Further Maths Critical path analysis Practice Questions

Free Further Maths Critical path analysis practice questions with full step-by-step worked solutions. Covers critical-path-analysis, activity-network, forward-pass, total-float. Practise exam-style problems and check your method.

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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.
Show worked solution

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 predecessorsA12noneB15noneC19A, BD10AE7noneF17B, D\begin{array}{c|c|c} \text{Activity} & \text{Duration (days)} & \text{Immediate predecessors} \\ \hline A & 12 & \text{none} \\ B & 15 & \text{none} \\ C & 19 & A,\,B \\ D & 10 & A \\ E & 7 & \text{none} \\ F & 17 & 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. The latest finish time of an activity is equal to the total duration of the project if that activity is an immediate predecessor of no activity, and is otherwise the smallest value of (latest finish time −- duration) taken over the activities that have it as an immediate predecessor. The total float of an activity is defined by (total float) == (latest finish time) −- (earliest start time) −- (duration), and an activity is critical if and only if its total float is zero. Which activity has the greatest total float?
Show worked solution

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+12,  0+15}=15,e(D)=max⁡{0+12}=12,e(E)=0,e(F)=max⁡{0+15,  12+10}=22e(A)=0,\quad e(B)=0,\quad e(C)=\max\left\{0+12,\;0+15\right\}=15,\quad e(D)=\max\left\{0+12\right\}=12,\quad e(E)=0,\quad e(F)=\max\left\{0+15,\;12+10\right\}=22

    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⁡{12,  15,  34,  22,  7,  39}=39T=\max\left\{12,\;15,\;34,\;22,\;7,\;39\right\}=39

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

  3. Backward pass: find the latest finish times of activities FF, EE, DD, CC, BB, AA

    l(F)=39,l(E)=39,l(D)=min⁡{39−17}=22,l(C)=39,l(B)=min⁡{39−19,  39−17}=20,l(A)=min⁡{39−19,  22−10}=12l(F)=39,\quad l(E)=39,\quad l(D)=\min\left\{39-17\right\}=22,\quad l(C)=39,\quad l(B)=\min\left\{39-19,\;39-17\right\}=20,\quad l(A)=\min\left\{39-19,\;22-10\right\}=12

    Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.

  4. Select the activity with the largest total float

    F(E)=32  >  F(B)=5F(E)=32\;>\;F(B)=5

    Activity EE has more spare time than any other activity in the project.

Answer
E
Question 3
4 marksintermediate
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 predecessorsA15noneB12AC14noneD7noneE19C, DF6B, C, DG11B, E\begin{array}{c|c|c} \text{Activity} & \text{Duration (days)} & \text{Immediate predecessors} \\ \hline A & 15 & \text{none} \\ B & 12 & A \\ C & 14 & \text{none} \\ D & 7 & \text{none} \\ E & 19 & C,\,D \\ F & 6 & B,\,C,\,D \\ G & 11 & B,\,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. The latest finish time of an activity is equal to the total duration of the project if that activity is an immediate predecessor of no activity, and is otherwise the smallest value of (latest finish time −- duration) taken over the activities that have it as an immediate predecessor. The total float of an activity is defined by (total float) == (latest finish time) −- (earliest start time) −- (duration), and an activity is critical if and only if its total float is zero. Which activity has the greatest total float?
Show worked solution

Worked solution

  1. Forward pass: find the earliest start times of activities AA, BB

    e(A)=0,e(B)=max⁡{0+15}=15e(A)=0,\quad e(B)=\max\left\{0+15\right\}=15

    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. Forward pass: find the earliest start times of activities CC, DD

    e(C)=0,e(D)=0e(C)=0,\quad e(D)=0

    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.

  3. Forward pass: find the earliest start times of activities EE, FF, GG

    e(E)=max⁡{0+14,  0+7}=14,e(F)=max⁡{15+12,  0+14,  0+7}=27,e(G)=max⁡{15+12,  14+19}=33e(E)=\max\left\{0+14,\;0+7\right\}=14,\quad e(F)=\max\left\{15+12,\;0+14,\;0+7\right\}=27,\quad e(G)=\max\left\{15+12,\;14+19\right\}=33

    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.

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

    T=max⁡{15,  27,  14,  7,  33,  33,  44}=44T=\max\left\{15,\;27,\;14,\;7,\;33,\;33,\;44\right\}=44

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

  5. Backward pass: find the latest finish times of activities GG, FF, EE, DD, CC, BB, AA

    l(G)=44,l(F)=44,l(E)=min⁡{44−11}=33,l(D)=min⁡{33−19,  44−6}=14,l(C)=min⁡{33−19,  44−6}=14,l(B)=min⁡{44−6,  44−11}=33,l(A)=min⁡{33−12}=21l(G)=44,\quad l(F)=44,\quad l(E)=\min\left\{44-11\right\}=33,\quad l(D)=\min\left\{33-19,\;44-6\right\}=14,\quad l(C)=\min\left\{33-19,\;44-6\right\}=14,\quad l(B)=\min\left\{44-6,\;44-11\right\}=33,\quad l(A)=\min\left\{33-12\right\}=21

    Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.

  6. Find the total float of every activity

    F(A)=21−0−15=6,  F(B)=33−15−12=6,  F(C)=14−0−14=0,  F(D)=14−0−7=7,  F(E)=33−14−19=0,  F(F)=44−27−6=11,  F(G)=44−33−11=0F(A)=21-0-15=6,\;F(B)=33-15-12=6,\;F(C)=14-0-14=0,\;F(D)=14-0-7=7,\;F(E)=33-14-19=0,\;F(F)=44-27-6=11,\;F(G)=44-33-11=0

    The total float of an activity is its latest finish time minus its earliest start time minus its duration.

  7. Select the activity with the largest total float

    F(F)=11  >  F(D)=7F(F)=11\;>\;F(D)=7

    Activity FF has more spare time than any other activity in the project.

Answer
F
Question 4
6 markshard
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 predecessorsA6noneB14noneC9A, BD10noneE8DF19EG11A, B, DH13C, FI17F\begin{array}{c|c|c} \text{Activity} & \text{Duration (days)} & \text{Immediate predecessors} \\ \hline A & 6 & \text{none} \\ B & 14 & \text{none} \\ C & 9 & A,\,B \\ D & 10 & \text{none} \\ E & 8 & D \\ F & 19 & E \\ G & 11 & A,\,B,\,D \\ H & 13 & C,\,F \\ I & 17 & F \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. The latest finish time of an activity is equal to the total duration of the project if that activity is an immediate predecessor of no activity, and is otherwise the smallest value of (latest finish time −- duration) taken over the activities that have it as an immediate predecessor. The total float of an activity is defined by (total float) == (latest finish time) −- (earliest start time) −- (duration), and an activity is critical if and only if its total float is zero. Which activity has the greatest total float?
Show worked solution

Worked solution

  1. Forward pass: find the earliest start time of activity AA

    e(A)=0e(A)=0

    Activity AA has no immediate predecessors, so it can start at time 00.

  2. Forward pass: find the earliest start time of activity BB

    e(B)=0e(B)=0

    Activity BB has no immediate predecessors, so it can start at time 00.

  3. Forward pass: find the earliest start time of activity CC

    e(C)=max⁡{0+6,  0+14}=14e(C)=\max\left\{0+6,\;0+14\right\}=14

    Activity CC cannot start until AA, BB have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  4. Forward pass: find the earliest start time of activity DD

    e(D)=0e(D)=0

    Activity DD has no immediate predecessors, so it can start at time 00.

  5. Forward pass: find the earliest start time of activity EE

    e(E)=max⁡{0+10}=10e(E)=\max\left\{0+10\right\}=10

    Activity EE cannot start until DD has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  6. Forward pass: find the earliest start time of activity FF

    e(F)=max⁡{10+8}=18e(F)=\max\left\{10+8\right\}=18

    Activity FF cannot start until EE has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  7. Forward pass: find the earliest start time of activity GG

    e(G)=max⁡{0+6,  0+14,  0+10}=14e(G)=\max\left\{0+6,\;0+14,\;0+10\right\}=14

    Activity GG cannot start until AA, BB, DD have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  8. Forward pass: find the earliest start time of activity HH

    e(H)=max⁡{14+9,  18+19}=37e(H)=\max\left\{14+9,\;18+19\right\}=37

    Activity HH cannot start until CC, FF have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  9. Forward pass: find the earliest start time of activity II

    e(I)=max⁡{18+19}=37e(I)=\max\left\{18+19\right\}=37

    Activity II cannot start until FF has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

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

    T=max⁡{6,  14,  23,  10,  18,  37,  25,  50,  54}=54T=\max\left\{6,\;14,\;23,\;10,\;18,\;37,\;25,\;50,\;54\right\}=54

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

  11. Backward pass: find the latest finish times of activities II, HH, GG, FF, EE, DD, CC, BB, AA

    l(I)=54,l(H)=54,l(G)=54,l(F)=min⁡{54−13,  54−17}=37,l(E)=min⁡{37−19}=18,l(D)=min⁡{18−8,  54−11}=10,l(C)=min⁡{54−13}=41,l(B)=min⁡{41−9,  54−11}=32,l(A)=min⁡{41−9,  54−11}=32l(I)=54,\quad l(H)=54,\quad l(G)=54,\quad l(F)=\min\left\{54-13,\;54-17\right\}=37,\quad l(E)=\min\left\{37-19\right\}=18,\quad l(D)=\min\left\{18-8,\;54-11\right\}=10,\quad l(C)=\min\left\{54-13\right\}=41,\quad l(B)=\min\left\{41-9,\;54-11\right\}=32,\quad l(A)=\min\left\{41-9,\;54-11\right\}=32

    Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.

  12. Find the total float of every activity

    F(A)=32−0−6=26,  F(B)=32−0−14=18,  F(C)=41−14−9=18,  F(D)=10−0−10=0,  F(E)=18−10−8=0,  F(F)=37−18−19=0,  F(G)=54−14−11=29,  F(H)=54−37−13=4,  F(I)=54−37−17=0F(A)=32-0-6=26,\;F(B)=32-0-14=18,\;F(C)=41-14-9=18,\;F(D)=10-0-10=0,\;F(E)=18-10-8=0,\;F(F)=37-18-19=0,\;F(G)=54-14-11=29,\;F(H)=54-37-13=4,\;F(I)=54-37-17=0

    The total float of an activity is its latest finish time minus its earliest start time minus its duration.

  13. Select the activity with the largest total float

    F(G)=29  >  F(A)=26F(G)=29\;>\;F(A)=26

    Activity GG has more spare time than any other activity in the project.

Answer
G
Question 5
9 markschallenging
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 predecessorsA8noneB18AC11BD13CE15DF5BG7DH3CI14GJ17C, FK10E, I\begin{array}{c|c|c} \text{Activity} & \text{Duration (days)} & \text{Immediate predecessors} \\ \hline A & 8 & \text{none} \\ B & 18 & A \\ C & 11 & B \\ D & 13 & C \\ E & 15 & D \\ F & 5 & B \\ G & 7 & D \\ H & 3 & C \\ I & 14 & G \\ J & 17 & C,\,F \\ K & 10 & E,\,I \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. The latest finish time of an activity is equal to the total duration of the project if that activity is an immediate predecessor of no activity, and is otherwise the smallest value of (latest finish time −- duration) taken over the activities that have it as an immediate predecessor. The total float of an activity is defined by (total float) == (latest finish time) −- (earliest start time) −- (duration), and an activity is critical if and only if its total float is zero. Which activity has the greatest total float?
Show worked solution

Worked solution

  1. Forward pass: find the earliest start time of activity AA

    e(A)=0e(A)=0

    Activity AA has no immediate predecessors, so it can start at time 00.

  2. Forward pass: find the earliest start time of activity BB

    e(B)=max⁡{0+8}=8e(B)=\max\left\{0+8\right\}=8

    Activity BB cannot start until AA has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  3. Forward pass: find the earliest start time of activity CC

    e(C)=max⁡{8+18}=26e(C)=\max\left\{8+18\right\}=26

    Activity CC cannot start until BB has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  4. Forward pass: find the earliest start time of activity DD

    e(D)=max⁡{26+11}=37e(D)=\max\left\{26+11\right\}=37

    Activity DD cannot start until CC has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  5. Forward pass: find the earliest start time of activity EE

    e(E)=max⁡{37+13}=50e(E)=\max\left\{37+13\right\}=50

    Activity EE cannot start until DD has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  6. Forward pass: find the earliest start time of activity FF

    e(F)=max⁡{8+18}=26e(F)=\max\left\{8+18\right\}=26

    Activity FF cannot start until BB has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  7. Forward pass: find the earliest start time of activity GG

    e(G)=max⁡{37+13}=50e(G)=\max\left\{37+13\right\}=50

    Activity GG cannot start until DD has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  8. Forward pass: find the earliest start time of activity HH

    e(H)=max⁡{26+11}=37e(H)=\max\left\{26+11\right\}=37

    Activity HH cannot start until CC has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  9. Forward pass: find the earliest start time of activity II

    e(I)=max⁡{50+7}=57e(I)=\max\left\{50+7\right\}=57

    Activity II cannot start until GG has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  10. Forward pass: find the earliest start time of activity JJ

    e(J)=max⁡{26+11,  26+5}=37e(J)=\max\left\{26+11,\;26+5\right\}=37

    Activity JJ cannot start until CC, FF have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

  11. Forward pass: find the earliest start time of activity KK

    e(K)=max⁡{50+15,  57+14}=71e(K)=\max\left\{50+15,\;57+14\right\}=71

    Activity KK cannot start until EE, II have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.

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

    T=max⁡{8,  26,  37,  50,  65,  31,  57,  40,  71,  54,  81}=81T=\max\left\{8,\;26,\;37,\;50,\;65,\;31,\;57,\;40,\;71,\;54,\;81\right\}=81

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

  13. Backward pass: find the latest finish times of activities KK, JJ, II

    l(K)=81,l(J)=81,l(I)=min⁡{81−10}=71l(K)=81,\quad l(J)=81,\quad l(I)=\min\left\{81-10\right\}=71

    Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.

  14. Backward pass: find the latest finish times of activities HH, GG, FF, EE

    l(H)=81,l(G)=min⁡{71−14}=57,l(F)=min⁡{81−17}=64,l(E)=min⁡{81−10}=71l(H)=81,\quad l(G)=\min\left\{71-14\right\}=57,\quad l(F)=\min\left\{81-17\right\}=64,\quad l(E)=\min\left\{81-10\right\}=71

    Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.

  15. Backward pass: find the latest finish times of activities DD, CC, BB, AA

    l(D)=min⁡{71−15,  57−7}=50,l(C)=min⁡{50−13,  81−3,  81−17}=37,l(B)=min⁡{37−11,  64−5}=26,l(A)=min⁡{26−18}=8l(D)=\min\left\{71-15,\;57-7\right\}=50,\quad l(C)=\min\left\{50-13,\;81-3,\;81-17\right\}=37,\quad l(B)=\min\left\{37-11,\;64-5\right\}=26,\quad l(A)=\min\left\{26-18\right\}=8

    Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.

  16. Find the total float of every activity

    F(A)=8−0−8=0,  F(B)=26−8−18=0,  F(C)=37−26−11=0,  F(D)=50−37−13=0,  F(E)=71−50−15=6,  F(F)=64−26−5=33,  F(G)=57−50−7=0,  F(H)=81−37−3=41,  F(I)=71−57−14=0,  F(J)=81−37−17=27,  F(K)=81−71−10=0F(A)=8-0-8=0,\;F(B)=26-8-18=0,\;F(C)=37-26-11=0,\;F(D)=50-37-13=0,\;F(E)=71-50-15=6,\;F(F)=64-26-5=33,\;F(G)=57-50-7=0,\;F(H)=81-37-3=41,\;F(I)=71-57-14=0,\;F(J)=81-37-17=27,\;F(K)=81-71-10=0

    The total float of an activity is its latest finish time minus its earliest start time minus its duration.

  17. Select the activity with the largest total float

    F(H)=41  >  F(F)=33F(H)=41\;>\;F(F)=33

    Activity HH has more spare time than any other activity in the project.

Answer
H

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