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10 FE practice problems: construction: CPM and earned value, with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

These ten original problems practice construction: CPM and earned value, a topic from the FE Civil exam specification, using the Civil Engineering, Earned-Value Analysis part of the FE Reference Handbook 10.6. Each problem gives the situation and the values with units. Work it with the handbook PDF open and commit to an answer before you open the solution. Every solution shows the handbook page, the equation, the substitution with units, a size check, and the mistake behind each wrong option, so a wrong pick tells you exactly what to fix.

How to use these problems

Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.

The problems

Problem 1 · FE Civil, Construction: CPM and earned value

A treatment-plant upgrade has a budget at completion (BAC) of $1,380,000. At the status date, the budgeted cost of work performed (BCWP) is $676,000, the actual cost of work performed (ACWP) is $839,000, and the budgeted cost of work scheduled (BCWS) is $647,000. Using the cost performance index, the estimate at completion (EAC) is most nearly:

  • A $1,710,000
  • B $1,750,000
  • C $1,540,000
  • D $1,110,000

Handbook: Civil Engineering, Earned-Value Analysis, FE Reference Handbook 10.6

Show the worked solution
Answer
A ($1,710,000)
Given
BAC = $1,380,000, BCWP = $676,000, ACWP = $839,000, BCWS = $647,000
Find
estimate at completion EAC ($)
Handbook
Civil Engineering, Construction, Earned-Value Analysis (indices and forecasting), pages 316 and 317
Equation
CPI = BCWP/ACWP; ETC = (BAC - BCWP)/CPI; EAC = ACWP + ETC
Substitute
  1. CPI = BCWP/ACWP = 676,000/839,000 = 0.8057
  2. ETC = (BAC - BCWP)/CPI = (1,380,000 - 676,000)/0.8057 = $873,800
  3. EAC = ACWP + ETC = 839,000 + $873,800 = $1,710,000
Result
$1,710,000, 3 significant figures
Check
CPI < 1, so EAC exceeds the budget; EAC = BAC/CPI = $1,710,000.
Why the others are wrong
  • B: used the planned value (BCWS) in place of the earned value
  • C: added the cost overrun to date to the budget (assumes the rest goes to plan)
  • D: multiplied the budget by the CPI instead of dividing

Problem 2 · FE Civil, Construction: CPM and earned value

A small project has these activities (duration, predecessors): A (5 days, no predecessor); B (11 days, after A); C (10 days, after A); D (3 days, after B); E (10 days, after C); F (4 days, after D and E). Using the critical path method, the total float of activity D is most nearly:

  • A 20 days
  • B 9 days
  • C 29 days
  • D 6 days

Handbook: Civil Engineering, CPM Precedence Relationships, FE Reference Handbook 10.6

Show the worked solution
Answer
D (6 days)
Given
dur = 5, 11, 10, 3, 10, 4, act = D
Find
total float of D (days)
Handbook
Civil Engineering, Construction, CPM Precedence Relationships (activity on node), page 316
Equation
ES = latest EF of predecessors; EF = ES + d; LF = earliest LS of successors; LS = LF - d; Float = LS - ES
Substitute
  1. Forward pass (ES = latest EF of predecessors; EF = ES + duration): A: ES 0, EF 5, B: ES 5, EF 16, C: ES 5, EF 15, D: ES 16, EF 19, E: ES 15, EF 25, F: ES 25, EF 29
  2. Project duration = 29 days
  3. Backward pass (LF = earliest LS of successors; LS = LF - duration): A: LS 0, LF 5, B: LS 11, LF 22, C: LS 5, LF 15, D: LS 22, LF 25, E: LS 15, LF 25, F: LS 25, LF 29
  4. Float of D = LS - ES = 22 - 16 = 6 days
Result
6 days, 3 significant figures
Check
activities with zero float form the critical path: A-C-E-F.
Why the others are wrong
  • A: is the length of the middle of the critical path, not a float
  • B: took LF - ES, which leaves the duration in
  • C: is the project duration, not the float

Problem 3 · FE Civil, Construction: CPM and earned value

A small project has these activities (duration, predecessors): A (10 days, no predecessor); B (7 days, after A); C (2 days, after A); D (3 days, after B); E (4 days, after C); F (2 days, after D and E). Using the critical path method, the total float of activity E is most nearly:

  • A 4 days
  • B 10 days
  • C 8 days
  • D 12 days

Handbook: Civil Engineering, CPM Precedence Relationships, FE Reference Handbook 10.6

Show the worked solution
Answer
A (4 days)
Given
dur = 10, 7, 2, 3, 4, 2, act = E
Find
total float of E (days)
Handbook
Civil Engineering, Construction, CPM Precedence Relationships (activity on node), page 316
Equation
ES = latest EF of predecessors; EF = ES + d; LF = earliest LS of successors; LS = LF - d; Float = LS - ES
Substitute
  1. Forward pass (ES = latest EF of predecessors; EF = ES + duration): A: ES 0, EF 10, B: ES 10, EF 17, C: ES 10, EF 12, D: ES 17, EF 20, E: ES 12, EF 16, F: ES 20, EF 22
  2. Project duration = 22 days
  3. Backward pass (LF = earliest LS of successors; LS = LF - duration): A: LS 0, LF 10, B: LS 10, LF 17, C: LS 14, LF 16, D: LS 17, LF 20, E: LS 16, LF 20, F: LS 20, LF 22
  4. Float of E = LS - ES = 16 - 12 = 4 days
Result
4 days, 3 significant figures
Check
activities with zero float form the critical path: A-B-D-F.
Why the others are wrong
  • B: is the length of the middle of the critical path, not a float
  • C: took LF - ES, which leaves the duration in
  • D: is the early start time, not the float

Problem 4 · FE Civil, Construction: CPM and earned value

A small project has these activities (duration, predecessors): A (6 days, no predecessor); B (9 days, after A); C (3 days, after A); D (11 days, after B); E (5 days, after C); F (7 days, after D and E). Using the critical path method, the total float of activity E is most nearly:

  • A 33 days
  • B 17 days
  • C 12 days
  • D 21 days

Handbook: Civil Engineering, CPM Precedence Relationships, FE Reference Handbook 10.6

Show the worked solution
Answer
C (12 days)
Given
dur = 6, 9, 3, 11, 5, 7, act = E
Find
total float of E (days)
Handbook
Civil Engineering, Construction, CPM Precedence Relationships (activity on node), page 316
Equation
ES = latest EF of predecessors; EF = ES + d; LF = earliest LS of successors; LS = LF - d; Float = LS - ES
Substitute
  1. Forward pass (ES = latest EF of predecessors; EF = ES + duration): A: ES 0, EF 6, B: ES 6, EF 15, C: ES 6, EF 9, D: ES 15, EF 26, E: ES 9, EF 14, F: ES 26, EF 33
  2. Project duration = 33 days
  3. Backward pass (LF = earliest LS of successors; LS = LF - duration): A: LS 0, LF 6, B: LS 6, LF 15, C: LS 18, LF 21, D: LS 15, LF 26, E: LS 21, LF 26, F: LS 26, LF 33
  4. Float of E = LS - ES = 21 - 9 = 12 days
Result
12 days, 3 significant figures
Check
activities with zero float form the critical path: A-B-D-F.
Why the others are wrong
  • A: is the project duration, not the float
  • B: took LF - ES, which leaves the duration in
  • D: is the late start time, not the float

Problem 5 · FE Civil, Construction: CPM and earned value

A small project has these activities (duration, predecessors): A (9 days, no predecessor); B (6 days, after A); C (11 days, after A); D (3 days, after B); E (7 days, after C); F (4 days, after D and E). Using the critical path method, the total float of activity D is most nearly:

  • A 12 days
  • B 31 days
  • C 18 days
  • D 9 days

Handbook: Civil Engineering, CPM Precedence Relationships, FE Reference Handbook 10.6

Show the worked solution
Answer
D (9 days)
Given
dur = 9, 6, 11, 3, 7, 4, act = D
Find
total float of D (days)
Handbook
Civil Engineering, Construction, CPM Precedence Relationships (activity on node), page 316
Equation
ES = latest EF of predecessors; EF = ES + d; LF = earliest LS of successors; LS = LF - d; Float = LS - ES
Substitute
  1. Forward pass (ES = latest EF of predecessors; EF = ES + duration): A: ES 0, EF 9, B: ES 9, EF 15, C: ES 9, EF 20, D: ES 15, EF 18, E: ES 20, EF 27, F: ES 27, EF 31
  2. Project duration = 31 days
  3. Backward pass (LF = earliest LS of successors; LS = LF - duration): A: LS 0, LF 9, B: LS 18, LF 24, C: LS 9, LF 20, D: LS 24, LF 27, E: LS 20, LF 27, F: LS 27, LF 31
  4. Float of D = LS - ES = 24 - 15 = 9 days
Result
9 days, 3 significant figures
Check
activities with zero float form the critical path: A-C-E-F.
Why the others are wrong
  • A: added the activity's own duration to its float
  • B: is the project duration, not the float
  • C: is the length of the middle of the critical path, not a float

Problem 6 · FE Civil, Construction: CPM and earned value

A treatment-plant upgrade has a budget at completion (BAC) of $4,230,000. At the status date, the budgeted cost of work performed (BCWP) is $2,926,000, the actual cost of work performed (ACWP) is $3,217,000, and the budgeted cost of work scheduled (BCWS) is $3,498,000. Using the cost performance index, the estimate at completion (EAC) is most nearly:

  • A $4,520,000
  • B $4,780,000
  • C $4,650,000
  • D $4,020,000

Handbook: Civil Engineering, Earned-Value Analysis, FE Reference Handbook 10.6

Show the worked solution
Answer
C ($4,650,000)
Given
BAC = $4,230,000, BCWP = $2,926,000, ACWP = $3,217,000, BCWS = $3,498,000
Find
estimate at completion EAC ($)
Handbook
Civil Engineering, Construction, Earned-Value Analysis (indices and forecasting), pages 316 and 317
Equation
CPI = BCWP/ACWP; ETC = (BAC - BCWP)/CPI; EAC = ACWP + ETC
Substitute
  1. CPI = BCWP/ACWP = 2,926,000/3,217,000 = 0.9095
  2. ETC = (BAC - BCWP)/CPI = (4,230,000 - 2,926,000)/0.9095 = $1,434,000
  3. EAC = ACWP + ETC = 3,217,000 + $1,434,000 = $4,650,000
Result
$4,650,000, 3 significant figures
Check
CPI < 1, so EAC exceeds the budget; EAC = BAC/CPI = $4,650,000.
Why the others are wrong
  • A: added the cost overrun to date to the budget (assumes the rest goes to plan)
  • B: used the schedule index instead of the cost index
  • D: used the planned value (BCWS) in place of the earned value

Problem 7 · FE Civil, Construction: CPM and earned value

A treatment-plant upgrade has a budget at completion (BAC) of $3,250,000. At the status date, the budgeted cost of work performed (BCWP) is $1,402,000, the actual cost of work performed (ACWP) is $1,234,000, and the budgeted cost of work scheduled (BCWS) is $1,634,000. Using the cost performance index, the estimate at completion (EAC) is most nearly:

  • A $3,080,000
  • B $3,690,000
  • C $2,860,000
  • D $2,660,000

Handbook: Civil Engineering, Earned-Value Analysis, FE Reference Handbook 10.6

Show the worked solution
Answer
C ($2,860,000)
Given
BAC = $3,250,000, BCWP = $1,402,000, ACWP = $1,234,000, BCWS = $1,634,000
Find
estimate at completion EAC ($)
Handbook
Civil Engineering, Construction, Earned-Value Analysis (indices and forecasting), pages 316 and 317
Equation
CPI = BCWP/ACWP; ETC = (BAC - BCWP)/CPI; EAC = ACWP + ETC
Substitute
  1. CPI = BCWP/ACWP = 1,402,000/1,234,000 = 1.136
  2. ETC = (BAC - BCWP)/CPI = (3,250,000 - 1,402,000)/1.136 = $1,627,000
  3. EAC = ACWP + ETC = 1,234,000 + $1,627,000 = $2,860,000
Result
$2,860,000, 3 significant figures
Check
CPI > 1, so EAC is under the budget; EAC = BAC/CPI = $2,860,000.
Why the others are wrong
  • A: added the remaining budget without dividing by the CPI
  • B: multiplied the budget by the CPI instead of dividing
  • D: used the planned value (BCWS) in place of the earned value

Problem 8 · FE Civil, Construction: CPM and earned value

A small project has these activities (duration, predecessors): A (12 days, no predecessor); B (4 days, after A); C (10 days, after A); D (9 days, after B); E (12 days, after C); F (7 days, after D and E). Using the critical path method, the total float of activity B is most nearly:

  • A 41 days
  • B 22 days
  • C 9 days
  • D 13 days

Handbook: Civil Engineering, CPM Precedence Relationships, FE Reference Handbook 10.6

Show the worked solution
Answer
C (9 days)
Given
dur = 12, 4, 10, 9, 12, 7, act = B
Find
total float of B (days)
Handbook
Civil Engineering, Construction, CPM Precedence Relationships (activity on node), page 316
Equation
ES = latest EF of predecessors; EF = ES + d; LF = earliest LS of successors; LS = LF - d; Float = LS - ES
Substitute
  1. Forward pass (ES = latest EF of predecessors; EF = ES + duration): A: ES 0, EF 12, B: ES 12, EF 16, C: ES 12, EF 22, D: ES 16, EF 25, E: ES 22, EF 34, F: ES 34, EF 41
  2. Project duration = 41 days
  3. Backward pass (LF = earliest LS of successors; LS = LF - duration): A: LS 0, LF 12, B: LS 21, LF 25, C: LS 12, LF 22, D: LS 25, LF 34, E: LS 22, LF 34, F: LS 34, LF 41
  4. Float of B = LS - ES = 21 - 12 = 9 days
Result
9 days, 3 significant figures
Check
activities with zero float form the critical path: A-C-E-F.
Why the others are wrong
  • A: is the project duration, not the float
  • B: is the length of the middle of the critical path, not a float
  • D: added the activity's own duration to its float

Problem 9 · FE Civil, Construction: CPM and earned value

A small project has these activities (duration, predecessors): A (8 days, no predecessor); B (4 days, after A); C (3 days, after A); D (3 days, after B); E (2 days, after C); F (3 days, after D and E). Using the critical path method, the total float of activity C is most nearly:

  • A 18 days
  • B 2 days
  • C 7 days
  • D 5 days

Handbook: Civil Engineering, CPM Precedence Relationships, FE Reference Handbook 10.6

Show the worked solution
Answer
B (2 days)
Given
dur = 8, 4, 3, 3, 2, 3, act = C
Find
total float of C (days)
Handbook
Civil Engineering, Construction, CPM Precedence Relationships (activity on node), page 316
Equation
ES = latest EF of predecessors; EF = ES + d; LF = earliest LS of successors; LS = LF - d; Float = LS - ES
Substitute
  1. Forward pass (ES = latest EF of predecessors; EF = ES + duration): A: ES 0, EF 8, B: ES 8, EF 12, C: ES 8, EF 11, D: ES 12, EF 15, E: ES 11, EF 13, F: ES 15, EF 18
  2. Project duration = 18 days
  3. Backward pass (LF = earliest LS of successors; LS = LF - duration): A: LS 0, LF 8, B: LS 8, LF 12, C: LS 10, LF 13, D: LS 12, LF 15, E: LS 13, LF 15, F: LS 15, LF 18
  4. Float of C = LS - ES = 10 - 8 = 2 days
Result
2 days, 3 significant figures
Check
activities with zero float form the critical path: A-B-D-F.
Why the others are wrong
  • A: is the project duration, not the float
  • C: is the length of the middle of the critical path, not a float
  • D: added the activity's own duration to its float

Problem 10 · FE Civil, Construction: CPM and earned value

A small project has these activities (duration, predecessors): A (10 days, no predecessor); B (9 days, after A); C (11 days, after A); D (7 days, after B); E (4 days, after C); F (2 days, after D and E). Using the critical path method, the total float of activity E is most nearly:

  • A 1 days
  • B 22 days
  • C 5 days
  • D 21 days

Handbook: Civil Engineering, CPM Precedence Relationships, FE Reference Handbook 10.6

Show the worked solution
Answer
A (1 days)
Given
dur = 10, 9, 11, 7, 4, 2, act = E
Find
total float of E (days)
Handbook
Civil Engineering, Construction, CPM Precedence Relationships (activity on node), page 316
Equation
ES = latest EF of predecessors; EF = ES + d; LF = earliest LS of successors; LS = LF - d; Float = LS - ES
Substitute
  1. Forward pass (ES = latest EF of predecessors; EF = ES + duration): A: ES 0, EF 10, B: ES 10, EF 19, C: ES 10, EF 21, D: ES 19, EF 26, E: ES 21, EF 25, F: ES 26, EF 28
  2. Project duration = 28 days
  3. Backward pass (LF = earliest LS of successors; LS = LF - duration): A: LS 0, LF 10, B: LS 10, LF 19, C: LS 11, LF 22, D: LS 19, LF 26, E: LS 22, LF 26, F: LS 26, LF 28
  4. Float of E = LS - ES = 22 - 21 = 1 days
Result
1 days, 3 significant figures
Check
activities with zero float form the critical path: A-B-D-F.
Why the others are wrong
  • B: is the late start time, not the float
  • C: added the activity's own duration to its float
  • D: is the early start time, not the float

A new problem posts every day inside the lab, with the full worked solution the same evening.

Frequently asked questions

Where is construction: CPM and earned value in the FE Reference Handbook?

Look in the Civil Engineering, Earned-Value Analysis part of FE Reference Handbook 10.6. Each solution gives the exact page, so you can practice finding it in the PDF the way you will on exam day.

Are these real FE exam questions?

No. They are original problems generated from the handbook formulas and checked by code. Real exam questions are confidential, and sharing them breaks the NCEES agreement every examinee accepts.

How do I check my answer before opening the solution?

Check the units of your result and whether its size makes sense for the situation. Each solution ends with the same kind of check, so you can compare your habit with ours.

Sources

  1. NCEES FE Civil CBT exam specifications (PDF). Retrieved October 3, 2026.
  2. NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.