FE Civil Construction Eng. practice problems

A Start-to-Start relationship ties the successor START to the predecessor START. With SS+2 lag, S can start 2 days after R starts. R starts on day 3, so $ES_S = 3 + 2 = 5$. The Day 3 choice ignored the 2-day lag. The Day 10 choice wrongly used R's early finish ($3 + 7 = 10$), treating it as Finish-to-Start. The Day 12 choice added the lag to R's finish, mixing FS with the lag.

1. R starts on day 3: $ES_R = 3$.
2. Start-to-Start with 2-day lag means S starts 2 days after R starts:
   $ES_S = ES_R + \text{lag} = 3 + 2 = 5$
3. The 7-day duration of R does not enter an SS relationship.

FE Reference Handbook 10.6, p. 316

In AOA notation, activities are arrows and events are nodes. When two activities share dependency logic but must remain distinct, a dummy activity (a dashed arrow with zero duration) preserves the correct precedence without implying real work. This is unique to AOA. The lag option adds a lag, which changes timing, not logic. The second predecessor node invents an extra node that does not fix the shared-dependency problem. The Finish-to-Finish constraint imposes a finish constraint between two activities that the problem says are independent.

1. AOA = Activity-on-Arrow: activities are arrows, events are nodes.
2. Shared dependencies that must stay logically distinct require a dummy activity.
3. A dummy has zero duration and only enforces precedence; it represents no real work.

FE Reference Handbook 10.6, p. 316

Forward pass: P(0,6). Q(6,10). R(6,15). S depends on Q and R, so $ES_S = \max(EF_Q, EF_R) = \max(10, 15) = 15$. S must wait for the slower path through R. The Day 10 option took $EF_Q$ (the minimum, the faster predecessor). The Day 6 option used $EF_P$ and ignored Q and R. The Day 20 option is the project finish ($EF_S$), not the start.

1. Forward pass: P(0,6), Q(6,10), R(6,15).
2. S depends on both Q and R:
   $ES_S = \max(EF_Q, EF_R) = \max(10, 15) = 15$
3. Check: $EF_S = 15 + 5 = 20$ equals the project duration. Critical path is P-R-S.

FE Reference Handbook 10.6, p. 316

Backward pass from $LF = 20$: S(15,20). Q feeds only S, so $LF_Q = LS_S = 15$. Then $LS_Q = 15 - 4 = 11$. The question asks for the late finish, which is 15. Day 10 is the early finish of Q from the forward pass, not the late finish. Day 11 is the late START of Q. Day 20 is the project end, the late finish of S.

1. Backward pass: $LF_S = 20$, $LS_S = 20 - 5 = 15$.
2. Q feeds only S:
   $LF_Q = LS_S = 15$
3. For reference, $LS_Q = LF_Q - D_Q = 15 - 4 = 11$, but the question asks for LF.

FE Reference Handbook 10.6, p. 316

Total float $= LF - EF$. First find $EF = ES + D = 6 + 8 = 14$. Then $TF = LF - EF = 20 - 14 = 6$ days. Equivalently, $LS = LF - D = 20 - 8 = 12$, and $TF = LS - ES = 12 - 6 = 6$. The 8-day option is just the duration. The 14-day option is $LF - ES = 20 - 6$, which double-counts the duration. The 2-day option subtracts incorrectly.

1. Early finish:
   $EF = ES + D = 6 + 8 = 14$
2. Total float:
   $TF = LF - EF = 20 - 14 = 6 \text{ days}$
3. Verify via late start: $LS = LF - D = 20 - 8 = 12$, $TF = LS - ES = 12 - 6 = 6$.

FE Reference Handbook 10.6, p. 316

Total float $= LF - EF$. If the late finish increases by 3 while the early finish is unchanged, total float becomes $0 + 3 = 3$ days. Any activity with positive total float is, by definition, NOT on the critical path. The critical path is the set of activities with zero total float. The "remains critical because its early dates did not change" choice confuses early dates with float, but float depends on the gap between early and late dates. The "negative float of 3 days" choice describes an over-allocated schedule, the opposite of gaining slack. The "free float increases but total float stays at zero" choice is impossible because free float can never exceed total float.

1. Originally critical: $TF = LF - EF = 0$.
2. LF increases by 3, EF unchanged:
   $TF = (LF + 3) - EF = 0 + 3 = 3 \text{ days}$
3. Positive total float means the activity is off the critical path.

FE Reference Handbook 10.6, p. 316

$CV = BCWP - ACWP = 1{,}050{,}000 - 980{,}000 = +70{,}000$. A positive cost variance means the work performed is worth more than what was spent, so the project is under budget. The $-150{,}000$ option computed the schedule variance ($BCWP - BCWS = 1{,}050{,}000 - 1{,}200{,}000$) instead of the cost variance. The $-70{,}000$ option flipped the sign by subtracting in the wrong order. The $+220{,}000$ option used $BCWS - ACWP$, which is not a standard variance.

1. Cost variance:
   $CV = BCWP - ACWP = 1{,}050{,}000 - 980{,}000 = +70{,}000$
2. Positive CV means earned value exceeds actual cost: under budget.

FE Reference Handbook 10.6, p. 316

Schedule variance $SV = BCWP - BCWS = 0$ forces $BCWP = BCWS$. Cost variance $CV = BCWP - ACWP = 0$ forces $BCWP = ACWP$. Combining both, all three values are equal: $BCWS = BCWP = ACWP$. The project is exactly on schedule and exactly on budget. The choice BCWP = 0 only happens before any work is earned. The choice BCWS = ACWP but BCWP may differ lets BCWP differ, which would make both variances nonzero. The choice ACWP > BCWP > BCWS describes a project both behind schedule and over budget.

1. Schedule variance zero:
   $SV = BCWP - BCWS = 0 \Rightarrow BCWP = BCWS$
2. Cost variance zero:
   $CV = BCWP - ACWP = 0 \Rightarrow BCWP = ACWP$
3. Therefore all three are equal:
   $BCWS = BCWP = ACWP$

FE Reference Handbook 10.6, p. 316

$SPI = BCWP / BCWS = 840{,}000 / 1{,}050{,}000 = 0.80$. An SPI below 1.0 means the project earned less value than planned, so it is behind schedule. The 1.25 option inverted the ratio to $BCWS / BCWP$. The other 0.80 (ahead) option has the right number but misreads the sign: below 1.0 is behind, not ahead. The $-210{,}000$ option is the schedule variance in dollars ($SV = BCWP - BCWS$), not the index.

1. Schedule Performance Index:
   $SPI = \frac{BCWP}{BCWS} = \frac{840{,}000}{1{,}050{,}000} = 0.80$
2. $SPI = 0.80 < 1.0$ means the project is behind schedule.

FE Reference Handbook 10.6, p. 317

First find $CPI = BCWP / ACWP = 1{,}500{,}000 / 1{,}800{,}000 = 0.8333$. Then $EAC = BAC / CPI = 4{,}000{,}000 / 0.8333 = 4{,}800{,}000$. Because the project is running over budget (CPI < 1.0), the forecast total exceeds the original budget. The 3,333,333 option inverted CPI, dividing by 1.20 instead. The 4,300,000 option added the overrun-to-date (ACWP − BCWP = 300,000) to BAC, ignoring that the trend continues. The 4,000,000 option is BAC with no adjustment.

1. Cost Performance Index:
   $CPI = \frac{BCWP}{ACWP} = \frac{1{,}500{,}000}{1{,}800{,}000} = 0.8333$
2. Estimate at completion:
   $EAC = \frac{BAC}{CPI} = \frac{4{,}000{,}000}{0.8333} = 4{,}800{,}000$
3. The forecast exceeds the 4,000,000 budget because the project is running over budget.

FE Reference Handbook 10.6, p. 317

The clues all point to Design-Build: fastest delivery (overlapping design and construction), single-point accountability (one firm liable), and acceptance of less design control. DB places design and construction under one contract. DBB is sequential and cannot overlap phases. CMAR keeps a separate designer, so accountability is split. Multiple-prime contracting fragments responsibility across several contractors, the opposite of single-point accountability.

1. Overlapping design and construction rules out the sequential DBB.
2. Single firm legally accountable rules out CMAR and multiple-prime.
3. Design-Build provides speed and single-point responsibility.

A GMP caps the owner's cost exposure: the owner never pays more than the GMP. When actual cost comes in under the GMP, a shared-savings (savings-split) clause divides the underrun between owner and CM by a pre-agreed ratio. This incentivizes the CM to control costs. The "owner keeps the entire savings" option ignores the explicit shared-savings clause. The "CM keeps the entire savings as profit" option is wrong because the CM does not pocket all savings under a shared clause. The "owner must pay the full GMP regardless of actual cost" option describes a different arrangement, not a GMP with a savings split, where the owner would pay the cap rather than the lower actual cost.

1. GMP = 8,000,000 is the maximum the owner pays.
2. Actual cost 7,600,000 is below the cap, creating a 400,000 underrun.
3. A shared-savings clause splits that underrun between owner and CM per the agreed ratio.

OSHA prohibits classifying previously disturbed soil as Type A. Previously disturbed, submerged, or seeping soils are treated as the weakest classification, Type C, which requires the flattest slope (1.5H:1V). The disturbance destroys the natural cohesion that would justify a steeper classification. The "Type A, because compacted backfill is the most stable" choice is exactly the forbidden assumption. The "Type B, because backfill is moderately stable" choice still overstates stability. The "it does not need classification if it was engineered fill" choice is wrong because all excavation soils must be classified by a competent person regardless of origin.

1. OSHA forbids classifying previously disturbed soil as Type A.
2. Disturbed, submerged, or seeping soils default to Type C (weakest).
3. Type C requires the flattest slope, 1.5H : 1V.

At 1H : 1V, each side slopes back 1 ft horizontally for every 1 ft of depth. At 16 ft deep, each side adds $1 \times 16 = 16$ ft of horizontal run. There are two sides, so total slope width = $2 \times 16 = 32$ ft. Add the 4 ft bottom: top width $= 32 + 4 = 36$ ft. The 20 ft option added only one side's run plus the bottom (16 + 4). The 28 ft option used a 0.75H : 1V ratio ($2 \times 12 + 4$). The 52 ft option used 1.5H : 1V ($2 \times 24 + 4$), the Type C slope, not Type B.

1. Horizontal run per side:
   $1 \times 16 = 16 \text{ ft}$
2. Two sides:
   $2 \times 16 = 32 \text{ ft}$
3. Top width = slope run + bottom width:
   $32 + 4 = 36 \text{ ft}$