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Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e

Fred L. Mannering, Scott S. Washburn, and

Walter P.

Preface

The solutions to the fourth edition of Principles of Highway Engineering and Traffic Analysis were prepared with the Mathcad1 software program. You will notice several notation conventions that you may not be familiar with if you are not a Mathcad user. Most of these notation conventions are self-explanatory or easily understood. The most common Mathcad specific notations in these solutions relate to the equals sign. You will notice the equals sign being used in three different contexts, and Mathcad uses three different notations to distinguish between each of these contexts. The differences between these equals sign notations are explained as follows.

• The ‘:=’ (colon-equals) is an assignment operator, that is, the value of the variable or expression on the left side of ‘:=’is set equal to the value of the expression on the right side. For example, in the statement, L := 1234, the variable ‘L’ is assigned (i.e., set equal to) the value of 1234. Another example is x := y + z. In this case, x is assigned the value of y + z.

• The ‘ = =’ (bold equals) is used when the Mathcad function solver was used to find the value of a variable in the equation. For example, in the equation , the = = is used to tell Mathcad that the value of the expression on the left side needs to equal the value of the expression on the right side. Thus, the Mathcad solver can be employed to find a value for the variable ‘t’ that satisfies this relationship. This particular example is from a problem where the function for arrivals at some time ‘t’ is set equal to the function for departures at some time ‘t’ to find the time to queue clearance.

• The ‘=’ (standard equals) is used for a simple numeric evaluation. For example, referring to the x := y + z assignment used previously, if the value of y was 10 [either by assignment (with :=), or the result of an equation solution (through the use of = =) and the value of z was 15, then the expression ‘x =’ would yield 25. Another example would be as follows: s := 1800/3600, with s = 0.5. That is, ‘s’ was assigned the value of 1800 divided by 3600 (using :=), which equals 0.5 (as given by using =).

Another symbol you will see frequently is ‘→’. In these solutions, it is used to perform an evaluation of an assignment expression in a single statement. For example, in the following statement, , Q(t) is assigned the value of Arrivals(t) – Departures(t), and this evaluates to 2.2t – 0.10t2

Finally, to assist in quickly identifying the final answer, or answers, for what is being asked in the problem statement, yellow highlighting has been used (which will print as light gray).

Solutions

to accompany

of Highway

Traffic

Manual

Principles

Engineering and

Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski.

1 www.mathcad.com

Determine the hourly volume.

BFFS65 :=

fLW 1.9 :=

fLC 0.80 :=

fN 0.0 :=

fID 0.0 :=

FFSBFFSfLW fLC fN fID :=

vp 1616 :=

P T 0.10 :=

E T 2.5 :=

fHV 1

1P T E T 1 () + :=

PHF0.90 :=

Vvp PHF N fHV fp :=

FFS62.3 =

fHV 0.87 =

N3 := fp 1.0 :=

veh/h V3794 =

Problem 6.1

(given)

(Table 6.3)

(Table 6.4)

(Table 6.5)

(Table 6.6)

(Eq. 6.2)

(Table 6.1, by interpolation)

(given)

(Table 6.8)

(Eq. 6.5)

(given)

(Eq. 6.3)

Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski.

Determine the grade length.

V5435 :=

FFS62.3 :=

vp 2323 :=

N3 := fp 1.0 := PHF0.90 :=

vp V PHFN ⋅ fHV ⋅ fp ⋅

fHV 0.867 =

P T 0.10 :=

fHV 1 1P T E T 1 () +

Problem 6.2

(given)

(From Problem 6.1)

(Table 6.1)

(given)

(Eq. 6.3)

(given)

(Eq. 6.5)

Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski.

to be between 0.75-1.0 mi

Using Table 6.8, the length is found

E T 2.54 =

Problem 6.3

Determine the maximum number of large trucks and buses.

first, determine the heavy vehicle factor assume urban freeway

BFFS70 := (given)

fLW 0.0 := (Table 6.3)

fLC 0.0 := (Table 6.4)

fN 4.5 := (Table 6.5)

fID 0.0 := (Table 6.6)

FFSBFFSfLW fLC fN fID := FFS65.5 = (Eq. 6.2)

PHF 1800 7004 ⋅ := PHF0.6429 = (Eq. 6.4)

fp 1.0 := N2 :=

(given)

vp 1800 PHFN fHV fp 1400 fHV (Eq. 6.3)

interpolate from Table 6.1 to find vp

vp 1680FFS65 () 17701680 7065 ⎛ ⎜

+ := vp 1689 =

fHV 1400 vp := fHV 0.8289 =

now, determine the number of trucks

E T 2.5 := (rolling terrain) (Table 6.7)

1800P T ⋅ 247.7143 = therefore 247 trucks and buses

Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski.

⎝ ⎞ ⎟ ⎠

⋅

fHV 1 1P T E T 1 () + 1 1P T 2.51 () + 1 11.5P T + (Eq. 6.5) 11.5P T + 1 fHV

T 0.1376 =

Determine the level of service

V1800 :=

P T 180 V :=

E T 4.0 :=

P T 0.1 =

fHV 1 1P T E T 1 () + :=

PHF0.643 :=

vp V PHFN ⋅ fHV ⋅ :=

fHV 0.769 =

N2 :=

vp 1819.6 =

Problem 6.4

(given)

(Table 6.9)

(Eq. 6.5)

(given)

(Eq. 6.3)

pc/h/ln

max service flows at 65 mi/h FFS from Table 6.1

LOS C = 1680

LOS D = 2090

therefore the LOS is D

Alternative solution:

by interpolation,

D V 64 :=

26.028.1 < 35.0 <

D28.125 =

pc/mi/ln

therefore the LOS is D

(Eq. 6.6)

Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski.

Problem 6.5

Determine the driver population factor.

(Table

Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski.

fp 0.867 = vp V PHFfHV N fp

6.1) vp 2250 := Freeway is operating at capacity, so for FFS = 55 mi/h

6.5) fHV 0.833 = fHV 1 1P T E T 1 () ⋅ + P R E R 1 () ⋅ + := Table 6.7 E R 3.0 := E T 2.0 := fHV Solve for (given) V3900 := N3 := P R 0.06 := P T 0.08 := PHF0.80 :=

(Eq.

Problem 6.6

Determine the new level of service.

BFFS60 :=

fLW 1.9 := (Table 6.3) fN 0.0 := (Table 6.5)

fLC 0.6 := (Table 6.4) fID 0.0 := (Table 6.6)

FFSBFFSfLW fLC fN fID := FFS57.5 = mi/h(Eq. 6.2)

calculate heavy vehicle adjustment

P T 0.12 := P R 0.06 :=

E T 4.5 := E R 4.0 := (Table 6.7)

fHV 1 1P T E T 1 () + P R E R 1 () + := fHV 0.625 = (Eq. 6.5)

calculate LOS before and after lane addition

N1 2 := N2 3 := fp 0.90 := PHF0.88 :=

23002250 + 2 2275 = vp 2275 := (Table 6.1 by interpolation)

vp V PHFfHV N1 fp V vp PHF fHV N1 fp := (Eq. 6.3)

V2252.25 =

vp2 V PHFfHV N2 fp := (Eq. 6.3)

vp2 1516.667 = pc/h/ln LOS D (Table 6.1, by interpolation)

max service flow:

1495 pc/h/ln for LOS C

1965 pc/h/ln for LOS D

Calculate FFS after lane addition

fLC 0.4 := FFS BFFSfLW fLC fN fID := FFS57.7 =

For vp of 1517, S FFS :=

D vp2 S := D26.285 = pc/mi/ln LOS D - 26 < 26.3 < 35(Eq. 6.6)

Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski.

Problem 6.7

Determine the LOS.

find the analysis flow rate

15000.03 ⋅ ()10000.04 ⋅ () + 2500 0.034 = (average grade)

2500 5280 0.473 = mi

P T 0.05 := (given)

E T 2.0 := (Table 6.8)

fHV 1 1P T E T 1 () + :=

fHV 0.952 = (Eq. 6.5)

N2.0 := PHF0.90 := fp 1.0 := V2000 := (given)

vp V PHFN fHV fp := vp 1166.667 = pc/h/ln

determine FFS

(Eq. 6.3)

BFFS65 := (given)

fLW 0.0 := (Table 6.3)

fLC 1.8 := (Table 6.4)

fN 4.5 := (Table 6.5)

fID 0.0 := (Table 6.6)

FFSBFFSfLW fLC fN fID := FFS58.7 = (Eq. 6.2)

find level of service for vp = 1167, S = FFS

D vp FFS := D19.88 = pc/mi/ln therefore LOS is C(Eq. 6.6)

Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski.

to accompany

of Highway

Traffic

Solutions Manual

Principles

Engineering and

Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski.

Problem 6.9

Determine density and level of service before and after the ban.

Before:

P T 0.06 := P B 0.05 := (given)

P TB P T P B + := P TB 0.11 =

E TB 2.5 := (Table 6.7)

fHVTB 1 1P TB E TB 1 () + := fHVTB 0.858 = (Eq. 6.5)

PHF0.95 := fp 1.0 := N 4 := V 5400 := (given)

vp V PHFfHVTB fp N := vp 1655.526 = (Eq. 6.3)

BFFS70 := (given)

fLW 1.9 := (Table 6.3)

fLC 0.4 := (Table 6.4)

fN 1.5 := (Table 6.5)

fID 3.7 := (Table 6.6)

FFSBFFSfLW fLC fN fID := FFS62.5 = (Eq. 6.2)

S 62.5 := (Figure 6.2)

D vp S := D26.5 = pc/mi/ln LOS D(Eq. 6.6)

After: V new V1P T () := V new 5076 =

NumBusesVP B := NumBuses270 =

Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski.

Solutions

Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by

All

reserved. P Bnew NumBuses V new := P Bnew 0.053 = E B 2.5 := (given) fHVB 1 1P Bnew E B 1 () + := fHVB 0.926 =

6.7)

6.5) vp V new PHFfHVB fp N := vp 1442.368 = (Eq. 6.3) S 62.5 :=

6.2) D vp S := D23.1 = pc/mi/ln LOS C (Eq. 6.6)

Fred L. Mannering, Scott

Washburn, and Walter P. Kilareski. Copyright

Inc.

rights

(Table

(Eq.

(Figure

Determine the

calculate free-flow speed

Problem 6.10

and after the strike.

BFFS65 := (given)

fLW 0.0 := (Table 6.3)

fLC 0.0 := (Table 6.4)

fN 3.0 := (Table 6.5)

fID 5.0 := (Table 6.6) (Eq. 6.7)

FFSBFFSfLW fLC fN fID := FFS57 = mi/h

calculate volume after bus strike

V 1 3800 := P

(given)

calculate heavy vehicle adjustments before and after

Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski.

V 2 V 1 P B V 1 6P B V1 () + := V 2 4560 =

E TB 4.0 := E T 5.0 :=

fHVTB 1 1P TB E TB 1 () + := fHVTB 0.847 = (Eq. 6.5) Calculate new P T P T V 1 P T V 2 := P T 0.017 = fHVT 1 1P T E T 1 () ⋅ + := fHVT 0.938 = N3 := fp 1.0 :=

vp1 V 1 PHFN ⋅ fHVTB ⋅ fp ⋅ := vp1 1660.741 = (Eq. 6.3)

bus strike

(Table 6.8)

(given)

LOS before

density, v/c ratio, and

PB 0.04 := PHF0.90

T 0.02 :=

:= P TB P T P B + :=

Solutions

Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski.

(Eq. 6.6) vp1 c 0.732 = D vp1 FFS := D29.14 = pc/mi/ln LOS D (Eq. 6.3) vp2 V 2 PHFN fHVT fp := vp2 1801.481 = vp2 c 0.794 = D vp2 FFS := D31.6 = pc/mi/ln LOS D(Eq. 6.6)