Chapter 6, Problem 20P

To determine

(a)

The maximum queue length that will be formed, total delay, number of vehicles that will be affected by the incident, the average individual delay for the expected demand flow of 70%of the capacity of the highway.

Answer to Problem 20P

For 70 % expected demand flow:

Themaximum queue length that will be formed is $q_{\max }=400\text{Veh}\text{.}$

The total delay is $d_r=445\text{hr}\text{.}$

The number of vehicles that will be affected by the incident is $10800\text{Veh}\text{.}$

The average individual delay is $0.0412\text{hr}\text{.}$

Explanation of Solution

Given:

We have been given the following information:

Total number of lanes = 3,

  $V=$$\text{90\%}\text{of}\text{6000,}$

  $t_{inc}=2\text{hr,}$

Mean free flow speed of the highway $=55\text{ mi/h}$,

Jam density $=135\text{ veh/mi/ln,}$

  $C_r=4000$.

Following is the lay out of the given highway section:

  

Calculation:

For the expected 70 % demand flow:

We have the following formula for the determination of maximum queue length that will be formed:

  $q_{\max }=\left(V-C_r\right)t_{inc}$

Where, $q_{\max }$ is the maximum queue length,

  $V$ is the total volume,

  $C_r$ is the reduced capacity,

  $t_{inc}$ is the duration of the incident.

Considering 70 percent of the flow and that the capacity of each lane is 2000 Veh/ h

Substituting the values in the following equation, we have

  $\begin{array}{l}{q}_{\max }=\left(V-C_r\right)t_{inc}\\ q_{\max }=\left(\left(70\%of6000\right)-4000\right)2\\ q_{\max }=\left(\left(6000\times \frac{70}{100}\right)-4000\right)2\end{array}$

  $\begin{array}{l}{q}_{\max }=\left(\left(60\times 70\right)-4000\right)2\\ q_{\max }=\left(4200-4000\right)2\\ q_{\max }=400\text{Veh}\text{.}\end{array}$

The maximum queue length that will be formed is $q_{\max }=400\text{Veh}\text{.}$

Now, the total delay, we have the following formula

  $d_r=\frac{t_{inc}^2\left(V-C_R\right)\left(C-C_R\right)}{2\left(C-V\right)}$

Where, $C_R$ is the reduced capacity and can be found as follows:

  $\begin{array}{l}{C}_R=Numberoflanesinuse\times capacityofeachlane.\\ C_R=2000\times (3-1).\\ C_R=2000\times 2=4000\text{Veh/h}\text{.}\end{array}$

And C is the total capacity and can be found as

  $\begin{array}{l}C=Totalnumberoflanesinuse\times capacityofeachlane.\\ C=2000\times 3.\\ C=6000\text{Veh/h}\text{.}\end{array}$

Now, substituting the values in the required equation, we have

  $\begin{array}{l}{d}_r=\frac{t_{inc}^2\left(V-C_R\right)\left(C-C_R\right)}{2\left(C-V\right)}\\ d_r=\frac{2^2\left(70\%\text{}\text{}\text{}of6000-4000\right)\left(6000-4000\right)}{2\left(6000-70\%\text{}\text{}\text{}of6000\right)}\end{array}$

  $d_r=\frac{4\left(4200-4000\right)\left(2000\right)}{2\left(6000-4200\right)}$

  $\begin{array}{l}{d}_r=\frac{4\left(100\right)\left(2000\right)}{(1800)}\\ d_r=\frac{4000}{9}\end{array}$

  $d_r=444.44=445\text{hr}\text{.}$

The total delay is $d_r=445\text{hr}\text{.}$

The number of vehicles that will be affected by the incident.

To calculate the number of vehicles that will be affected by the incident can be found using the following formula:

Number of vehicles affected = $\text{V}\times {\text{t}}_{\text{inc}}.$

Substituting the values, we have

  $\begin{array}{l}\text{=}5400\times 2.\\ \text{=}10800\text{Veh}\text{.}\end{array}$

The number of vehicles that will be affected by the incident is $10800\text{Veh}\text{.}$

To calculate the average individual delay, we have the following formula:

Average individual delay $=\frac{d_r}{\text{Number of vehicles affected }}$

Substituting the values, we have

  $\begin{array}{l}\text{Average}\text{individual delay}=\frac{d_r}{\text{Number of vehicles affected }}\\ =\frac{445}{10800}.\\ =0.0412\text{hr}\text{.}\end{array}$

The average individual delay is $0.0412\text{hr}\text{.}$

Conclusion:

Therefore, for 70 % expected demand flow:

Themaximum queue length that will be formed is $q_{\max }=400\text{Veh}$, the total delay is $d_r=445\text{hr,}$

The number of vehicles that will be affected by the incident is $10800\text{Veh}$ and the average individual delay is $0.0412\text{hr}\text{.}$

To determine

(b)

The maximum queue length that will be formed, total delay, number of vehicles that will be affected by the incident, the average individual delay for the expected demand flow of 75%of the capacity of the highway.

Answer to Problem 20P

For 75 % expected demand flow:

Themaximum queue length that will be formed is $q_{\max }=1000\text{Veh}\text{.}$

The total delay is $d_r=1334\text{hr}\text{.}$

The number of vehicles that will be affected by the incident is $10800\text{Veh}\text{.}$

The average individual delay is $0.1235\text{hr}\text{.}$

Explanation of Solution

Given:

We have been given the following information:

Total number of lanes = 3,

  $V=$$\text{90\%}\text{of}\text{6000,}$

  $t_{inc}=2\text{hr,}$

Mean free flow speed of the highway $=55\text{ mi/h}$,

Jam density $=135\text{ veh/mi/ln,}$

  $C_r=4000$.

Following is the lay out of the given highway section:

  

Calculation:

For the expected 75 % demand flow:

We have the following formula for the determination of maximum queue length that will be formed:

  $q_{\max }=\left(V-C_r\right)t_{inc}$

Where, $q_{\max }$ is the maximum queue length,

  $V$ is the total volume,

  $C_r$ is the reduced capacity,

  $t_{inc}$ is the duration of the incident.

Considering 75 percent of the flow and that the capacity of each lane is 2000 Veh/ h

Substituting the values in the following equation, we have

  $\begin{array}{l}{q}_{\max }=\left(V-C_r\right)t_{inc}\\ q_{\max }=\left(\left(75\%of6000\right)-4000\right)2\\ q_{\max }=\left(\left(6000\times \frac{75}{100}\right)-4000\right)2\\ q_{\max }=\left(\left(60\times 75\right)-4000\right)2\\ q_{\max }=\left(4500-4000\right)2\\ q_{\max }=1000\text{Veh}\text{.}\end{array}$

Themaximum queue length that will be formed is $q_{\max }=1000\text{Veh}\text{.}$

Now, the total delay, we have the following formula

  $d_r=\frac{t_{inc}^2\left(V-C_R\right)\left(C-C_R\right)}{2\left(C-V\right)}$

Where, $C_R$ is the reduced capacity and can be found as follows:

  $\begin{array}{l}{C}_R=Numberoflanesinuse\times capacityofeachlane.\\ C_R=2000\times (3-1).\\ C_R=2000\times 2=4000\text{Veh/h}\text{.}\end{array}$

And C is the total capacity and can be found as

  $\begin{array}{l}C=Totalnumberoflanesinuse\times capacityofeachlane.\\ C=2000\times 3.\\ C=6000\text{Veh/h}\text{.}\end{array}$

Now, substituting the values in the required equation, we have

  $\begin{array}{l}{d}_r=\frac{t_{inc}^2\left(V-C_R\right)\left(C-C_R\right)}{2\left(C-V\right)}\\ d_r=\frac{2^2\left(75\%\text{}\text{}\text{}of6000-4000\right)\left(6000-4000\right)}{2\left(6000-75\%\text{}\text{}\text{}of6000\right)}\\ d_r=\frac{4\left(4500-4000\right)\left(2000\right)}{2\left(6000-4500\right)}\\ d_r=\frac{4\left(500\right)\left(2000\right)}{3000}\\ d_r=\frac{8\left(500\right)}{3}\\ d_r=1333.33=1334\text{hr}\text{.}\end{array}$

The total delay is $d_r=1334\text{hr}\text{.}$

The number of vehicles that will be affected by the incident.

To calculate the number of vehicles that will be affected by the incident can be found using the following formula:

Number of vehicles affected = $\text{V}\times {\text{t}}_{\text{inc}}.$

Substituting the values, we have

  $\begin{array}{l}\text{=}5400\times 2.\\ \text{=}10800\text{Veh}\text{.}\end{array}$

The number of vehicles that will be affected by the incident is $10800\text{Veh}\text{.}$

To calculate the average individual delay, we have the following formula:

Average individual delay $=\frac{d_r}{\text{Number of vehicles affected }}$

Substituting the values, we have

  $\begin{array}{l}\text{Average}\text{individual delay}=\frac{d_r}{\text{Number of vehicles affected }}\\ =\frac{1334}{10800}.\\ =0.1235\text{hr}\text{.}\end{array}$

The average individual delay is $0.1235\text{hr}\text{.}$

Conclusion:

For 75 % expected demand flow:

Themaximum queue length that will be formed is $q_{\max }=1000\text{Veh,}$ the total delay is $d_r=1334\text{hr,}$ the number of vehicles that will be affected by the incident is $10800\text{Veh}$ and the average individual delay is $0.1235\text{hr}\text{.}$

To determine

(c)

The maximum queue length that will be formed, total delay, number of vehicles that will be affected by the incident, the average individual delay for the expected demand flow of 80% of the capacity of the highway.

Answer to Problem 20P

For 80 % expected demand flow:

Themaximum queue length that will be formed is $q_{\max }=1600\text{Veh}\text{.}$

The total delay is $d_r=2667\text{hr}\text{.}$

The number of vehicles that will be affected by the incident is $10800\text{Veh}\text{.}$

The average individual delay is $0.247\text{hr}\text{.}$

Explanation of Solution

Given:

We have been given the following information:

Total number of lanes = 3,

  $V=$$\text{90\%}\text{of}\text{6000,}$

  $t_{inc}=2\text{hr,}$

Mean free flow speed of the highway $=55\text{ mi/h}$,

Jam density $=135\text{ veh/mi/ln,}$

  $C_r=4000$.

Following is the lay out of the given highway section:

  

Calculation:

For the expected 80 % demand flow:

We have the following formula for the determination of maximum queue length that will be formed:

  $q_{\max }=\left(V-C_r\right)t_{inc}$

Where, $q_{\max }$ is the maximum queue length,

  $V$ is the total volume,

  $C_r$ is the reduced capacity,

  $t_{inc}$ is the duration of the incident.

Considering 80 percent of the flow and that the capacity of each lane is 2000 Veh/ h

Substituting the values in the following equation, we have

  $\begin{array}{l}{q}_{\max }=\left(V-C_r\right)t_{inc}\\ q_{\max }=\left(\left(80\%of6000\right)-4000\right)2\\ q_{\max }=\left(\left(6000\times \frac{80}{100}\right)-4000\right)2\\ q_{\max }=\left(\left(60\times 80\right)-4000\right)2\\ q_{\max }=\left(4800-4000\right)2\\ q_{\max }=1600\text{Veh}\text{.}\end{array}$

The maximum queue length that will be formed is $q_{\max }=1600\text{Veh}\text{.}$

Now, the total delay, we have the following formula

  $d_r=\frac{t_{inc}^2\left(V-C_R\right)\left(C-C_R\right)}{2\left(C-V\right)}$

Where, $C_R$ is the reduced capacity and can be found as follows:

  $\begin{array}{l}{C}_R=Numberoflanesinuse\times capacityofeachlane.\\ C_R=2000\times (3-1).\\ C_R=2000\times 2=4000\text{Veh/h}\text{.}\end{array}$

And C is the total capacity and can be found as

  $\begin{array}{l}C=Totalnumberoflanesinuse\times capacityofeachlane.\\ C=2000\times 3.\\ C=6000\text{Veh/h}\text{.}\end{array}$

Now, substituting the values in the required equation, we have

  $\begin{array}{l}{d}_r=\frac{t_{inc}^2\left(V-C_R\right)\left(C-C_R\right)}{2\left(C-V\right)}\\ d_r=\frac{2^2\left(4800-4000\right)\left(6000-4000\right)}{2\left(6000-4800\right)}\\ d_r=\frac{4\left(800\right)\left(2000\right)}{2\left(1200\right)}\\ d_r=\frac{4\left(800\right)\left(20\right)}{24}\\ d_r=2666.66\text{hr}\text{.}\\ d_r=2667\text{hr}\text{.}\end{array}$

The total delay is $d_r=2667\text{hr}\text{.}$

The number of vehicles that will be affected by the incident.

To calculate the number of vehicles that will be affected by the incident can be found using the following formula:

Number of vehicles affected = $\text{V}\times {\text{t}}_{\text{inc}}.$

Substituting the values, we have

  $\begin{array}{l}\text{=}5400\times 2.\\ \text{=}10800\text{Veh}\text{.}\end{array}$

Therefore, the number of vehicles that will be affected by the incident is $10800\text{Veh}\text{.}$

To calculate the average individual delay, we have the following formula:

Average individual delay $=\frac{d_r}{\text{Number of vehicles affected }}$

Substituting the values, we have

  $\begin{array}{l}\text{Average}\text{individual delay}=\frac{d_r}{\text{Number of vehicles affected }}\\ =\frac{2667}{10800}.\\ =0.247\text{hr}\text{.}\end{array}$

The average individual delay is $0.247\text{hr}\text{.}$

Conclusion:

For 80 % expected demand flow:

Themaximum queue length that will be formed is $q_{\max }=1600\text{Veh,}$ the total delay is $d_r=2667\text{hr,}$ the number of vehicles that will be affected by the incident is $10800\text{Veh}$ and

the average individual delay is $0.247\text{hr}\text{.}$

To determine

(d)

The maximum queue length that will be formed, total delay, number of vehicles that will be affected by the incident, the average individual delay for the expected demand flowof 85% of the capacity of the highway.

Answer to Problem 20P

For 85 % expected demand flow:

Themaximum queue length that will be formed is $q_{\max }=2200\text{Veh}\text{.}$

The total delay is $d_r=4889\text{hr}\text{.}$

The number of vehicles that will be affected by the incident is $10800\text{Veh}\text{.}$

The average individual delay is $0.453\text{hr}\text{.}$

Explanation of Solution

Given:

We have been given the following information:

Total number of lanes = 3,

  $V=$$\text{90\%}\text{of}\text{6000,}$

  $t_{inc}=2\text{hr,}$

Mean free flow speed of the highway $=55\text{ mi/h}$,

Jam density $=135\text{ veh/mi/ln,}$

  $C_r=4000$.

Following is the lay out of the given highway section:

  

Calculation:

For the expected 85 % demand flow:

We have the following formula for the determination of maximum queue length that will be formed:

  $q_{\max }=\left(V-C_r\right)t_{inc}$

Where, $q_{\max }$ is the maximum queue length,

  $V$ is the total volume,

  $C_r$ is the reduced capacity,

  $t_{inc}$ is the duration of the incident.

Considering 85 percent of the flow and that the capacity of each lane is 2000 Veh/ h

Substituting the values in the following equation, we have

  $\begin{array}{l}{q}_{\max }=\left(V-C_r\right)t_{inc}\\ q_{\max }=\left(\left(85\%of6000\right)-4000\right)2\\ q_{\max }=\left(\left(6000\times \frac{85}{100}\right)-4000\right)2\\ q_{\max }=\left(\left(60\times 85\right)-4000\right)2\\ q_{\max }=\left(5100-4000\right)2\\ q_{\max }=2200\text{Veh}\text{.}\end{array}$

Therefore, the maximum queue length that will be formed is $q_{\max }=2200\text{Veh}\text{.}$

Now, the total delay, we have the following formula

  $d_r=\frac{t_{inc}^2\left(V-C_R\right)\left(C-C_R\right)}{2\left(C-V\right)}$

Where, $C_R$ is the reduced capacity and can be found as follows:

  $\begin{array}{l}{C}_R=Numberoflanesinuse\times capacityofeachlane.\\ C_R=2000\times (3-1).\\ C_R=2000\times 2=4000\text{Veh/h}\text{.}\end{array}$

And C is the total capacity and can be found as

  $\begin{array}{l}C=Totalnumberoflanesinuse\times capacityofeachlane.\\ C=2000\times 3.\\ C=6000\text{Veh/h}\text{.}\end{array}$

Now, substituting the values in the required equation, we have

  $\begin{array}{l}{d}_r=\frac{t_{inc}^2\left(V-C_R\right)\left(C-C_R\right)}{2\left(C-V\right)}\\ d_r=\frac{2^2\left(85\%\text{}\text{}\text{}of6000-4000\right)\left(6000-4000\right)}{2\left(6000-5100\right)}\\ d_r=\frac{4\left(5100-4000\right)\left(2000\right)}{2\left(900\right)}\\ d_r=\frac{4\left(1100\right)\left(20\right)}{18}\\ d_r=4888.88\text{hr}\text{.}\\ d_r=4889\text{hr}\text{.}\end{array}$

Therefore, the total delay is $d_r=4889\text{hr}\text{.}$

The number of vehicles that will be affected by the incident.

To calculate the number of vehicles that will be affected by the incident can be found using the following formula:

Number of vehicles affected = $\text{V}\times {\text{t}}_{\text{inc}}.$

Substituting the values, we have

  $\begin{array}{l}\text{=}5400\times 2.00.\\ \text{=}10800\text{Veh}\text{.}\end{array}$

The number of vehicles that will be affected by the incident is $10800\text{Veh}\text{.}$

To calculate the average individual delay, we have the following formula:

Average individual delay $=\frac{d_r}{\text{Number of vehicles affected }}$

Substituting the values, we have

  $\begin{array}{l}\text{Average}\text{individual delay}=\frac{d_r}{\text{Number of vehicles affected }}\\ =\frac{4889}{10800}.\\ =0.453\text{hr}\text{.}\end{array}$

the average individual delay is $0.453\text{hr}\text{.}$

The graph of average individual delay versus the expected demand flow is as follows:

  

Conclusion:

For 85 % expected demand flow:

Themaximum queue length that will be formed is $q_{\max }=2200\text{Veh,}$ the total delay is $d_r=4889\text{hr,}$ the number of vehicles that will be affected by the incident is $10800\text{Veh}$ and

the average individual delay is $0.453\text{hr}\text{.}$