Chapter 6, Problem 19P

To determine

The graph of average individual delay versus the repair period and use this graph to discuss the effect of the expected repair time on the average delay.

Explanation of Solution

Given:

We have been given the following information:

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 repair period of 1 hour.

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

Substituting the values in the following equation, we have

  $\begin{array}{l}{q}_{\max }=\left(V-C_r\right)t_{inc}\\ q_{\max }=\left(\left(90\%of6000\right)-4000\right)2\\ q_{\max }=\left(\left(6000\times \frac{90}{100}\right)-4000\right)2\\ q_{\max }=\left(\left(60\times 90\right)-4000\right)2\\ q_{\max }=\left(5400-4000\right)2\\ q_{\max }=2800\text{Veh}\text{.}\end{array}$

Themaximum queue length that will be formed is $q_{\max }=2800\text{Veh}\text{.}$

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

  $\begin{array}{l}{d}_r=\frac{t_{inc}^2\left(V-C_R\right)\left(C-C_R\right)}{2\left(C-V\right)}\\ d_r=\frac{1^2\left(90\%\text{}\text{}\text{}of6000-4000\right)\left(6000-4000\right)}{2\left(6000-5400\right)}\end{array}$

  $d_r=\frac{\left(5400-4000\right)\left(2000\right)}{2\left(600\right)}$

  $\begin{array}{l}{d}_r=\frac{\left(1400\right)\left(20\right)}{12}\\ d_r=2333.33=2334\text{hr}\text{.}\end{array}$

The total delay is $d_r=2334\text{hr}\text{.}$

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 foundusing the following formula:

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

Substituting the values, we have

  $\begin{array}{l}\text{=}5400\times 1.\\ \text{=}5400\text{Veh}\text{.}\end{array}$

The number of vehicles that will be affected by the incident is $5400\text{Veh}\text{.}$

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

  $\begin{array}{l}\text{Average}\text{individual delay}=\frac{d_r}{\text{Number of vehicles affected }}\\ =\frac{2334}{5400}.\\ =0.432\text{hr}\text{.}\end{array}$

The average individual is $0.432\text{hr}\text{.}$

For the expected repair period of 1.5 hour.

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 and,

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

Considering 90 percent of the flow and that the capacity of each lane is 2000 Veh/ h

Substituting the values in the following equation, we have

  $\begin{array}{l}{q}_{\max }=\left(V-C_r\right)t_{inc}\\ q_{\max }=\left(\left(90\%of6000\right)-4000\right)2\\ q_{\max }=\left(\left(6000\times \frac{90}{100}\right)-4000\right)2\\ q_{\max }=\left(\left(60\times 90\right)-4000\right)2\\ q_{\max }=\left(5400-4000\right)2\\ q_{\max }=2800\text{Veh}\text{.}\end{array}$

Themaximum queue length that will be formed is $q_{\max }=2800\text{Veh}\text{.}$

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

  $\begin{array}{l}{d}_r=\frac{t_{inc}^2\left(V-C_R\right)\left(C-C_R\right)}{2\left(C-V\right)}\\ d_r=\frac{{1.5}^2\left(90\%\text{}\text{}\text{}of6000-4000\right)\left(6000-4000\right)}{2\left(6000-5400\right)}\\ d_r=\frac{2.25\left(5400-4000\right)\left(2000\right)}{2\left(600\right)}\\ d_r=\frac{2.25\left(1400\right)\left(20\right)}{12}\\ d_r=5250\text{hr}\text{.}\end{array}$

The total delay is $d_r=5250\text{hr}\text{.}$

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

  $\begin{array}{l}\text{=}5400\times 1.5.\\ \text{=}8100\text{Veh}\text{.}\end{array}$

The number of vehicles that will be affected by the incident is $8100\text{Veh}\text{.}$

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

  $\begin{array}{l}\text{Average}\text{individual delay}=\frac{d_r}{\text{Number of vehicles affected }}\\ =\frac{5250}{8100}.\\ =0.648\text{hr}\text{.}\end{array}$

Theaverage individualdelay is $0.648\text{hr}\text{.}$

For the expected repair period of 2.5 hour.

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 and,

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

Considering 90 percent of the flow and that the capacity of each lane is 2000 Veh/ h

Substituting the values in the following equation, we have

  $\begin{array}{l}{q}_{\max }=\left(V-C_r\right)t_{inc}\\ q_{\max }=\left(\left(90\%of6000\right)-4000\right)2\\ q_{\max }=\left(\left(6000\times \frac{90}{100}\right)-4000\right)2\\ q_{\max }=\left(\left(60\times 90\right)-4000\right)2\\ q_{\max }=\left(5400-4000\right)2\\ q_{\max }=2800\text{Veh}\text{.}\end{array}$

The maximum queue length that will be formed is $q_{\max }=2800\text{Veh}\text{.}$

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

  $\begin{array}{l}{d}_r=\frac{t_{inc}^2\left(V-C_R\right)\left(C-C_R\right)}{2\left(C-V\right)}\\ d_r=\frac{{2.5}^2\left(90\%\text{}\text{}\text{}of6000-4000\right)\left(6000-4000\right)}{2\left(6000-5400\right)}\\ d_r=\frac{6.25\left(5400-4000\right)\left(2000\right)}{2\left(600\right)}\\ d_r=\frac{6.25\left(1400\right)\left(20\right)}{12}\\ d_r=14583.33\text{hr}\text{.}\\ d_r=14584\text{hr}\text{.}\end{array}$

The total delay is $d_r=14584\text{hr}\text{.}$

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

  $\begin{array}{l}\text{=}5400\times 2.5.\\ \text{=}13500\text{Veh}\text{.}\end{array}$

Therefore, the number of vehicles that will be affected by the incident is $13500\text{Veh}\text{.}$

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

  $\begin{array}{l}\text{Average}\text{individual delay}=\frac{d_r}{\text{Number of vehicles affected }}\\ =\frac{14584}{13500}.\\ =1.0803\text{hr}\text{.}\end{array}$

The average individualdelay is $1.0803\text{hr}\text{.}$

For the expected repair period of 2.75 hour.

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

Substituting the values in the following equation, we have

  $\begin{array}{l}{q}_{\max }=\left(V-C_r\right)t_{inc}\\ q_{\max }=\left(\left(90\%of6000\right)-4000\right)2\\ q_{\max }=\left(\left(6000\times \frac{90}{100}\right)-4000\right)2\\ q_{\max }=\left(\left(60\times 90\right)-4000\right)2\\ q_{\max }=\left(5400-4000\right)2\\ q_{\max }=2800\text{Veh}\text{.}\end{array}$

Therefore, the maximum queue length that will be formed is $q_{\max }=2800\text{Veh}\text{.}$

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

  $\begin{array}{l}{d}_r=\frac{t_{inc}^2\left(V-C_R\right)\left(C-C_R\right)}{2\left(C-V\right)}\\ d_r=\frac{{2.75}^2\left(90\%\text{}\text{}\text{}of6000-4000\right)\left(6000-4000\right)}{2\left(6000-5400\right)}\\ d_r=\frac{7.5625\left(5400-4000\right)\left(2000\right)}{2\left(600\right)}\\ d_r=\frac{7.5625\left(1400\right)\left(20\right)}{12}\\ d_r=17645.83\text{hr}\text{.}\\ d_r=17646\text{hr}\text{.}\end{array}$

Therefore, the total delay is $d_r=17646\text{hr}\text{.}$

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

  $\begin{array}{l}\text{=}5400\times 2.75.\\ \text{=}14850\text{Veh}\text{.}\end{array}$

The number of vehicles that will be affected by the incident is $14850\text{Veh}\text{.}$

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

  $\begin{array}{l}\text{Average}\text{individual delay}=\frac{d_r}{\text{Number of vehicles affected }}\\ =\frac{17646}{14850}.\\ =1.188\text{hr}\text{.}\end{array}$

the average individual delay is $1.188\text{hr}\text{.}$

For the expected repair period of 3.0 hour.

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 and,

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

Considering 90 percent of the flow and that the capacity of each lane is 2000 Veh/ h

Substituting the values in the following equation, we have

  $\begin{array}{l}{q}_{\max }=\left(V-C_r\right)t_{inc}\\ q_{\max }=\left(\left(90\%of6000\right)-4000\right)2\\ q_{\max }=\left(\left(6000\times \frac{90}{100}\right)-4000\right)2\\ q_{\max }=\left(\left(60\times 90\right)-4000\right)2\\ q_{\max }=\left(5400-4000\right)2\\ q_{\max }=2800\text{Veh}\text{.}\end{array}$

The maximum queue length that will be formed is $q_{\max }=2800\text{Veh}\text{.}$

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

  $\begin{array}{l}{d}_r=\frac{t_{inc}^2\left(V-C_R\right)\left(C-C_R\right)}{2\left(C-V\right)}\\ d_r=\frac{{3.0}^2\left(90\%\text{}\text{}\text{}of6000-4000\right)\left(6000-4000\right)}{2\left(6000-5400\right)}\\ d_r=\frac{9.0\left(5400-4000\right)\left(2000\right)}{2\left(600\right)}\\ d_r=\frac{9.0\left(1400\right)\left(20\right)}{12}\\ d_r=21000.00\text{hr}\text{.}\end{array}$

The total delay is $d_r=21000.00\text{hr}\text{.}$

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

  $\begin{array}{l}\text{=}5400\times 3.00.\\ \text{=}16200\text{Veh}\text{.}\end{array}$

The number of vehicles that will be affected by the incident is $16200\text{Veh}\text{.}$

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

  $\begin{array}{l}\text{Average}\text{individual delay}=\frac{d_r}{\text{Number of vehicles affected }}\\ =\frac{21000}{16200}.\\ =1.30\text{hr}\text{.}\end{array}$

The average individual delay is $1.30\text{hr}\text{.}$

Plot the graph of average individual delay versus the repair period is as follows:

  

Conclusion:

Therefore, for 1.0 hour : Themaximum queue length that will be formed is $q_{\max }=2800\text{Veh}$, total delay is $d_r=2334\text{hr}$ ,number of vehicles that will be affected by the incident is $5400\text{Veh}$ and theaverage individual is $0.432\text{hr}\text{.}$

For 1.5hour:

Themaximum queue length that will be formed is $q_{\max }=2800\text{Veh}$ ,total delay is $d_r=5250\text{hr}$ ,number of vehicles that will be affected by the incident is $8100\text{Veh}$ and the average individual is $0.648\text{hr}\text{.}$

For 2.50hour:

Themaximum queue length that will be formed is $q_{\max }=2800\text{Veh}\text{.}$

The total delay is $d_r=14584\text{hr}\text{.}$

The number of vehicles that will be affected by the incident is $13500\text{Veh}\text{.}$

The average individual delayis $1.0803\text{hr}\text{.}$

For 2.75hour:

Themaximum queue length that will be formed is $q_{\max }=2800\text{Veh}\text{.}$

The total delay is $d_r=9334\text{hr}\text{.}$

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

The average individual delay is $1.188\text{hr}\text{.}$

For 3.0 hour:

Themaximum queue length that will be formed is $q_{\max }=2800\text{Veh}\text{.}$

The total delay is $d_r=21000.00\text{hr}\text{.}$

The number of vehicles that will be affected by the incident is $16200\text{Veh}\text{.}$

The average individual delay is $1.30\text{hr}\text{.}$