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Question 1
ప్రశ్న 1
Under the classical single-critical-path PERT approximation, which expression gives the standard normal variable for an assumed project completion target T, when the selected critical path has expected duration mu and positive variance V?
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Explanation:
Classical PERT first estimates the expected duration and variance of a selected controlling path. The standard deviation is the positive square root of the path variance. A target time is standardized by subtracting the mean and dividing by that standard deviation. Therefore Z equals (T - mu)/sqrt(V). The cumulative standard normal function Phi(Z) gives the approximate probability of finishing by the target. For networks with several competing paths, this one-path probability need not equal the true project-completion probability.
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Question 2
ప్రశ్న 2
A pipeline-crossing activity has PERT time estimates a = 2 days, m = 8 days and b = 14 days. Using the conventional three-time formulas, what is its activity VARIANCE?
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Explanation:
Conventional PERT approximates activity variance as [(b - a)/6] squared. The pessimistic time is fourteen days and the optimistic time is two days. Their difference is twelve days. Dividing by six gives an estimated standard deviation of two days. Squaring two gives an activity variance of four days squared. The most likely time affects the expected duration but does not enter this conventional variance expression.
వివరణ:
Question 3
ప్రశ్న 3
Three activities form the ONLY sequential path of a project. Their estimated duration variances are 4, 9 and 3 days squared, and their durations are statistically independent. What is the standard deviation of the total project duration?
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Explanation:
For independent sequential activity durations, their variances are additive. The total project variance is 4 + 9 + 3 = 16 days squared. The standard deviation is the square root of the total variance. Taking the square root of sixteen gives four days. Adding the three individual standard deviations would be incorrect. The independence assumption is needed for the simple variance-addition rule.
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Question 4
ప్రశ్న 4
A project has ONLY three sequential activities with PERT estimates A: (2, 5, 8), B: (3, 6, 15) and C: (4, 7, 10) days, in optimistic-most likely-pessimistic order. Their random durations are independent. Which pair gives the project's expected duration and duration variance using conventional PERT?
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Explanation:
The expected durations are 5 days for A, 7 days for B and 7 days for C. Their sum is a project expected duration of nineteen days because all activities are sequential. The conventional variances are 1, 4 and 1 days squared respectively. With independent durations, their sum is six days squared. The corresponding project standard deviation is the square root of six days. Variance addition would not be justified without checking dependence among activity durations.
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Question 5
ప్రశ్న 5
A single-path PERT project is approximated by a normal completion time with mean 24 days and standard deviation 4 days. For target completion time T = 28 days, what is the approximate probability of finishing BY the target? Use Phi(1.00) = 0.8413.
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Explanation:
The standardized target is Z = (28 - 24)/4. This gives Z = 1.00, one standard deviation above the expected completion time. The supplied cumulative normal value is Phi(1.00) = 0.8413. Hence the approximate probability of completing by day 28 is 84.13 percent. The complement, 15.87 percent, represents finishing after the target under the assumed normal model. This is an approximation based on the stated completion-time distribution.
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Question 6
ప్రశ్న 6
A single-path PERT project has approximately normal duration with mean 35 days and standard deviation 5 days. What is the probability of completing BY day 30? Use Phi(-1.00) = 0.1587.
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Explanation:
The target is five days earlier than the expected completion. The standardized variable is Z = (30 - 35)/5 = -1. The supplied standard normal cumulative value at -1 is 0.1587. Thus the approximate probability of finishing by day 30 is 15.87 percent. The probability of finishing later is the complement, 84.13 percent. A negative Z value indicates a target earlier than the assumed mean, not a negative activity duration.
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Question 7
ప్రశ్న 7
Under a symmetric normal approximation, a single-path project's expected completion time is 40 days, and its standard deviation is 6 days. What is the probability of completion BY day 40?
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Explanation:
The specified target equals the normal distribution's mean. Standardization gives Z = (40 - 40)/6 = 0. A normal distribution has half its probability at or below its mean. Therefore Phi(0) = 0.5 and the completion probability is fifty percent. An expected date is not the same as a guaranteed deadline. This interpretation applies to the normal model assumed in the question.
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Question 8
ప్రశ్న 8
For a single-path project, the classical PERT expected duration is 30 days and estimated project variance is 16 days squared. Under the normal approximation, what is the completion probability BY day 38? Use Phi(2.00) = 0.9772.
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Explanation:
The standard deviation equals the square root of project variance. It is therefore four days. Standardizing day 38 gives Z = (38 - 30)/4 = 2. The supplied standard normal cumulative value is 0.9772. Hence the approximate probability of completing by day 38 is 97.72 percent. The chance of exceeding the target is about 2.28 percent under the same assumptions.
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Question 9
ప్రశ్న 9
A single-path project's completion duration is approximated as normal with mean 28 days and standard deviation 3 days. What is the probability that completion takes MORE THAN 25 days? Use Phi(1.00) = 0.8413 and normal symmetry.
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Explanation:
The target day 25 is one standard deviation below the mean of day 28. Its standardized score is Z = (25 - 28)/3 = -1. Normal symmetry gives Phi(-1) = 1 - Phi(1) = 0.1587. This is the probability of completion on or before day 25. The question instead asks for completion after day 25. Its probability is 1 - 0.1587 = 0.8413, or 84.13 percent.
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Question 10
ప్రశ్న 10
A project with one sequential PERT path has assumed normally distributed completion time with mean 20 days and standard deviation 2 days. What is the probability of completing BY day 16? Use Phi(-2.00) = 0.0228.
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Explanation:
The specified deadline is four days shorter than the expected project duration. Standardization gives Z = (16 - 20)/2 = -2. The corresponding cumulative normal probability is 0.0228. Therefore completion by day 16 has an approximate probability of 2.28 percent. The complementary probability of missing this deadline is 97.72 percent. This computation assumes that the stated single-path normal approximation is appropriate.
వివరణ:
Question 11
ప్రశ్న 11
A project has a single assumed normal completion-time distribution with mean 30 days and standard deviation 4 days. What target time corresponds to approximately 90% probability of completing BY that target? Use the supplied 90th-percentile normal score Z = 1.28.
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Explanation:
A target percentile can be converted from its standard normal score. The required relation is T = mu + Z times sigma. The supplied score for approximately ninety percent reliability is 1.28. Substitution gives T = 30 + 1.28 × 4 = 35.12 days. This is the approximate deadline corresponding to the desired one-sided completion probability. The calculation does not introduce any extra contingency beyond the specified normal model.
వివరణ:
Question 12
ప్రశ్న 12
A single-path PERT project has a normally approximated completion duration with mean 26 days and standard deviation 3 days. The engineer must select a WHOLE-NUMBER day target that provides at least 95% probability of completion under this model. Using one-sided Z = 1.645 for 95%, what is the minimum acceptable integer-day target?
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Explanation:
The continuous ninety-fifth percentile is T = mu + 1.645 sigma. Substituting the stated mean and standard deviation gives 26 + 1.645 × 3 = 30.935 days. A target of thirty days would fall below this percentile and provide less than ninety-five percent probability. Since targets must be whole-number days, the smallest acceptable integer is the next higher day. Thus a thirty-one-day target satisfies the stated probability requirement. This is a one-sided percentile, not the two-sided normal interval value of 1.96.
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Question 13
ప్రశ్న 13
Under the normal approximation, a single-path project has standard deviation 4 days. A target of 36 days corresponds to approximately 84.13% probability of completion BY that target. Given Phi(1.00) = 0.8413, what is the implied expected project duration?
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Explanation:
A cumulative probability of 0.8413 corresponds to Z = 1.00. Standardization gives 1 = (36 - mu)/4. Multiplying by four yields 4 = 36 - mu. Solving gives an expected project duration of 32 days. This means the target lies one standard deviation above the mean. A normal completion-time interpretation is required to infer the mean from this probability.
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Question 14
ప్రశ్న 14
A single-path PERT schedule has expected completion of 24 days. The approximate probability of completion BY day 30 is 97.72%. Using Phi(2.00) = 0.9772, what are the corresponding PROJECT STANDARD DEVIATION and VARIANCE under the normal approximation?
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Explanation:
A completion probability of 97.72 percent corresponds to a standard normal score of two. The target exceeds the mean by 30 - 24 = 6 days. Therefore Z = 6/sigma = 2. Solving gives sigma = 3 days. The project variance is the square of the standard deviation, or nine days squared. This inference assumes the provided normal completion-time model is appropriate.
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Question 15
ప్రశ్న 15
A project contains ONLY three independent sequential activities with PERT estimates A: (2, 5, 8), B: (1, 4, 13), and C: (3, 6, 15) days. With conventional PERT expected time and variance formulas, estimate the completion probability BY day 20 using a normal approximation. Use Phi(1.00) = 0.8413.
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Explanation:
The expected activity durations are five, five and seven days respectively. Their sum is the expected serial project duration of seventeen days. The corresponding variances are one, four and four days squared. Their sum is nine days squared, giving a project standard deviation of three days. The target score is Z = (20 - 17)/3 = 1. The given normal cumulative probability is therefore 84.13 percent.
వివరణ:
Question 16
ప్రశ్న 16
An AON PERT project has A and B starting independently, C following A, D following B, and E following BOTH C and D. Its (optimistic, most likely, pessimistic) estimates in days are A: (2,5,8), B: (1,4,7), C: (1,4,13), D: (2,5,8), and E: (3,6,15). Assuming independent durations, what is the CLASSICAL SINGLE-CRITICAL-PATH PERT approximation for completing the project BY day 20? Use Phi(1.00) = 0.8413.
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Explanation:
Expected durations are A = 5, B = 4, C = 5, D = 5 and E = 7 days. The path A-C-E totals seventeen days, exceeding the B-D-E total of sixteen days. The selected expected-duration critical path has variances 1, 4 and 4 days squared, giving variance nine and standard deviation three days. For target twenty days, Z = (20 - 17)/3 = 1. The classical one-path PERT result is Phi(1) = 84.13 percent. Because the other near-critical path is uncertain, this result must not be presented as an exact probability for the entire merged network.
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Question 17
ప్రశ్న 17
Two INDEPENDENT parallel paths each have a normally distributed completion time with mean 20 days and standard deviation 2 days. Both begin at time zero and BOTH must finish for project completion. Under these stated distributional assumptions, what is the exact probability that the PROJECT completes BY day 20?
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Explanation:
For either normal path, a deadline equal to its mean has completion probability fifty percent. Project completion requires both parallel paths to finish by the deadline. The question explicitly states that the two path durations are independent. Their joint success probability is therefore 0.5 multiplied by 0.5. This gives 0.25, or twenty-five percent. Treating the project as only one representative path would incorrectly give fifty percent and overlook merge risk.
వివరణ:
Question 18
ప్రశ్న 18
A project consists of only TWO independent required parallel work packages. Their probabilities of completion BY the same specified deadline are 0.80 and 0.90 respectively. What is the probability that the WHOLE PROJECT finishes by the deadline?
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Explanation:
Project completion requires both work packages to meet the target. The two required completion events are independent according to the question. Therefore the joint probability equals the product of their separate probabilities. Multiplication gives 0.80 × 0.90 = 0.72. The whole-project success probability is seventy-two percent. Averaging the two probabilities would not correctly represent a project that needs both packages.
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Question 19
ప్రశ్న 19
Two sequential activity durations X and Y have variances 4 and 9 days squared, and covariance Cov(X,Y) = +3 days squared. What is the variance of their TOTAL duration, without assuming independence?
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Explanation:
The variance of a sum includes covariance when durations are dependent. The general formula is Var(X + Y) = Var(X) + Var(Y) + 2 Cov(X,Y). Substituting the given values gives 4 + 9 + 2 × 3. The resulting variance is nineteen days squared. Simply adding four and nine would ignore positive correlation. This is why the usual PERT addition of path variances requires a suitable independence assumption.
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Question 20
ప్రశ్న 20
An uncertain network has multiple parallel paths converging at a required completion milestone, with several near-critical paths. Which conclusion about the CLASSICAL single-expected-critical-path PERT probability estimate is most defensible?
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Explanation:
The project finishes only when every required branch leading to its finish is complete. Different uncertain paths may control in different realizations. Classical PERT often approximates completion risk using only the path longest in expected duration. That practice may overlook a shorter expected path that runs late and delays the merge. The resulting one-path completion probability can be larger than the whole-project probability. Simulation or a suitable network-wide probabilistic model can examine the changing controlling path more realistically.
వివరణ:
Question 21
ప్రశ్న 21
The completion duration of a single-path project is approximately normal with mean 18 days and standard deviation 3 days. What is the probability that completion occurs BETWEEN days 15 and 21, inclusive in the continuous-model sense? Use Phi(1.00) = 0.8413 and Phi(-1.00) = 0.1587.
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Explanation:
The lower target is one standard deviation below the mean, giving Z = -1. The upper target is one standard deviation above the mean, giving Z = 1. The probability between the two values is Phi(1) minus Phi(-1). Using the supplied figures gives 0.8413 - 0.1587 = 0.6826. Therefore the approximate probability of completion between days 15 and 21 is 68.26 percent. In the continuous normal model, including or excluding the exact endpoints does not change the probability.
వివరణ:
Question 22
ప్రశ్న 22
A single-path project's duration is approximately normal with mean 24 days and standard deviation 3 days. What is the probability that it finishes OUTSIDE the interval from day 21 to day 27? Use Phi(-1.00) = 0.1587 and Phi(1.00) = 0.8413.
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Explanation:
The interval from day 21 to day 27 lies one standard deviation either side of the mean. Standardization gives lower score -1 and upper score +1. The probability within this interval is 0.8413 - 0.1587 = 0.6826. The probability outside the interval is its complement, 1 - 0.6826 = 0.3174. Therefore the required probability is 31.74 percent. It includes both unusually early and unusually late completion under the model.
వివరణ:
Question 23
ప్రశ్న 23
A single-path project has approximately normal completion duration with mean 40 days and standard deviation 5 days. Using the supplied approximate 80th-percentile standard normal score Z = 0.84, which TARGET duration gives about 80% completion reliability?
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Explanation:
A target at a chosen normal percentile is obtained from T = mu + Z sigma. The question supplies mean forty days, standard deviation five days and score 0.84. The required addition to the mean is 0.84 × 5 = 4.20 days. Therefore the target is 40 + 4.20 = 44.20 days. This provides approximately eighty percent completion probability under the assumed normal distribution. A different desired probability would require a different standard normal percentile.
వివరణ:
Question 24
ప్రశ్న 24
A project consisting of one activity originally has PERT estimates (a,m,b) = (2,8,14) days. A revised assessment gives (5,8,11) days. Using the conventional PERT formulas and normal approximation, what happens to the predicted probability of completion BY day 10? Use Phi(1.00) = 0.8413 and Phi(2.00) = 0.9772.
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Explanation:
The original expected time is (2 + 4 × 8 + 14)/6 = 8 days. The revised expected time is also eight days, so the mean does not change. Originally the standard deviation is (14 - 2)/6 = 2 days, making the day-ten Z score equal to one. The revised standard deviation is (11 - 5)/6 = 1 day, making the Z score equal to two. The supplied cumulative probabilities are 84.13 and 97.72 percent respectively. The narrower activity-time range produces a higher predicted chance of meeting a target above the unchanged mean.
వివరణ:
Question 25
ప్రశ్న 25
An activity is modelled with identical PERT estimates a = m = b = 12 days, and it is the ONLY activity in the project. Under the conventional PERT formulas, what is the proper interpretation of completion probabilities for deadlines of 11 and 12 days?
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Explanation:
The PERT expected duration is (12 + 4 × 12 + 12)/6 = 12 days. Its conventional variance is [(12 - 12)/6] squared, which equals zero. The standard deviation is therefore zero, making division by sigma in the ordinary Z formula undefined. With the resulting fixed-duration model, completion is always at time twelve days. Consequently completion by day eleven has probability zero and completion by day twelve has probability one. A nondegenerate normal approximation should not be imposed when the estimated variance is zero.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. Z = (T - mu)/sqrt(V), with approximate probability of completion by T equal to Phi(Z). సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: A. 4 days squared. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: D. 4 days. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: B. 19 days and 6 days squared. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: A. 84.13%. సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: C. 15.87%. సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: B. Exactly 50% under the stated normal model. సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: D. 97.72%. సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: C. 84.13%. సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: A. 2.28%. సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: D. Approximately 35.12 days. సమాధానం: D.
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Question 12 ప్రశ్న 12Answer: B. 31 days. సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: A. 32 days. సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: C. 3 days and 9 days squared. సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: D. 84.13%. సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: B. 84.13%, based on expected-duration path A-C-E of 17 days and its variance of 9 days squared. సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: C. 25%. సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. 72%. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: D. 19 days squared. సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: B. It can be optimistic because another path may become controlling; network-wide simulation can assess the competing-path risk. సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: A. 68.26%. సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: C. 31.74%. సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: D. 44.20 days. సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: B. It increases from 84.13% to 97.72% because the mean stays 8 days and the estimated standard deviation falls from 2 to 1 day. సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: C. Variance is zero, so the usual Z ratio is undefined; completion by day 11 has probability 0 and by day 12 has probability 1 in the resulting fixed-duration model. సమాధానం: C.