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Networks, STD2 N3 EQ-Bank 30

A project involves nine activities, `A` to `I`.

The immediate predecessor(s) of each activity is shown in the table below.

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

A directed network for this project will require a dummy activity.

Sketch the network diagram, clearly identifying the dummy activity.   (3 marks)

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Show Worked Solution

`text(Sketch network diagram:)`
 

`text(The dummy activity needs to be drawn)`

`text(from the end of activity)\ B\ text(to the start)`

`text(of activity)\ E.`

Networks, STD2 N3 EQ-Bank 32

An engineering project requires activities `A` to `G` to be completed, as shown in the table.
 


 

The minimum completion time for the project is 25 days and the critical path includes activities `B, D, E` and `F`. The float for activity `G` is two days and the float for activity `C` is four days.

Find the possible duration for each of the activities `A, C, F` and `G`. Include a network diagram in your answer.   (5 marks)

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`text(Duration of)\ F = 1\ text(day)`

`text(Duration of)\ G = 4\ text(days)`

`text(*Understand why there are many possibilities for the duration of)`

`A and C, text(provided they add up to 20 days and)\ A\ text(is not longer)`

`text{than 7 days (or a new critical path is created)}.`

Show Worked Solution

`text(Sketch the network:)`
 

`text{Critical path information included in the network (where EST = LST)}`

`F\ text(is on critical path)rightarrow text(no float)`

`:.\ text(Duration of)\ F=  25-24 = 1\ text(day)`

 

`text(Duration of)\ G` `=\ text(LST of next activity − EST of)\ G-text(float)`
  `= 25-19-2= 4\ text(days)`

 

`text(The float of)\ C\ text{is 4 days (given)}`

`:.\ text(Duration of)\ C` `=\ text(LST of)\ F-text(EST of)\ C-4`
  `= 24-text(EST of)\ C-4`
  `= 20-text(EST of)\ C`

 

`text(EST of)\ C =\ text(Duration of)\ A\ \ (A\ text(has no prerequisites))`

`=>\ text(Duration of)\ A +\ text(Duration of)\ C = 20\ text(days)`

`:.\ text(Possible durations of)\ A\ text(and)\ C\ text(are:)`

`A = 5\ text(days), C = 15\ text(days)`
 

`text(*Understand why there are many possibilities for the duration of)`

`A and C, text(provided they add up to 20 days and)\ A\ text(is not longer)`

`text{than 7 days (or a new critical path is created)}.`

Networks, STD2 N3 EQ-Bank 31

An engineering project requires activities `A` to `G` to be completed, as shown in the table.
 


 

The minimum completion time for the project is 40 weeks and the critical path includes activities `A, C, E` and `G`. The float for activity `F` is six weeks and the float for activity `D` is 9 weeks.

Find the possible duration for each of the activities `B, D, F` and `G`. Include a network diagram in your answer.   (5 marks)

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`text(Duration of)\ G = 3\ text(weeks)`

`text(Duration of)\ F = 9\ text(weeks)`

`text(There are many possibilities for the duration of)\ B and D`

`text(provided they add up to 28 weeks and)\ B\ text(is not longer)`

`text{than 10 weeks (new critical path)}.`

Show Worked Solution

`text(Sketch network:)`

`text{Critical path added to network (where EST = LST)}`

`G\ text(is on critical path ⇒ no float)`

`:.\ text(Duration of)\ G = 40-37 = 3\ text(weeks)`

 

`text(Duration of)\ F` `=40- text(EST of)\ F-text(float)`
  `= 40-25-6= 9\ text(weeks)`

 

`text(Float of)\ D = 9\ text{weeks (given)}`

`:.\ text(Duration of)\ D` `=\ text(LST of)\ G-\ text(EST of)\ D-9`
  `= 37-\ text(EST of)\ D-9`
  `= 28-\ text(EST of)\ D`

 

`text(EST of)\ D =\ text(Duration of)\ B\ \ \ (B\ text(has no prerequisites))`

`text(Duration of)\ B +\ text(Duration of)\ D = 28\ text(weeks)`

`:.\ text(Possible durations of)\ B\ text(and)\ D\ text(are:)`

`B = 1\ text(week), D = 27\ text(weeks)`

 

`text(*Understand why there are many possibilities for the duration of)`

`B and D, text(provided they add up to 28 weeks and)\ B\ text(is not longer)`

`text{than 10 weeks (or a new critical path is created)}.`

Networks, STD2 N3 2012 FUR2 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.
 


 

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

  2. An activity with zero duration starts at the end of activity `B`.

     

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

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

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

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

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a.    `12\ text(days)`

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 zero)`

`text{duration activity is drawn as an extension of}\ B\ text(to)`

`text(show that it is also a predecessor of)\ G\ text(and)\ H.`

c.    `15\ text(days)`

d.    `ABHILM`

e.    `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 zero)`

`text{duration activity is drawn as an extension of}\ B\ text(to)`

`text(show that it is also a predecessor of)\ G\ text(and)\ H.`
 

♦♦ “Few students” were able to correctly deal with the zero duration activity in part (c).

c.    `text(Scanning forwards:)`
 

  
`text(EST for)\ H = 15\ text(days)`
   

d.   `text(The critical path is)\ \ ABHILM`
 


 

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)` 

Networks, STD2 N3 2006 FUR1 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 optimised by reducing the completion time of one activity only.

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

  1. 1 hour 
  2. 3 hours
  3. 4 hours
  4. 5 hours
Show Answers Only

`C`

Show Worked Solution

`text(Scanning forward:)`
 


 

`text(Critical path:)`

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

`=> BDCEHJ\ text{(19 hours)}`
 

`text(Other routes not through)\ B,`

`ACEHJ\ text{(15 hours),}\ AFJ\ text{(14 hours)}`
 

`:.\ text(Activity)\ B\ text(could be reduced by 4 hours without)`

`text(a new critical path emerging.)`
 

`=> C`