/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A queue, or line, of traffic can... [FREE SOLUTION] | 91Ó°ÊÓ

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A queue, or line, of traffic can form when a feeder road meets a main road with a high volume of traffic. In this exercise we assume that gaps in traffic on the main road (allowing cars to enter from the feeder road) appear randomly and that cars on the feeder road arrive at the intersection randomly also. We let \(s\) be the average rate at which a gap in traffic appears and \(a\) the average arrival rate, both per minute. Assume that \(s\) is greater than \(a\). a. The average length \(L\) of a queue is given by the rational function $$ L=\frac{a^{2}}{s(s-a)} . $$ Explain what happens to average queue length if arrival rate and gap rate are nearly the same. b. The average waiting time \(w\) (the time in minutes spent in the queue) is given by $$ w=\frac{a}{s(s-a)} . $$ If the gap rate is 3 per minute, what arrival rate will result in a driver's average waiting time of 2 minutes?

Short Answer

Expert verified
Near equal rates make queue long; arrival rate for 2 min wait is about 2.57 cars/min.

Step by step solution

01

Understanding Queue Length Equation

The average queue length is given by \( L = \frac{a^2}{s(s-a)} \). When the arrival rate \( a \) approaches the gap rate \( s \), the denominator \( s-a \) becomes very small, which causes \( L \) to become very large. This leads to a significant increase in the average queue length.
02

Analyzing Waiting Time Equation

The waiting time \( w \) is given by \( w = \frac{a}{s(s-a)} \). We need to determine \( a \) for \( w = 2 \) when \( s = 3 \).
03

Substitute Known Values

Substitute \( w = 2 \) and \( s = 3 \) into the waiting time formula: \[ 2 = \frac{a}{3(3-a)} \].
04

Solve for Arrival Rate

First, multiply both sides by \( 3(3-a) \): \[ 2 \cdot 3(3-a) = a \].Next, simplify and rearrange the equation:\[ 6(3-a) = a \]\[ 18 - 6a = a \]Combine like terms:\[ 18 = 7a \]Finally, solve for \( a \):\[ a = \frac{18}{7} \approx 2.57 \] (cars per minute).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Traffic Flow
Traffic flow refers to the movement of vehicles or pedestrians through intersections and roadways. In queue theory, especially where feeder roads meet busy highways, traffic flow can considerably influence the formation of queues. This concept is particularly important in understanding how vehicles merge into main traffic streams.

When considering traffic flow in terms of probability, both vehicles arriving at an intersection and the gaps available in the main traffic flow behave randomly. The feeder road vehicles attempt to enter traffic whenever there is a sufficient gap. If these gaps are infrequent or short, queues can quickly build up.
  • Gap rate \( s \): This is the rate at which gaps occur in the main road traffic. More frequent gaps mean more chances for cars from a feeder road to join the traffic, thus optimizing traffic flow.
  • Arrival rate \( a \): This is the frequency of cars arriving at the intersection from the feeder road. When this rate is higher than the gap rate, it results in congestion and longer queues.
Understanding these two rates helps in designing better intersection controls and traffic lights, aiming to balance the flow and reduce congestion.
Rational Functions
In the context of traffic, a rational function is utilized to understand how queue length and waiting time react to changes in arrival and gap rates. A rational function is simply a ratio of two polynomials. In this case, the functions take into account how many cars arrive and how often gaps occur on the main road.

The formula for the average queue length \( L \) is a classic example: \[ L = \frac{a^2}{s(s-a)} \] This illustrates a rational function because it divides two polynomials of arrival rate \( a \) and gap rate \( s \).
  • If \( a \) (arrival rate) increases relative to \( s \) (gap rate), the denominator \( s-a \) shrinks, which drastically increases \( L \), implying longer queues.
  • The form of these equations helps visualize how sensitive the system is to small changes, especially as the arrival rate nears the gap rate.
Rational functions are crucial in predicting and optimizing systems prone to congestion, such as traffic flow, by showing potential issues before they occur.
Waiting Time Calculation
Waiting time is an essential metric in traffic management, representing the average duration vehicles spend in a queue before merging into the main road. Calculating this time helps in assessing and improving the efficiency of traffic systems. In our exercise, the average waiting time \( w \) is given by:\[ w = \frac{a}{s(s-a)} \]
The formula helps traffic engineers determine how changes in arrival and gap rates impact waiting times. This is especially useful in planning signal timings or road expansions.
  • To find the arrival rate \( a \) that corresponds to a specific waiting time, such as 2 minutes when \( s = 3 \), you plug these values into the equation to solve for \( a \).
  • This results in setting up the equation: \[ 2 = \frac{a}{3(3-a)} \] Solving it can help find the critical rate \( a \) where significant waiting is reduced.
In this problem, the solution showed that an arrival rate of approximately \( 2.57 \) cars per minute leads to the desired average waiting time of 2 minutes. This shows the delicate balance required in traffic management to minimize delays.

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Most popular questions from this chapter

For traffic moving along a highway, we use \(q\) to denote the mean flow rate. That is the average number of vehicles per hour passing a certain point. We let \(q_{m}\) denote the maximum flow rate, \(k\) the mean traffic density (that is, the average number of vehicles per mile), and \(k_{m}\) the density at which flow rate is a maximum (that is, the value of \(k\) when \(q=q_{m}\) ). a. An important measurement of traffic on a highway is the relative density \(R\), which is defined as $$ R=\frac{k}{k_{m}}. $$ i. What does a value of \(R<1\) indicate about traffic on a highway? ii. What does a value of \(R>1\) indicate about traffic on a highway? b. Let \(u\) denote the mean speed of vehicles on the road and \(u_{f}\) the free speed -that is, the speed when there is no traffic congestion at all. One study \({ }^{43}\) proposes the following relation between density and speed: $$ u=u_{f} e^{-0.5 R^{2}}. $$ Use function composition to find a formula that directly relates mean speed to mean traffic density. c. Make a graph of mean speed versus mean traffic density, assuming that \(k_{m}\) is 122 cars per mile and \(u_{f}\) is 75 miles per hour. (Include values of mean traffic density up to 250 vehicles per mile.) Paying particular attention to concavity, explain the significance of the point \(k=122\) on the graph. d. Traffic is considered to be seriously congested if the mean speed drops to 35 miles per hour. Use the graph from part c to determine what density will result in serious congestion.

If you invest \(P\) dollars (the present value of your investment) in a fund that pays an interest rate of \(r\), as a decimal, compounded yearly, then after \(t\) years your investment will have a value \(F\) dollars, which is known as the future value. The discount rate \(D\) for such an investment is given by $$ D=\frac{1}{(1+r)^{t}}, $$ where \(t\) is the life, in years, of the investment. The present value of an investment is the product of the future value and the discount rate. Find a formula that gives the present value in terms of the future value, the interest rate, and the life of the investment.

By comparing the surface area of a sphere with its volume and assuming that air resistance is proportional to the square of velocity, it is possible to make a heuristic argument to support the following premise: For similarly shaped objects, terminal velocity varies in proportion to the square root of length. Expressed in a formula, this is $$ T=k L^{0.5}, $$ where \(L\) is length, \(T\) is terminal velocity, and \(k\) is a constant that depends on shape, among other things. This relation can be used to help explain why small mammals easily survive falls that would seriously injure or kill a human. a. A 6-foot man is 36 times as long as a 2-inch mouse (neglecting the tail). How does the terminal velocity of a man compare with that of a mouse? b. If the 6-foot man has a terminal velocity of 120 miles per hour, what is the terminal velocity of the 2 -inch mouse? c. Neglecting the tail, a squirrel is about 7 inches long. Again assuming that a 6-foot man has a terminal velocity of 120 miles per hour, what is the terminal velocity of a squirrel?

The number of seconds \(n\) for the yellow light is critical to safety at a traffic signal. One study recommends a formula for setting the time that permits a driver who sees the yellow light shortly before entering the intersection either to stop the vehicle safely or to cross the intersection at the current approach speed before the end of the yellow light. \({ }^{69}\) For a street of width 70 feet under standard conditions, the formula is $$ n=1+\frac{v}{30}+\frac{90}{v} . $$ Here \(v\) is the approach speed in feet per second. (See Figure 5.127.) a. Make a graph of \(n\) as a function of \(v\). Include speeds from 30 to 80 feet per second (roughly 20 to 55 miles per hour). b. Express using functional notation the length of the yellow light when the approach speed is 45 feet per second, and then calculate that value. c. Explain in practical terms the behavior of the graph near the pole at \(v=0\). d. What is the minimum length of time for a yellow light?

Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let \(A\) be the amount of litter present, in grams per square meter, as a function of time \(t\) in years. If the litter falls at a constant rate of \(L\) grams per square meter per year, and if it decays at a constant proportional rate of \(k\) per year, then the limiting value of \(A\) is \(R=L / k\). For this exercise and the next, we suppose that at time \(t=0\), the forest floor is clear of litter. a. If \(D\) is the difference between the limiting value and \(A\), so that \(D=R-A\), then \(D\) is an exponential function of time. Find the initial value of \(D\) in terms of \(R\). b. The yearly decay factor for \(D\) is \(e^{-k}\). Find a formula for \(D\) in terms of \(R\) and \(k\). Reminder: \(\left(a^{b}\right)^{c}=a^{b c}\). c. Explain why \(A=R-R e^{-k t}\).

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