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Question 1
ప్రశ్న 1
For an isolated fixed-time signalized intersection operating within the usual range of Webster’s method, which expression gives the approximate minimum-delay cycle length C0 in seconds when L is the total lost time per cycle and Y is the sum of critical flow ratios?
Explanation:
Webster’s classical approximate optimum-cycle expression is C0 = (1.5L + 5)/(1 - Y). The quantity L is the sum of the effective lost times of the relevant sequential phases in seconds. The quantity Y is the sum of their critical demand-to-saturation-flow ratios. The formula estimates a cycle associated with low average delay under its modelling assumptions. Practical design must also check pedestrian service, queues, clearance intervals and coordination constraints.
వివరణ:
Question 2
ప్రశ్న 2
For a two-phase signal, each phase serves two compatible lane groups concurrently without shared-lane constraints. How should the critical flow ratio for each phase generally be identified for a basic Webster calculation?
Explanation:
Critical-movement analysis identifies the lane group that places the greatest normalized demand on each relevant phase. Concurrent compatible lane groups share their green rather than automatically requiring separate green allocations. Consequently the largest applicable flow ratio governs a simple phase-critical calculation in the stated arrangement. The critical ratios for sequential phases are then added to obtain Y. Shared lanes, overlaps and complex phasing can require more detailed critical-path analysis.
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Question 3
ప్రశ్న 3
A signalized traffic lane has an observed saturation discharge headway of 2.25 seconds per vehicle after start-up. What is its corresponding saturation flow rate, assuming the headway remains uniform?
Explanation:
Saturation flow is the equivalent hourly discharge rate during sustained queued flow. When saturation headway h is measured in seconds per vehicle, the rate is 3600/h. Substituting h = 2.25 seconds gives 3600/2.25 = 1600 vehicles per hour. The result applies to one lane under the stated discharge conditions. Initial start-up lost time is accounted for separately rather than being added to every saturation headway.
వివరణ:
Question 4
ప్రశ్న 4
Two compatible lane groups are served simultaneously during one phase, with no shared-lane restrictions. Group P carries 540 vehicles/h at a saturation flow of 1800 vehicles/h, and group Q carries 400 vehicles/h at a saturation flow of 1600 vehicles/h. What is the phase-critical flow ratio?
Explanation:
The demand-to-saturation-flow ratio for group P is 540/1800 = 0.30. The corresponding ratio for group Q is 400/1600 = 0.25. Both compatible groups receive green simultaneously under the stated assumption. The larger ratio of 0.30 therefore governs this phase for a simple critical-group calculation. Adding the two ratios would incorrectly allocate separate phase time to movements that are served together.
వివరణ:
Question 5
ప్రశ్న 5
A three-phase Webster analysis gives phase-critical flow ratios of 0.25, 0.30 and 0.20 for three sequential phases. What is the sum of critical flow ratios Y?
Explanation:
The total critical flow ratio Y is the sum of the critical flow ratios for the sequential phases. The three values are 0.25, 0.30 and 0.20. Their sum is 0.25 + 0.30 + 0.20 = 0.75. This total enters the denominator of Webster’s optimum-cycle expression. It is not the volume-to-capacity ratio of any one approach.
వివరణ:
Question 6
ప్రశ్న 6
A signal has three sequential phases, each with 2 seconds of start-up lost time and 2 seconds of clearance lost time. What is the total lost time L per cycle for the adopted Webster calculation?
Explanation:
Lost time for one phase is the sum of its start-up and clearance losses. Each phase therefore contributes 2 + 2 = 4 seconds. There are three sequential phases within the cycle. The total is L = 3 times 4 = 12 seconds. Displayed amber and all-red periods should not be added again if their unusable portions have already been represented by clearance lost time.
వివరణ:
Question 7
ప్రశ్న 7
Using Webster’s expression C0 = (1.5L + 5)/(1 - Y), determine the optimum cycle length for L = 10 seconds and Y = 0.60.
Explanation:
The numerator of Webster’s formula is 1.5 times 10 plus 5. This equals 20 seconds. The denominator is 1 - 0.60 = 0.40. Therefore C0 = 20/0.40 = 50 seconds. The computation assumes that the specified Y and lost time are appropriate for the proposed phasing arrangement.
వివరణ:
Question 8
ప్రశ్న 8
A proposed signal has total lost time L = 12 seconds and sum of critical flow ratios Y = 0.70. Under Webster’s formula, what is the approximate optimum cycle length to one decimal place?
Explanation:
The numerator is 1.5 times 12 plus 5 = 23 seconds. The denominator is 1 - 0.70 = 0.30. Dividing gives C0 = 23/0.30 = 76.6667 seconds. To one decimal place the result is 76.7 seconds. The computed value remains a preliminary optimum estimate subject to operational and pedestrian checks.
వివరణ:
Question 9
ప్రశ్న 9
A three-phase fixed-time signal has critical lane-group flows of 300, 450 and 750 vehicles/h, with a saturation flow of 2000 vehicles/h for each critical group. Each phase has 4 seconds of lost time. Applying Webster’s optimum-cycle formula and splitting effective green in proportion to the critical flow ratios, which complete design result is correct?
Explanation:
The phase-critical ratios are 300/2000 = 0.15, 450/2000 = 0.225 and 750/2000 = 0.375, giving Y = 0.75. Three phases with 4 seconds lost per phase give L = 12 seconds. Webster’s formula yields C0 = (1.5 times 12 + 5)/(1 - 0.75) = 92 seconds. Available effective green is 92 - 12 = 80 seconds, which is allocated in the ratio 0.15:0.225:0.375. The resulting phase-effective greens are 16, 24 and 40 seconds, whose sum is 80 seconds.
వివరణ:
Question 10
ప్రశ్న 10
A fixed-time signal has an adopted cycle length of 84 seconds and a total effective lost time of 12 seconds per cycle. Under the usual sequential-phase timing convention, how much total effective green time is available?
Explanation:
The usable effective green time over a cycle is C - L for the specified sequential phases. The cycle length is 84 seconds. Total lost time is 12 seconds. Consequently the total effective green is 84 - 12 = 72 seconds. That effective time must be shared among the required phases while respecting safety and minimum service constraints.
వివరణ:
Question 11
ప్రశ్న 11
A two-phase signal has an adopted, not necessarily Webster-optimum, cycle length of 60 seconds, lost time L = 10 seconds and critical flow ratios y1 = 0.24 and y2 = 0.36. Using proportional Webster green splitting, what are the two effective green times?
Explanation:
The sum of critical ratios is Y = 0.24 + 0.36 = 0.60. Available effective green is 60 - 10 = 50 seconds. Phase 1 receives (0.24/0.60) times 50 = 20 seconds. Phase 2 receives (0.36/0.60) times 50 = 30 seconds. Both effective greens add to the available 50 seconds rather than the full 60-second cycle.
వివరణ:
Question 12
ప్రశ్న 12
A three-phase signal has an adopted cycle of 80 seconds, total lost time of 8 seconds and critical ratios of 0.15, 0.25 and 0.20. What are the proportional effective green allocations, in phase order?
Explanation:
The available effective green is 80 - 8 = 72 seconds. The critical ratios sum to 0.15 + 0.25 + 0.20 = 0.60. Phase 1 receives 72 times 0.15/0.60 = 18 seconds. Phase 2 receives 72 times 0.25/0.60 = 30 seconds, and Phase 3 receives 72 times 0.20/0.60 = 24 seconds. The allocated values sum to 72 seconds as required.
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Question 13
ప్రశ్న 13
For a signal phase, effective green g = 24 seconds, amber Yc = 3 seconds, all-red R = 1 second, start-up lost time l1 = 2 seconds and clearance lost time l2 = 3 seconds. Using g = G + Yc + R - l1 - l2, what displayed green G is required?
Explanation:
The given relationship defines effective green using the displayed intervals and phase lost times. Rearranging gives G = g - Yc - R + l1 + l2. Substitution gives G = 24 - 3 - 1 + 2 + 3. This yields a displayed green of 25 seconds. Effective and displayed green need not be numerically equal because the end-of-phase intervals and lost times can differ.
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Question 14
ప్రశ్న 14
For one phase the start-up lost time is 2.5 seconds and clearance lost time is 1.5 seconds. Its amber indication lasts 3 seconds and its all-red indication lasts 2 seconds. What phase lost time should be included in Webster’s total L using the stated lost-time convention?
Explanation:
Phase lost time is the sum of start-up and clearance losses under the stated convention. The two components are 2.5 seconds and 1.5 seconds. Their sum is 4.0 seconds. Amber and all-red display times are relevant to the signal timing, but their unused components are already represented in clearance lost time. Adding all displayed clearance intervals again would double-count part of the loss.
వివరణ:
Question 15
ప్రశ్న 15
For a signal with fixed total lost time L = 10 seconds, Webster’s expression is used first with Y = 0.50 and then with Y = 0.75. What are the respective optimum cycles?
Explanation:
With L = 10 seconds, the common Webster numerator is 1.5 times 10 plus 5 = 20 seconds. For Y = 0.50 the denominator is 0.50, giving C0 = 40 seconds. For Y = 0.75 the denominator is 0.25, giving C0 = 80 seconds. The resulting cycle doubles because the denominator halves. This comparison demonstrates the sensitivity of the classical formula to increasing critical flow ratios.
వివరణ:
Question 16
ప్రశ్న 16
A proposed isolated fixed-time intersection has Y = 1.02 and a positive total lost time L. Which engineering interpretation of Webster’s optimum-cycle expression is most defensible?
Explanation:
Webster’s formula has the denominator 1 - Y. If Y = 1.02, that denominator is negative. A negative optimum cycle is not a physically useful signal setting. The critical demand exceeds the available saturation-flow fraction in the adopted simplified phasing framework. The engineer should reassess demand, lane-group capacity, staging or other treatment instead of accepting the numerical output.
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Question 17
ప్రశ్న 17
For a signal with Y = 0.75, total lost time L = 12 seconds and adopted cycle C = 60 seconds, compute the illustrative critical degree-of-saturation measure X = Y/[1 - L/C].
Explanation:
Effective-green fraction available across the critical phases is 1 - L/C. With L = 12 seconds and C = 60 seconds, this fraction is 1 - 12/60 = 0.80. The stated critical degree-of-saturation measure is therefore 0.75/0.80. This equals 0.9375, or approximately 0.94. It differs from Y because the intersection cannot use the entire cycle for effective green.
వివరణ:
Question 18
ప్రశ్న 18
A signalized lane group has saturation flow s = 1800 vehicles/h, allocated effective green g = 24 seconds and cycle length C = 60 seconds. If its demand is 600 vehicles/h, which capacity and volume-to-capacity ratio are correct using c = s(g/C)?
Explanation:
The effective green ratio is 24/60 = 0.40. Multiplying this ratio by saturation flow 1800 gives capacity c = 720 vehicles per hour. The demand is 600 vehicles per hour. Therefore v/c = 600/720 = 0.8333, approximately 0.83. This capacity check is necessary because a computed green split does not by itself confirm acceptable queueing or operational performance.
వివరణ:
Question 19
ప్రశ్న 19
For sequential phases, assume the available effective green is C - L and phase greens can be allocated in the proportions y_i/Y. If total lost time is L = 12 seconds and critical sum Y = 0.75, what is the minimum theoretical cycle length ensuring the summed critical demand does not exceed the available effective-green share?
Explanation:
The maximum effective-green fraction available is (C - L)/C. A necessary capacity condition in the stated simplified model is Y = L and hence C >= L/(1 - Y). Substituting L = 12 and Y = 0.75 gives 12/0.25 = 48 seconds. This capacity-feasibility bound is not Webster’s minimum-delay optimum, which uses a different numerator.
వివరణ:
Question 20
ప్రశ్న 20
Two compatible approaches can operate concurrently in a proposed signal phase without opposing or pedestrian conflicts. Their critical flow ratios are 0.22 and 0.31. Which treatment is appropriate when constructing a basic Webster critical-path analysis for that shared phase?
Explanation:
Compatible lane groups can discharge during the same green interval. They need not receive separate nonoverlapping green portions solely because they are on different approaches. In the specified arrangement the larger ratio of 0.31 governs the required phase share. Summing 0.22 and 0.31 would overstate the demand for sequential phase time. Shared lanes, overlaps and other constraints must be checked before using this simple rule.
వివరణ:
Question 21
ప్రశ్న 21
Webster’s calculation produces a phase-effective green that is shorter than the required pedestrian crossing service time for a movement scheduled with that phase. Which design response is appropriate?
Explanation:
Webster’s result provides an approximate traffic-delay-based timing starting point. It does not override pedestrian safety and crossing-clearance requirements. If a phase needs more time than its proportional allocation, the timing plan must be adjusted. Other phases and the total cycle may also need revision to preserve service and clearance. The revised plan should be checked for capacity, delay, queue storage and safe operation.
వివరణ:
Question 22
ప్రశ్న 22
Two sequential signal phases have critical lane-group flows and saturation flows as follows: Phase 1 has 450 and 1500 vehicles/h; Phase 2 has 480 and 2400 vehicles/h. What is the correct sum of critical flow ratios Y?
Explanation:
Each phase-critical flow ratio uses its own corresponding saturation flow. Phase 1 has y1 = 450/1500 = 0.30. Phase 2 has y2 = 480/2400 = 0.20. Therefore Y = 0.30 + 0.20 = 0.50. Combining all demands and dividing by the sum of saturation flows would not give the required sum of phase-critical ratios.
వివరణ:
Question 23
ప్రశ్న 23
A two-phase signal has start-up lost times of 2.0 and 1.5 seconds and clearance lost times of 2.5 and 2.0 seconds, respectively. What total cycle lost time L should be used in Webster’s expression?
Explanation:
Total cycle lost time is the sum of the relevant lost-time components of both sequential phases. Phase 1 contributes 2.0 + 2.5 = 4.5 seconds. Phase 2 contributes 1.5 + 2.0 = 3.5 seconds. Adding these gives L = 4.5 + 3.5 = 8.0 seconds. The phase component values should not be confused with their displayed amber or all-red durations.
వివరణ:
Question 24
ప్రశ్న 24
A two-phase signal has phase-critical flows of 450 and 630 vehicles/h, each with saturation flow of 1800 vehicles/h. The total lost time is 10 seconds per cycle. Using Webster’s optimum-cycle expression followed by proportional effective-green allocation, which is the correct result to one decimal place?
Explanation:
The critical ratios are y1 = 450/1800 = 0.25 and y2 = 630/1800 = 0.35, so Y = 0.60. Webster’s cycle is C0 = (1.5 times 10 + 5)/(1 - 0.60) = 50 seconds. Available effective green is 50 - 10 = 40 seconds. Phase 1 receives 40 times 0.25/0.60 = 16.6667 seconds. Phase 2 receives 40 times 0.35/0.60 = 23.3333 seconds, giving approximately 16.7 and 23.3 seconds.
వివరణ:
Question 25
ప్రశ్న 25
A Webster calculation for a heavily loaded isolated junction gives a very long optimum cycle, while the site also has pedestrian crossing requirements, short turning-bay storage and coordination constraints. Which engineering judgement is most appropriate?
Explanation:
Webster’s optimum-cycle equation is an approximate method developed for specified operating assumptions. It becomes very sensitive when the sum of critical flow ratios approaches unity. Longer cycles can create large queues and may conflict with pedestrian or network requirements. The calculated timing should therefore be checked against safety, capacity, storage and coordination constraints. If the scheme is infeasible, alternative phase arrangements, geometric improvements or demand-management measures may be needed.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. C0 = (1.5L + 5)/(1 - Y). సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. Use the largest demand-to-saturation-flow ratio among lane groups served concurrently in each phase, then sum the two phase-critical ratios. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. 1600 vehicles/h/lane. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 0.30. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 0.75. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. 12 seconds. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: D. 50 seconds. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: C. 76.7 seconds. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: A. C0 = 92 seconds; effective greens = 16, 24 and 40 seconds. సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: B. 72 seconds. సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: D. 20 seconds and 30 seconds. సమాధానం: D.
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Question 12 ప్రశ్న 12Answer: C. 18, 30 and 24 seconds. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. 25 seconds. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: A. 4.0 seconds. సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: D. 40 seconds and 80 seconds. సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: B. Do not treat the resulting nonphysical cycle as a valid design; review demand, capacity, geometry or phasing because Y exceeds unity. సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: C. Approximately 0.94. సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. Capacity 720 vehicles/h and v/c approximately 0.83. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: C. 48 seconds. సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: D. Use the governing concurrent critical ratio of 0.31, subject to verifying the actual lane-group and movement compatibility. సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: A. Apply the pedestrian minimum service constraint and revise the green split or cycle as needed, then recheck the whole timing plan. సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: C. 0.50. సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: B. 8.0 seconds. సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: D. C0 = 50 seconds; g1 = 16.7 seconds and g2 = 23.3 seconds. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: A. Treat the Webster cycle as a preliminary estimate and compare feasible timings and geometric or operational alternatives against capacity, queueing, pedestrian safety and coordination needs. సమాధానం: A.
