`:. 4\ text(activities can’t be delayed.)`

`=> C`

Filed Under: Critical Path Analysis Tagged With: Band 4, Band 5, smc-621-20-Critical Paths/EST

NETWORKS, FUR2 2016 VCAA 3

A new skateboard park is to be built in Beachton.

This project involves 13 activities, `A` to `M`.

The directed network below shows these activities and their completion times in days.
 


 

  1. Determine the earliest start time for activity `M`.   (1 mark)

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

  2. The minimum completion time for the skateboard park is 15 days.

     

    Write down the critical path for this project.   (1 mark)

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

  3. Which activity has a float time of two days?   (1 mark)

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

  4. The completion times for activities `E, F, G, I` and `J` can each be reduced by one day.

     

    The cost of reducing the completion time by one day for these activities is shown in the table below.
     

     

       

     

    What is the minimum cost to complete the project in the shortest time possible?   (1 mark)

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

  5. The original skateboard park project from part (a), before the reduction of time in any activity, will be repeated at another town named Campville, but with the addition of one extra activity.

     

    The new activity, `N`, will take six days to complete and has a float time of one day.

     

    Activity `N` will finish at the same time as the project.

     

     i.  Add activity `N` to the network below.   (1 mark) 

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


      
    ii.  What is the latest start time for activity `N`?   (1 mark)

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

Show Answers Only
  1. `11\ text(days)`
  2. `AEIK`
  3. `text(Activity)\ H`
  4. `text(Minimum cost of $2000 when activity)\ I\ text(is reduced by 1 day.)`

    1. `9\ text(days from the start)`
Show Worked Solution
a.    `text(EST)` `= 1 + 4 + 6`
    `= 11\ text(days)`

  
b.
  `text(Critical Path:)\ AEIK`

♦♦ Mean mark part (c) 37%, part (d) 21%.
MARKER’S COMMENT: In part (d), `ADK` cannot be crashed, therefore shortest duration is 14 days. Activity `I` is cheapest to reduce.
  

c.   `text(Activity)\ H`
  

d.   `text(Minimum days to complete is 14 days by reducing)`

`text(either)\ E\ text(or)\ I\ text(by 1 day.)`

`:. text(Minimum cost of $2000 when activity)\ I\ text(is reduced)`

`text(by 1 day.)`
  

e.i.   

♦♦ Mean mark part (e)(i) 21%, (e)(ii) 27%.
MARKER’S COMMENT: In (e)(ii), activity `N` must have an arrow on it.
  

e.ii.    `text(LST)` `=\ text(critical path time − 6 days)`
    `= 15-6`
    `= 9\ text(days from the start.)`

Filed Under: Critical Path Analysis Tagged With: Band 4, Band 5, Band 6, smc-621-20-Critical Paths/EST, smc-621-30-Float time/LST, smc-621-40-Crashing/Reduce completion time, smc-621-45-Adding activities

NETWORKS, FUR1 2016 VCAA 6-7 MC

The directed graph below shows the sequence of activities required to complete a project.

All times are in hours.
 

 
Part 1

The number of activities that have exactly two immediate predecessors is

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4

 
Part 2

There is one critical path for this project.

Three critical paths would exist if the duration of activity

  1. I were reduced by two hours.
  2. E were reduced by one hour.
  3. G were increased by six hours.
  4. K were increased by two hours.
  5. F were increased by two hours. 
Show Answers Only

`text(Part 1:)\ C`

`text(Part 1:)\ B`

Show Worked Solution

`text(Part 1)`

`I\ text(and)\ J`

`=> C`

 

`text(Part 2)`

♦ Mean mark of Part 2: 45%.

`text(The possible paths are:)`

`ADIL – 19\ text(mins)`

`BEIL – 20\ text(mins)`

`BFJL – 17\ text(mins)`

`CGJL – 13\ text(mins)`

`CHKL – 19\ text(mins)`

`text(If)\ E\ text(were reduced by 1 hour,)\ BEIL\ text(will)`

`text{take 19 minutes (i.e. 3 critical paths of 19}`

`text{minutes would exist).}`

`=> B`

Filed Under: Critical Path Analysis Tagged With: Band 4, Band 5, smc-621-20-Critical Paths/EST

NETWORKS, FUR1 2008 VCAA 8-9 MC

The network below shows the activities that are needed to finish a particular project and their completion times (in days).

 

networks-fur1-2008-vcaa-8-mc

Part 1

The earliest start time for Activity `K`, in days, is

A.     7

B.   15

C.   16

D.   19

E.   20

 

Part 2

This project currently has one critical path.

A second critical path, in addition to the first, would be created by

A.   increasing the completion time of `D` by 7 days.

B.   increasing the completion time of `G` by 1 day.

C.   increasing the completion time of `I` by 2 days.

D.   decreasing the completion time of `C` by 1 day.

E.   decreasing the completion time of `H` by 2 days.

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ A`

Show Worked Solution

`text(Part 1)`

`text(EST for Activity)\ K`

`=\ text(Duration)\ ACFI`

`= 2 + 5 + 6 + 3`

`= 16`

`=> C`

 

`text(Part 2)`

♦♦ Mean mark of Part 2 was 35%.

`text(Original critical path is)`

`ACFHJL\ text{(22 days)}`

`text(Consider option)\ A,`

`text(New critical path is created)`

`ABDJL\ text{(22 days)}`

`=> A`

Filed Under: Critical Path Analysis Tagged With: Band 4, Band 5, smc-621-20-Critical Paths/EST

NETWORKS, FUR1 2010 VCAA 8 MC

A project has 12 activities. The network below gives the time (in hours) that it takes to complete each activity.
 


 

The critical path for this project is

A.   `ADGK`

B.   `ADGIL`

C.   `BHJL`

D.   `CEGIL`

E.   `CEHJL`

Show Answers Only

`D`

Show Worked Solution

`text(Critical path is)\ \ CEGIL`

♦ Mean mark 41%.

`=>  D`

 

Filed Under: Critical Path Analysis Tagged With: Band 6, smc-621-20-Critical Paths/EST, smc-621-50-Dummy activities

NETWORKS, FUR1 2006 VCAA 9 MC

The network below shows the activities and their completion times (in hours) that are needed to complete a project.
 


 

The project is to be crashed by reducing the completion time of one activity only.

This will reduce the completion time of the project by a maximum of

A.   1 hour 

B.   2 hours

C.   3 hours

D.   4 hours

E.   5 hours

Show Answers Only

`D`

Show Worked Solution

`text(The critical path is)\ BDCEHJ\ text{(19 hours)}`

♦♦♦ Mean mark 17%.
MARKER’S COMMENT: When choosing an activity to crash, take care that a new critical path is not created.

 
`text(Also,)\ ACEHJ\ text{(15 hours)}`

`:.\ text(Activity)\ B\ text(could be crashed by 4 hours)`

`text(without a new critical path emerging.)`

`rArr D`

Filed Under: Critical Path Analysis Tagged With: Band 6, smc-621-20-Critical Paths/EST, smc-621-50-Dummy activities

NETWORKS, FUR1 2007 VCAA 5-6 MC

The following network shows the activities that are needed to complete a project and their completion times (in hours).
 


 

Part 1

Which one of the following statements regarding this project is false?

A.   Activities `A, B` and `C` all have the same earliest start time.

B.   There is only one critical path for this project.

C.   Activity `J` may start later than activity `H.`

D.   The shortest path gives the minimum time for project completion.

E.   Activity `L` must be on the critical path.

 

Part 2

The earliest start time for activity `L`, in hours, is

A.   11

B.   12

C.   14

D.   15

E.   16

Show Answers Only

`text (Part 1:)\ D`

`text (Part 2:)\ E`

Show Worked Solution

`text (Part 1)`

♦ Mean mark 43%.

`A, B, C,\ text(and)\ E\ text(can be shown to be true.)`

`=>  D`

 

`text (Part 2)`

`text(Critical path is)\ CDFKL`

`:.\ text(EST for)\ L` `= 5 + 0 + 4 + 7`
  `= 16\ text(hours)`

  
`=>  E`

Filed Under: Critical Path Analysis Tagged With: Band 5, smc-621-20-Critical Paths/EST, smc-621-50-Dummy activities

NETWORKS, FUR1 2011 VCAA 8 MC

The diagram shows the tasks that must be completed in a project.

Also shown are the completion times, in minutes, for each task.
 

 
The critical path for this project includes activities

A.   `B and I.`

B.   `C and H.`

C.   `D and E.`

D.   `F and K.`

E.   `G and J.`

Show Answers Only

`D`

Show Worked Solution

`text(The critical path is)\ \ ACFIK.`

`=>  D`

Filed Under: Critical Path Analysis Tagged With: Band 4, smc-621-20-Critical Paths/EST, smc-621-50-Dummy activities

NETWORKS, FUR1 2012 VCAA 9 MC

John, Ken and Lisa must work together to complete eight activities, `A, B, C, D, E, F, G` and `H`, in minimum time.

The directed network below shows the activities, their completion times in days, and the order in which they must be completed.

net9

Several activities need special skills. Each of these activities may be completed only by a specified person.

  • Activities `A` and `F` may only be completed by John.
  • Activities `B` and `C` may only be completed by Ken.
  • Activities `D` and `E` may only be completed by Lisa.
  • Activities `G` and `H` may be completed by any one of John, Ken or Lisa.

With these conditions, the minimum number of days required to complete these eight activities is

A.  14

B.  17

C.  20

D.  21

E.  24

Show Answers Only

`D`

Show Worked Solution

`text(The minimum days require the following path:)`

♦♦♦ Mean mark 18%.

`text(John completes)\ A.`

`text(Lisa completes)\ D\ text(then)\ E\ text{(restriction)}.`

`H\ text(is completed.)`

`text(All other activities can be completed)`

`text(during this time.)`

`text(Minimum days)` `= 3 + 7 + 6 + 5`
  `= 21`

`rArr D`

Filed Under: Critical Path Analysis Tagged With: Band 6, smc-621-20-Critical Paths/EST

NETWORKS, FUR2 2007 VCAA 4

A community centre is to be built on the new housing estate.

Nine activities have been identified for this building project.

The directed network below shows the activities and their completion times in weeks.
 

NETWORKS, FUR2 2007 VCAA 4
 

  1. Determine the minimum time, in weeks, to complete this project.   (1 mark)

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

  2. Determine the slack time, in weeks, for activity `D`.   (2 marks)

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

The builders of the community centre are able to speed up the project.

Some of the activities can be reduced in time at an additional cost.

The activities that can be reduced in time are `A`, `C`, `E`, `F` and `G`.

  1. Which of these activities, if reduced in time individually, would not result in an earlier completion of the project?   (1 mark)

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

The owner of the estate is prepared to pay the additional cost to achieve early completion.

The cost of reducing the time of each activity is $5000 per week.

The maximum reduction in time for each one of the five activities, `A`, `C`, `E`, `F`, `G`, is 2 weeks.

  1. Determine the minimum time, in weeks, for the project to be completed now that certain activities can be reduced in time.   (1 mark)

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

  2. Determine the minimum additional cost of completing the project in this reduced time.   (1 mark)

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

Show Answers Only
  1. `19\ text(weeks)`
  2. `5\ text(weeks)`
  3. `A, E, G`
  4. `text(15 weeks)`
  5. `$25\ 000`
Show Worked Solution

a.   `B-C-F-H-I\ \ text(is the critical path.)`

♦ Mean mark of all parts (combined) 40%.
`:.\ text(Minimum time)` `= 4 + 3 + 4 + 2 + 6`
  `= 19\ text(weeks)`

 

b.    `text(EST of)\ D` `= 4`
  `text(LST of)\ D` `= 9`
`:.\ text(Slack time of)\ D` `= 9-4`
  `= 5\ text(weeks)`

  
c.
   `A, E,\ text(and)\ G\ text(are not currently on)`

`text(the critical path, therefore crashing)`

`text(them will not affect the completion)`

`text(time.)`
  

d.   `text(Reduce)\ C\ text(and)\ F\ text(by 2 weeks.)`

`text(However, a new critical path)`

`B-E-H-I\ text(takes 16 weeks.)`

`:.\ text(Also reduce)\ E\ text(by 1 week.)`

`:.\ text(Minimum time = 5 weeks)`

 

e.    `text(Additional cost)` `= 5 xx $5000`
    `= $25\ 000`

Filed Under: Critical Path Analysis Tagged With: Band 4, Band 5, Band 6, smc-621-20-Critical Paths/EST, smc-621-30-Float time/LST, smc-621-40-Crashing/Reduce completion time

NETWORKS, FUR1 2014 VCAA 8 MC

Which one of the following statements about critical paths is true?

  1. There can be only one critical path in a project.
  2. A critical path always includes at least two activities.
  3. A critical path will always include the activity that takes the longest time to complete.
  4. Reducing the time of any activity on a critical path for a project will always reduce the minimum completion time for the project.
  5. If there are no other changes, increasing the time of any activity on a critical path will always increase the completion time of a project.
Show Answers Only

`E`

Show Worked Solution

`=>  E` 

♦ Mean mark 40%.

 

Filed Under: Critical Path Analysis Tagged With: Band 5, smc-621-20-Critical Paths/EST

NETWORKS, FUR2 2009 VCAA 4

A walkway is to be built across the lake.

Eleven activities must be completed for this building project.

The directed network below shows the activities and their completion times in weeks.
 

NETWORKS, FUR2 2009 VCAA 4
 

  1. What is the earliest start time for activity `E`?   (1 mark)

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

  2. Write down the critical path for this project.   (1 mark)

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

  3. The project supervisor correctly writes down the float time for each activity that can be delayed and makes a list of these times.

     

    Determine the longest float time, in weeks, on the supervisor’s list.   (1 mark)

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

A twelfth activity, `L`, with duration three weeks, is to be added without altering the critical path.

Activity `L` has an earliest start time of four weeks and a latest start time of five weeks.

 

NETWORKS, FUR2 2009 VCAA 4

  1. Draw in activity `L` on the network diagram above.   (1 mark)

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

  2. Activity `L` starts, but then takes four weeks longer than originally planned.

     

    Determine the total overall time, in weeks, for the completion of this building project.   (1 mark)

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

Show Answers Only
  1. `7`
  2. `BDFGIK`
  3. `H\ text(or)\ J\ text(can be delayed for)`
    `text(a maximum of 3 weeks.)`
  4.  
    NETWORKS, FUR2 2009 VCAA 4 Answer
  5. `text(25 weeks)`
Show Worked Solution

a.   `7\ text(weeks)`

♦ Mean mark of all parts (combined): 44%.
  

b.   `BDFGIK`
  

c.   `H\ text(or)\ J\ text(can be delayed for a maximum)`

`text(of 3 weeks.)`
  

d.    NETWORKS, FUR2 2009 VCAA 4 Answer

  
e.
  `text(The new critical path is)\ BLEGIK.`

`L\ text(now takes 7 weeks.)`

`:.\ text(Time for completion)`

`= 4 + 7 + 1 + 5 + 2 + 6`

`= 25\ text(weeks)`

Filed Under: Critical Path Analysis Tagged With: Band 4, Band 5, Band 6, smc-621-20-Critical Paths/EST, smc-621-40-Crashing/Reduce completion time, smc-621-45-Adding activities

NETWORKS, FUR2 2010 VCAA 4

In the final challenge, each of four teams has to complete a construction project that involves activities `A` to `I`.
 

NETWORKS, FUR2 2010 VCAA 4
 

Table 1 shows the earliest start time (EST), latest start time (LST) and duration, in minutes, for each activity. 

The immediate predecessor is also shown. The earliest start time for activity `F` is missing.

  1. What is the least number of activities that must be completed before activity `F` can commence?   (1 mark)

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

  2. What is the earliest start time for activity `F`?   (1 mark)

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

  3. Write down all the activities that must be completed before activity `G` can commence.   (1 mark)

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

  4. What is the float time, in minutes, for activity `G`?   (1 mark)

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

  5. What is the shortest time, in minutes, in which this construction project can be completed?   (1 mark)

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

  6. Write down the critical path for this network.   (1 mark)

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

Show Answers Only
  1. `2`
  2. `9\ text(minutes)`
  3. `A\ text(and)\ C`
  4. `4\ text(minutes)`
  5. `16\ text(minutes)`
  6. `A-B-D-H`
Show Worked Solution

a.   `2`
  

b.    `text(EST for)\ F` `= 5 + 4`
    `= 9\ text(minutes)`

  
c.
   `A\ text(and)\ C`
  

d.    `text(Float time for)\ G` `= 13-9`
    `= 4\ text(minutes)`

 

e.   `text(Shortest construction time)`

`= 5 + 6 + 2 + 3`

`= 16\ text(minutes)`

 

f.   `A-B-D-H`

Filed Under: Critical Path Analysis Tagged With: Band 4, Band 5, smc-621-10-Network table, smc-621-20-Critical Paths/EST

NETWORKS, FUR2 2011 VCAA 3

A section of the Farnham showgrounds has flooded due to a broken water pipe. The public will be stopped from entering the flooded area until repairs are made and the area has been cleaned up.

The table below shows the nine activities that need to be completed in order to repair the water pipe. Also shown are some of the durations, Earliest Start Times (EST) and the immediate predecessors for the activities.
 

NETWORKS, FUR2 2011 VCAA 3 

  1. What is the duration of activity `B`?   (1 mark)

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

  2. What is the Earliest Start Time (EST) for activity `D`?   (1 mark)

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

  3. Once the water has been turned off (Activity `B`), which of the activities `C` to `I` could be delayed without affecting the shortest time to complete all activities?   (1 mark)

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

It is more complicated to replace the broken water pipe (Activity `E`) than expected. It will now take four hours to complete instead of two hours.

  1. Determine the shortest time in which activities `A` to `I` can now be completed.   (1 mark)

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

Turning on the water to the showgrounds (Activity `H`) will also take more time than originally expected. It will now take five hours to complete instead of one hour.

  1. With the increased duration of Activity `H` and Activity `E`, determine the shortest time in which activities `A` to `I` can be completed.   (1 mark)

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

Show Answers Only
  1. `text(2 hours)`
  2. `text(3 hours)`
  3. `text(Activities)\ F and H`
  4. `13\ text(hours)`
  5. `14\ text(hours)`
Show Worked Solution
a.    `text(Duration of)\ B` `= text(EST of)\ C`
    `= 2\ text(hours)`
♦ Mean mark of all parts (combined) was 42%.

  
b.
  `text(EST of)\ C = 3\ text(hours)`

  
c.
   `text(Activities)\ F and H`

MARKER’S COMMENT: Many students incorrectly included `G` in this answer (note that `G` is not on the critical path).

  
d.
  `text(Shortest time)\ (A\ text(to)\ I)`

`= 2 + 1 + 1 + 4 + 4 + 1`

`= 13\ text(hours)`
  

e.   `text(New shortest time)`

`= 2 + 1 + 1 + 4 + 5 + 1`

`= 14\ text(hours)`

Filed Under: Critical Path Analysis Tagged With: Band 4, Band 5, smc-621-10-Network table, smc-621-20-Critical Paths/EST, smc-621-40-Crashing/Reduce completion time

NETWORKS, FUR2 2012 VCAA 2

Thirteen activities must be completed before the produce grown on a farm can be harvested. 

The directed network below shows these activities and their completion times in days.

 

NETWORKS, FUR2 2012 VCAA 2
  

  1. Determine the earliest starting time, in days, for activity `E`.   (1 mark)

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

  2. A dummy activity starts at the end of activity `B`.

     

    Explain why this dummy activity is used on the network diagram.   (1 mark)

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

  3. Determine the earliest starting time, in days, for activity `H`.   (1 mark)

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

  4. In order, list the activities on the critical path.   (1 mark)

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

  5. Determine the latest starting time, in days, for activity `J`.   (1 mark)

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

Show Answers Only
  1. `12\ text(days)`
  2. `F\ text(has)\ B\ text(as a predecessor while)\ G\ text(and)\ H`
    `text(have)\ B\ text(and)\ C\ text(as predecessors.)`
    `text(S)text(ince there cannot be 2 activities called)\ B,`
    `text{a dummy activity is drawn as an extension of}`
    `B\ text(to show that it is also a predecessor of)\ G\ text(and)`
    `H\ text{(with zero time).}`
  3. `15\ text(days)`
  4. `A-B-H-I-L-M`
  5. `25\ text(days)`
Show Worked Solution
a.    `text(EST of)\ E` `= 10 + 2`
    `= 12\ text(days)`
♦ Mean mark of all parts (combined) 47%.

 

b.   `F\ text(has)\ B\ text(as a predecessor while)\ G\ text(and)\ H`

`text(have)\ B\ text(and)\ C\ text(as predecessors.)`

`text(S)text(ince there cannot be 2 activities called)\ B,`

`text{a dummy activity is drawn as an extension of}`

`B\ text(to show that it is also a predecessor of)\ G\ text(and)`

`H\ text{(with zero time).}`

 

♦♦ Exact data unavailable but “few students” were able to correctly deal with the dummy activity in this question.
c.    `text(EST of)\ H` `= 10 + 5`
    `= 15\ text(days)`

 

d.   `text(The critical path is)`

`A-B-H-I-L-M`

 

e.   `text(The shortest time to complete all the activities)`

MARKER’S COMMENT: A correct calculation based on an incorrect critical path in part (d) gained a consequential mark here. Show your working!

`= 10 + 5 + 4 + 3  + 4 + 2`

`= 28\ text(days)`

 

`:.\ text(LST of)\ J` `= 28-3`
  `= 25\ text(days)`

Filed Under: Critical Path Analysis Tagged With: Band 3, Band 4, Band 5, smc-621-20-Critical Paths/EST, smc-621-30-Float time/LST, smc-621-50-Dummy activities

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