Critical path analysis Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Critical path analysis questions. See exactly how to solve problems on critical-path-analysis, activity-network, forward-pass, total-float.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A project is made up of the activities in the following table, which gives the duration of each activity, in days, and the immediate predecessors of each activity. ActivityDuration (days)Immediate predecessorsA10noneB4noneC15noneD3A, CE14A, B, CF19E\begin{array}{c|c|c} \text{Activity} & \text{Duration (days)} & \text{Immediate predecessors} \\ \hline A & 10 & \text{none} \\ B & 4 & \text{none} \\ C & 15 & \text{none} \\ D & 3 & A,\,C \\ E & 14 & A,\,B,\,C \\ F & 19 & E \end{array} The earliest start time of an activity is 00 if it has no immediate predecessors, and is otherwise the largest value of (earliest start time ++ duration) taken over its immediate predecessors. The total duration of the project is the largest value of (earliest start time ++ duration) taken over all of the activities. Find the total duration of the project.

Worked solution

  1. Forward pass: find the earliest start times of activities AA, BB, CC, DD, EE, FF

    e(A)=0,e(B)=0,e(C)=0,e(D)=max⁡{0+10,  0+15}=15,e(E)=max⁡{0+10,  0+4,  0+15}=15,e(F)=max⁡{15+14}=29e(A)=0,\quad e(B)=0,\quad e(C)=0,\quad e(D)=\max\left\{0+10,\;0+15\right\}=15,\quad e(E)=\max\left\{0+10,\;0+4,\;0+15\right\}=15,\quad e(F)=\max\left\{15+14\right\}=29

    Each of these activities starts as soon as all of its immediate predecessors are complete, so its earliest start time is the largest of the earliest finish times of those predecessors.

  2. Find the total duration of the project from the earliest finish times

    T=max⁡{10,  4,  15,  18,  29,  48}=48T=\max\left\{10,\;4,\;15,\;18,\;29,\;48\right\}=48

    The project is finished when the last activity finishes, so the total duration is the largest earliest finish time.

  3. State the total duration of the project

    total duration=48 days\text{total duration}=48\text{ days}

    The project cannot be completed in fewer than 4848 days.

Answer
4848
Question 2
2 markseasy
A project is made up of the activities in the following table, which gives the duration of each activity, in days, and the immediate predecessors of each activity. ActivityDuration (days)Immediate predecessorsA2noneB16noneC12noneD14A, CE15A, CF4B, D, E\begin{array}{c|c|c} \text{Activity} & \text{Duration (days)} & \text{Immediate predecessors} \\ \hline A & 2 & \text{none} \\ B & 16 & \text{none} \\ C & 12 & \text{none} \\ D & 14 & A,\,C \\ E & 15 & A,\,C \\ F & 4 & B,\,D,\,E \end{array} The earliest start time of an activity is 00 if it has no immediate predecessors, and is otherwise the largest value of (earliest start time ++ duration) taken over its immediate predecessors. The total duration of the project is the largest value of (earliest start time ++ duration) taken over all of the activities. Find the total duration of the project.

Worked solution

  1. Forward pass: find the earliest start times of activities AA, BB, CC, DD, EE, FF

    e(A)=0,e(B)=0,e(C)=0,e(D)=max⁡{0+2,  0+12}=12,e(E)=max⁡{0+2,  0+12}=12,e(F)=max⁡{0+16,  12+14,  12+15}=27e(A)=0,\quad e(B)=0,\quad e(C)=0,\quad e(D)=\max\left\{0+2,\;0+12\right\}=12,\quad e(E)=\max\left\{0+2,\;0+12\right\}=12,\quad e(F)=\max\left\{0+16,\;12+14,\;12+15\right\}=27

    Each of these activities starts as soon as all of its immediate predecessors are complete, so its earliest start time is the largest of the earliest finish times of those predecessors.

  2. Find the total duration of the project from the earliest finish times

    T=max⁡{2,  16,  12,  26,  27,  31}=31T=\max\left\{2,\;16,\;12,\;26,\;27,\;31\right\}=31

    The project is finished when the last activity finishes, so the total duration is the largest earliest finish time.

  3. State the total duration of the project

    total duration=31 days\text{total duration}=31\text{ days}

    The project cannot be completed in fewer than 3131 days.

Answer
3131
Question 3
2 markseasy
A project is made up of the activities in the following table, which gives the duration of each activity, in days, and the immediate predecessors of each activity. ActivityDuration (days)Immediate predecessorsA4noneB12noneC10noneD7A, CE6B, DF15D\begin{array}{c|c|c} \text{Activity} & \text{Duration (days)} & \text{Immediate predecessors} \\ \hline A & 4 & \text{none} \\ B & 12 & \text{none} \\ C & 10 & \text{none} \\ D & 7 & A,\,C \\ E & 6 & B,\,D \\ F & 15 & D \end{array} The earliest start time of an activity is 00 if it has no immediate predecessors, and is otherwise the largest value of (earliest start time ++ duration) taken over its immediate predecessors. The total duration of the project is the largest value of (earliest start time ++ duration) taken over all of the activities. Find the total duration of the project.

Worked solution

  1. Forward pass: find the earliest start times of activities AA, BB, CC, DD, EE, FF

    e(A)=0,e(B)=0,e(C)=0,e(D)=max⁡{0+4,  0+10}=10,e(E)=max⁡{0+12,  10+7}=17,e(F)=max⁡{10+7}=17e(A)=0,\quad e(B)=0,\quad e(C)=0,\quad e(D)=\max\left\{0+4,\;0+10\right\}=10,\quad e(E)=\max\left\{0+12,\;10+7\right\}=17,\quad e(F)=\max\left\{10+7\right\}=17

    Each of these activities starts as soon as all of its immediate predecessors are complete, so its earliest start time is the largest of the earliest finish times of those predecessors.

  2. Find the total duration of the project from the earliest finish times

    T=max⁡{4,  12,  10,  17,  23,  32}=32T=\max\left\{4,\;12,\;10,\;17,\;23,\;32\right\}=32

    The project is finished when the last activity finishes, so the total duration is the largest earliest finish time.

  3. State the total duration of the project

    total duration=32 days\text{total duration}=32\text{ days}

    The project cannot be completed in fewer than 3232 days.

Answer
3232
Question 4
2 markseasy
A project is made up of the activities in the following table, which gives the duration of each activity, in days, and the immediate predecessors of each activity. ActivityDuration (days)Immediate predecessorsA18noneB10noneC15BD17noneE8A, CF14A, C\begin{array}{c|c|c} \text{Activity} & \text{Duration (days)} & \text{Immediate predecessors} \\ \hline A & 18 & \text{none} \\ B & 10 & \text{none} \\ C & 15 & B \\ D & 17 & \text{none} \\ E & 8 & A,\,C \\ F & 14 & A,\,C \end{array} The earliest start time of an activity is 00 if it has no immediate predecessors, and is otherwise the largest value of (earliest start time ++ duration) taken over its immediate predecessors. The total duration of the project is the largest value of (earliest start time ++ duration) taken over all of the activities. Find the total duration of the project.

Worked solution

  1. Forward pass: find the earliest start times of activities AA, BB, CC, DD, EE, FF

    e(A)=0,e(B)=0,e(C)=max⁡{0+10}=10,e(D)=0,e(E)=max⁡{0+18,  10+15}=25,e(F)=max⁡{0+18,  10+15}=25e(A)=0,\quad e(B)=0,\quad e(C)=\max\left\{0+10\right\}=10,\quad e(D)=0,\quad e(E)=\max\left\{0+18,\;10+15\right\}=25,\quad e(F)=\max\left\{0+18,\;10+15\right\}=25

    Each of these activities starts as soon as all of its immediate predecessors are complete, so its earliest start time is the largest of the earliest finish times of those predecessors.

  2. Find the total duration of the project from the earliest finish times

    T=max⁡{18,  10,  25,  17,  33,  39}=39T=\max\left\{18,\;10,\;25,\;17,\;33,\;39\right\}=39

    The project is finished when the last activity finishes, so the total duration is the largest earliest finish time.

  3. State the total duration of the project

    total duration=39 days\text{total duration}=39\text{ days}

    The project cannot be completed in fewer than 3939 days.

Answer
3939
Question 5
2 markseasy
A project is made up of the activities in the following table, which gives the duration of each activity, in days, and the immediate predecessors of each activity. ActivityDuration (days)Immediate predecessorsA8noneB4noneC11noneD5A, CE13B, DF2B, D\begin{array}{c|c|c} \text{Activity} & \text{Duration (days)} & \text{Immediate predecessors} \\ \hline A & 8 & \text{none} \\ B & 4 & \text{none} \\ C & 11 & \text{none} \\ D & 5 & A,\,C \\ E & 13 & B,\,D \\ F & 2 & B,\,D \end{array} The earliest start time of an activity is 00 if it has no immediate predecessors, and is otherwise the largest value of (earliest start time ++ duration) taken over its immediate predecessors. The total duration of the project is the largest value of (earliest start time ++ duration) taken over all of the activities. Find the earliest start time of activity DD.

Worked solution

  1. Forward pass: find the earliest start times of activities AA, BB, CC, DD, EE, FF

    e(A)=0,e(B)=0,e(C)=0,e(D)=max⁡{0+8,  0+11}=11,e(E)=max⁡{0+4,  11+5}=16,e(F)=max⁡{0+4,  11+5}=16e(A)=0,\quad e(B)=0,\quad e(C)=0,\quad e(D)=\max\left\{0+8,\;0+11\right\}=11,\quad e(E)=\max\left\{0+4,\;11+5\right\}=16,\quad e(F)=\max\left\{0+4,\;11+5\right\}=16

    Each of these activities starts as soon as all of its immediate predecessors are complete, so its earliest start time is the largest of the earliest finish times of those predecessors.

  2. Find the total duration of the project from the earliest finish times

    T=max⁡{8,  4,  11,  16,  29,  18}=29T=\max\left\{8,\;4,\;11,\;16,\;29,\;18\right\}=29

    The project is finished when the last activity finishes, so the total duration is the largest earliest finish time.

  3. State the earliest start time of activity DD

    e(D)=11e(D)=11

    Activity DD cannot begin until all of AA, CC are complete, and the last of them finishes at time 1111.

Answer
1111

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