3 Example Problem 1

3.1 Level, Straight, Passing Constrained Segment

This example problem illustrates the computation of the LOS in a single direction on a straight, 0.75-mi-long Passing Constrained segment in level terrain. This example problem follows the flowchart of analysis steps outlined in Exhibit 15-9.

The Facts

The segment has the following known characteristics:

• Segment length = 3,960 ft (0.75 mi);

• Segment type = Passing Constrained;

• No upstream passing lanes;

• Vehicle count in the analysis direction = 752 veh/h;

• PHF = 0.94;

• Posted speed limit: 50 mi/h;

• Percent heavy vehicles (%HV) = 5%;

• Percent grade = 0%;

• Horizontal curvature = none;

• Lane width = 12 ft;

• Shoulder width = 6 ft; and

• Access points = 0.

Objective

Estimate the LOS in the subject direction on the two-lane highway segment as described.

Step 1: Identify Facility Study Boundaries and Segmentation

The limits of the segment were identified following the guidance given in Step 1 on page 15-15. The characteristics of this segment were determined by examination to be essentially homogenous. These characteristics included the ability to pass, lane geometry, grades, lane and shoulder widths, posted speed limit, traffic demands, adjacent land uses, and driveways.

A field examination of the segment determined that it met the definition of a Passing Constrained segment, being a segment in which “passing in the oncoming lane is either prohibited or is effectively negligible due to geometric or sight distance limitations.”

Step 2: Determine Demand Flow Rates, Capacity, and d/c Ratio

In this step, the hourly demand volume at the upstream entrance of the directional segment being evaluated is converted to a peak 15-min flow rate by applying the peak hour factor (PHF) using Equation 15-1:

\[v_{d} = \frac{V_{d}}{\text{PHF}} = \frac{752}{0.94} = 800\ \text{veh/h}\]

The capacity of a Passing Constrained segment is 1,700 veh/h, as stated in the description of Step 2 on page 15-18. The demand flow rate is less than capacity; therefore the calculation process proceeds to Step 3.Note that it is only necessary to compute the actual opposing flow rate for Passing Zone segments.

Step 3: Determine Vertical Alignment Classification

The segment is assigned a vertical alignment classification of 1, based on Exhibit 15-11 for a level (0% grade), 0.75-mi-long segment. From Exhibit 15-10, the segment length of 0.75 mi is between the minimum (0.25 mi) and maximum (3.0 mi) lengths for a Passing Constrained segment of vertical class 1, and therefore no adjustment is needed to the segment length.

Step 4: Determine the Free-Flow Speed

As stated on page 15-18, direct field measurement of FFS is preferred. In this case it is not feasible to measure the FFS, so it will be estimated using the procedure given in Step 4.

Because this is a Passing Constrained segment, the opposing flow rate v_o is set at 1,500 veh/h in Equation 15-4 for the purposes of computing FFS, regardless of the actual opposing flow rate. First, the base free-flow speed BFFS is estimated using Equation 15-2. Next, Equation 15-4 through Equation 15-6 are used to determine factors relating to lane and shoulder width, access-point density, and heavy-vehicle percentage, which are used in the estimation of FFS. Finally, the FFS is estimated by Equation 15-3.

\[BFFS = 1.14 \times S_{\text{pl}} = 1.14 \times 50 = 57.0\ \text{mi/h}\]

\[f_{LS} = 0.6 \times (12 - LW) + 0.7 \times (6 - SW)\]

\[f_{LS} = 0.6 \times (12 - 12) + 0.7 \times (6 - 6) = 0\]

\[f_{A} = \text{min}\left( \frac{ {APD}}{4},10 \right) = \text{min}\left( \frac{0}{4},10 \right) = 0\]

\[a = \text{max}\lbrack 0.0333,\ 0\rbrack = 0.0333\]

\[FFS = BFFS - a(HV\%)\]

\[FFS = 57.0 - (0.0333)(5) - 0 - 0 = 56.83 \text{ mi/h}\]

There are no geometry-related FFS adjustments for 12-ft lanes and 6-ft shoulders. There are also no adjustments for access points (driveways or streets), because this segment has no access points.

Step 5: Estimate the Average Speed

Because the demand flow rate in the subject direction is greater than 100 veh/h, the equations given in Step 5 are used to estimate the average speed.

Step 5a: Calculate the Slope Coefficient

The slope coefficient m determines how rapidly the average speed is estimated to decrease as a function of the entering flow rate. It is computed as a function of six coefficients b₀ to b₅, which are obtained from Exhibit 15-13 for a Passing Constrained segment. For a segment of vertical class 1, these coefficients are 0.0558, 0.0542, 0.3278, 0.1029, 0, and 0, respectively. Equation 15-8 is then used to determine the slope coefficient.

\[m = \text{max}\left\lbrack b_{5},b_{0} + b_{1} \times FFS + b_{2} \times \sqrt{\frac{v_{o}}{1,000}} + \text{max}\left( 0,b_{3} \right) \times \sqrt{L} + \text{max}\left( 0,b_{4} \right) \times \sqrt{HV\%} \right\rbrack\]

\[m = \text{max}\left\lbrack 0,0.0558 + 0.0542 \times 56.83 + 0.3278 \times \sqrt{\frac{1,500}{1,000}} + \text{max}(0,\ 0.1029) \\ \times \sqrt{0.75} + \text{max}(0,\ 0) \times \sqrt{5}\right\rbrack\]

\[m = 3.626\]

Step 5b: Calculate the Power Coefficient

The power coefficient p is used to estimate how fast the average speed decreases at higher flow rates. The equation uses nine coefficients f₀ to f₈, which are obtained from Exhibit 15-19 for a Passing Constrained segment. For a segment of vertical class 1, all of these coefficients take on values of 0, except for f₀ (0.67576), f₃ (0.12060), and f₄ (−0.35919). Equation 15-11 is then used to determine the power coefficient.

\[p = \text{max}\left\lbrack f_{8},f_{0} + f_{1} \times FFS + f_{2} \times L + f_{3} \times \frac{v_{o}}{1,000} + f_{4} \times \sqrt{\frac{v_{o}}{1,000}} \\ + f_{5} \times HV\% + f_{6} \times \sqrt{HV\%} + f_{7} \times (L \times HV\%) \right\rbrack\]

\[p = \text{max}\left\lbrack 0,0.67576 + 0 \times 56.83 + 0 \times 0.75 + 0.12060 \times \frac{1,500}{1,000} - 0.35919 \times \sqrt{\frac{1,500}{1,000}} \\ + 0 \times 5 + 0 \times \sqrt{5} + 0 \times (0.75 \times 5) \right\rbrack\]

\[p = 0.41676\]

Step 5c: Calculate Average Speed for the Segment

The average speed for the given entry flow rate is computed in Step 5-3 using Equation 15-7. The previously computed flow rate v_d, slope coefficient m, and power coefficient p are used in this equation.

\[S = FFS - {m\left( \frac{v_{d}}{1,000} - 0.1 \right)}^{p}\]

\[S = 56.83 - {3.626\left( \frac{800}{1,000} - 0.1 \right)}^{0.41676}\]

\[S = 53.7\ \text{mi/h}\]

Step 5d: Adjust Speed for Horizontal Alignment

Because this is a straight segment, no adjustment to the speed estimate is required for horizontal alignment.

Assessment of Speed Results

The average speed at a directional flow rate of 800 veh/h is estimated to be approximately 3 mi/h (about 5%) lower than the FFS, but about 4 mi/h (about 7%) higher than the posted speed limit.

Step 6: Estimate the Percent Followers

The service measure percent followers is estimated in Step 6 for the given flow rate v_d. under the prevailing geometric conditions and percent heavy vehicles. First, the percent followers at 100% of capacity is estimated. Next, the percent followers at 25% of capacity is estimated. Then, the slope and power coefficients for an exponential curve fitting those two points (percent followers at 25% and 100% capacity) are estimated. Finally, the fitted curve is used to estimate the percent followers for the given flow rate.

Step 6a: Compute Percent Followers at Capacity

Percent followers at capacity is calculated using Equation 15-18, applying eight parameters b₀ to b₇ obtained from Exhibit 15-24.

\({PF}_{cap} = b_{0} + b_{1}(L) + b_{2}\left(\sqrt{L} \right) + b_{3}\left({FFS} \right) \\+ b_{4}\left( \sqrt{FFS} \right) + b_{5}(HV\%) + \ b_{6}\left( FFS \times \frac{v_{o}}{1,000} \right) + b_{7}\left( \sqrt{\frac{v_{0}}{1,000}} \right)\)

\({PF}_{cap} = 37.68080 + 3.05089(0.75) - 7.90866\left( \sqrt{0.75} \right) - 0.94321(56.83) \\ + 13.64266\left( \sqrt{56.83} \right) - 0.00050(5) - 0.05500\left(56.83 \times \frac{1,500}{1,000} \right) + 7.1376\left(\sqrt{\frac{1,500}{1,000}} \right)\)

\[{PF}_{cap} = 86.41\%\]

Step 6b: Compute Percent Followers at 25% Capacity

Percent followers at 25 percent of capacity is calculated using Equation 15-20, applying eight parameters c₀ to c₇ obtained from Exhibit 15-26.

\[{PF}_{25cap} = c_{0} + c_{1}(L) + c_{2}\left( \sqrt{L} \right) + c_{3}\left({FFS} \right) + c_{4}\left( \sqrt{\text{FFS}} \right) + c_{5}(HV\%) + c_{6}\left( FFS \times \frac{v_{o}}{1,000} \right) \\ + c_{7}\left( \sqrt{\frac{v_{0}}{1,000}} \right)\]

\[{PF}_{25cap} = 18.01780 + 10.00000(0.75) - 21.60000\left( \sqrt{0.75} \right) - 0.97853(56.83) \\ + 12.05214\left( \sqrt{56.83} \right) - 0.00750(5) - 0.06700\left( 56.83 \times \frac{1,500}{1,000} \right) + 11.6041\left( \sqrt{\frac{1,500}{1,000}} \right)\]

\[{PF}_{25cap} = 50.52\%\]

Step 6c: Calculate the Slope Coefficient

Equation 15-22 is used to compute the slope coefficient m for an exponential curve fitted between percent following at capacity and percent following at 25% capacity. It employs two parameters d₁ and d₂ obtained from Exhibit 15-28; the parameters for Passing Constrained segments are used.

\[m = d_{1}\left( \frac{0 - \text{ln}\left\lbrack 1 - \frac{{PF}_{25cap}}{100} \right\rbrack}{0.25\left\lbrack \frac{{cap}}{1,000} \right\rbrack} \right) + d_{2}\left( \frac{0 - \text{ln}\left\lbrack 1 - \frac{{PF}_{\text{cap}}}{100} \right\rbrack}{\left\lbrack \frac{{cap}}{1,000} \right\rbrack} \right)\]

\[m = - 0.29764\left( \frac{0 - \text{ln}\left\lbrack 1 - \frac{50.52}{100} \right\rbrack}{0.25\left\lbrack \frac{1,700}{1,000} \right\rbrack} \right) - 0.71917\left( \frac{0 - \text{ln}\left\lbrack 1 - \frac{86.41}{100} \right\rbrack}{\left\lbrack \frac{1,700}{1,000} \right\rbrack} \right)\]

\[m = - 1.337\]

Step 6d: Calculate the Power Coefficient

Equation 15-23 is used to compute the power coefficient p for an exponential curve fitted between percent following at capacity and percent following at 25% capacity. It employs five parameters e₀ through e₄ obtained from Exhibit 15-29; the parameters for Passing Constrained segments are used.

\[p = e_{0} + e_{1}\left( \frac{0 - \text{ln}\left\lbrack 1 - \frac{{PF}_{25cap}}{100} \right\rbrack}{0.25\left\lbrack \frac{cap}{1,000} \right\rbrack} \right) + e_{2}\left( \frac{0 - \text{ln}\left\lbrack 1 - \frac{{PF}_{{cap}}}{100} \right\rbrack}{\left\lbrack \frac{cap}{1,000} \right\rbrack} \right) + \ e_{3}\sqrt{\frac{0 - \text{ln}\left( 1 - \frac{{PF}_{25cap}}{100} \right)}{0.25\left\lbrack \frac{cap}{1,000} \right\rbrack}} \\ + e_{4}\sqrt{\frac{0 - \text{ln}\left( 1 - \frac{{PF}_{cap}}{100} \right)}{\left\lbrack \frac{{cap}}{1,000} \right\rbrack}}\]

\[p = 0.81165 + 0.37920\left( \frac{0 - \text{ln}\left\lbrack 1 - \frac{50.52}{100} \right\rbrack}{0.25\left\lbrack \frac{1,700}{1,000} \right\rbrack} \right) - 0.49524\left( \frac{0 - \text{ln}\left\lbrack 1 - \frac{86.41}{100} \right\rbrack}{\left\lbrack \frac{1,700}{1,000} \right\rbrack} \right) \\ - 2.11289\sqrt{\frac{0 - \text{ln}\left( 1 - \frac{50.52}{100} \right)}{0.25\left\lbrack \frac{1,700}{1,000} \right\rbrack}} + 2.41146\sqrt{\frac{0 - \text{ln}\left( 1 - \frac{86.41}{100} \right)}{\left\lbrack \frac{1,700}{1,000} \right\rbrack}}\]

\[p = 0.7524\]

Step 6e: Calculate Percent Followers

Equation 15-17 is used to compute percent followers PF.

\[PF = 100 \times \left\lbrack 1 - e^{\left( m\ \times \left\{ \ \frac{v_{d}}{1,000} \right\}^{p} \right)} \right\rbrack\]

\[PF = 100 \times \left\lbrack 1 - e^{\left( - 1.337\ \times \left\{ \ \frac{800}{1,000} \right\}^{0.7524} \right)} \right\rbrack\]

\[PF = 67.7\%\]

Step 7: Calculate Additional Performance Measure Values for a Passing Lane Segment

This step is only applicable to passing lane segments. Therefore, Step 7 is skipped.

Step 8: Calculate Follower Density

Follower density FD is estimated using Equation 15-35.

\[FD = \frac{\text{PF}}{100} \times \frac{v_{d}}{S} = \frac{67.7}{100} \times \frac{800}{53.7} = 10.1\ \text{followers/mi/ln}\]

Step 9: Determine Potential Adjustment to Follower Density

There is no passing lane upstream of the analysis segment. Therefore, no adjustment is needed to follower density and Step 9 is skipped.

Step 10: Determine LOS

The segment’s LOS is determined from Exhibit 15-6, using the column for a higher-speed highway (posted speed limit equal to or greater than 50 mi/h). With 10.1 followers/mi, the subject direction of travel on the segment operates at LOS D.

Discussion

The estimated FFS and average speed for a flow rate of 800 veh/h are both above the posted speed limit. This result is reasonable for a flat, straight segment in this volume range. However, the follower density produces LOS D operations. This flow rate is large enough to produce fairly high levels of platooning, but not so high as to cause significant reductions in speed. The combination of a moderately high flow rate and moderately high level of platooning will result in travelers perceiving a relatively poor level of service.