smc-916-50-Dummy Activity

Networks, STD2 N3 SM-Bank 44

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

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

	Activity	Immediate predecessor(s)
	`A`	`-`
	`B`	`A`
	`C`	`A`
	`D`	`B`
	`E`	`B, C`
	`F`	`D`
	`G`	`D`
	`H`	`E, F`
	`I`	`G, H`

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

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

Show Answers Only

`B`

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 19

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 ⇒ 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 18

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`	`=\ text(40 − 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.

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

--- 1 WORK AREA LINES (style=lined) ---
An activity with zero duration starts at the end of activity `B`.

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

--- 1 WORK AREA LINES (style=lined) ---
Determine the earliest starting time, in days, for activity `H`. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
In order, list the activities on the critical path. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Determine the latest starting time, in days, for activity `J`. (1 mark)

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

Show Answers Only

`12\ text(days)`
`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)`
`B\ text(show that it is also a predecessor of)\ G\ text(and)\ H.`
`15\ text(days)`
`ABHILM`
`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

A. 1 hour

B. 3 hours

C. 4 hours

D. 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.)`

`rArr C`