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Question 1
ప్రశ్న 1
A construction planner is estimating the duration of an unfamiliar tunnelling operation using PERT. Which statement correctly distinguishes the THREE activity-time estimates?
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Explanation:
PERT uses three estimates to represent uncertainty in activity duration. Optimistic time is an estimate under favourable operating conditions. Most likely time is the duration considered most probable under the assumed conditions. Pessimistic time represents a longer duration associated with unfavourable but plausible circumstances. These estimates need not be equally spaced. They are combined through the conventional PERT weighted-average formula to estimate activity duration.
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Question 2
ప్రశ్న 2
A project engineer submits optimistic time a, most likely time m, and pessimistic time b for a pipeline-crossing activity. Which set of estimates is logically consistent with the conventional PERT ordering?
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Explanation:
The conventional PERT estimates must satisfy optimistic time less than or equal to most likely time. Most likely time must also be less than or equal to pessimistic time. The first option satisfies the ordering 3 ≤ 8 ≤ 14. The second option places the most likely time below the optimistic time. The remaining options also violate the required ordering. Such inconsistencies should be corrected before calculating the expected activity duration.
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Question 3
ప్రశ్న 3
A sewer-construction activity has optimistic duration 3 days, most likely duration 6 days and pessimistic duration 15 days. Using the conventional PERT weighted-average formula, what is its EXPECTED duration?
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Explanation:
The conventional PERT expected-duration formula is te = (a + 4m + b)/6. The optimistic duration is three days and the most likely duration is six days. The pessimistic duration is fifteen days. Substitution gives (3 + 4 × 6 + 15)/6 = 42/6. The expected activity duration is therefore seven days. This value differs from the most likely duration because the pessimistic estimate is relatively high.
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Question 4
ప్రశ్న 4
The duration estimates for installing a large-diameter valve are optimistic 4 days, most likely 7 days and pessimistic 16 days. What is the conventional PERT EXPECTED activity duration?
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Explanation:
PERT gives four times the weight to the most likely estimate compared with either extreme estimate. The weighted duration numerator is 4 + 4 × 7 + 16. This equals 48 days in the numerator. Dividing by six gives an expected duration of eight days. The weighted estimate is greater than the seven-day most likely duration. The difference arises from the relatively long pessimistic estimate.
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Question 5
ప్రశ్న 5
A reservoir-foundation activity has optimistic time 4 days and pessimistic time 16 days. Under the conventional PERT approximation, what is its activity STANDARD DEVIATION?
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Explanation:
The conventional PERT approximation for activity standard deviation is sigma = (b - a)/6. The pessimistic and optimistic estimates are sixteen and four days respectively. Their difference is twelve days. Dividing twelve by six gives two days. The result measures the estimated spread of the activity-duration distribution. Standard deviation has the same unit as the activity duration.
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Question 6
ప్రశ్న 6
A transmission-main activity has optimistic time 2 days and pessimistic time 14 days. What is its activity VARIANCE under the standard PERT approximation?
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Explanation:
Activity variance in conventional PERT is calculated as [(b - a)/6] squared. The time range is 14 - 2 = 12 days. The estimated standard deviation is therefore 12/6 = 2 days. Squaring this value gives a variance of four days squared. Variance and standard deviation are different measures of duration uncertainty. The most likely estimate is not required for this conventional variance formula.
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Question 7
ప్రశ్న 7
Two independent construction activities have PERT estimates X: (a = 2, m = 6, b = 10) days and Y: (a = 4, m = 6, b = 14) days. Which statement correctly compares their EXPECTED durations?
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Explanation:
The two activities have the same most likely duration of six days. For X, the weighted calculation is (2 + 4 × 6 + 10)/6 = 6 days. For Y, the corresponding calculation is (4 + 4 × 6 + 14)/6 = 7 days. Therefore the activities do not have identical expected durations. The optimistic and pessimistic estimates influence the weighted average along with the most likely estimate. This example shows why the most likely time alone is insufficient for PERT estimation.
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Question 8
ప్రశ్న 8
A pumping-station activity has optimistic time 3 days and pessimistic time 13 days. Its conventional PERT expected duration is calculated as 8 days. What MOST LIKELY duration must have been used?
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Explanation:
The conventional PERT formula is te = (a + 4m + b)/6. The expected time is eight days, so the numerator must equal 48. The known optimistic and pessimistic estimates contribute 3 + 13 = 16. Therefore four times the most likely duration equals 48 - 16 = 32. Dividing by four gives a most likely duration of eight days. This result satisfies the required ordering between optimistic and pessimistic times.
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Question 9
ప్రశ్న 9
A pipeline-testing activity has optimistic time 3 days, most likely time 9 days and expected PERT duration 9 days. What is its PESSIMISTIC time?
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Explanation:
The PERT expected-duration equation is 9 = (3 + 4 × 9 + b)/6. Multiplying both sides by six gives a weighted total of 54. The known optimistic and most likely contributions add to 3 + 36 = 39. The pessimistic estimate must therefore equal 54 - 39 = 15 days. This is greater than the most likely estimate, as required. Substitution of fifteen days back into the equation confirms the expected duration of nine days.
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Question 10
ప్రశ్న 10
Two activities have PERT estimates X: (2, 6, 10) days and Y: (0, 6, 12) days, where each tuple gives optimistic, most likely and pessimistic times. Which statement about their EXPECTED durations and VARIANCES is correct?
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Explanation:
The expected duration of X is (2 + 4 × 6 + 10)/6 = 6 days. The expected duration of Y is (0 + 4 × 6 + 12)/6 = 6 days. The standard deviation of X is (10 - 2)/6 = 4/3 days. Its variance is therefore 16/9 days squared. The standard deviation of Y is twelve divided by six, giving a variance of four days squared. Thus both expected durations are equal, but Y has greater estimated uncertainty.
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Question 11
ప్రశ్న 11
For a construction activity, the optimistic and pessimistic estimates remain unchanged, but the most likely duration is revised upward by 3 days. Under the conventional PERT formula, by how much does the EXPECTED duration increase?
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Explanation:
The conventional PERT expected duration is (a + 4m + b)/6. The most likely estimate carries a weighting factor of four. Increasing m by three days increases the numerator by 4 × 3 = 12 days. The denominator remains six. Therefore the expected duration increases by twelve divided by six, or two days. The conventional variance formula remains unchanged because the optimistic and pessimistic estimates have not changed.
వివరణ:
Question 12
ప్రశ్న 12
A construction activity originally has PERT estimates (a = 2, m = 5, b = 8) days. Only its pessimistic estimate increases from 8 to 14 days. What are its REVISED expected duration and variance?
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Explanation:
The revised expected duration is calculated as (2 + 4 × 5 + 14)/6. This gives 36/6 = 6 days. The revised optimistic-to-pessimistic range is 14 - 2 = 12 days. Hence the revised standard deviation is 12/6 = 2 days. Its variance is therefore four days squared. Both the expected duration and variance increase compared with the original estimates, which gave five days and one day squared respectively.
వివరణ:
Question 13
ప్రశ్న 13
A sewerage project contains THREE sequential activities A, B and C, with PERT estimates A: (2, 4, 6) days, B: (1, 5, 9) days and C: (3, 6, 15) days. Each activity must finish before the next starts, and there are no other activities or lags. What is the EXPECTED project duration?
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Explanation:
The expected time of A is (2 + 4 × 4 + 6)/6 = 4 days. The expected time of B is (1 + 4 × 5 + 9)/6 = 5 days. The expected time of C is (3 + 4 × 6 + 15)/6 = 7 days. All three activities are sequential and comprise the entire project. The expected total duration is the sum of their expected times, giving 4 + 5 + 7 = 16 days. This expectation is additive for a serial sequence even without assuming statistical independence.
వివరణ:
Question 14
ప్రశ్న 14
A PERT network has A and B starting independently; C follows A, D follows B, and E follows BOTH C and D. Their three-point time estimates (optimistic, most likely, pessimistic) are A: (2, 5, 8), B: (1, 3, 11), C: (2, 4, 6), D: (2, 5, 14), and E: (1, 3, 5), all in days. Using expected activity times and zero-lag finish-to-start links, what are the estimated project duration and controlling path?
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Explanation:
The expected activity durations are A = 5, B = 4, C = 4, D = 6 and E = 3 days. The A-C branch reaches E after 5 + 4 = 9 days. The B-D branch reaches E after 4 + 6 = 10 days. E must wait for both branches and therefore starts at estimated time 10. Adding E gives the conventional PERT project-duration estimate of 10 + 3 = 13 days. The controlling expected-duration path is B-D-E.
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Question 15
ప్రశ్న 15
For an AON PERT network, A and B begin at project time zero, C follows A, D follows B, and E follows BOTH C and D. Their PERT estimates are A: (1, 4, 7), B: (2, 5, 8), C: (2, 5, 8), D: (2, 5, 14), and E: (3, 6, 9), all in days. Using conventional expected activity durations, what is the EARLIEST START of E?
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Explanation:
The expected durations are A = 4, B = 5, C = 5 and D = 6 days. The branch A-C reaches the merge after 4 + 5 = 9 days. The branch B-D reaches the merge after 5 + 6 = 11 days. E requires both predecessors and cannot begin before the longer branch finishes. Its earliest start in the expected-duration network is therefore eleven days. The six-day expected duration of E is added only when its earliest finish or project completion is required.
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Question 16
ప్రశ్న 16
Which statement correctly distinguishes the BASIC time-estimation approach of traditional PERT from deterministic CPM?
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Explanation:
Traditional PERT was developed to address uncertainty in activity durations. It combines three estimates into a weighted expected activity time. Deterministic CPM generally performs network calculations using one specified duration per activity. Both techniques depend on meaningful activity definitions and valid precedence relationships. Both can identify controlling paths after durations have been assigned. The distinction concerns how duration uncertainty is represented, rather than whether network logic is required.
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Question 17
ప్రశ్న 17
A project has only TWO independent parallel activities A and B, both starting at time zero, and project completion requires BOTH to finish. A has PERT estimates (6, 7, 8) days, while B has estimates (2, 6, 22) days. Which statement is correct when the standard PERT expected-duration network is analysed?
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Explanation:
The expected duration of A is (6 + 4 × 7 + 8)/6 = 7 days. The expected duration of B is (2 + 4 × 6 + 22)/6 = 8 days. The project requires completion of both parallel activities. In the conventional expected-duration calculation, B is the longer branch and controls the eight-day estimate. The result differs from simply comparing most likely durations. The true expected duration of the stochastic project need not equal eight days because either branch may finish last in an actual realization.
వివరణ:
Question 18
ప్రశ్న 18
A project consists of THREE sequential activities on its only path. Their PERT estimates are A: (2, 5, 8), B: (3, 6, 15) and C: (4, 7, 10), all in days. Assuming statistically independent activity durations, what is the estimated VARIANCE of the project duration under conventional PERT?
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Explanation:
Activity A has standard deviation (8 - 2)/6 = 1 day and variance 1 day squared. Activity B has standard deviation (15 - 3)/6 = 2 days and variance 4 days squared. Activity C has standard deviation (10 - 4)/6 = 1 day and variance 1 day squared. The project has a single serial path containing all three activities. Because their durations are assumed independent, the variance of the total duration is the sum of their variances. Therefore project-duration variance is 1 + 4 + 1 = 6 days squared.
వివరణ:
Question 19
ప్రశ్న 19
For a project with only TWO sequential activities, their PERT duration variances are 9 days squared and 16 days squared. Assume their durations are statistically independent. What is the STANDARD DEVIATION of the total project duration?
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Explanation:
For independent activity durations, the variance of their sum equals the sum of individual variances. The first activity has variance nine days squared. The second has variance sixteen days squared. Their total-duration variance is therefore 9 + 16 = 25 days squared. Standard deviation is the positive square root of variance. Consequently the project-duration standard deviation is five days, not the sum of the variances.
వివరణ:
Question 20
ప్రశ్న 20
Two sequential activities have duration standard deviations of 2 days and 3 days. Suppose their random duration deviations are PERFECTLY POSITIVELY CORRELATED, so both respond proportionately to the same uncertainty. What is the variance of their total duration?
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Explanation:
The variance of a sum includes covariance when random variables are correlated. For perfect positive correlation, the covariance equals the product of the two standard deviations. The standard deviation of the total is therefore 2 + 3 = 5 days. Squaring gives a total-duration variance of 25 days squared. Simply adding the individual variances would give 4 + 9 = 13 days squared and would ignore the positive covariance. This demonstrates why statistical independence is important in the usual PERT path-variance calculation.
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Question 21
ప్రశ్న 21
A PERT estimate reports the expected duration of an activity as 12 days and its activity variance as 9 days squared. Which interpretation of the UNITS is correct?
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Explanation:
Expected duration is a measure of central tendency and retains the original time unit. Variance is formed from squared deviations in duration. It therefore carries squared time units, such as days squared. Standard deviation is the square root of variance. It consequently returns to the original unit of days. For the stated variance of nine days squared, the standard deviation is three days.
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Question 22
ప్రశ్న 22
Which statement most accurately describes the relationship between conventional PERT three-point estimates and the BETA-distribution approximation?
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Explanation:
Traditional PERT commonly models uncertain activity durations using a beta-like bounded distribution. Its familiar expected-duration equation gives four times the weight to the most likely estimate. This weighted average is an approximation adopted for scheduling calculations. The three elicited duration values do not uniquely determine every possible probability-distribution shape without assumptions. Different distribution models can yield different means and variances. Therefore a PERT estimate should be treated as a modelling result rather than an exact guarantee.
వివరణ:
Question 23
ప్రశ్న 23
An activity's three PERT estimates are identical: optimistic = 4 days, most likely = 4 days and pessimistic = 4 days. Under the conventional PERT formulas, which pair is correct?
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Explanation:
The expected-duration formula becomes (4 + 4 × 4 + 4)/6. Its numerator is twenty-four, so the expected duration is four days. The pessimistic and optimistic estimates have no difference. The conventional standard deviation is therefore (4 - 4)/6 = 0. Squaring gives zero variance. These particular inputs model a fixed duration, although real-world certainty would still require appropriate evidence.
వివరణ:
Question 24
ప్రశ్న 24
A PERT project has A and B starting independently; C follows A, D follows B, and E follows BOTH C and D. The three-point estimates in days are A: (2, 5, 8), B: (2, 4, 6), C: (3, 6, 9), D: (3, 5, 7), and E: (2, 5, 8). Under conventional PERT, assume independent activity durations and calculate project variance using the critical path identified from expected times. Which result is correct?
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Explanation:
The expected durations are A = 5, B = 4, C = 6, D = 5 and E = 5 days. The A-C-E path has expected duration 5 + 6 + 5 = 16 days. The B-D-E path has expected duration 4 + 5 + 5 = 14 days. Hence A-C-E is the controlling path in the expected-duration network. Activities A, C and E each have variance [(6)/6] squared = 1 day squared, giving a critical-path variance of three days squared. This is the conventional PERT path-based approximation and is not necessarily the exact variance of the stochastic project's completion time.
వివరణ:
Question 25
ప్రశ్న 25
A PERT project has TWO parallel required paths with uncertain activity durations. A planner calculates the expected duration of each path and selects the longer expected-duration path as the controlling path. Which limitation must be recognised when interpreting the resulting project-duration estimate?
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Explanation:
In a network with parallel required paths, project completion occurs when the last required path finishes. The path that is longest in an individual realization can differ from the path with the largest expected duration. Conventional PERT commonly identifies one critical path using expected activity times. That path-based calculation does not fully model changes in which path governs completion. Mathematically, the expected maximum of uncertain path durations can exceed the maximum of their separate expected durations. Therefore the conventional estimate may understate the actual expected project completion time when competing paths can become controlling.
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Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. Optimistic time represents favourable conditions, most likely time represents the most probable duration, and pessimistic time represents adverse but plausible conditions. సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: A. a = 3 days, m = 8 days, b = 14 days. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: D. 7 days. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: B. 8 days. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: D. 2 days. సమాధానం: D.
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Question 6 ప్రశ్న 6Answer: C. 4 days squared. సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: B. X has an expected duration of 6 days and Y has an expected duration of 7 days. సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: A. 8 days. సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: C. 15 days. సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: D. Both have expected duration 6 days, but Y has variance 4 days squared, greater than X's 16/9 days squared. సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: B. 2 days. సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: A. Expected duration 6 days; variance 4 days squared. సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: D. 16 days. సమాధానం: D.
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Question 14 ప్రశ్న 14Answer: C. 13 days; B-D-E. సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: A. 11 days. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: D. Traditional PERT uses optimistic, most likely and pessimistic duration estimates to account for uncertainty, whereas deterministic CPM commonly uses a specified duration for each activity. సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: B. B controls with an expected-duration path of 8 days, although its most likely duration is smaller than A's. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. 6 days squared. సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: A. 5 days. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: B. 25 days squared, because the standard deviations add under perfect positive correlation. సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: D. Expected duration is measured in days, variance in days squared, and standard deviation in days. సమాధానం: D.
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Question 22 ప్రశ్న 22Answer: A. The formula (a + 4m + b)/6 is a conventional weighted approximation informed by beta-like duration modelling, not a guarantee of an exact mean for every possible beta distribution. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: C. Expected duration = 4 days; variance = 0 days squared. సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: B. Critical path A-C-E; expected duration 16 days; path variance 3 days squared. సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: D. The longest expected-duration path provides a conventional PERT scheduling approximation, but uncertainty and path switching can make the true expected completion time larger. సమాధానం: D.