Hard Further Maths Critical path analysis Questions

Challenging, exam-style Further Maths Critical path analysis questions with worked solutions. Stretch yourself on the hardest critical-path-analysis, activity-network, forward-pass, total-float problems.

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Further Maths34 questionsStep-by-step solutions

Question 1

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.

\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

0

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

Forward pass: find the earliest start time of activity $A$
$e(A)=0$
Activity $A$ has no immediate predecessors, so it can start at time $0$ .
Forward pass: find the earliest start time of activity $B$
$e(B)=\max\left\{0+8\right\}=8$
Activity $B$ cannot start until $A$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $C$
$e(C)=\max\left\{8+18\right\}=26$
Activity $C$ cannot start until $B$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $D$
$e(D)=\max\left\{26+11\right\}=37$
Activity $D$ cannot start until $C$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $E$
$e(E)=\max\left\{37+13\right\}=50$
Activity $E$ cannot start until $D$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $F$
$e(F)=\max\left\{8+18\right\}=26$
Activity $F$ cannot start until $B$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $G$
$e(G)=\max\left\{37+13\right\}=50$
Activity $G$ cannot start until $D$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $H$
$e(H)=\max\left\{26+11\right\}=37$
Activity $H$ cannot start until $C$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $I$
$e(I)=\max\left\{50+7\right\}=57$
Activity $I$ cannot start until $G$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $J$
$e(J)=\max\left\{26+11,\;26+5\right\}=37$
Activity $J$ cannot start until $C$ , $F$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $K$
$e(K)=\max\left\{50+15,\;57+14\right\}=71$
Activity $K$ cannot start until $E$ , $I$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Find the total duration of the project from the earliest finish times
$T=\max\left\{8,\;26,\;37,\;50,\;65,\;31,\;57,\;40,\;71,\;54,\;81\right\}=81$
The project is finished when the last activity finishes, so the total duration is the largest earliest finish time.
Backward pass: find the latest finish times of activities $K$ , $J$ , $I$
$l(K)=81,\quad l(J)=81,\quad l(I)=\min\left\{81-10\right\}=71$
Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.
Backward pass: find the latest finish times of activities $H$ , $G$ , $F$ , $E$
$l(H)=81,\quad l(G)=\min\left\{71-14\right\}=57,\quad l(F)=\min\left\{81-17\right\}=64,\quad l(E)=\min\left\{81-10\right\}=71$
Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.
Backward pass: find the latest finish times of activities $D$ , $C$ , $B$ , $A$
$l(D)=\min\left\{71-15,\;57-7\right\}=50,\quad l(C)=\min\left\{50-13,\;81-3,\;81-17\right\}=37,\quad l(B)=\min\left\{37-11,\;64-5\right\}=26,\quad l(A)=\min\left\{26-18\right\}=8$
Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.
Find the total float of every activity
$F(A)=8-0-8=0,\;F(B)=26-8-18=0,\;F(C)=37-26-11=0,\;F(D)=50-37-13=0,\;F(E)=71-50-15=6,\;F(F)=64-26-5=33,\;F(G)=57-50-7=0,\;F(H)=81-37-3=41,\;F(I)=71-57-14=0,\;F(J)=81-37-17=27,\;F(K)=81-71-10=0$
The total float of an activity is its latest finish time minus its earliest start time minus its duration.
Select the activity with the largest total float
$F(H)=41\;>\;F(F)=33$
Activity $H$ has more spare time than any other activity in the project.

Answer

Question 2

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.

\begin{array}{c|c|c} \text{Activity} & \text{Duration (days)} & \text{Immediate predecessors} \\ \hline A & 9 & \text{none} \\ B & 12 & \text{none} \\ C & 14 & B \\ D & 16 & A,\,C \\ E & 3 & A,\,B \\ F & 8 & E \\ G & 5 & F \\ H & 2 & E \\ I & 13 & C,\,G \\ J & 18 & H,\,I \\ K & 15 & I \end{array}

The earliest start time of an activity is

0

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 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 of the following lists all of the critical activities?

Show worked solution

Worked solution

Forward pass: find the earliest start time of activity $A$
$e(A)=0$
Activity $A$ has no immediate predecessors, so it can start at time $0$ .
Forward pass: find the earliest start time of activity $B$
$e(B)=0$
Activity $B$ has no immediate predecessors, so it can start at time $0$ .
Forward pass: find the earliest start time of activity $C$
$e(C)=\max\left\{0+12\right\}=12$
Activity $C$ cannot start until $B$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $D$
$e(D)=\max\left\{0+9,\;12+14\right\}=26$
Activity $D$ cannot start until $A$ , $C$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $E$
$e(E)=\max\left\{0+9,\;0+12\right\}=12$
Activity $E$ cannot start until $A$ , $B$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $F$
$e(F)=\max\left\{12+3\right\}=15$
Activity $F$ cannot start until $E$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $G$
$e(G)=\max\left\{15+8\right\}=23$
Activity $G$ cannot start until $F$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $H$
$e(H)=\max\left\{12+3\right\}=15$
Activity $H$ cannot start until $E$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $I$
$e(I)=\max\left\{12+14,\;23+5\right\}=28$
Activity $I$ cannot start until $C$ , $G$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $J$
$e(J)=\max\left\{15+2,\;28+13\right\}=41$
Activity $J$ cannot start until $H$ , $I$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $K$
$e(K)=\max\left\{28+13\right\}=41$
Activity $K$ cannot start until $I$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Find the total duration of the project from the earliest finish times
$T=\max\left\{9,\;12,\;26,\;42,\;15,\;23,\;28,\;17,\;41,\;59,\;56\right\}=59$
The project is finished when the last activity finishes, so the total duration is the largest earliest finish time.
Backward pass: find the latest finish times of activities $K$ , $J$ , $I$
$l(K)=59,\quad l(J)=59,\quad l(I)=\min\left\{59-18,\;59-15\right\}=41$
Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.
Backward pass: find the latest finish times of activities $H$ , $G$ , $F$ , $E$
$l(H)=\min\left\{59-18\right\}=41,\quad l(G)=\min\left\{41-13\right\}=28,\quad l(F)=\min\left\{28-5\right\}=23,\quad l(E)=\min\left\{23-8,\;41-2\right\}=15$
Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.
Backward pass: find the latest finish times of activities $D$ , $C$ , $B$ , $A$
$l(D)=59,\quad l(C)=\min\left\{59-16,\;41-13\right\}=28,\quad l(B)=\min\left\{28-14,\;15-3\right\}=12,\quad l(A)=\min\left\{59-16,\;15-3\right\}=12$
Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.
Find the total float of every activity
$F(A)=12-0-9=3,\;F(B)=12-0-12=0,\;F(C)=28-12-14=2,\;F(D)=59-26-16=17,\;F(E)=15-12-3=0,\;F(F)=23-15-8=0,\;F(G)=28-23-5=0,\;F(H)=41-15-2=24,\;F(I)=41-28-13=0,\;F(J)=59-41-18=0,\;F(K)=59-41-15=3$
The total float of an activity is its latest finish time minus its earliest start time minus its duration.
Select the list of activities whose total float is zero
$\text{critical}:\;B,\;E,\;F,\;G,\;I,\;J$
These are the activities that cannot be delayed at all without delaying the whole project.

Answer

B,\;E,\;F,\;G,\;I,\;J

Question 3

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.

\begin{array}{c|c|c} \text{Activity} & \text{Duration (days)} & \text{Immediate predecessors} \\ \hline A & 3 & \text{none} \\ B & 18 & \text{none} \\ C & 16 & \text{none} \\ D & 15 & A \\ E & 10 & C \\ F & 13 & B \\ G & 9 & D \\ H & 12 & D,\,E \\ I & 19 & A,\,C \\ J & 14 & F,\,H,\,I \\ K & 4 & J \end{array}

The earliest start time of an activity is

0

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 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. The critical path is the chain of critical activities that runs from an activity with no immediate predecessors to an activity that is an immediate predecessor of no activity, each activity in the chain being an immediate predecessor of the next one. State the critical path for this project. Which of the following is the critical path?

Show worked solution

Worked solution

Forward pass: find the earliest start time of activity $A$
$e(A)=0$
Activity $A$ has no immediate predecessors, so it can start at time $0$ .
Forward pass: find the earliest start time of activity $B$
$e(B)=0$
Activity $B$ has no immediate predecessors, so it can start at time $0$ .
Forward pass: find the earliest start time of activity $C$
$e(C)=0$
Activity $C$ has no immediate predecessors, so it can start at time $0$ .
Forward pass: find the earliest start time of activity $D$
$e(D)=\max\left\{0+3\right\}=3$
Activity $D$ cannot start until $A$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $E$
$e(E)=\max\left\{0+16\right\}=16$
Activity $E$ cannot start until $C$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $F$
$e(F)=\max\left\{0+18\right\}=18$
Activity $F$ cannot start until $B$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $G$
$e(G)=\max\left\{3+15\right\}=18$
Activity $G$ cannot start until $D$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $H$
$e(H)=\max\left\{3+15,\;16+10\right\}=26$
Activity $H$ cannot start until $D$ , $E$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $I$
$e(I)=\max\left\{0+3,\;0+16\right\}=16$
Activity $I$ cannot start until $A$ , $C$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $J$
$e(J)=\max\left\{18+13,\;26+12,\;16+19\right\}=38$
Activity $J$ cannot start until $F$ , $H$ , $I$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $K$
$e(K)=\max\left\{38+14\right\}=52$
Activity $K$ cannot start until $J$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Find the total duration of the project from the earliest finish times
$T=\max\left\{3,\;18,\;16,\;18,\;26,\;31,\;27,\;38,\;35,\;52,\;56\right\}=56$
The project is finished when the last activity finishes, so the total duration is the largest earliest finish time.
Backward pass: find the latest finish times of activities $K$ , $J$ , $I$
$l(K)=56,\quad l(J)=\min\left\{56-4\right\}=52,\quad l(I)=\min\left\{52-14\right\}=38$
Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.
Backward pass: find the latest finish times of activities $H$ , $G$ , $F$ , $E$
$l(H)=\min\left\{52-14\right\}=38,\quad l(G)=56,\quad l(F)=\min\left\{52-14\right\}=38,\quad l(E)=\min\left\{38-12\right\}=26$
Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.
Backward pass: find the latest finish times of activities $D$ , $C$ , $B$ , $A$
$l(D)=\min\left\{56-9,\;38-12\right\}=26,\quad l(C)=\min\left\{26-10,\;38-19\right\}=16,\quad l(B)=\min\left\{38-13\right\}=25,\quad l(A)=\min\left\{26-15,\;38-19\right\}=11$
Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.
Find the total float of every activity
$F(A)=11-0-3=8,\;F(B)=25-0-18=7,\;F(C)=16-0-16=0,\;F(D)=26-3-15=8,\;F(E)=26-16-10=0,\;F(F)=38-18-13=7,\;F(G)=56-18-9=29,\;F(H)=38-26-12=0,\;F(I)=38-16-19=3,\;F(J)=52-38-14=0,\;F(K)=56-52-4=0$
The total float of an activity is its latest finish time minus its earliest start time minus its duration.
Select the chain of activities with zero total float
$C\to E\to H\to J\to K\quad\left(16+10+12+14+4=56\right)$
Every activity on this chain has zero total float, and its total length is the total duration of the project.

Answer

C\to E\to H\to J\to K

Question 4

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.

\begin{array}{c|c|c} \text{Activity} & \text{Duration (days)} & \text{Immediate predecessors} \\ \hline A & 15 & \text{none} \\ B & 6 & \text{none} \\ C & 19 & B \\ D & 14 & C \\ E & 17 & B \\ F & 7 & \text{none} \\ G & 12 & E \\ H & 8 & D,\,F \\ I & 9 & H \\ J & 18 & A,\,I \\ K & 2 & A,\,E,\,F \end{array}

The earliest start time of an activity is

0

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 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)

-

Show worked solution

Worked solution

Forward pass: find the earliest start time of activity $A$
$e(A)=0$
Activity $A$ has no immediate predecessors, so it can start at time $0$ .
Forward pass: find the earliest start time of activity $B$
$e(B)=0$
Activity $B$ has no immediate predecessors, so it can start at time $0$ .
Forward pass: find the earliest start time of activity $C$
$e(C)=\max\left\{0+6\right\}=6$
Activity $C$ cannot start until $B$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $D$
$e(D)=\max\left\{6+19\right\}=25$
Activity $D$ cannot start until $C$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $E$
$e(E)=\max\left\{0+6\right\}=6$
Activity $E$ cannot start until $B$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $F$
$e(F)=0$
Activity $F$ has no immediate predecessors, so it can start at time $0$ .
Forward pass: find the earliest start time of activity $G$
$e(G)=\max\left\{6+17\right\}=23$
Activity $G$ cannot start until $E$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $H$
$e(H)=\max\left\{25+14,\;0+7\right\}=39$
Activity $H$ cannot start until $D$ , $F$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $I$
$e(I)=\max\left\{39+8\right\}=47$
Activity $I$ cannot start until $H$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $J$
$e(J)=\max\left\{0+15,\;47+9\right\}=56$
Activity $J$ cannot start until $A$ , $I$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $K$
$e(K)=\max\left\{0+15,\;6+17,\;0+7\right\}=23$
Activity $K$ cannot start until $A$ , $E$ , $F$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Find the total duration of the project from the earliest finish times
$T=\max\left\{15,\;6,\;25,\;39,\;23,\;7,\;35,\;47,\;56,\;74,\;25\right\}=74$
The project is finished when the last activity finishes, so the total duration is the largest earliest finish time.
Backward pass: find the latest finish times of activities $K$ , $J$ , $I$
$l(K)=74,\quad l(J)=74,\quad l(I)=\min\left\{74-18\right\}=56$
Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.
Backward pass: find the latest finish times of activities $H$ , $G$ , $F$ , $E$
$l(H)=\min\left\{56-9\right\}=47,\quad l(G)=74,\quad l(F)=\min\left\{47-8,\;74-2\right\}=39,\quad l(E)=\min\left\{74-12,\;74-2\right\}=62$
Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.
Backward pass: find the latest finish times of activities $D$ , $C$ , $B$ , $A$
$l(D)=\min\left\{47-8\right\}=39,\quad l(C)=\min\left\{39-14\right\}=25,\quad l(B)=\min\left\{25-19,\;62-17\right\}=6,\quad l(A)=\min\left\{74-18,\;74-2\right\}=56$
Working backwards, an activity must finish early enough for every one of its successors to start at its own latest start time.
Find the total float of every activity
$F(A)=56-0-15=41,\;F(B)=6-0-6=0,\;F(C)=25-6-19=0,\;F(D)=39-25-14=0,\;F(E)=62-6-17=39,\;F(F)=39-0-7=32,\;F(G)=74-23-12=39,\;F(H)=47-39-8=0,\;F(I)=56-47-9=0,\;F(J)=74-56-18=0,\;F(K)=74-23-2=49$
The total float of an activity is its latest finish time minus its earliest start time minus its duration.
Select the chain of activities with zero total float
$B\to C\to D\to H\to I\to J\quad\left(6+19+14+8+9+18=74\right)$
Every activity on this chain has zero total float, and its total length is the total duration of the project.

Answer

B\to C\to D\to H\to I\to J

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.

\begin{array}{c|c|c} \text{Activity} & \text{Duration (days)} & \text{Immediate predecessors} \\ \hline A & 4 & \text{none} \\ B & 9 & A \\ C & 11 & \text{none} \\ D & 6 & A \\ E & 3 & \text{none} \\ F & 14 & B,\,D \\ G & 10 & F \\ H & 17 & E \\ I & 13 & C,\,G \\ J & 19 & B,\,C,\,E \\ K & 8 & H,\,I,\,J \end{array}

The earliest start time of an activity is

0

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. Every activity requires exactly one worker for the whole of its duration, no worker may work on two activities at the same time, and every activity is scheduled to start at its earliest start time. An activity with earliest start time

a

and duration

d

then occupies the time interval from

a

a+d

. A lower bound for the number of workers needed to complete the project in its minimum time is given by

\left\lceil \dfrac{\text{sum of all the activity durations}}{\text{total duration of the project}} \right\rceil

. Find this lower bound.

Show worked solution

Worked solution

Forward pass: find the earliest start time of activity $A$
$e(A)=0$
Activity $A$ has no immediate predecessors, so it can start at time $0$ .
Forward pass: find the earliest start time of activity $B$
$e(B)=\max\left\{0+4\right\}=4$
Activity $B$ cannot start until $A$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $C$
$e(C)=0$
Activity $C$ has no immediate predecessors, so it can start at time $0$ .
Forward pass: find the earliest start time of activity $D$
$e(D)=\max\left\{0+4\right\}=4$
Activity $D$ cannot start until $A$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $E$
$e(E)=0$
Activity $E$ has no immediate predecessors, so it can start at time $0$ .
Forward pass: find the earliest start time of activity $F$
$e(F)=\max\left\{4+9,\;4+6\right\}=13$
Activity $F$ cannot start until $B$ , $D$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $G$
$e(G)=\max\left\{13+14\right\}=27$
Activity $G$ cannot start until $F$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $H$
$e(H)=\max\left\{0+3\right\}=3$
Activity $H$ cannot start until $E$ has finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $I$
$e(I)=\max\left\{0+11,\;27+10\right\}=37$
Activity $I$ cannot start until $C$ , $G$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $J$
$e(J)=\max\left\{4+9,\;0+11,\;0+3\right\}=13$
Activity $J$ cannot start until $B$ , $C$ , $E$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Forward pass: find the earliest start time of activity $K$
$e(K)=\max\left\{3+17,\;37+13,\;13+19\right\}=50$
Activity $K$ cannot start until $H$ , $I$ , $J$ have all finished, so its earliest start time is the largest earliest finish time among its immediate predecessors.
Find the total duration of the project from the earliest finish times
$T=\max\left\{4,\;13,\;11,\;10,\;3,\;27,\;37,\;20,\;50,\;32,\;58\right\}=58$
The project is finished when the last activity finishes, so the total duration is the largest earliest finish time.
Add the durations of all the activities to find the total amount of work
$4+9+11+6+3+14+10+17+13+19+8=114$
This is the total number of worker-days of work in the project.
Check the total duration by adding the durations along the longest chain of activities
$A\to B\to F\to G\to I\to K:\;4+9+14+10+13+8=58$
The total duration of the project is the length of the longest chain of activities, so the two calculations must agree.
Evaluate the lower bound
$\left\lceil \dfrac{114}{58} \right\rceil=2$
The $114$ worker-days of work must be squeezed into $58$ days, so at least $2$ workers are needed.

Answer

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