Principle of Highway Engineering and Traffic Analysis, 4 Edition, Solution Manual

Solutions Manual to accompany

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

Chapter 2 Road Vehicle Performance U.S. Customary Units

Copyright © 2008, 2008, by John Wiley & Sons, Inc. All rights reserved.

Preface The solutions to the fourth edition of Principles of Highway Engineering and Traffic Analysis 1 were prepared with the Mathcad software program. 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. 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 sign notations are explained as follows. 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 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 in the equation. For example, in the 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 from a problem where the function 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 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 2 Arrivals(t) – Departures(t), and this evaluates to 2.2t – 0.10t .

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). 1

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Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski. Copyright © 2008, by John Wiley Wiley & Sons, Inc. All rights reserved. reserved.





Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski. Copyright © 2008, by John Wiley & Sons, Inc. All rights reserved.





Problem 2.11

Determine the m aximum g rade. 3500 ne := 60

i := 0.035

ε o := 3.2

r :=

14 12

W

:= 2500 lb

(given)

assume F=F e calculate velocity

V

:=

2 ⋅ π ⋅ r ⋅ ne ⋅ ( 1

− i) V

εo

= 128.9

(Eq. 2.18)

ft/s

calculate aerodynamic resistance

ρ := 0.002378

CD := 0.35

ρ

Ra := ⋅ CD ⋅ Af ⋅ V 2

2

Af := 25 (Eq. 2.3)

Ra

= 172.99

lb

calculate rolling resistance V ⎞ ⎛ f rl := 0.01 1 + ⎝ 147 ⎠ Rrl := f rl ⋅ W

(Eq. 2.5)

Rrl

= 46.93

lb

calculate engine-generated tractive effort Me := 200 Fe :=

nd := 0.90

Me ⋅ ε o ⋅ nd

Fe

r

= 493.71

lb

(Eq. 2.17)

calculate grade resistance Rg := Fe Rg

− Ra − Rrl

(Eq. 2.2)

= 273.79

solve for G G :=

G

Rg

(Eq. 2.9)

W

= 0.1095

therefore G = 11.0%


Al ter nat iv e cal cu lat io n f or gr ade, u si ng tr ig rel ati on sh ip s

θg := asin

⎛ Rg ⎞ ⎝ W ⎠

θg = 0.1097 degθg := θg ⋅ degθg

radians 180

π

convert from radians to degrees

= 6.287

tan deg = opposite side/adjacent side

G := tan

(θg) ⋅ 100

G

= 11.02

%

Thus, error is minimal when assuming G = sin

θg for small to medium grades


Problem 2.12 Determine the torqu e the engine is produ cing and the engine speed.

− ΣR

Fe

γ m ⋅m ⋅a

at top speed, acceleration = 0; thus, F e V

:= 124⋅

5280

V

3600

− ΣR

0

= 181.867 ft/s

(given)

calculate aerodynamic resistance

ρ := 0.00206

CD := 0.28

ρ

Ra := ⋅ CD ⋅ Af ⋅ V 2

2

Af := 19.4 Ra

= 185.056

(Eq. 2.3)

calculate rolling resistance

⎛ + ⎝

⎞ 147 ⎠ V

f rl := 0.01 ⋅ 1 W

f rl

= 0.022

(Eq. 2.5)

:= 2700

(given)

Rrl := f rl ⋅ W

Rrl

(Eq. 2.6)

= 60.404

Rg := 0 sum of resistances is equal to engine-generated tractive effort, solve for Me Fe := Ra + Rrl i := 0.03

Fe Me

+ Rg

ηd := 0.90

Me ⋅ ε 0 ⋅ ηd

ε 0 := 2.5

r :=

Me :=

r

= 114.548

(Eq. 2.2)

Fe = 245.46 12.6 12

Fe ⋅ r

(Eq. 2.17)

ε 0 ⋅ η d

ft-lb

Knowing velocity, solve for ne

V

ne

2 ⋅ π ⋅ r ⋅ ne ⋅ ( 1

− i)

ε0 = 71.048

rev s

V ⋅ε 0

ne := 2 ⋅ π ⋅ r ⋅ ( 1 ne ⋅ 60 = 4263

(Eq. 2.18)

− i) rev min




Rear - wheel drive

μ := 0.2

f rl := 0.011

h := 20

L := 120

lf := 60








2

Sal

V2 :=

2

γ b ⋅ V1 − V2 2 ⋅ g ⋅ ( η b ⋅ μ m + f rl − 0.03) 2

V1

−

2 ⋅ Sal ⋅ g ⋅

(Eq. 2.43)

(η b ⋅ μ m + f rl − 0.03) γb


Problem 2.24 Determine if th e driver shoul d appeal the ticket.

μ := 0.6 (for good, wet pavement, and slide value because of skidding) γ b := 1.04 V2

:= 40 ⋅

g := 32.2

5280

V2

3600

ηb := 0.95

= 58.667

S := 200

(given)

f rl := 0.015

Solve for the initial velocity of the car using theoretical stopping distance

2

S

V1

2

γ b ⋅ V1 − V2 2 ⋅ g ⋅ ( ηb μ ⋅ + f rl − 0.04)

V1

:=

2⋅S ⋅g⋅

(ηb ⋅ μ + f rl − 0.04) γb

+ V2

2

(Eq. 2.43)

= 100.952

V1 ⋅

3600 5280

= 68.83

mi h

No, the driver should not appeal the ticket as the initial velocity was higher than the speed limit, in addition to the road being wet.





G

= 0.059

G ⋅ 100 = 5.93

%



Multiple Choice Problems Determine the minimum t ractive effort.

CD := 0.35

2

A f := 20 ft

ρ := 0.002045

Problem 2.31 slugs ft

V := 70⋅

⎛ 5280 ⎞ ⎝ 3600 ⎠

ft s

3

(given)

W := 2000 lb

G := 0.05

grade resistance R g := 2000G ⋅

R g

= 100

R a

= 75.44

(Eq. 2.9)

lb

aerodynamic resistance

ρ

2

R a := ⋅ CD⋅ A f ⋅ V 2

lb

(Eq. 2.3)

rolling resistance

⎛ ⎝

f rl := 0.01⋅ 1 +

V ⎞ 147 ⎠

R rl := f rl⋅ W

f rl = 0.02

R rl = 33.97

(Eq. 2.5)

lb

(Eq. 2.6)

summation of resistances

F := R a

+

R rl + R g

F = 209.41 lb

(Eq. 2.2)


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Using mi/h instead of ft/s for velocity mi

V := 70

h

⎛ ⎝

f rl := 0.01⋅ 1 +

V ⎞ 147 ⎠

R rl := f rl⋅ W R a :=

ρ 2

F := R a

2

f rl = 0.01

R rl = 29.52

lb

= 35.07

lb

⋅ CD⋅ A f ⋅ V

R a

+

F = 164.6

R rl + R g

lb

2) not including aerodynamic resistance V := 70⋅

5280 3600

F := R rl + R g

F

= 129.52

lb

3) not including rolling resistance F := R a

+

R g

F

= 135.07

lb


Determine the acceleration.

Problem 2.32 V := 20⋅

h := 20

Cd := 0.3

5280 3600

V = 29.33

ft

A f := 20

s

ρ := 0.002045

μ := 1

M e := 95 r :=

ft

W

2

slugs ft

3

L := 110

in

lf := 50

in

lb

(given)

ηd := 0.90

ft

12

:= 2500

ε o := 4.5

ft-lb

14

in

aerodynamic resistnace

ρ

2

R a := ⋅ Cd ⋅ A f ⋅ V 2

R a

= 5.28

(Eq. 2.3)

lb

rolling resistance f rl := 0.01⎛ 1 + ⎝

(Eq. 2.5)

V ⎞ 147 ⎠

R rl := 0.01⋅ ⎛ 1 +

⎝

V ⎞ 147 ⎠

⋅ 2500

R rl = 29.99

(Eq. 2.6)

lb

engine-generated tractive effort

Fe :=

M e⋅ ε o ⋅ ηd

Fe

r

= 329.79

(Eq. 2.17)

lb

mass factor

γ m := 1.04 +

0.0025ε o

2

(Eq. 2.20)

γ m = 1.09

lr := 120 − lf

acceleration

μ⋅ W⋅ ( lr + f rl⋅ h) Fmax :=

a :=

Fe

L

1+

μ⋅h

lb

(Eq. 2.15)

L

− R a − R rl

γ m⋅ ⎛

Fmax = 1350.77

2500 ⎞

⎝ 32.2 ⎠

a

= 3.48

ft s

(Eq. 2.19)

2


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Use a mass factor of 1.04

2) Use Fmax instead of Fe

γ m := 1.04

a :=

γ m := 1.091

Fe

a :=

− R a − R rl

ft a

2500 ⎞ γ m⋅ ⎛ ⎝ 32.2 ⎠ Fmax − R a

− R rl

2500 ⎞ γ m⋅ ⎛ ⎝ 32.2 ⎠

= 3.65

s

2

ft a

= 15.53

a

= 15.9

s

2

3) Rear wheel instead of front wheel drive

μ⋅ W⋅ ( lf − f rl⋅ h) Fmax :=

L

1−

μ⋅ h L

Fmax = 1382.22

a :=

Fmax − R a

γ m⋅ ⎛

− R rl

2500 ⎞

⎝ 32.2 ⎠

ft s

2


Determine the percentage of braking force. V := 65⋅

5280

ft

3600

s

Problem 2.33

μ := 0.90

L := 120

in

lf := 50 in

h := 20

in

lr := L − lf

(given) in

determine the coefficient of rolling resistance

⎛ ⎝

f rl := 0.01⋅ 1 +

V ⎞ 147 ⎠

f rl = 0.02

(Eq. 2.5)

determine the brake force ratio lr + h ⋅ BFR frmax:= lf − h ⋅

(μ + f rl) (μ + f rl)

BFR frmax = 2.79

(Eq. 2.30)

calculate percentage of braking force allocated to rear axle PBFr :=

100 1 + BFRfrmax

PBFr = 26.39 %

(Eq. 2.32)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Use front axle equation 100 PBFf := 100 − 1 + BFRfrmax

PBFf = 73.61 %

(Eq 2.31)

2) Use incorrect brake force ratio equation

BFR frmax:=

PBFr :=

(μ + f rl) lf + h⋅ ( μ + f rl) lr − h ⋅

100 1 + BFRfrmax

BFR frmax = 0.76

PBFr = 56.94 %

3) Switch l f and l r in brake force ratio equation

BFR frmax:=

PBFr :=

(μ + f rl) lr − h ⋅ (μ + f rl)

lf + h⋅

100 1 + BFRfrmax

BFR frmax = 1.32

PBFr = 43.06 %


Determine the theoretical stoppi ng dis tance on level grade. CD := 0.59

V := 80⋅ 2

A f := 26 ft

5280

ft

3600

s

Problem 2.34 (given)

μ := 0.7

γ b := 1.04

η b := 0.75

(assumed values)

Coefficient of Rolling Resistance

⎛

⎞

V 2

f rl := 0.01⋅ 1 +

⎝

f rl = 0.014

147 ⎠

(Eq. 2.5)

Theoretical Stopping Distance

S :=

2

γ b ⋅

V1

2

− V2

(

2⋅ g ⋅ η b⋅ μ

S = 412.8

)

+

f rl

s

2

(Eq. 2.43)

ft

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Not dividing the velocity by 2 for the coeffeicent of rolling resistance

⎛ ⎝

f rl := 0.01⋅ 1 +

S :=

γ b⋅

2

V1

V ⎞ 147 ⎠ 2

− V2

(

2⋅ g⋅ η b⋅ μ

+

S

)

f rl

= 409.8

s

2

ft

2) Using mi/h instead of ft/s for the velocity V1 := 80

V2 := 0

⎛ ⎝

S :=

γ b⋅

2

V1

(

⎞

V

f rl := 0.01⋅ 1 +

2

147 ⎠ 2

− V2

2⋅ g⋅ η b⋅ μ

V := 80

+

)

f rl

S

= 192.4

s

2

ft


3) Using γ = 1.0 value

γ b := 1.0 S :=

2

γ b ⋅

V1

2

− V2

(

2⋅ g ⋅ η b⋅ μ

+

S = 397.9

)

f rl

s

2

ft

Problem 2.35

Determine the stopping si ght dis tance. V := 45⋅

5280

ft/s

(given)

3600 ft

a := 11.2

s

tr := 2.5

2

g := 32.2

s

ft s

2

( assumed )

Braking Distance 2

d :=

V 2⋅ g⋅

d

⎛ a ⎞ ⎝ g ⎠

= 194.46

ft

(Eq. 2.47)

Perception/Reaction Distance d r := V⋅ tr

d r = 165.00 ft

(Eq. 2.49)

Total Stopping Distance d s

:= d +

d r

d s

= 359.46

ft

(Eq. 2.50)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) just the braking distance value d

= 194.46

ft

2) just the perception/reaction distance value d r = 165.00

ft


3) use the yellow signal interval deceleration rate a := 10.0 2

d :=

d s

V

⎛ a ⎞ 2⋅ g⋅ ⎝ g ⎠

:= d +

d r

d

= 217.80

ft

d s

= 382.80

ft

Problem 2.36

Determine the vehicle sp eed.

CD := 0.35

G := 0.04 2

A f := 16 ft W

:= 2500

V1 := 88⋅

S := 150

γ b := 1.04 η b := 1

ft slugs

lb

ρ := 0.002378

5280

ft

3600

s

g := 32.2

ft

ρ 2

⋅ CD⋅ A f

K a

μ := 0.8

3

f rl := 0.017

ft s

K a :=

(given)

2

= 0.007

Given V2 := 0

S

⎡ η ⋅ μ ⋅ W + K ⋅ V 2 + f ⋅ W + W⋅ G ⎤ b a 1 rl ⋅ ln 2⋅ g⋅ K a ⎢ ⎥ 2 ⎣ η b⋅ μ ⋅ W + K a⋅ V2 + f rl⋅ W + W⋅ G⎦ γ b⋅ W

(Eq. 2.39)

( )

V2 := Find V2

V2 = 91.6

V2 1.467

= 62.43

mi h

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answers:

1) Use 0% grade G := 0.0 Given V2 := 0

⎡ η ⋅ μ ⋅ W + K ⋅ V 2 + f ⋅ W + W⋅ G ⎤ b a 1 rl ⋅ ln 2⋅ g⋅ K a ⎢ ⎥ 2 W + K a⋅ V2 + f rl⋅ W + W ⋅ G η ⋅ μ ⋅ b ⎣ ⎦ γ b⋅ W

S

( )

V2 := Find V2

V2

V2 = 93.6

1.467

mi

= 63.78

h

2) Ignoring aerodynamic resistance G := 0.04

V2 :=

V2 1.467

2

V1

−

(

S⋅ 2⋅ g ⋅ η b⋅ μ

+

)

f rl + G

V2 = 93.3

γ b

(Eq 2.43) rearranged to solve for V2

mi

= 63.6

h

3) Ignoring aerodynamic resistance and using G = 0

G := 0

V2 :=

V2 1.467

2

V1

= 64.9

−

(

S⋅ 2⋅ g⋅ η b ⋅ μ

γ b

+

)

f rl + G

V2 = 95.2

mi h



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

Chapter 3 Geometric Design of Highways U.S. Customary Units

Copyright © 2008, by John Wiley & Sons, Inc. All rights reserved.

Preface The solutions to the fourth edition of Principles of Highway Engineering and Traffic Analysis 1 were prepared with the Mathcad 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 =).



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Problem 3.6 Compute the difference in design curve lengths for 2005 and 2025 designs. G1 := 1

G2 := −2

A := G2 − G1

A = 3

(given)

find required L for 70 mi/h design speed K := 247

(Table 3.2)

L2005 := K ⋅ A

L2005 = 741

(Eq. 3.17)

SSD2005 := 730 V := 70⋅

(Table 3.1)

5280

V = 102.667

3600

H1 := 3

H2 := 1

g := 32.2

G := 0

(given)

For 2025 values, a increases by 25% and t r increases by 20% a2025 := 11.2⋅ 1.25

a2025 = 14

t r2025 := 2.5 ⋅ 1.2

tr2025 = 3

Calculate required stopping sight distance in 2025 V

S2025 := 2⋅g⋅

2

⎡⎛ a2025 ⎞ ⎣⎝

g

⎠

⎤

+ V ⋅ t r2025

S2025 = 684.444

(Eq. 3.12)

−G

⎦

Using this distance, calculate required minimum curve length in 2025 2

A ⋅ S2025

L2025 :=

200⋅

(

(Eq. 3.13)

)2

H1 +

H2

L2025 = 941.435

Diff := L2025 − L2005

Diff = 200.43

ft

Al ter nat iv e Sol ut io n 2

L2002 :=

A ⋅ SSD2005 2158

L2005 = 741

L2025 − L2005 = 200.435












)2

(

K c ⋅ 4 + Gcon 200

−

(

)

4.0 ⋅ Kc ⋅ 4 + Gcon 100

+

(

Gcon ⋅ Ltotal − Kc ⋅ 4 + Gcon 100

)

− Ks ⋅ Gcon

2

+

K s ⋅ Gcon 200

Δ elev


(

G con − G 1s 200

)

2

⋅K s

+

G con

(

)

⋅ L total − G con − G 1s ⋅ K s − G con − G 2c ⋅ K c + 100

(

G con − G 2c 200

)

2

⋅K c

elev diff +

(

⎡

G con − G 1s ⋅ K s ⋅ G 1s

⎣

100

)

+

G con − G 2c ⋅ K c ⋅ G 2c

)⎤

100

(

⎦










Problem 3.24 Determine the station of th e PT. sta PC := 12410 e := 0.06

sta PI := 13140

V := 70

(given)

g := 32.2

f s := 0.10

(Table 3.5)

calculate radius Rv :=

( V ⋅ 1.467)

(

g ⋅ e + f s

2

)

Rv = 2046.8

since road is single-lane,

(Eq. 3.34)

R := Rv

R = 2046.8 T := sta PI − sta PC

T = 730

knowing tangent length and radius, solve for central angle T

R ⋅ tan

⎛ Δ ⎞ ⎝ 2 ⎠

Δ := 2 ⋅ atan

⎛ T ⎞ ⎝ R ⎠

Δ = 39.258deg

Δ := 39

(Eq. 3.36)

calculate length L :=

π 180

⋅ R⋅ (Δ )

sta PT := sta PC + L

L = 1393.2 sta PT = 13803.229

(Eq. 3.39) sta PT = 138 + 03.23




Problem 3.28 Determine the radius and s tationing of the PC and PT. sta PI := 25050 e := 0.08

g := 32.2

V := 65

(given)

Δ := 35

f s := 0.11

(Table 3.5)

calculate radius

Rv :=

( V ⋅ 1.467)

(

2

Rv = 1486.2

)

g ⋅ f s + e

(Eq. 3.34)

ft

since the road is two-lane with 12-ft lanes R := Rv + 6

R = 1492.2

ft

calculate length and tangent length of curve L :=

π 180

⋅R⋅Δ

T := R ⋅ tan

L = 911.534

⎡⎛ Δ ⎞ ⋅ deg⎤ ⎣⎝ 2 ⎠ ⎦

(Eq. 3.39)

T = 470.489

(Eq. 3.36)

sta PC := sta PI − T

sta PC = 24579.511

sta PC = 245+79.51

sta PT := sta PC + L

sta PT = 25491.044

sta PT = 254+91.04

Problem 3.29 Give the radius, degree of curvature, and length of cur ve that you would recommend.

Δ := 40

2 10-ft lanes

(given)

for a 70 mph design speed with e restricted to 0.06, R := Rv + L :=

D :=

π 180

5 2

⋅ R⋅ Δ

18000

π ⋅R

R = 2052.5

L = 1432.92

D = 2.79

Rv := 2050 ft

(Table 3.5)

ft

ft

degrees

(Eq. 3.39)

(Eq. 3.35)




Problem 3.32 Determine a maximum safe speed to the nearest 5 mi/h.

Δ := 34

(given)

e := 0.08

PT := 12934 L := PT − PC

PC := 12350 L = 584

12 since this is a two-lane road with 12-ft lanes, M := 20.3 + s 2 L

π 180

⋅ R⋅ Δ

Rv := R − 6

R :=

L ⋅ 180

π ⋅Δ

R = 984.139

Ms = 26.3 (Eq. 3.39)

Rv = 978.139


Problem 3.33 Determine the distance that mu st be cl eared from the inside edge of t he inside l ane to pro vide adequate SSD. V is 70 mi/h SSD := 730

(Table 3.1)

Rv := 2050

⎛

Ms := Rv⋅ 1 − cos

⎝

(Prob. 3.29)

⎛ 90⋅ SSD ⋅ deg ⎞ ⎞ ⎝

π ⋅ Rv

(Eq. 3.42)

⎠ ⎠

Ms = 32.41 To inside edge of inside lane (subtracting 1/2 of lane width) Ms − 5 = 27.41

ft



Problem 3.35 Determine the length of th e horizontal curve. G1 := 1

G2 := 3

Ls := 420

Δ := 37

As := G2 − G1 Ks :=

Ls As

(given) e := 0.06

As = 2 Ks = 210

(Eq. 3.10)

safe design speed is 75 mi/h (K = 206 for 75 mi/h) Rv1 := 2510

or

V := 75

(Table 3.5)

f s := 0.09 ( V ⋅ 1.467)

2

Rv2 := 32.2 ⋅ fs + e

(

)

Rv2 = 2506.31

since the road is two-lane with 12-ft lanes, L :=

π 180

⋅ R⋅ Δ

(Table 3.3)

L = 1624.76

ft

R := Rv1 +

(Eq. 3.34) 12 2

R = 2516 (Eq. 3.39)


Problem 3.36 Determine Determine the station of th e PT. PT. G1 := −2.5

G2 := 1.5

Δ := 38

e := 0.08

L := K ⋅ A

L = 824

A := G2 − G1

A = 4

K := 206

(given)

PVT := 2510 (Eq. 3.10)

PVC := PVT − L

PVC = 1686

PC := PVC − 292

PC = 1394

Rv := 2215

(Table 3.5)

since the road is two-lane, 12-ft lanes R := Rv + L :=

12 2

π ⋅ R⋅ Δ 180

PT := PC + L

R = 2221

ft

L = 1473.02

ft

PT = 2867.02

(Eq. 3.39) sta PT = 28 + 67.02




Multiple Choice Problems Determine Determine the elevation elevation of the lowest poin t of t he curve. G1 := −4.0 c := 500

G2 := 2.5

Probl em 3.38 3.38

L := 4 stations

(given)

ft

stationing and elevation for lowest point on the curve dy dx

:= ( 2a⋅ x + b)

(Eq. 3.1)

0

(Eq. 3.3)

b := −4.0

a :=

x :=

G2 − G1 2⋅ L

− b 2⋅ a

(Eq. 3.6)

a = 0.813

(Eq. 3.1)

x = 2.462 stations

Lowe Lowest st Poi Point nt sta stati tion onin ing: g:

(100 (100 + 00) 00) + (2 (2 + 46) 46) = 102 102 + 46 46

Lowest Point elevation:

y := a⋅ x

( 2) + b⋅ x + c

y = 495.077 ft

(Eq. 3.1)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Miscalculation

y := 492.043

2) Miscalculate "a"

a :=

G1 − G2 2⋅ L

ft

a = −0.813

( 2) + b⋅ x + c

y := a⋅ x

3) Assume lowest point at L/2

Station = 102 + 46

y = 485.231

ft

Station = 102 + 00 x := 2

( 2) + b⋅ x + c

y := a⋅ x

y = 495.25

ft


Determine Determine the station of

T := 1200

PT.

Δ :=

ft

Problem 3.39 ⋅ 0.5211180

(given)

π

Calculate radius T

R :=

⎡⎛ Δ ⎞

tan [mc]

⎣⎝ 2 ⎠

⋅ deg⎤

R = 4500.95 ft

(Eq 3.36)

L = 2345.44 ft

(Eq 3.39)

⎦

Solve for length of curve L :=

π 180

⋅ R ⋅ Δ

Calculate stationing of PT stationing PC = 145 + 00 minus 12+00 = 133 + 00 stationing PT = stationing PC + L = 133 + 00 plus 23 + 45.43 = 156 + 45.43

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Add length of curve to stationing PI stationing PT = 145 + 000 plus 23 + 45.43 = 168 + 45.43 2) Use radians instead of degrees T

R := tan

L :=

⎛ 0.5211 ⎞ ⎝

π 180

2

R = 4500.95 ft

⎠

⋅ R ⋅ 0.5211

L = 40.94

ft

stationing PT = 133 + 00 plus 40 + 94 = 173 + 94 3) add half of length to stationing PI stationong PT = 145 + 00 plus 11 +72.72 = 156 + 72.72


Problem 3.40

Determine Determine the o ffset.

G1 := 5.5 %

G2 := 2.5 %

x := 750 ft

L := 1600 ft

(given)

determine the absolute value of the difference of grades A := G1 − G2

A=3

determine offset at 750 feet from the PVC

Y :=

A

2

200 200⋅ L

⋅x

Y = 5.273 ft

(Eq 3.7)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: Answers: 1) Use Ym equation. Ym :=

A⋅L 800

Ym = 6

ft

(Eq 3.8)

Yf = 24

ft

(Eq 3.9)

2) Use Yf equation. Yf :=

A⋅L 200

3) Use 0.055 and 0.025 for grades. G1 := 0.055

G2 := 0.025

A := G1 − G2 Y :=

A 200L ⋅

2

⋅x

A = 0.03 Y = 0.053 ft


Determine the minimum length of curve.

V := 65⋅

⎛ 5280 ⎞ ⎝ 3600 ⎠

ft

Problem 3.41

G1 := 1.5

s

G2 := −2.0

(given)

ignoring the effect of grades using Table 3.1, SSD for 65 mi/h would be 645 ft (assuming L > SSD) SSD := 645

(Table 3.1)

ft

A := G1 − G2

A = 3.50

2

Lm :=

A ⋅ SSD 2158

Lm = 674.74

ft

(Eq. 3.15)

674.74 > 645

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers 1) assume L < SSD Lm := 2⋅ SSD −

2158

Lm = 673.43 ft

A

(Eq. 3.16)

2) Misinterpret chart for 70 mi/h 2

SSD := 730

Lm :=

A ⋅ SSD 2158

Lm = 864.30 ft

3) Assume SSD is equivalent to L m

SSD := 645

Lm := SSD

therefore

Lm = 645.00 ft


Determine the stopping si ght dis tance.

V1 := 35⋅

a := 11.2

5280

ft

3600

s

ft s

g := 32.2

2

Problem 3.42 G :=

ft s

2

3

(given)

100

tr := 2.5 s

(assumed)

Determine stopping sight distance

2

V1

SSD := 2⋅ g ⋅

⎛ a

− G ⎞

⎝ g

+ V1⋅ tr

SSD = 257.08 ft

(Eq 3.12)

⎠

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Assume grade is positive (uphill)

2

SSD :=

V1

⎛ a + G ⎞ 2⋅ g ⋅ ⎝ g ⎠

+ V1⋅ tr

SSD = 236.63 ft

2) Use g = 9.81 m/s 2 instead of g = 32.2 ft/s 2 g := 9.81

m s

2 2

SSD :=

V1

a 2⋅ g ⋅ ⎛ − G ⎞ ⎝ g ⎠

+ V1⋅ tr

SSD = 249.15 ft

3) Miscalculation SSD := 254.23 ft


Determine the minimum length o f the vertical curve. G1 := 4.0

G2 := −2.0

H1 := 6.0 ft

Problem 3.43

H2 := 4.0 ft

(given) S := 450 ft

V := 40⋅

5280

ft

3600

s

Calculate the minim length of vertical curve A := G1 − G2

200⋅

Lm := 2⋅ S −

(

)

H1 +

H2

A

2

Lm = 240.07 ft

(Eq 3.14)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Use equation 3.13 2

A⋅ S

Lm := 200⋅

(

H1 +

)

H2

Lm = 306.85 ft

2

(Eq 3.13)

2) Use AASHTO guidelines for heights and equation 3.13 H1 := 3.5 ft

H2 := 2.0 ft 2

A⋅ S

Lm := 200⋅

(

H1 +

)

H2

Lm = 562.94 ft

2

3) Solve for S and not Lm

Lm := 450 ft S :=

Lm + 200⋅

(

H1 + 2

)

H2

2

S = 1304.15 ft




Chapter 4 Pavement Design U.S. Customary Units


Preface The solutions to the fourth edition of Principles of Highway Engineering and Traffic Analysis 1 were prepared with the Mathcad 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.

•


•


•




www.mathcad.com


Problem 4.1 Determine radial-horizontal st ress. P := 5000

p := 100

(given)

E := 43500 a :=

P

a

p ⋅ π z := 0

at z a

and r

=0

A := 1.0

a

= 3.989

(Eq. 4.3)

r := 0.8

= 0.201

H := 1.97987

(Table 4.1)

Solving for Poisson's ratio

Δ z := 0.016

(given)

Find μ:

Δz

p⋅ ( 1 +

μ) ⋅ a

⋅ ⎡ ⋅ A + (1 − μ) ⋅ H⎤ z

⎣a

E

⎦

(Eq. 4.6)

μ = 0.345 function C := 0

σr := p⋅ 2μ ⋅ A + σr = 84.47

and function F := 0.5 C + ( 1 − 2μ) F

(Eq. 4.5)

lb 2

in


Determine Modulus of Elasticity. contactarea

P := 6700

in

lb

(given)

contactarea

a :=

a

2

:= 80

Problem 4.2

π

= 5.046

inches

P

p := a

(Eq. 4.3)

2

⋅π lb

p

= 83.75

2

in

z := 2 z a

r := 0

and

r

= 0.3963

=0

a

Using linear interpolation: A := 0.62861

H := 1.35407

and

Δ z := 0.035

Δz E :=

μ := 0.5

p ⋅ ( 1 +

μ) ⋅ a

z

⎣a

E p ⋅ ( 1 +

⋅⎡ ⋅A +

μ) ⋅ a

⋅⎡ ⋅A + z

⎣a

Δz 4

E = 1.678 × 10

(Table 4.1) (given)

⎤ ⎦

(1 − μ )⋅ H

(Eq. 4.6)

⎤ ⎦

( 1 − μ)⋅ H

lb 2

in


Problem 4.3 Determine load applied to th e wheel.

z

:= 0 z

=

a A

r := 0

and

r

0

a

:= 1

=

a := 3.5

and

(given)

0

C := 0

F

:= 0.5

H := 2

(Table 4.1)

σ r := p [ 2μ ⋅ A + C + ( 1 − 2μ )F]

(Eq. 4.5)

lb

σ r := 87

2

in

Therefore,

p

:=

σ r 2μ

p

+ 0.5 − μ

:=

σ r μ + 0.5

Find μ by substituting p into Equation 4.6

Δ z := 0.0165

p ⋅ ( 1 + μ ) ⋅ a

Δz

z

(Eq. 4.6)

⎦

⎛ 87 ⎞ ⋅( 1 + μ ) ⋅ 3.5 ⎝ μ + 0.5 ⎠ ⋅[ 0 + ( 1 − μ ) ⋅2.0] 43500

μ =

:=

⋅⎡ ⋅A + ( 1 − μ ) ⋅H⎤

⎣a

E

Δz

p

(given)

0.281 87

p

μ + 0.5

:=

P

:= a2( p π ⋅ )

P

:= 12.25⋅( p ⋅ π )

P

=

lb 2

(Eq. 4.3)

p π ⋅

4285.20

111.349

in

P

a

=

lb


Problem 4.4

Determine the vertical stress, radial-horizontal stress, and deflection.

:= 25

z z a

=

r := 50

and

r

1.969

a

A

:= 0.0116

p

:= 101.5

=

and

a

(given)

:= 12.7

3.937

B := − 0.0041 C := 0.01527

F

:= −0.005465

H := 0.22418

(Table 4.1)

μ := 0.4

σ z := p ⋅( A + B)

(Eq. 4.4)

lb

σz =

0.761

2

in

σ r := p ⋅ 2μ ⋅A + C + (1 − 2μ )F σ r =

(Eq. 4.5)

lb

2.381

2

in E := 36250

Δ z :=

p ⋅ ( 1 +

Δz =

μ ) ⋅a

⋅⎡ ⋅A + (1 − μ ) ⋅ H⎤ z

⎣a

E

−3

7.833 × 10

⎦

(Eq. 4.6)

inches


Problem 4.5 Determine which truck will c ause more pavement damage. a1 := 0.44

a2 := 0.2

a3 := 0.11

D1 := 3

D2 := 6

D3 := 8

(given)

M 3 := 1.0

(given)

M2 := 1.0

SN := a1⋅ D1 + a2⋅ D2⋅ M 2 + a3⋅ D3⋅ M 3

(Table 4.6)

(Eq. 4.9)

SN = 3.4

Using linear interpolation for Truck A

(Table 4.2, 4.3, and 4.4)

single12kip := 0.2226 single23kip := 2.574 Total18kipESALA := single12kip + single23kip Total18kipESALA

= 2.7966

Using linear interpolation for Truck B single8kip := 0.0046 tandem43kip := 2.706 Total18kipESALB := single8kip + tandem43kip Total18kipESALB = 2.7106

Therefore, Truck A causes more damage.


Problem 4.6 How many 25-kip singl e-axle loads can be c arried before the pavement reaches its TSI? a1 := 0.44

a2 := 0.18

a3 := 0.11

D1 := 4

D2 := 7

D3 := 10

(given)

M2 := 0.9

M3 := 0.8

(given)

(Table 4.6)

SN := a1⋅ D1 + a2⋅ D2⋅ M2 + a3⋅ D3⋅ M 3

(Eq. 4.9)

SN = 3.774 ZR := −1.282 So := 0.4

(Table 4.5)

ΔPSI := 2.0

MR := 5000

lb

(given)

2

in

Using Eq. 4.7:

⎡⎡ x := ⎢⎢ZR ⋅ So + 9.36(log( SN + 1) ) − 0.20 + ⎢⎢ ⎣⎣

⎤ ⎤ + 2.32⋅ log( MR )⎥ − 8.07⎥ 1094 ⎤ ⎥ ⎥ 0.40 + ⎡ 5.19 ⎣ (SN + 1) ⎦ ⎦ ⎦ log

⎛ ΔPSI ⎞ ⎝ 2.7 ⎠

x = 5.974 x

5

W 18 := 10

W 18 = 9.4204645× 10

Interpolating from Table 4.2 to determine equivalent 25-kip axle load equivalency factor: TableValue1 :=

TableValue2 :=

3.09 + 4.31

TableValue1 = 3.7

2 2.89 + 3.91

TableValue2 = 3.4

2

EquivFactor := 3.7 −

⎛ TableValue1 − TableValue2 ⎞ ⋅ ( SN − 3) 1 ⎝ ⎠

EquivFactor = 3.4678 W 18 EquivFactor

= 2.716554×

5

10

25-kip loads


Problem 4.7 Determine minimum acceptable soil resilient modu lus. a1 := 0.35

a2 := 0.20

a3 := 0.11

D1 := 4

D2 := 6

D3 := 7

(given)

M2 := 1

M3 := 1

(given)

(Table 4.6)

SN := a1⋅ D1 + a2⋅ D2⋅ M2 + a3⋅ D3⋅ M 3

(Eq. 4.9)

SN = 3.37

Axle Loads (interpolating):

(Table 4.2, 4.3, and 4.4)

single10kip := 0.1121300 ⋅ single18kip := 1⋅ 120 single23kip := 2.578100 ⋅ tandem32kip := 0.8886100 ⋅ single32kip := 9.87130 ⋅ triple40kip := 0.546100 ⋅ TotAxleEqv := single10kip + single18kip + single23kip + tandem32kip + single32kip + triple40kip W 18 := TotAxleEqv⋅ 10⋅ 365 6

W 18 = 3.106 × 10 ZR := −1.036

ΔPSI := 2.2

(Table 4.5)

So := 0.30

⎡ log ( W 18) ⎢ZR ⋅ So + 9.36(log( SN + 1) ) − 0.20 + ⎢ ⎣ 3

MR = 9.0098× 10

⎤ + 2.32⋅ log( MR )⎥ − 8.07 1094 ⎤ ⎥ 0.40 + ⎡ 5.19 ⎣ (SN + 1) ⎦ ⎦ log

⎛ ΔPSI ⎞ ⎝ 2.7 ⎠

lb 2

in


Determine minimum acceptable soil resilient modu lus. a1 := 0.35

a2 := 0.20

D1 := 4

D2 := 6

a3 := 0.11

M2 := 1.0

Problem 4.8 (Table 4.6)

D3 := 7

(given)

M3 := 1.0

(given)

SN := a1⋅ D1 + a2⋅ D2⋅ M2 + a3⋅ D3⋅ M 3

(Eq. 4.9)

SN = 3.37


(Table 4.2, 4.3, and 4.4)

single10kip := 0.1121300 ⋅ single18kip := 1⋅ 20 single23kip := 2.682100 ⋅ tandem32kip := 0.8886100 ⋅ single32kip := 9.87190 ⋅ triple40kip := 0.546100 ⋅

TotAxleEqv := single10kip + single18kip + single23kip + tandem32kip + single32kip + triple40kip W 18 := TotAxleEqv⋅ 10⋅ 365 ZR := −1.036

ΔPSI := 2.2

(Table 4.5)

So := 0.30

(given)

Using Eq. 4.7:

⎡ log ( W 18) ⎢ZR ⋅ So + 9.36(log( SN + 1) ) − 0.20 + ⎢ ⎣ 4

M R = 1.1005× 10

⎤ + 2.32⋅ log( MR )⎥ − 8.07 1094 ⎤ ⎥ 0.40 + ⎡ 5.19 ⎣ (SN + 1) ⎦ ⎦ log

⎛ ΔPSI ⎞ ⎝ 2.7 ⎠

lb 2

in


Determine the soil resilient modul us of th e soil used in design.

Problem 4.9

a1 := 0.44

a2 := 0.40

(Table 4.6)

D1 := 4

D2 := 4

D3 := 8

(given)

M2 := 1.0

M3 := 1.0

(given)

a3 := 0.11

SN := a1⋅ D1 + a2⋅ D2⋅ M2 + a3⋅ D3⋅ M 3

(Eq. 4.9)

SN = 4.24


(Tables 4.2 and 4.3)

single8kip := 0.039251300 ⋅ = 51.025 tandem15kip := 0.0431900 ⋅ = 38.79 single40kip := 22.16420 ⋅ = 443.28 tandem40kip := 2.042200 ⋅ = 408.4 TotAxleEqv := single8kip + tandem15kip + single40kip + tandem40kip TotAxleEqv = 941.495 6

W 18 := TotAxleEqv⋅ 12⋅ 365 = 4.1237× 10 ZR := −0.524 So := 0.5

(Table 4.5)

ΔPSI := 2.0

(given)

Using Eq. 4.7:

⎡ log( W 18) ⎢ZR ⋅ So + 9.36(log( SN + 1) ) − 0.20 + ⎢ ⎣ MR = 5250.1

⎛ ΔPSI ⎞ ⎝ 2.7 ⎠

⎤ + 2.32⋅ log( MR )⎥ − 8.07 1094 ⎡ ⎤ ⎥ 0.40 + 5.19 ⎣ ( SN + 1) ⎦ ⎦ log

lb 2

in


Problem 4.10 Determine the reduction i n pavement design life. SN := 3.8

So := 0.40

PSI := 4.7

W 18 := 1800⋅ 365⋅ N

TSI := 2.5

ZR := −1.645

(given)

(for 95% reliability)

(Table 4.5)

CBR := 9

Current Design Life:

ΔPSI := PSI − TSI

ΔPSI = 2.2

MR := 1500⋅ CBR

MR = 1.35 × 10

(given) 4

Using Eq. 4.7:

⎡ log ( W 18) ⎢ZR ⋅ So + 9.36(log( SN + 1) ) − 0.20 + ⎢ ⎣

⎤ + 2.32⋅ log( MR )⎥ − 8.07 1094 ⎤ ⎥ 0.40 + ⎡ 5.19 ⎣ (SN + 1) ⎦ ⎦ log

⎛ ΔPSI ⎞ ⎝ 2.7 ⎠

6

W 18 = 8.0730705× 10 N before :=

W 18 1800⋅ 365

N before

= 12.288

3

1800⋅ 1.3 = 2.34 × 10

Nafter :=

W 18 2340⋅ 365

Δyears := N before − Nafter

Nafter = 9.452

Δyears = 2.836


Problem 4.11 How many years would yo u be 95% sure the pavement will last? ZR := −1.645

(Table 4.5)

MR := 5000

ΔPSI := 1.9

So := 0.45

(given)

a1 := 0.35

a2 := 0.20

a3 := 0.11

D1 := 6

D2 := 9

D3 := 10

(given)

M2 := 1.0

M3 := 1.0

(given)

(Table 4.6)

SN := a1⋅ D1 + a2⋅ D2⋅ M2 + a3⋅ D3⋅ M3

(Eq. 4.9)

SN = 5


(Table 4.2, 4.3, and 4.4)

single2kip := 0.000420000 ⋅ single10kip := 0.088200 ⋅ tandem22kip := 0.18⋅ 200 single12kip := 0.189410 ⋅ tandem18kip := 0.077410 ⋅ triple50kip := 1.22⋅ 410 TotAxleEqv := single2kip + single10kip + tandem22kip + single12kip + tandem18kip + triple50kip TotAxleEqv = 670.86

Using Eq. 4.7:

⎡ log ( W 18) ⎢ZR ⋅ So + 9.36(log( SN + 1) ) − 0.20 + ⎢ ⎣

⎤ + 2.32⋅ log( MR )⎥ − 8.07 1094 ⎡ ⎤ ⎥ 0.40 + 5.19 ⎣ (SN + 1) ⎦ ⎦ log

⎛ ΔPSI ⎞ ⎝ 2.7 ⎠

6

W 18 = 3.545637× 10 N :=

W 18 TotAxleEqv⋅ 365

N = 14.48

years


Problem 4.12 Determine the Struc tural Number. N := 15

(given)

Assume: SN := 4.0

Axle Loads: single10kip := 0.1025000 ⋅ single24kip := 2.89⋅ 400 tandem30kip := 0.6951000 ⋅ tandem50kip := 4.64⋅ 100 TotAxleEqv := single10kip + single24kip + tandem30kip + tandem50kip 3

TotAxleEqv = 2.825× 10

7

DirW18 := TotAxleEqv⋅ 365⋅ N

DirW18 = 1.5466875× 10

PDL := 0.8

(Table 4.11)

DesignLaneW 18 := 0.8⋅ DirW18 7

DesignLaneW 18 = 1.23735× 10 ZR := −1.282

(Table 4.5)

So := 0.4

ΔPSI := 1.8 MR := 13750

Using Eq. 4.7:

⎡ log( DesignLaneW 18) ⎢ZR ⋅ So + 9.36(log( SN + 1) ) − 0.20 + ⎢ ⎣ SN = 4

⎤ + 2.32⋅ log( MR )⎥ − 8.07 1094 ⎡ ⎤ ⎥ 0.40 + 5.19 ⎣ ( SN + 1) ⎦ ⎦ log

⎛ ΔPSI ⎞ ⎝ 2.7 ⎠

Therefore the assumed value of 4 is good.


Problem 4.13 Determine the probablilit y (reliability) that this pavement will last 20 years before reaching it s terminal s erviceability . a1 := 0.44

a2 := 0.44

a3 := 0.11

D1 := 5

D2 := 6

D3 := 10

(given)

M2 := 1.0

M3 := 1.0

(given)

(Table 4.6)

SN := a1⋅ D1 + a2⋅ D2⋅ M2 + a3⋅ D3⋅ M 3

(Eq. 4.9)

SN = 5.94 N := 20

years


(Tables 4.2, 4.3, 4.4)

single20kip := 1.538200 ⋅ tandem40kip := 2.122200 ⋅ single22kip := 2.26480 ⋅ TotAxleEqv := single20kip + tandem40kip + single22kip TotAxleEqv = 913.12 W 18 := TotAxleEqv⋅ N⋅ 365 So := 0.6

M R := 3000

6

W 18 = 6.665776× 10

ΔPSI := 2.0

(given)

Using Eq. 4.7

⎡ log ( W 18) ⎢ZR ⋅ So + 9.36(log( SN + 1) ) − 0.20 + ⎢ ⎣

⎤ + 2.32⋅ log( MR )⎥ − 8.07 1094 ⎡ ⎤ ⎥ 0.40 + 5.19 ⎣ (SN + 1) ⎦ ⎦ log

⎛ ΔPSI ⎞ ⎝ 2.7 ⎠

ZR = −0.92773 R := 82.3

%

(interpolating from Table 4.5)


Problem 4.14 Determine the probabili ty that t he pavement wil l have a PSI greater th an 2.5 after 20 years. PSI := 4.5

(given)

TSI := 2.5


ΔPSI = 2

MR := 12000 So := 0.40 a1 := 0.35 D1 := 4

a2 := 0.18

a3 := 0.11

D2 := 6

D3 := 8

(given)

M2 := 1.0

M3 := 1.0

(given)

(Table 4.6)

SN := a1⋅ D1 + a2⋅ D2⋅ M2 + a3⋅ D3⋅ M 3

(Eq. 4.9)

SN = 3.36 N := 20

years

W 18 := 1290⋅ N⋅ 365

6

W 18 = 9.417 × 10

Using Eq. 4.7:

⎡ log ( W 18) ⎢ZR ⋅ So + 9.36(log( SN + 1) ) − 0.20 + ⎢ ⎣

⎤ + 2.32⋅ log( MR )⎥ − 8.07 1094 ⎡ ⎤ ⎥ 0.40 + 5.19 ⎣ (SN + 1) ⎦ ⎦ log

⎛ ΔPSI ⎞ ⎝ 2.7 ⎠

ZR = −0.1612 R := 56.40

%

(interpolating from Table 4.5)


Problem 4.15 Determine probabilit y of pavement lastin g 25 years. a1 := 0.35

a2 := 0.20

a3 := 0.11

D1 := 4

D2 := 10

D3 := 10

M2 := 1.0

M3 := 1.0

SN := a1⋅ D1 + a2⋅ D2 M 2 + a3⋅ D3⋅ M 3 N := 25

(Table 4.6) (given) (given)

SN = 4.5

(Eq. 4.9)

years

Axle load equivalency factors from Tables 4.2 and 4.3, interpolating for SN = 4.5 single20kip := 1.49⋅ 400 tandem35kip := 1.24⋅ 900 TotAxleEqv := single20kip + tandem35kip 3


W 18 := TotAxleEqv⋅ 365⋅ 25

So := 0.45

CBR := 8

M R := 1500⋅ CBR

PSI := 4.2

TSI := 2.5


4

MR = 1.2 × 10

(given)

Using Eq. 4.7: log

(

)

log W 18

ZR ⋅ So

+

9.36(log( SN + 1) )

− 0.20 + 0.40 +

⎛ ΔPSI ⎞ ⎝ 2.7 ⎠

⎡ ⎤ 5.19 ⎣ ( SN + 1) ⎦ 1094

+

(

) − 8.07

2.32⋅ log MR

ZR = −1.265

Probability is approximately 89.7% (interpolating from Table 4.5), that PSI will equal 2.5 after 25 years; thus, probabilities less than this will correspond to PSIs above 2.5 after 25 years


Problem 4.16

Determine the struc tural numb er to be used s o the PCC and flexible pavements have the same life expectancy. D := 10

So := 0.35

R = 90%

PSI := 4.6

Sc := 700

TSI := 2.5 6

Ec := 4.5⋅ 10

CBR := 2

J := 3.0

Cd := 1.0


(given)

ΔPSI = 2.1

k := 100

(Table 4.10)

ZR := −1.282

(Table 4.5)

Rigid Pavement Design Using Eq. 4.19:

⎛ ΔPSI ⎞ ⎝ 3.0 ⎠

0.75 ⎡ ⎤ Sc Cd ⋅ ( D − 1.132) x := ZR ⋅ So + 7.35⋅ ( log ( D + 1) ) − 0.06 + + ( 4.22 − 0.32⋅T SI) ⋅ log ⎢ 7 ⎤ 0.75 ⎡ 18.42 ⎤ ⎤ ⎥ ⎡ 1.62410 ⎡ ⋅ 215.63 J D ⋅ − ⎢ ⎥ 1+ 0.25 ⎢ ⎢ ⎢ E ⎥ ⎥ ⎥ 8.46 ⎛ c ⎞ ⎣ ( D + 1) ⎦ ⎣ ⎣ ⎣ ⎝ k ⎠ ⎦ ⎦ ⎦

log

x = 7.156 x

W 18 := 10

7

W 18 = 1.433 × 10

PDL := 0.80

For 3 lanes and a conservative design, 6

(

RemainingW18 := W 18 − CurrentW 18

)

7

CurrentW 18 := 13⋅ 10 ⋅ PDL = 1.04 × 10

6

RemainingW18 = 3.928 × 10

Rigid Pavement Axle load equivalency factors from Tables 4.7 and 4.8, for D = 10 single12kip := 0.175 tandem24kip := 0.441


TotAxleEqvrigid := single12kip + tandem24kip RemainingTrucks :=

RemainingW18

TotAxleEqvrigid = 0.616 6

RemainingTrucks = 6.376× 10

TotAxleEqvrigid

Flexible Pavement Design Flexible Pavement Axle load equivalency factors from Tables 4.2 and 4.3 Assume SN = 5 single12kip := 0.189 tandem24kip := 0.260 TotAxleEqvflex := single12kip + tandem24kip RemainingW18 := RemainingTrucks⋅ TotAxleEqvflex

TotAxleEqvflex = 0.449 6

RemainingW18 = 2.863 × 10

3

MR := 1500⋅ CBR

MR = 3 × 10

W 18 := RemainingW18

Using Eq. 4.7:

log

(

)

log W 18

ZR ⋅ So

+

9.36(log( SN + 1) )

− 0.20 +

⎛ ΔPSI ⎞ ⎝ 2.7 ⎠

1094 ⎤ 0.40 + ⎡ 5.19 ⎣ (SN + 1) ⎦

+

(

) − 8.07

2.32⋅ log MR

SN = 5.07


Problem 4.17 Determine pavement slab thic kness. P := 10000 a :=

p := 90

P

= 5.947

a

p ⋅ π

E := 4200000

(given) (Eq. 4.3)

in

μ := 0.25

k := 150

(given)

σe := 218.5 3 ⎞ P ⎞ ⎛ ⎛ E⋅ h ⎞ ⎛ 0.529⋅ ( 1 + 0.54μ) ⋅ ⋅ log − 0.71 2 4 ⎝ h ⎠ ⎝ ⎝ k⋅ a ⎠ ⎠

σe

h

= 10

inches

Problem 4.18

Determine the interior st ress. E := 3500000

μ := 0.30

Δ i := 0.008195

⎡ l

E⋅ h

l := 30.106

⎤

3

h := 8

P := 12000

γ := 0.577215

(given)

Euler's constant

0.25

(Eq. 4.13)

⎣ 12⋅ (1 − μ )⋅ k ⎦ 2

lb

k = 199.758

3

in

Δi

a

2 1 ⎞ ⎛ ⎛ a ⎞ 5 ⎞ ⎛ a ⎞ ⎤ ⎛ ⋅ 1+ ⋅ ln +γ− ⋅ 4 ⎠ ⎝ l ⎠ ⎦ 2 ⎣ 2⋅ π ⎠ ⎝ ⎝ 2⋅ l ⎠ ⎝ 8⋅ k ⋅ l

P

= 4.32

σi :=

⎡

(Eq. 4.12)

inches

3⋅ P⋅ ( 1 + 2⋅ π⋅ h

σi = 297.87

2

μ)

2 2⋅ l ⎞ 3⋅ P⋅ ( 1 + μ ) ⎛ a ⎞ ⎛ ⎛ ⎞ ⋅ ln + 0.5 − γ + ⋅ 2 ⎝ ⎝ a ⎠ ⎠ ⎝ l ⎠ 64⋅ h

(Eq. 4.11)

lb 2

in


Problem 4.19 Determine the modulus of elasticity o f the pavement. h := 10

P := 17000

μ := 0.36

al := 7

k := 250

Δ c := 0.05

⎡ ⎛ al ⎞ ⎤ 1.205 − 0.69⋅ 2 ⎣ ⎝ l ⎠ ⎦ k⋅ l P

Δc

⋅

(given)

(Eq. 4.18)

l = 38.306

⎡ l

E⋅ h

(

⎣ 12⋅ 1 −

⎤

3

0.25

(Eq. 4.13)

)

2

μ ⋅ k ⎦

6

E = 5.6219543× 10

lb 2

in


Problem 4.20 Determine the interior and edge stresses, as well as th e interior and edge slab deflections. h := 12

k := 300

E := 4000000

P := 9000

μ := 0.40

a := 5

l :=

⎡

E⋅ h

⎤

3

l = 38.883

⎣ 12⋅ (1 − μ )⋅ k ⎦

σi :=

2⋅ π⋅ h

μ)

2

(given)

0.25

2

3⋅ P⋅ ( 1 +

γ := 0.577215 Euler's constant

2 2⋅ l ⎞ 3⋅ P⋅ ( 1 + μ ) ⎛ a ⎞ ⎛ ⎛ ⎞ ⋅ ln + 0.5 − γ + ⋅ 2 ⎝ ⎝ a ⎠ ⎠ ⎝ l ⎠ 64⋅ h

(Eq. 4.13)

(Eq. 4.11)

lb

σi = 111.492

2

in

2 1 ⎞ ⎛ ⎛ a ⎞ 5 ⎞ ⎛ a ⎞ ⎤ ⎛ Δ i := ⋅ 1+ ⋅ ln +γ− ⋅ 2 ⎣ 4 ⎠ ⎝ l ⎠ ⎦ 2⋅ π ⎠ ⎝ ⎝ 2⋅ l ⎠ ⎝ 8⋅ k ⋅ l

P

⎡

Δ i = 2.45809×

−3

10

(Eq. 4.12)

inches

3 ⎞ P ⎞ ⎛ ⎛ E⋅ h ⎞ ⎛ σe := 0.529⋅ ( 1 + 0.54⋅ μ ) ⋅ ⋅ log − 0.71 2 4 ⎝ h ⎠ ⎝ ⎝ k⋅ a ⎠ ⎠

(Eq. 4.15)

lb

σe = 155.051

2

in

Δ e := 0.408⋅ ( 1 +

Δ e = 9.391 ×

0.4μ) ⋅

−3

10

⎛ P ⎞ 2 ⎝ k⋅ l ⎠

(Eq. 4.16)

inches


Problem 4.21 Considering Ex. 4.5, determine which tr uck wi ll cause mor e pavement damage? Axle load equivalency factors for truck A: single12kip := 0.175 single23kip := 2.915 TotAxleEqvA := single12kip + single23kip TotAxleEqvA

= 3.09

18 kip ESAL

Axle load equivalency factors for truck B: single8kip := 0.032 tandem23kip := 5.245 TotAxleEqvB := single8kip + tandem23kip TotAxleEqvB = 5.277

18 kip ESAL

Therefore, Truck B causes more damage.


Problem 4.22

Determine Estimated Daily Truck Traffic. D := 11

N := 20

ZR := −1.282

Sc := 600

So := 0.35

Cd := 0.8

PSI := 4.8

J := 2.8

TSI := 2.5

(given)

6

Ec := 4⋅ 10


k := 150

ΔPSI = 2.3 Using Eq. 4.19:

⎛ ΔPSI ⎝ 3.0 ⎠

0.75 ⎡ ⎤ − 1.132) Sc Cd ⋅ ( D + ( 4.22 − 0.32⋅ TSI) ⋅ log⎢ log( W 18) ZR ⋅ So + 7.35⋅ ( log( D + 1) ) − 0.06 + 7 ⎤ ⎡ 1.62410 ⎡ 0.75 − ⎡ 18.42 ⎤ ⎤ ⎥⎥ ⋅ 215.63J⋅ D ⎢ 1+ 8.46 ⎢ ⎢ ⎢ ⎛ Ec ⎞ 0.25 ⎥ ⎥ ⎥ ⎣ (D + 1) ⎦ ⎣ ⎣ ⎣ ⎝ k ⎦⎦ ⎦

log

7

W 18 = 1.162 × 10

Axle load equivalency factors from Tables 4.7, 4.8, and 4.9, for D = 11 single20kip := 1.58 tandem26kip := 0.619 triple34kip := 0.593 TotAxleEqv := single20kip + tandem26kip + triple34kip TotAxleEqv = 2.792

DailyDesignLaneTraffic :=

W 18 365⋅ N⋅ TotAxleEqv

DailyDesignLaneTraffic = 570.2

trucks/day (for design lane)

For 3 lanes and a conservative design, TotalTraffic :=

DailyDesignLaneTraffic PDL

PDL := 0.80 TotalTraffic = 712.8

trucks/day (total for 3 lanes)


Problem 4.23 Determine how long p avement will last with new loading and the additional lane. D := 11

Cd1 := 1.0 6

Ec := 5⋅ 10

So := 0.3

Sc := 700

J := 3.0

PSI := 4.7

(given)

TSI := 2.5 N1 := 20 years

CBR := 25 k := 290

(Table 4.10)

ZR := −1.645

(Table 4.5)


ΔPSI = 2.2

Before Using Eq. 4.19:

⎡

0.75 ⎡ ⎤⎤ Sc Cd1⋅ ( D − 1.132) x1 := ⎢ ZR ⋅ So + 7.35⋅ ( log( D + 1) ) − 0.06 + + ( 4.22 − 0.32⋅T SI) ⋅ log⎢ ⎥ 7 ⎤ 0.75 ⎡ 18.42 ⎤ ⎤ ⎥ ⎡ ⎡ ⎢ 1.624⋅ 10 215.63 J D ⋅ − ⎢ ⎥⎥ 1+ 0.25 ⎢ ⎢ ⎢ ⎢ ⎛ Ec ⎞ ⎥ ⎥ ⎥ ⎥ 8.46 ( D 1 ) + ⎣ ⎦ ⎢ ⎥ ⎣ ⎣ ⎣ ⎣ ⎝ k ⎦⎦ ⎦⎦

log

⎛ ΔPSI ⎝ 3.0 ⎠

x1 = 7.514 x1

W 18before := 10

7

W 18before = 3.265 × 10

Axle load equivalency factors from Tables 4.7 and 4.8, for D = 11 single18kip := 1.0 tandem28kip := 0.850 TotAxleEqv1 := single18kip + tandem28kip DailyDesignLaneTraffic :=

TotAxleEqv1 = 1.85

W 18before 365⋅ N1⋅ TotAxleEqv1 3

DailyDesignLaneTraffic = 2.418× 10

trucks/day


For 2 lanes designed conservatively: PDL1 := 1.00

(Table 4.11)

TotalDailyTraffic :=

DailyDesignLaneTraffic

3

TotalDailyTraffic = 2.418× 10

PDL1

trucks/day

After Cd2 := 0.8

(given)

Using Eq. 4.19:

⎡

0.75 ⎡ ⎤⎤ Sc Cd2⋅ ( D − 1.132) x2 := ⎢ ZR ⋅ So + 7.35⋅ ( log( D + 1) ) − 0.06 + + ( 4.22 − 0.32⋅T SI) ⋅ log⎢ ⎥ 7 ⎤ 0.75 ⎡ 18.42 ⎤ ⎤ ⎥ ⎡ ⎡ ⎢ 1.624⋅ 10 215.63 J ⋅ D − ⎢ ⎥⎥ 1+ 0.25 ⎢ ⎢ ⎢ ⎢ ⎛ Ec ⎞ ⎥ ⎥ ⎥ ⎥ 8.46 ( D + 1 ) ⎣ ⎦ ⎢ ⎥ ⎣ ⎣ ⎣ ⎣ ⎝ k ⎦⎦ ⎦⎦

log

⎛ ΔPSI ⎝ 3.0 ⎠

x2 = 7.182 x2

7

W 18after := 10

W 18after = 1.522 × 10

Axle load equivalency factors from Tables 4.7 and 4.8 for D = 11 single20kip := 1.58 tandem34kip := 1.96 TotAxleEqv2 := single20kip + tandem34kip 3

TotalDailyTraffic = 2.418× 10

With addition of third lane,

TotAxleEqv2 = 3.54

trucks/day

(from Before condition)

PDL2 := 0.80

DailyDesignLaneTraffic2 := TotalDailyTraffic⋅ PDL2 3

DailyDesignLaneTraffic2 = 1.934 × 10 N2 :=

W 18after 365⋅ TotAxleEqv2 DailyDesignLaneTraffic2

N2 = 6.1

years


Problem 4.24 Determine the required slab th ickness i f a 20-year design life is used. N := 20

(given)

Assume: D := 11

Axle loads:

(Tables 4.7, 4.8 and 4.9)

single22kip := 2.67580 ⋅ tandem25kip := 0.5295570 ⋅ tandem39kip := 3.55⋅ 50 triple48kip := 2.58⋅ 80 TotAxleEqv := single22kip + tandem25kip + tandem39kip + triple48kip TotAxleEqv = 899.715 ZR := −1.645

(Table 4.5)

W 18 := TotAxleEqv⋅ N⋅ 365 6

W 18 = 6.5679195× 10 Sc := 600

Cd := 0.9

ΔPSI := 1.7

Ec := 5⋅ 10

J := 3.2

TSI := 2.5

So := 0.4

k := 200

6

(given)

Using Eq. 4.19:

⎛ ΔPSI ⎝ 3.0 ⎠

0.75 ⎡ ⎤ − 1.132) Sc Cd ⋅ ( D + ( 4.22 − 0.32⋅ TSI) ⋅ log⎢ log( W 18) ZR ⋅ So + 7.35⋅ ( log( D + 1) ) − 0.06 + 7 ⎤ ⎡ 1.62410 ⎡ 0.75 − ⎡ 18.42 ⎤ ⎤ ⎥⎥ ⋅ 215.63J⋅ D ⎢ 1+ 8.46 ⎢ ⎢ ⎢ ⎛ Ec ⎞ 0.25 ⎥ ⎥ ⎥ + ( D 1 ) ⎣ ⎦ ⎣ ⎣ ⎣ ⎝ k ⎦⎦ ⎦

log

D = 11.43

inches

So assumption was close.


Problem 4.25 Determine the design modulus o f ruptu re. k := 300

Cd := 1.0

ΔPSI := 2.0

D := 8.5

Ec := 4⋅ 10

TSI := 2.5

J := 3.0

So := 0.5

N := 12

6

ZR := −0.524

(given)

(Table 4.5)

Axle loads:

(Tables 4.7, 4.8 and 4.9)

single8kip := 0.03251300 ⋅ tandem15kip := 0.06575900 ⋅ single40kip := 26⋅ 20 tandem40kip := 3.645200 ⋅ TotAxleEqv := single8kip + tandem15kip + single40kip + tandem40kip 3


W 18 := TotAxleEqv⋅ N⋅ 365 6

W 18 = 5.9148615× 10

Using Eq. 4.19

⎛ ΔPSI ⎝ 3.0 ⎠


log

Sc

= 575.343

lb 2

in


Problem 4.26 Determine the assumed soil resilient modulu s. D := 10

PSI := 4.7

So := 0.35

Ec := 6⋅ 10

TSI := 2.5

k := 190

Sc := 432


Cd := 0.8

J := 3.0

ΔPSI = 2.2

6

ZR := −1.282

(given)

(Table 4.5)

Rigid Pavement: Axle loads from Tables 4.7, 4.8 and 4.9: single20kip1 := 1.58⋅ 100 tandem42kip1 := 4.74⋅ 100 TotAxleEqvrig := single20kip1 + tandem42kip1 TotAxleEqvrig = 632

Flexible Pavement: SN := 4

Axle loads from Tables 4.2, 4.3 and 4.4: single20kip2 := 1.47⋅ 100 tandem42kip2 := 2.43⋅ 100 TotAxleEqvflex := single20kip2 + tandem42kip2 TotAxleEqvflex = 390

Using Eq. 4.19:

⎛ ΔPSI ⎝ 3.0 ⎠

0.75 ⎡ ⎤ − 1.132) Sc Cd ⋅ ( D + ( 4.22 − 0.32⋅ TSI) ⋅ log⎢ log( W 18) ZR ⋅ So + 7.35⋅ ( log( D + 1) ) − 0.06 + 7 ⎤ ⎡ 1.62410 ⎡ 0.75 − ⎡ 18.42 ⎤ ⎤ ⎥⎥ ⋅ 215.63J⋅ D ⎢ 1+ 8.46 ⎢ ⎢ ⎢ ⎛ Ec ⎞ 0.25 ⎥ ⎥ ⎥ ⎣ (D + 1) ⎦ ⎣ ⎣ ⎣ ⎝ k ⎦⎦ ⎦

log

6

W 18 = 1.4723512× 10


N :=

W 18 TotAxleEqvrig⋅ 365

N = 6.383

years

Now applying Eq. 4.7 with: W 18 := N⋅ TotAxleEqvflex⋅ 365

⎡ log ( W 18) ⎢ZR ⋅ So + 9.36(log( SN + 1) ) − 0.20 + ⎢ ⎣ 3

MR = 3.67 × 10

5

W 18 = 9.08571× 10

⎤ + 2.32⋅ log( MR )⎥ − 8.07 1094 ⎡ ⎤ ⎥ 0.40 + 5.19 ⎣ (SN + 1) ⎦ ⎦ log

⎛ ΔPSI ⎞ ⎝ 2.7 ⎠

lb 2

in


What slab thickness sho uld have been used? D := 8

Sc := 700

R = 90%

Cd := 1.0

So := 0.3

J := 3.0

PSI := 4.6 TSI := 2.5 CBR := 25


6

Ec := 5⋅ 10 Nyears


:= 20

ΔPSI = 2.1

ZR := −1.282

(Table 4.5)

k := 290

(Table 4.10)

PDL := 0.75

(Table 4.11, for # Lanes = 4)

Using Eq. 4.19:

⎛ ΔPSI ⎝ 3.0 ⎠


log

6

W 18 = 5.7402× 10

Ignored Traffic in Design Axle load equivalency factors from Tables 4.7, and 4.8. D = 11 is assumed as a result of an iterative process. single22kip := 2.40 tandem30kip := 1.14 TotAxleEqv := single22kip + tandem30kip TotAxleEqv := 3.54


AddW 18 := 1000⋅ TotAxleEqv⋅ PDL⋅ 365⋅ Nyears 7

AddW 18 = 1.938 × 10

NewW18 := W 18 + AddW 18

W18 not originally included in design

NewW18

= 2.512 ×

7

10

Using Eq. 4.19: log⎛

ΔPSI ⎞

⎡ S C ⋅ (D0.75 − 1.132) ⎤ c d log( NewW18) ZR ⋅ So + 7.35⋅ ( log( D + 1) ) − 0.06 + + ( 4.22 − 0.32⋅ T SI) ⋅ log⎢ 7 ⎤ 0.75 ⎡ 18.42 ⎤⎤ ⎥ ⎡ 1.62410 ⋅ 215.63J⋅ ⎡D − ⎢ ⎥ 1+ 0.25 8.46 ⎢ ⎢ ⎢ ⎛ Ec ⎞ ⎥⎥ ⎥ ⎣ ( D + 1) ⎦ ⎣ ⎣ ⎣ ⎝ k ⎠ ⎦⎦ ⎦ ⎝

D = 10.216

3.0

⎠

inches


Problem 4.28 Determine the number o f years the pavement will l ast.

6

k := 200

Ec := 5⋅ 10

PSI := 4.7

So := 0.30

D := 8

Sc := 600

TSI := 2.5

ZR := −1.036

J := 3.2

Cd := 1.0

(given) (Table 4.5)

ΔPSI := PSI − TSI ΔPSI = 2.2

Axle loads:

(Tables 4.7, 4.8 and 4.9)

single10kip := 0.084300 ⋅ single18kip := 1.00⋅ 200 single23kip := 2.75⋅ 100 tandem32kip := 1.47⋅ 100 single32kip := 10.1⋅ 30 tandem40kip := 1.16⋅ 100 TotAxleEqv := single10kip + single18kip + single23kip + tandem32kip + single32kip + tandem40kip 3


Using Eq. 4.19:

⎛ ΔPSI ⎝ 3.0 ⎠

0.75 ⎡ ⎤ − 1.132) Sc Cd ⋅ ( D + ( 4.22 − 0.32⋅ TSI) ⋅ log⎢ log( W 18) ZR ⋅ So + 7.35⋅ ( log( D + 1) ) − 0.06 + 7 ⎤ ⎡ 1.62410 ⎡ 0.75 − ⎡ 18.42 ⎤ ⎤ ⎥⎥ ⋅ 215.6 3J⋅ D ⎢ 1+ 8.46 ⎢ ⎢ ⎢ ⎛ Ec ⎞ 0.25 ⎥ ⎥ ⎥ + ( D 1 ) ⎣ ⎦ ⎣ ⎣ ⎣ ⎝ k ⎦⎦ ⎦

log

6

W 18 = 2.8833862× 10 N :=


N = 7.409

years


Problem 4.29 Determine the number o f years the pavement will l ast with a 95% reliability. 6

k := 200

Ec := 5⋅ 10

PSI := 4.7

So := 0.30

D := 8

Sc := 600

TSI := 2.5

ZR := −1.645

J := 3.2

Cd := 1.0

(given) (Table 4.5)

ΔPSI := PSI − TSI ΔPSI = 2.2

Axle loads:

(Tables 4.7, 4.8 and 4.9)

single10kip := 0.084300 ⋅ single18kip := 1.00⋅ 200 single23kip := 2.75⋅ 100 tandem32kip := 1.47⋅ 100 single32kip := 10.1⋅ 30 tandem40kip := 1.16⋅ 100 TotAxleEqv := single10kip + single18kip + single23kip + tandem32kip + single32kip + tandem40kip 3


Using Eq. 4.19:

⎛ ΔPSI ⎝ 3.0 ⎠

0.75 ⎡ ⎤ − 1.132) Sc Cd ⋅ ( D + ( 4.22 − 0.32⋅ TSI) ⋅ log⎢ log( W 18) ZR ⋅ So + 7.35⋅ ( log( D + 1) ) − 0.06 + 7 ⎤ ⎡ 1.62410 ⎡ 0.75 − ⎡ 18.42 ⎤ ⎤ ⎥⎥ ⋅ 215.6 3J⋅ D ⎢ 1+ 8.46 ⎢ ⎢ ⎢ ⎛ Ec ⎞ 0.25 ⎥ ⎥ ⎥ ⎣ (D + 1) ⎦ ⎣ ⎣ ⎣ ⎝ k ⎦⎦ ⎦

log

6

W 18 = 1.893227× 10 N :=


N = 4.865

years


Problem 4.30

Determine the design life. D := 10

MR := 5000

Ec := 4500000

Cd1 := 0.8

ΔPSI := 1.9

Sc := 900

Cd2 := 0.6

So := 0.45

J := 3.2

k := 300

(given)

TSI := 2.5

ZR := −1.645

(Table 4.5)


(Tables 4.7, 4.8, and 4.9)

single2kip := 0.000220000 ⋅ = 4.0

single12kip := 0.175410 ⋅ = 71.75

single10kip := 0.081200 ⋅ = 16.2

tandem18kip := 0.132410 ⋅ = 54.12

tandem22kip := 0.305200 ⋅ = 61.0

triple50kip := 3.02410 ⋅ = 1238.2

TotAxleEqv := single2kip + single10kip + tandem22kip + single12kip + tandem18kip + triple50kip TotAxleEqv = 1445.27

Using Eq. 4.19 with Cd=0.8:

⎛ ΔPSI ⎝ 3.0 ⎠

0.75 ⎡ ⎤ − 1.132) Sc Cd1⋅ ( D + ( 4.22 − 0.32⋅ TSI) ⋅ log⎢ log( W 18) ZR ⋅ So + 7.35⋅ ( log( D + 1) ) − 0.06 + 7 ⎤ ⎡ 1.62410 ⎡ 0.75 − ⎡ 18.42 ⎤ ⎤ ⎥⎥ ⋅ 215.6 3J⋅ D ⎢ 1+ 8.46 ⎢ ⎢ ⎢ ⎛ Ec ⎞ 0.25 ⎥ ⎥ ⎥ + ( D 1 ) ⎣ ⎦ ⎣ ⎣ ⎣ ⎝ k ⎦⎦ ⎦

log

6

W 18 = 8.1485079× 10 N :=

W 18

N

TotAxleEqv⋅ 365

= 15.45

years

Using Eq. 4.19 with Cd=0.6:

⎛ ΔPSI ⎝ 3.0 ⎠

0.75 ⎡ ⎤ − 1.132) Sc Cd2⋅ ( D log( W 18) ZR ⋅ So + 7.35⋅ ( log( D + 1) ) − 0.06 + + ( 4.22 − 0.32⋅ TSI) ⋅ log⎢ 7 ⎤ ⎡ 1.62410 ⎡ 0.75 − ⎡ 18.42 ⎤ ⎤ ⎥⎥ ⋅ 215.6 3J⋅ D ⎢ 1+ 8.46 ⎢ ⎢ ⎢ ⎛ Ec ⎞ 0.25 ⎥ ⎥ ⎥ ⎣ (D + 1) ⎦ ⎣ ⎣ ⎣ ⎝ k ⎦⎦ ⎦

log

6

W 18 = 3.0464048× 10

N :=


N

= 5.77

years


Problem 4.31 Determine how the design life of the soi l changes.

k := 150

Sc := 800

So := 0.45

D := 12

J := 3.0

Cd := 1.0

6

Ec := 6⋅ 10

ΔPSI := 2.0

TSI := 2.5

(given)

N1 := 20

ZR := −1.645

(Table 4.5)

DailyTruckTraffic := 1227.76

trucks day

Axle loads:

(Tables 4.7, 4.8 and 4.9)

single16kip := 0.599⋅ DailyTruckTraffic single20kip := 1.590⋅ DailyTruckTraffic tandem35kip := 2.245⋅ DailyTruckTraffic TotAxleEqv := single16kip + single20kip + tandem35kip 3


Using Eq. 4.19:

⎛ ΔPSI ⎝ 3.0 ⎠

0.75 ⎡ ⎤ − 1.132) Sc Cd ⋅ ( D + ( 4.22 − 0.32⋅ TSI) ⋅ log⎢ log( W 18) ZR ⋅ So + 7.35⋅ ( log( D + 1) ) − 0.06 + 7 ⎤ 0.75 ⎡ 18.42 ⎤ ⎤ ⎥ ⎡ 1.62410 ⋅ 215.6 3J⋅ ⎡ D − ⎢ ⎥ 1+ 0.25 E ⎢ ⎢ ⎢ ⎥ ⎥ ⎥ 8.46 ⎛ c ⎞ ⎣ (D + 1) ⎦ ⎣ ⎣ ⎣ ⎝ k ⎦⎦ ⎦

log

7

W 18 = 3.7684418× 10 W 18

N2 := TotAxleEqv⋅ 365

N2 = 18.965

Δ N := N1 − N2 Δ N = 1.03

year reduction


Determine how the design life of the soi l changes. k := 190

Sc := 600

So := 0.45

D := 12

J := 3.0

Cd := 1.0

6

Ec := 6⋅ 10

ΔPSI := 2.0

TSI := 2.5


N1 := 20

ZR := −1.645

(Table 4.5)

DailyTruckTraffic := 1227.76

trucks day

Axle loads:

(Tables 4.7, 4.8 and 4.9)

single16kip := 0.599⋅ DailyTruckTraffic single20kip := 1.590⋅ DailyTruckTraffic tandem35kip := 2.245⋅ DailyTruckTraffic TotAxleEqv := single16kip + single20kip + tandem35kip 3


Using Eq. 4.19:

⎛ ΔPSI ⎝ 3.0 ⎠

0.75 ⎡ ⎤ − 1.132) Sc Cd ⋅ ( D + ( 4.22 − 0.32⋅ TSI) ⋅ log⎢ log( W 18) ZR ⋅ So + 7.35⋅ ( log( D + 1) ) − 0.06 + 7 ⎤ 0.75 ⎡ 18.42 ⎤ ⎤ ⎥ ⎡ 1.62410 ⋅ 215.6 3J⋅ ⎡ D − ⎢ ⎥ 1+ 0.25 E 8.46 ⎢ ⎢ ⎢ ⎥ ⎥ ⎥ ⎛ c ⎞ ⎣ (D + 1) ⎦ ⎣ ⎣ ⎣ ⎝ k ⎦⎦ ⎦

log

7

W 18 = 1.4857318× 10 N2 :=


N2 = 7.477

years

Δ N := N1 − N2 Δ N = 12.52

year reduction


Problem 4.33

Determine Determine the deflection at a point at a depth of 20 inches. lb

p := 90

μ := 0.45

2

a :=

in z := 20

r := 20

in

10 2

E := 45000

lb 2

in

(given)

in

use Table 4.1 z a

= 4.0000

r

and

= 4.0000

a

and C := 0.00492 and

therefore A := 0.01109 and

F := 0.00209 and

B := 0.00595

⎤ ⎦

Δ z = 0.0021

H := 0.17640

deflection

Δ z :=

p ⋅ ( 1 +

μ)⋅a

⋅⎡ ⋅A + z

⎣a

E

( 1 − μ) ⋅ H

inches

(Eq 4.6)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Solve for radial-horizontal stress

σr := p ⋅ [ 2⋅ μ ⋅ A +

C + ( 1 − 2⋅ μ ) ⋅ F]

σr = 1.3599

inches

(Eq 4.5)

2) Solve for vertical stress

σz := p⋅ ( A +

σz = 1.5336

B)

inches

(Eq 4.4)

3) Careless with rounding A := 0.01

Δ z :=

H := 0.18

and

p ⋅ ( 1 + E

μ) ⋅ a

⋅⎡ ⋅A + z

⎣a

⎤ ⎦

( 1 − μ) ⋅ H

Δ z = 0.0020

inches


Determine Determine the d eflection. lb

p := 100

a :=

2

in z := 0 in

r := 0

Problem 4.34

12

μ := 0.50

in

2

E := 55000

lb

(given)

2

in

in

(assumed since on pavement surface)

Use Table 4.1 z a

r

=0

a

A := 1.0

=0

B := 0

C := 0

F := 0.5

H := 2.0

(Table 4.1)

deflection

Δ z :=

p ⋅ ( 1 +

μ)⋅a

E

⋅⎡ ⋅A + z

⎣a

⎤ ⎦

( 1 − μ) ⋅ H

Δ z = 0.016 in

(Eq 4.6)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers 1) Use stress equation

σr := p ⋅ [ 2⋅ μ ⋅ A +

C + ( 1 − 2μ ) ⋅ F]

σr = 100

(Eq. 4.5)

in

2) Use diameter instead of radius a := 12 in

Δ z :=

p ⋅ ( 1 +

μ) ⋅ a

⋅⎡ ⋅A + z

⎣a

E

( 1 − μ ) ⋅ H⎤

⎦

Δ z = 0.033

3) Use F instead of H in deflection equation

a := 6

Δ z :=

p ⋅ ( 1 + E

μ) ⋅ a

⋅⎡ ⋅A + z

⎣a

⎤ ⎦

( 1 − μ) ⋅ F

Δ z = 4.091 ×

−3

10


Problem 4.35

Deteremine Deteremine the structur al number of the pavement. pavement. a1 := 0.35

a2 := 0.18

a3 := 0.11

(Table 4.6)

D1 := 5 in

D2 := 9 in

D3 := 10 in

(given)

M2 := 0.90

M 3 := 1.0

(given)

SN := a1⋅ D1 + a2⋅ D2⋅ M2 + a3⋅ D3⋅ M3

(Eq. 4.9)

SN = 4.31

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers 1) Use M2 = 1.0 M2 := 1.0

SN := a1⋅ D1 + a2⋅ D2⋅ M2 + a3⋅ D3⋅ M3

SN = 4.47

2) Use Hot-mix asphaltic concrete Structural-Layer Coefficient M2 := 0.90

a1 := 0.44

SN := a1⋅ D1 + a2⋅ D2⋅ M2 + a3⋅ D3⋅ M3

SN = 4.76

3) Use crushed stone Structural-Layer Coefficient a1 := 0.35

a2 := 0.14

SN := a1⋅ D1 + a2⋅ D2⋅ M2 + a3⋅ D3⋅ M3

SN = 3.98


Problem 4.36

Calculate Calculate the edge stress (in lb/in 2 ). P := 25000 lb

h := 11.0 in

p := 120

lb

E := 4500000

2

in

μ := 0.20

k := 180

lb 2

in

(given)

lb 3

in

radius of relative stiffness

l :=

⎡

E⋅ h

⎤

3

0.25

l = 41.23

⎣ 12⋅ ( 1 − μ )⋅ k ⎦ 2

(Eq. 4.13)

radius of tire footprint

a :=

P p⋅ π

a

= 8.14

in

(Eq. 4.14)

stress 3 ⎞ P ⎞ ⎛ ⎛ E⋅ h ⎞ ⎛ σe := 0.529⋅ ( 1 + 0.54⋅ μ) ⋅ ⋅ log − 0.71 2 4 ⎝ h ⎠ ⎝ ⎝ k⋅ a ⎠ ⎠

(Eq. 4.15)

σe = 383.76 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answers 1) Use al for radius of tire footprint

3 ⎞ P ⎞ ⎛ ⎛ E⋅ h ⎞ ⎛ σe := 0.529⋅ ( 1 + 0.54⋅ μ) ⋅ ⋅ log − 0.71 2 4 ⎝ h ⎠ ⎝ ⎝ k⋅ al ⎠ ⎠

al := a⋅ 2

σe = 310.85 2) Solve for deformation instead of stress P Δ e := 0.408⋅ ( 1 + 0.4μ) ⋅ ⎛ ⎞ Δ e = 0.036 2 ⎝ k⋅ l ⎠

(Eq. 4.16)

3) Solve for corner loading stress

⎡

⎛ al ⎞ σc := ⋅ 1− 2 ⎝ l ⎠ h ⎣ 3⋅ P

⎤

0.72

⎦

(Eq. 4.17)

σc = 372.37


Problem 4.37

Determine Determine the corner deflection fr om the load.

P := 10000 lb

p := 110

lb

h := 9.5 in

2

μ := 0.18

in

(given) lb

E := 4100000

k := 175

2

in

lb 3

in

radius of relative stiffness

l :=

⎡

E⋅ h

⎤

3

0.25

⎣ 12⋅ ( 1 − μ )⋅ k ⎦ 2

l = 36.267

(Eq 4.13)

= 5.379

(Eq 4.14)

radius of tire footprint

a :=

P

a

p⋅ π

al := a⋅ 2

al = 7.608

deformation

Δ c :=

⎡ ⎛ al ⎞⎤ 1.205 − 0.69⋅ 2 ⎣ ⎝ l ⎠⎦ k⋅ l P

⋅

Δ c = 0.046 in

(Eq 4.18)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answers 1) Use a instead of al

Δ c :=

P

⋅ ⎡1.205 − 0.69⋅ ⎛

⎣ k⋅ l 2

a ⎞⎤

⎝ l ⎠⎦

Δ c = 0.048 in

2) Solve for stress instead of deformation

⎡

⎛ al ⎞ σc := ⋅ 1− 2 ⎣ ⎝ l ⎠ h 3⋅ P

⎤

0.72

⎦

σc = 224.435

in

(Eq 4.17)

3) Solve for total interior deflection 2 ⎡ ⎛ 1 ⎞ ⎛ ⎛ a ⎞ 5 ⎞ ⎛ a ⎞ ⎤ Δ i := ⋅ 1+ ⋅ ln +γ − ⋅ 2π ⎠ ⎝ ⎝ 2l ⎠ 4 ⎠ ⎝ l ⎠ ⎦ 2 ⎣ ⎝ 8⋅ k ⋅ l

P

Δ i = 0.005

in

(Eq 4.12)


Determine the corner deflection fro m the load. P := 12000 lb

lb

p := 100

h := 12 in

2

(given)

in

μ := 0.16

k := 250

lb 3

Problem 4.38

E := 4650000

in

lb 2

in

Determine al P

a :=

a

p⋅ π

al := a⋅ 2

= 6.1804

al = 8.7404

(Eq 4.14)

(For corner of slab)

Determine radius of relative stiffness

l :=

⎡

E⋅ h

⎤

3

0.25

⎣ 12⋅ ( 1 − μ )⋅ k ⎦ 2

l = 40.7178

(Eq 4.13)

Δ c = 0.0306 in

(Eq 4.18)

Calculate the deflection

Δ c :=

⎡ ⎛ al ⎞⎤ 1.205 − 0.69⋅ 2 ⎣ ⎝ l ⎠⎦ k⋅ l P

⋅

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answers: 1) Solve for corner stress

⎡

⎛ al ⎞ σc := ⋅ 1− 2 ⎣ ⎝ l ⎠ h 3⋅ P

⎤

0.72

lb

σc = 167.4348

⎦

(Eq 4.17)

2

in

2) Solve for edge loading deflection

Δ e := 0.408⋅ ( 1 +

0.4⋅ μ) ⋅

⎛ P ⎞ 2 ⎝ k⋅ l ⎠

Δ e = 0.0126 in

(Eq 4.16)

3) Use a in place of a l

Δ c :=

P

⋅ ⎡1.205 − 0.69⋅ ⎛

⎣ k⋅ l 2

a ⎞⎤

⎝ l ⎠⎦

Δ c = 0.0319 in




Chapter 5 Fundamentals of Traffic Flow and Queuing Theory U.S. Customary Units



•


•


•




www.mathcad.com


Determine space-mean speed at capacity and free-flow speed.

⎡

u

q

3.5 k ⎞ ⎤ ⎛ uf ⋅ 1 − ⎣ ⎝ k j ⎠ ⎦

k ⋅u

k j := 225

qcap := 3800

(given)

⎡

3.5 k ⎞ ⎤ ⎛ uf ⋅ k − k ⋅ ⎣ ⎝ k j ⎠ ⎦

at capacity,

d

q

Problem 5.1

(Eq. 5.14)

0

dk

⎡ 0

1

⎢ ⎣

k cap qcap

− 4.5 ⋅

⎛ k

cap

⎜ ⎝

k j

⎞⎤

3.5 3.5

⎟⎥ ⎠⎦

= 146.4 k cap ⋅ ucap

(Eq. 5.14)

qcap ucap := k cap

ucap

qcap

uf := k cap

− kcap ⋅

⎛ kcap ⎞ ⎝

k j

3.5

= 25.96

mi h

uf = 33.37

mi h

⎠


a

= −3.222

b := −2 ⋅ a ⋅ u b

= 193.333

now, when

u2

= 51.58

q := 1400

mi/h

at q = 0, the free flow speed is as follows q := 0

u

= 60

mi/h


Problem 5.3 Determine the capacity, speed at capacity, and densit y at o ne quarter capacity.

(

q ( k ) := 50 ⋅ k d

q (k)

− 0.156⋅ k

2

)

(given)

→ 50 − .312 ⋅ k

dk 50 − .312 ⋅ kcap

0

k cap := 160

(

)

qcap := q k cap

qcap

ucap := k cap

qcap

= 4006.4

ucap

= 25.04

veh h mi h

at 25% of capacity

q := q

qcap 4

= 1001.6

k1

= 21.5

k2

= 299

veh h

veh m veh m


Problem 5.4

Calculate the flo w, average speed, and d ensity of the traffic stream in this lane. hbar := 3 q :=

q

k

(Eq. 5.4)

hbar q ⋅ 3600 = 1200

veh/hr/ln

1 s bar

= 0.007

u :=

(given)

1

= 0.333

k :=

s bar := 150

q ⋅ 3600 k ⋅ 5280

(Eq. 5.13) k ⋅ 5280 = 35.2

u

= 34.09

veh/mi/ln

mi/hr

Also

q :=

k :=

u :=

3600 hbar 5280 sbar s bar hbar

q

= 1200

veh/hr/ln

k

= 35.2

veh/mi/ln

u

= 50

ft/s

u⋅

3600 5280

= 34.09

mi/hr


Problem 5.5

Determine the time-mean speed and space-mean speed. Lane 1 30 mi/h (44 ft/s)

Lane 2 45 mi/h 66 (ft/s)

Lane 3 60 mi/h (88 ft/s)

Spacing := 2640 ft 2640 Hdwy1 := 44

= 60

2640 Hdwy2 := 66

sec

2640 Hdwy3 := 88

= 40

= 30

For 30 minutes t := 30

⎛ 60 ⎞ Freq1 := t ⋅ ⎝ Hdwy1 ⎠

Freq1 = 30

⎛ 60 ⎞ Freq2 := t ⋅ ⎝ Hdwy2 ⎠

Freq2 = 45

⎛ 60 ⎞ Freq3 := t ⋅ ⎝ Hdwy3 ⎠

Freq3 = 60

Speed1 := 30 mi/h

ut_bar :=

Speed2 := 45

Speed1 ⋅ Freq1 + Speed2 ⋅ Freq2 Freq1

ut_bar = 48.33

vehicles in 30 minutes

+

Freq2

+

+

Speed3 := 60 Speed3 ⋅ Freq3

Freq3

mi/h

1

us_bar :=

1 Freq1 + Freq2

us_bar = 45.0

+

Freq3

⋅

⎡Freq ⋅ ⎛ 1 ⎞ + Freq ⋅ ⎛ 1 ⎞ + Freq ⋅ ⎛ 1 ⎞⎤ 1 Speed 2 Speed 3 Speed 1 ⎠ 2 ⎠ 3 ⎠⎦ ⎣ ⎝ ⎝ ⎝

mi/h


Problem 5.6 Determine the time-mean speed and space-mean speed.

u1 := 195⋅

5280 3600

5280 u3 := 185⋅ 3600

= 271.333

2.5 ⋅ 5280 u2 2.5 ⋅ 5280

secperlap3 :=

u3 2.5 ⋅ 5280

secperlap4 :=

laps3 :=

u3

u1

secperlap2 :=

laps1 :=

= 286

2.5 ⋅ 5280

secperlap1 :=

total_sec

u1

u4

:= 30⋅ 60

secperlap1 total_sec secperlap3

TMS :=

laps1 ⋅ u1

ft/s

= 47.368

secperlap3

= 48.649

secperlap4

= 50

u4 = 264

= 1800

laps2 :=

laps3

= 37

laps4 :=

+

laps3

+

+

laps3 ⋅ u3 + laps4 ⋅ u4

laps4

total_sec secperlap2 total_sec secperlap4

total_laps

laps2

= 38

laps4

= 36

= 150

TMS⋅

3600 5280

= 187.67

mi/h

1

:=

1 total_laps

SMS

secperlap2

= 39

laps2 ⋅ u2

u2 = 278.667

3600

total_laps

TMS = 275.244

SMS

+

= 46.154

laps1

total_laps := laps1 + laps2

5280

5280 u4 := 180⋅ 3600

secperlap1

total_sec

total_sec

u2 := 190⋅

ft/s

= 275

ft/s

⋅

⎛ laps ⋅ 1 + laps ⋅ 1 + laps ⋅ 1 + laps ⋅ 1 ⎞ 1 u 2 u 3 u 4 u 1 2 3 4 ⎠ ⎝ SMS ⋅

3600 5280

= 187.5

mi/h


Problem 5.7 Determine the space-mean speed. 195 + 190 + 185 + 180 4

= 187.5

mi/h

Problem 5.8 Estimate the probability of having 4 cars in an interval. t := 13

sec

− q⋅ ( t) P ( h ≥ 13)

(Eq. 5.26)

3600

− 0.0036 ⋅ q

0.6

q :=

e

e

ln ( 0.6)

−0.0036

qsec

:=

q 3600

q

= 141.9

qsec

= 0.039

veh/h

veh/s

during 30 sec intervals,

x := 30 ⋅ qsec

P ( n) :=

x

n

x

= 1.182

−x

⋅e n!

P ( 4)

= 0.025


Determine how m any of t hese 120 intervals have 3 cars arriving. t := 20

Problem 5.9

sec (Given)

18

P ( 0)

120 ( λ ⋅ t)

P ( n)

⋅e n!

λ = 0.095

P ( n) := P ( 3)

− λ⋅ t

n

(Eq. 5.23)

veh/s

( λ ⋅ t)

n

− λ⋅ t

⋅e n!

= 0.1707

P ( 3) ⋅ 120 = 20.48

intervals

Extra

λ ⋅ 20 = 1.897

veh/20 s

λ ⋅ 60 = 5.691

veh/min

λ ⋅ 3600 = 341.5

veh/h

− qt P ( 0)

e

(Eq. 5.26)

3600

− q( 20 ) 18 20 0.15

− q( 0.00556 )

e

ln ( 0.15) q := −

3600

e

−q ( 0.00556)

ln ( 0.15) 0.00556

q

= 341.2


Problem 5.10 Determine percentages for headways. vehicles := 1.9

time := 20

vehicles

q :=

sec

(given)

q = 0.095

time

tgreat := 10

tless

:= 6

(given) (Eq. 5.26)

− q⋅ t

P ( t) := e

P ( h ≥ 10)

(

) = 0.387

P(h >= 10) = 0.387

P tgreat P (h

≥ 6)

(

P tless P (h 1

) = 0.566

< 6)

1 − P (h

≥ 6)

− P ( tless ) = 0.434

P(h < 6) = 0.434

Problem 5.11 Determine the probability of an accident. q := 280

t := 1.5

+

2.5 → 4.0

− q⋅ t P (h

≥ 4)

e

(given) (Eq. 5.26)

3600

− q⋅ t e

3600

P (h

= 0.733

< 4)

1 − P ( h ≥ 4)

− q⋅ t 1

−e

3600

= 0.267

Probability of an accident = 0.267


Problem 5.12 Determine the driver reaction times th at would make the probabili ty of an accident o ccurin g equal to 0.15. q := 280 we want P (h

t

≥ t)

= 2.09

(given) P (h

< 4)

0.15

so,

0.85

s

so the driver reaction time should be, t

− 1.5 = 0.59


Problem 5.14 Determine the t otal vehicl e delay.

λ1 := 8

veh/min

for t <= 20

λ2 := 0

veh/min

for t > 20 to t <= 30

λ3 := 2

veh/min

for t > 30

μ := 4

veh/min

for all t

Queue at t = 30 min Q30 := λ1 ⋅ 20 + λ2 ⋅ 10 − μ ⋅ 30 Q30 = 40 for queue dissipation after 30 minutes, note that:

t

= 20

min

so queue dissipation occurs at, t

+

30 = 50

min

using areas, the total delay can be calculated as,

Dt :=

20 ⋅ 80 2

+

( 80 + 40) ⋅ 10 2

+

40 ⋅ 20

Dt

2

= 1800

veh − min

Also t1 := 20 t2 := 10 t3 := 20 D t :=

Dt

1 2

⋅t 1⋅

( t 1 ⋅ λ 1) − ( t 1 μ⋅ )

+

1 2

⋅t 2⋅

( λ 1 ⋅t 1) − (μ ⋅ t 1)

+

(λ 1 ⋅t 1 + λ 2 ⋅ t 2) − μ ⋅( t 1 + t 2)

+

1 2

⋅ t 3⋅ (λ 1 ⋅ t 1 + λ 2 ⋅ t 2) − μ ⋅ (t 1 +

= 1800


)

t2


Problem 5.16 Determine when the queue will dissip ate and the to tal delay.

λ ( t) := 4.1 +

0.01t

Arrivals( t) :=

⌠ ⎮ λ ( t) dt → 4.10 ⋅ t + ⎮ ⌡

Departures( t) :=

= 15.34

veh/min

(given)

-3 2

5.00⋅ 10

⋅t

⌠ ⎮ μ dt → 12 ⋅ t ⎮ ⌡

Departures( t) := 12⋅ ( t

t

μ := 12

veh/min

− 10)

for service starting 10 min after arrivals

min

Check that total arrivals equals total departures after 15.34 min Arrivals( t)

= 64.065 = 64.065

Departures( t)

Calculate total delay (using triangular area below departures function)

⌠ t Dt := ⎮ ⌡0

4.1 ⋅ t

+

0.005 ⋅ t dt

⌠ t Dt := ⎮ ⌡0

4.1 ⋅ t

+

0.005 ⋅ t dt

Dt

= 317.32

2

− 0.5 ⋅ b ⋅ h

2

− [ ( 0.5) ⋅ ( t − 10) ⋅ ( Departures( t) ) ]

veh-min

Alternative total delay calculation (taking integrals for both arrival and departure functions Q ( t) := Arrivals( t)

⌠ ⎮ Dt := ⎮ ⌡ Dt

− Departures( t) ⌠ t ⌠ t Dt := ⎮ Arrivals( t ) dt − ⎮ ⌡0 ⌡10

Q ( t ) dt

= 317.32

Departures( t) dt

veh-min


Problem 5.17 Determine time when qu eue clears.

λ ( t) := 5.2 − 0.01t

μ ( t ) := 3.3 +

2.4 ⋅ t

(given)

Integrate to obtain # of arrivals at specific time Arrivals( t) :=

⌠ ⎮ λ ( t) dt → 5.20 ⋅ t − 5.00⋅ 10-3 ⋅ t2 ⎮ ⌡

Integrate to obtain # of departures at specific time

⌠ ⎮ μ ( t) dt → 3.30 ⋅ t + Departures( t ) := ⎮ ⌡

1.20 ⋅ t

2

Since vehicle service begins (i.e., toll booth opens) 10 minutes after vehicles begin to arrive, Departures( t ) := 3.3( t

− 10) +

1.2 ⋅ ( t − 10)

2

Time to queue clearance: At what time does arrival rate equal constant 10 veh/min?

t

= 2.792

tcon := t

+

10

(

Departures tcon

t

tcon

= 12.792

time at which service becomes constant at 10 veh/min

) = 18.565

= 22.265

queue clears 22 minutes and 15.9 seconds after 7:50 AM

Check that total arrivals equals total departures Arrivals( 22.265)

= 113.3

Departures 10( 22.265 − 12.792)

= 94.73

94.73 + 18.568 = 113.3

arrivals after arrival rate becomes 10 veh/mi add arrivals before constant arrival rate


Problem 5.18 Determine the total delay and the lo ngest queue length .

λ := 6

μ ( t) := 2 +

veh/min

0.5 ⋅ t

veh/min

(given)

Integrate to obtain # of arrivals and departures at specific time

⌠ ⎮ λ dt → 6 ⋅ t Arrivals( t) := ⎮ ⌡ ⌠ ⎮ μ ( t) dt → 2. ⋅ t + Departures( t) := ⎮ ⌡

.250 ⋅ t

2

Time to queue clearance Q ( t) := Arrivals( t)

− Departures( t) → 4. ⋅ t − .250 ⋅ t

2

or Q ( t) :=

⌠ ⌠ ⎮ λ dt − ⎮ μ ( t) dt → 4. ⋅ t − .250 ⋅ t2 ⎮ ⎮ ⌡ ⌡

When q(t) = 0, queue is cleared

= 16

t

Total delay is area between arrival and departure curve

⌠ 16 Delay := ⎮ ⌡0

Q ( t) dt

→ 170.667

veh − min

Maximum queue length d

Q ( t)

4 − 0.5 ⋅ t

0

dt t :=

4

t

0.5

Q ( t) := 4 ⋅ t

=8

− 0.25 ⋅ t

2

Q ( 8)

= 16

cars

Check Arrivals( 8)

= 48

Departures( 8)

= 32

Q := Arrivals( 8)

− Departures( 8) → 16.000


Problem 5.19 Determine the delay and queue information .

λ ( t) := 5.2 − 0.20t

μ := 3

veh/min

veh/min

(given)

Integrate to obtain # of arrivals and departures at specific time

⌠ ⎮ λ ( t) dt → 5.20 ⋅ t − .100 ⋅ t2 Arrivals( t) := ⎮ ⌡ Departures( t) :=

⌠ ⎮ μ dt → 3 ⋅ t ⎮ ⌡

Time to queue clearance Q ( t) := Arrivals( t)

− Departures( t) → 2.20 ⋅ t − .100 ⋅ t

2

or

⌠ ⌠ 2 ⎮ λ ( t ) dt − ⎮ μ dt → 2.20 ⋅ t − .100 ⋅ t Q ( t) := ⎮ ⎮ ⌡ ⌡ When Q(t)=0, queue is cleared

t

= 22

Total delay is area between arrival and departure curve

⌠ 22 Delay := ⎮ ⌡0

Q ( t) dt

→ 177.467

veh − min

Maximum queue length d

Q ( t)

2.2

− .2 ⋅ t

t :=

0

dt Q ( 11)

= 12.1

2.2 0.2

t

= 11

veh

Wait time for 20th vehicle

− 20 0 t = 6.667

3t

time when 20th vehicle departs

wait := 6.667 − 4.183

wait

= 2.48

min



Problem 5.21 Determine delay and maximum queue length.

λ ( t) := 1.2 +

μ := 12

veh/min

0.3t

veh/min

(given)

Integrate to obtain # of arrivals at specific time

Arrivals( t) :=

⌠ ⎮ λ ( t) dt → 1.20 ⋅ t + ⎮ ⌡

Departures( t) :=

.150 ⋅ t

2

⌠ ⎮ μ dt → 12 ⋅ t ⎮ ⌡

Departures( t) := 12⋅ ( t

− 10)

for service starting 10 min after arrivals

For time to queue clearance, set arrival rate = to departure rate

1.2 ⋅ t

t

+

0.15 ⋅ t

= 13.729

2

12 ⋅ ( t

− 10)

min

⌠ t ⌠ t Dt := ⎮ Arrivals( t) dt − ⎮ ⌡0 ⌡10 Dt

= 159.04

Departures( t) dt

veh-min

By inspection, maximum queue is at t = 10 (point at which service begins)

Qmax ( t ) := Arrivals( t ) Qmax ( 10)

= 27

at t=10

veh


Determine the average length of queue, average time spent in the system, and average waiting ti me spent in the queue.

Problem 5.23

M/M/1

μ := 5 ρ :=

λ := 4

λ μ

(given)

ρ = 0.8

(Eq. 5.27)

2

ρ Qbar := 1−ρ tbar := wbar :=

1

μ−λ λ μ ⋅ ( μ − λ)

Qbar = 3.2

veh

(Eq. 5.31)

tbar = 1

min/veh

(Eq. 5.33)

wbar = 0.8

min/veh

(Eq. 5.32)


Problem 5.24

Determine the average length of queue, average time spent in the system, and average waiting ti me spent in the queue. M/D/1

μ := 3 λ ρ := μ

λ := 2 ρ = 0.667 ρ

Qbar := 2⋅ (1

− ρ) ρ

− ρ)

−ρ 2 ⋅ μ ⋅ ( 1 − ρ ) 2

(Eq. 5.27)

2

wbar := ⋅ ⋅ (1 2 μ tbar :=

(given)

Qbar = 0.667

veh

(Eq. 5.28)

wbar = 0.333

min/veh

(Eq. 5.30)

tbar = 0.667

min/veh

(Eq. 5.29)


Problem 5.26 Determine the new distrib ution r ate required to c lear the queue.

λ ( t) := 120 + 8 ⋅ ( t − 30)

(from Problem 5.25)

μ ( t)

(from Problem 5.25, with an unknown service rate)

μ ⋅ ( t − 30)

for dissipation at t = 60 (8:45 - 9:45): 120 + 8 ⋅ ( t

− 30)

μ ⋅ ( t − 30)

μ = 12

veh/min

Problem 5.27 Determine the arrival rate.

μ := 4

veh/min

tc := 30

min

lane_cap := 30

(time to queue clearance) vehicles

at queue clearance

λ ⋅ tc

μ ⋅ ( tc − tp)

Where t p is the time until processing begins (i.e., time at which queuing lane becomes full) tp

lane_cap

λ

substituting expression for t p and solving for arrival rate gives,

λ ⋅ tc λ=2

μ⋅

⎛ tc ⎞ t − ⎝ c λ ⎠ veh/min


Problem 5.28 Develop an expression for determining p rocessing rates in terms of x.

λ ⋅t

x

λ

and since x

t

2

at queue dissipation

2

λ ⋅ t μ ⋅ ( t − 13) substituting gives x

μ

x 2

− 13

Problem 5.29 Determine the time until queue dissipation. 1 2

⋅b⋅h

Delay is calculated by finding area of triangle formed by uniform arrival and departure rates

3600

b := 30

h :=

3600⋅ 2

μ := 4 h

μ

time at which processing begins

= 60

h

b

= 240

veh/min

min

(given)

time until queue to clearance after arrival of first vehicle


Problem 5.30 Determine when th e queue dissipates, the total delay, and t he length of the long est queue.

λ ( t) := 4.3 − 0.22t μ := 2

veh/min

(given)

veh/min

Arrivals( t) :=

⌠ ⎮ λ ( t) dt → 4.30 ⋅ t − .110 ⋅ t2 ⎮ ⌡

⌠ ⎮ μ dt → 2 ⋅ t Departures( t) := ⎮ ⌡ At time of queue clearance, Arrivals( t)

4.3 ⋅ t

t

Departures( t)

− 0.11 ⋅ t

= 20.91

2

2⋅ t

min

for total delay,

⌠ t Dt := ⎮ Arrivals( t) dt − ⌡0 Dt

= 167.59

1 2

⋅ t ⋅ ( μ ⋅ t)

second term is triangular area formed by uniform departures

veh − min

length of queue at time t is, Q ( t) := Arrivals( t)

− Departures( t) → 2.30 ⋅ t − .110 ⋅ t

2

for maximum length of queue, d

Q ( t)

0

0.22t

− 2.3

dt t :=

t

2.3 0.22

= 10.455

Q ( t)

= 12.02

veh


Problem 5.31 Determine how many veh/min sho uld be pro cessed. length of queue at time t is

Q ( t)

⌠ t ⎮ ⌡0

Q (t )

3.3 ⋅ t

⌠ ⎮ μ dt 3.3 − 0.1 ⋅ t dt − ⎮ ⌡ 2

− 0.05 ⋅ t − μ ⋅ t

for maximum d

Q ( t)

0

3.3

−μ

3.3

− 0.1 ⋅ t − μ

dt

t

0.1

first substitute for t, then find departure rate, 2

Q(t)

⎛ 3.3 − μ ⎞ − 0.05 ⋅ ⎛ 3.3 − μ ⎞ − μ ⋅ ⎛ 3.3 − μ ⎞ 3.3 ⋅ ⎝ 0.1 ⎠ ⎝ 0.1 ⎠ ⎝ 0.1 ⎠

μ = 2.41

4

veh/min


Problem 5.32 Determine the probability that the number of trucks wil l exceed 5.

λ := 1.5 ρ :=

μ := 2

λ μ

veh/min

(given)

ρ = 0.75

(Eq. 5.27)

N := 1 1

P0 :=

+

1

P0

ρ 1−ρ

(Eq. 5.34)

= 0.25

n

Pn

ρ ⋅ P0 n− N

N

(Eq. 5.36)

⋅ N!

1

P1 :=

ρ ⋅ P0 1− N

N

⋅ N!

P1

= 0.1875

P2

= 0.1406

P3

= 0.1055

P4

= 0.0791

P5

= 0.0593

2

P2 :=

ρ ⋅ P0 2− N

N

⋅ N!

3

P3 :=

ρ ⋅ P0 3− N

N

⋅ N!

4

P4 :=

ρ ⋅ P0 4− N

N

⋅ N!

5

P5 :=

ρ ⋅ P0 5− N

N

⋅ N!

Psum := P0 + P1 + P2 Pgreater_than_5 := 1

+

P3

+

− Psum

P4

+

P5

Psum

= 0.822

Pgreater_than_5 = 0.178


Problem 5.33 Determine the number o f spaces.

1

P0 :=

1

1+

P0

2

1!

2

+

2

2!

3

+

2

3!

(Eq. 5.34) 4

+

2

4!

5

2

+

5!

6

+

2

( 6!) ⋅ 0.5

= 0.134

Pn_greater_than_N

P0 ⋅ ρ

N+ 1

⎛ − ⎝ N ⎠

N! ⋅ N ⋅ 1

ρ := 2

ρ ⎞

(Eq. 5.37)

(from Example 5.12)

N := 6

P6 :=

P0 ⋅ ρ

N+ 1

⎛ − ρ ⎞ ⎝ N ⎠

N! ⋅ N ⋅ 1

P6

= 0.006

(this is less than 1%, but also check 5 spaces)

N := 5

P5 :=

P0 ⋅ ρ

N+ 1

⎛ − ρ ⎞ ⎝ N ⎠

N! ⋅ N ⋅ 1

P5

= 0.024

so 6 spaces must be provided, because with only 5 spaces, the probability of 2.4% exceeds 1%


Problem 5.34 Determine the total time spent in system by all vehicles in a one hour period.

λ := 430⋅

μ :=

1 10

1 60

⋅ 60

λ μ

ρ :=

λ = 7.167

veh/min

μ=6

veh/min

(given)

ρ = 1.194

(Eq. 5.27)

N := 2

ρ N

= 0.597 1

P0 := 1+

Qbar :=

tbar :=

ρ + 1!

P0 ⋅ ρ

ρ

P0

= 0.252

(Eq. 5.34)

⎛ − ρ ⎞ ⎝ N ⎠

2! ⋅ 1

N+ 1

N! ⋅ N

ρ+

2

⋅

⎡ 1 ⎤ ⎢ ⎛ ρ ⎞ 2 ⎥ 1− ⎣ ⎝ N ⎠ ⎦

Qbar

λ

Qbar = 0.662

veh

(Eq. 5.38)

t bar = 0.259

min/veh

(Eq. 5.40)

so total time spent in one hour: 430⋅ tbar = 111.4

min/h


Problem 5.35 Determine the minimum number of boo ths needed.

λ := 500⋅ μ := ρ :=

1 15

1 60

⋅ 60

λ μ

λ = 8.333

veh/min

μ=4

veh/min

(given)

ρ = 2.083

(Eq. 5.27)

N must be greater than 2 for the equations to apply, try 3 booths N := 3 1

P0 :=

2

1

Qbar :=

wbar :=

+

ρ ρ + + 1! 2!

P 0 ⋅ ρ

N+ 1

⋅

N! ⋅ N

ρ+

Qbar

λ

⎛ ⎝

3! ⋅ 1 −

1

(Eq. 5.34)

N ⎠

Qbar = 1.101

wbar = 0.132

μ

= 0.098

ρ ⎞

⎡ 1 ⎤ ⎢ ⎛ ρ ⎞ 2 ⎥ 1− ⎣ ⎝ N ⎠ ⎦

−

wbar ⋅ 60 = 7.924

ρ

P0

3

veh

(Eq. 5.38)

(Eq. 5.39)

min/veh

sec/veh

try 4 booths N := 4 1

P0 :=

2

1

Qbar :=

wbar :=

+

3

ρ ρ ρ + + + 1! 2! 3!

P 0 ⋅ ρ

N+ 1

N! ⋅ N

ρ+

Qbar

λ

wbar ⋅ 60 = 1.526

⋅

ρ

1

μ

sec/veh

= 0.119

(Eq. 5.34)

veh

(Eq. 5.38)

⎛ − ρ ⎞ ⎝ N ⎠

4! ⋅ 1

⎡ 1 ⎤ ⎢ ⎛ ρ ⎞ 2 ⎥ 1− ⎣ ⎝ N ⎠ ⎦

−

P0

4

Qbar = 0.212

wbar = 0.025

min/veh

(Eq. 5.39)

so a minimum of 4 booths must be open


Determine the tim e-mean speed of the miniv ans.

Problem 5.36

t1 := 98 s l := 3

t2 := 108 s

t3 := 113 s

t4 := 108 s

t5 := 102 s

(given)

n := 5 minivans

miles

velocities l

V1 :=

t1 l

V3 :=

t3 l

V5 :=

t5

⋅ 3600

mi

⋅ 3600

mi

⋅ 3600

mi

h

h

V2 :=

V4 :=

l t2 l t4

⋅ 3600

mi

⋅ 3600

mi

h

h

h

time-mean speed n

∑ ut

u t :=

i

(Eq 5.5)

ui

=1 n

V1 + V2 + V3 + V4 + V5 n

ut

= 102.332

mi h

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answers 1) Use space-mean speed equation us

:=

1 1 n

⋅ ⎛

1

⎝ V1

+

1 V2

+

1 V3

+

1 V4

+

us

1 ⎞

= 102.079

mi

(Eq 5.9)

h

V5 ⎠

2) Use all vehicles and not just minivans V6 :=

u t :=

3 101

⋅ 3600

mi

V7 :=

h

3 85

⋅ 3600

mi

V8 :=

h

V1 + V2 + V3 + V4 + V5 + V6 + V7 + V8

ut

8

3 95

= 107.417

⋅ 3600

mi h

mi h

3) Inattention to rounding V1 := 110

u t :=

mi

V2 := 100

h

V1 + V2 + V3 + V4 + V5 n

mi h

V3 := 96

mi h

V4 := 100

ut

mi h

= 102.4

V5 := 106

mi h

mi h


Determine the probability that more than five vehicles wil l arrive in a one-minute interval.

λ :=

400

veh

3600

s

n 0 := 0

n1 := 1

t := 60 s

n2 := 2

Problem 5.37

(given)

n3 := 3

n 4 := 4

n 5 := 5

solve for the probabilities of 5 and fewer vehicles

P0 :=

P1 :=

P2 :=

P3 :=

P4 :=

P5 :=

( λ ⋅ t)

( λ ⋅ t)

( λ ⋅ t)

( λ ⋅ t)

( λ ⋅ t)

( λ ⋅ t)

n0

⋅e n0!

n1

⋅e n1!

n2

⋅e n2!

n3

⋅e n3!

n4

⋅e n4!

n5

⋅e n5!

− λ⋅ t P0 = 0.0013

(Eq 5.23)

− λ⋅ t P1 = 0.0085

− λ⋅ t P2 = 0.0283

− λ⋅ t P3 = 0.0628

− λ⋅ t P4 = 0.1047

− λ⋅ t P5 = 0.1397

solve for 6 or more vehicles P := 1 − P0 − P1 − P2 − P3 − P4 − P5

P

= 0.6547

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answers 1) Do not account for P(0) P := 1 − P1 − P2 − P3 − P4 − P5

P

= 0.656

P

= 0.7944

2) Do not account for P(5) P := 1 − P0 − P1 − P2 − P3 − P4

3) Solve just for P(6)

n6 := 6

P6 :=

( λ ⋅ t)

n6

⋅e n6!

− λ⋅ t P6 = 0.1552


Problem 5.38

At wh at t im e do es t he m axi mu m q ueu e len gt h o cc ur .

λ ( t) := 1.8 +

0.25⋅ t

− 0.003t ⋅

2

μ ( t) := 1.4 +

0.11⋅ t

t := 60 min

(given)

set the integral arrival and departure rates equal to each other

⌠ t ⎮ 1.8 + ⌡0

0.25⋅ t

1.8⋅ t

0.125t

+

− 0.003t⋅

2

2

− 0.001t⋅

Q( t ) := −0.001⋅ t

3

+

⌠ t ⎮ 1.4 + ⌡0

d t

3

0.07⋅ t

1.4⋅ t 2

+

+

0.11⋅ t d t

0.055t ⋅

2

0.4⋅ t

solve for time at which maximum queue length occurs d d t

Q( t)

2

−.003⋅ t +

0.14⋅ t

+

0.4

0

use quadratic equation

t1 :=

t2 :=

−0.14 +

(0.142) − 4⋅ −0.003⋅ 0.4 2⋅ −0.003

−0.14 −

(0.142) − 4⋅ −0.003⋅ 0.4 2⋅ −0.003

t1 = −2.7

min

t2 = 49.4 min

(this is the reasonable answer of the two provided by the quadratic equation)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answers 1) Solve for total vehicle delay

⌠ 60 ⎮ 1.8 + ⌡0

⌠ 60 0.25⋅ t − 0.003t ⋅ d t − ⎮ 1.4 + ⌡0 2

0.11⋅ t d t

= 60

min

2) Take the t1 value as positive t := 2.701 min

3) Set Q(t) = 0 and assume t is in hours Q( t ) := −0.001⋅ t 3

−0.001⋅ t +

t :=

−0.07 +

3

+

0.07⋅ t

2

+

0.07⋅ t

2

0.4⋅ t

+

0.4⋅ t 0

(0.072) − 4⋅ −0.001⋅ 0.4 2 − 0.001

t

=0

t⋅ 3660 = 19.4

min


Determine the average waiting time for t his qu euing system.

λ :=

200

veh

60

min

μ :=

60

veh

15

min


calculate traffic intensity

λ μ

ρ :=

ρ = 0.833

(Eq 5.27)

calculate average waiting time in queue (M/M/1)

w :=

λ μ⋅ ( μ − λ )

w

= 1.25

min

(Eq 5.32)

veh

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 1) Assume M/D/1 w :=

ρ 2⋅ μ ⋅ ( 1 − ρ)

w

= 0.625

min

(Eq 5.29)

veh

2) Find average time spent in the system t :=

1

t

μ−λ

= 1.5

min

(Eq 5.30)

veh

3) Use average queue length equation 2

ρ Q := ( 1 − ρ)

Q = 4.167

min

(Eq 5.28)

veh

(note: the units for the average queue length equation are actually in "veh" so that should be an obvious sign this is the inocrrect equation to use.)


Determine the average time spent in the syst em.

λ :=

60

veh

60

min

μ :=

60

veh

30

min


compute the average time spent in the system t :=

1

t

μ−λ

=1

min

(Eq 5.33)

veh

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers 1) Assume M/D/1 queuing

ρ :=

λ μ

t :=

2−

ρ 2⋅ μ ⋅ ( 1 − ρ)

t

= 0.75

min

(Eq 5.30)

veh

2) Use average waiting time equation w :=

λ μ⋅ (μ − λ )

w

= 0.5

(Eq 5.32)

min veh

3) Keep units in veh/hr

λ := 60

μ := 120

t :=

1

μ−λ

t

= 0.017

min veh


How would the probability of waiting i n a queue change if a fourth tol l both w ere opened? 850

λ :=

μ :=

60

Problem 5.41

60 12

(given)

N := 3

determine probability of having no vehicles in system

ρ :=

λ μ

P0 :=

ρ = 2.833

1 2

ρ ρ 1+ + + 1! 2!

P0 = 0.013

3

ρ

⎛ ⎝

3!⋅ 1 −

(Eq 5.34)

ρ ⎞

N ⎠

determine probability of having to wait in a queue N + 1

P3 :=

P0⋅ ρ

⎛ ⎝

N !⋅ N ⋅ 1 −

P3 = 0.847

ρ ⎞

(Eq 5.37)

N ⎠

determine probability of having to wait in queue for four toll booths N := 4 1

P0 :=

2

1+

3

ρ ρ ρ + + + 1! 2! 3!

4

ρ

⎛ ⎝

4!⋅ 1 −

P0 = 0.048

(Eq 5.34)

P4 = 0.313

(Eq 5.37)

ρ ⎞

N ⎠

N + 1

P4 :=

P0⋅ ρ

⎛ ρ ⎞ N !⋅ N ⋅ 1 − ⎝ N ⎠

calculate difference

P3 − P4 = 0.534

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answers 1) Solve for probability for just 3 toll booths P3 = 0.847

2) Solve for probability for just 4 toll booths P4 = 0.313

3) Use Equation 5.36 for 4 toll booths n := 15.247

(approximate value for average length of queue

N := 3 n

ρ ⋅ P0

Pn3 := N

n− N

⋅ N !

Pn3

= 0.09

Pn4

= 0.003

N := 4

n

ρ ⋅ P0

Pn4 := N Pn3

n− N

⋅ N !

− Pn4 = 0.088




Chapter 6 Highway Capacity and Level of Service Analysis U.S. Customary Units


Preface The solutions to the fourth edition of Principles of Highway Engineering and Traffic Analysis 1 were prepared with the Mathcad 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. •


•


•



Finally, to assist in quickly identifying the final answer, or answers, for what is being asked in the problem statement, yellow highlighting has be en used (which will print as light gray). 1

www.mathcad.com


Problem 6.1 Determine the hourly volume. BFFS := 65

(given)

f LW := 1.9

(Table 6.3)

f LC := 0.80

(Table 6.4)

f N := 0.0

(Table 6.5)

f ID := 0.0

(Table 6.6)

FFS := BFFS − fLW − f LC − f N − f ID

FFS = 62.3

vp := 1616

(Eq. 6.2)

(Table 6.1, by interpolation)

PT := 0.10

(given)

ET := 2.5

(Table 6.8)

1 f HV := 1 + PT⋅ ET − 1

(

PHF := 0.90

V := vp ⋅ PHF ⋅ N ⋅ f HV ⋅ f p

) f p := 1.0

f HV = 0.87

(Eq. 6.5)

N := 3

(given)

V = 3794

veh/h

(Eq. 6.3)


Problem 6.2 Determine Determine the g rade length. V := 5435

(given)

FFS := 62.3

(From Problem 6.1)

vp := 2323

(Table 6.1)

PHF := 0.90

vp

f p := 1.0

N := 3

(given)

V (Eq. 6.3)

PHF ⋅ N ⋅ f HV ⋅ f p

f HV = 0.867 PT := 0.10

f HV

(given)

1

(

)

1 + PT⋅ ET − 1

(Eq. 6.5)

ET = 2.54 Using Table 6.8, the length is found to be between 0.75-1.0 mi


Problem 6.3 Determine Determine the maximum numb er of large trucks and buses. first, determine the heavy vehicle factor assume urban freeway BFFS := 70

(given)

f LW := 0.0

(Table 6.3)

f LC := 0.0

(Table 6.4)

f N := 4.5

(Table 6.5)

f ID := 0.0

(Table 6.6)

FFS := BFFS − fLW − f LC − f N − f ID PHF :=

1800

PHF = 0.6429

700⋅ 4

f p := 1.0 vp

FFS = 65.5

(Eq. 6.2)

(Eq. 6.4)

N := 2

(given)

1800

1400

PHF ⋅ N ⋅ f HV ⋅ f p

f HV

(Eq. 6.3)

interpolate from Table 6.1 to find v p

vp := 1680 + ( FFS − 65) ⋅

f HV :=

1400

⎛ 1770 − 1680 ⎞ ⎝ 70 − 65 ⎠

vp = 1689

f HV = 0.8289

vp

now, determine the number of trucks ET := 2.5 f HV

1

(

)

1 + PT⋅ ET − 1

1 + 1.5 ⋅ PT

(Table 6.7)

(rolling terrain)

1

1

1 + PT⋅ ( 2.5 − 1)

1 + 1.5PT

(Eq. 6.5)

1 f HV

PT = 0.1376 1800⋅ PT = 247.7143

therefore 247 trucks and buses


Problem 6.4 Determine Determine the level of servic e V := 1800 PT :=

(given)

180

PT = 0.1

V

ET := 4.0

(Table 6.9)

1 f HV := 1 + PT⋅ ET − 1

(

PHF := 0.643 vp :=

V PHF ⋅ N ⋅ f HV

f HV = 0.769

)

(given)

N := 2 vp = 1819.6

(Eq. 6.5)

pc/h/ln

(Eq. 6.3)

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

26.0 < 28.1 < 35.0

D = 28.125

pc/mi/ln

(Eq. 6.6)

therefore the LOS is D


Probl em 6.5 6.5 Determine Determine the driver population factor . PHF := 0.80

PT := 0.08

PR := 0.06

N := 3

V := 3900

(given)

Solve for f HV ET := 2.0

ER := 3.0

Table 6.7

1 f HV := 1 + PT⋅ ET − 1 + PR ⋅ ER − 1

(

)

(

)

f HV = 0.833

Freeway is operating at capacity, so for FFS = 55 mi/h

vp

(Eq. 6.5)

vp := 2250

(Table 6.1)

V PHF ⋅ fHV ⋅ N ⋅ f p

f p = 0.867


Probl em 6.6 6.6 Determine Determine the n ew level of servic e.

BFFS := 60 f LW := 1.9

(Table 6.3)

f N := 0.0

(Table 6.5)

f LC := 0.6

(Table 6.4)

f ID := 0.0

(Table 6.6)


FFS = 57.5

mi/h

(Eq. 6.2)

calculate heavy vehicle adjustment PT := 0.12

PR := 0.06

ET := 4.5

ER := 4.0

(Table 6.7)

1 f HV := 1 + PT⋅ ET − 1 + PR ⋅ ER − 1

(

)

(

)

f HV = 0.625

(Eq. 6.5)

calculate LOS before and after lane addition N1 := 2

N2 := 3

2300 + 2250 2 vp

f p := 0.90

= 2275

V PHF ⋅ fHV ⋅ N1 ⋅ f p

PHF := 0.88

vp := 2275

(Table 6.1 by interpolation) interpolation)

V := vp ⋅ PHF ⋅ f HV ⋅ N1 ⋅ f p

(Eq. 6.3)

V = 2252.25 V vp2 := PHF ⋅ fHV ⋅ N2 ⋅ f p vp2 = 1516.667

(Eq. 6.3)

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 f LC := 0.4 For vp of 1517,

D :=

vp2 S


FFS = 57.7

S := FFS

D = 26.285

pc/mi/ln

LOS D - 26 < 26.3 < 35

(Eq. 6.6)


Problem 6.7 Determine the LOS. find the analysis flow rate ( 1500 1500⋅ 0.03 0.03) + ( 1000⋅ 0.04) 2500 2500 5280

= 0.473

= 0.034

(average grade)

mi

PT := 0.05

(given)

ET := 2.0

(Table 6.8)

1 f HV := 1 + PT⋅ ET − 1

(

N := 2.0

f HV = 0.952

)

PHF := 0.90

V vp := PHF ⋅ N ⋅ f HV ⋅ f p

f p := 1.0 vp = 1166.667

V := 2000

pc/h/ln

(Eq. 6.5)

(given)

(Eq. 6.3)

determine FFS BFFS := 65

(given)

f LW := 0.0

(Table 6.3)

f LC := 1.8

(Table 6.4)

f N := 4.5

(Table 6.5)

f ID := 0.0

(Table 6.6)


FFS = 58.7

(Eq. 6.2)

pc/mi/ln therefore LOS is C

(Eq. 6.6)

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

D :=

vp FFS

D = 19.88



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

Problem 6.9

Before: PT := 0.06

PB := 0.05

PTB := PT + PB

(given)

PTB = 0.11

ETB := 2.5

(Table 6.7)

1 f HVTB := 1 + PTB⋅ ETB − 1

(

PHF := 0.95

f HVTB = 0.858

)

f p := 1.0

V vp := ⋅ f p ⋅ N PHF ⋅ fHVTB

N := 4

V := 5400

vp = 1655.526

(Eq. 6.5)

(given)

(Eq. 6.3)

BFFS := 70

(given)

f LW := 1.9

(Table 6.3)

f LC := 0.4

(Table 6.4)

f N := 1.5

(Table 6.5)

f ID := 3.7

(Table 6.6)


FFS = 62.5

S := 62.5

D :=

(Eq. 6.2) (Figure 6.2)

vp

D = 26.5

S

pc/mi/ln

LOS D

(Eq. 6.6)

After:

(

)

Vnew := V ⋅ 1 − PT

Vnew = 5076

NumBuses := V ⋅ PB

NumBuses = 270


PBnew :=

NumBuses PBnew = 0.053

Vnew

EB := 2.5

(given)

1 f HVB := 1 + PBnew⋅ EB − 1

(

)

f HVB = 0.926

(Table 6.7) (Eq. 6.5)

vp :=

Vnew PHF ⋅ fHVB ⋅ f p ⋅ N

vp = 1442.368 (Eq. 6.3)

S := 62.5

D :=

vp S

D = 23.1

pc/mi/ln

LOS C

(Figure 6.2) (Eq. 6.6)


Problem 6.10 Determine the density, v/c ratio, and LOS before and after the strike. calculate free-flow speed BFFS := 65

(given)

f LW := 0.0

(Table 6.3)

f LC := 0.0

(Table 6.4)

f N := 3.0

(Table 6.5)

f ID := 5.0

(Table 6.6)


FFS = 57 mi/h

(Eq. 6.7)

calculate volume after bus strike V 1 := 3800

PT := 0.02

PB := 0.04

(

)

V2 := V1 − PB ⋅ V1 + 6 ⋅ P B ⋅ V1

PHF := 0.90

PTB := PT + PB

(given)

V2 = 4560

calculate heavy vehicle adjustments before and after bus strike ETB := 4.0

ET := 5.0

(Table 6.8)

1 f HVTB := 1 + P TB⋅ E TB − 1

(

)

PT :=

Calculate new P T 1 f HVT := 1 + P T⋅ E T − 1

(

N := 3

)

f HVTB = 0.847 V1 ⋅ PT V2

PT = 0.017

f HVT = 0.938

f p := 1.0 V1

vp1 := PHF ⋅ N⋅ f HVTB⋅ f p

(Eq. 6.5)

(given)

vp1 = 1660.741

(Eq. 6.3)


vp1 c

= 0.732

D = 29.14

D :=

vp1 FFS

pc/mi/ln

LOS D

V2 vp2 := PHF ⋅ N ⋅ f HVT⋅ f p vp2 c

= 0.794

D :=

(Eq. 6.6)

(Eq. 6.3)

vp2 = 1801.481

vp2 FFS

D = 31.6

pc/mi/ln

LOS D

(Eq. 6.6)


Problem 6.11 Determine level of service. PT := 0.06

PR := 0.02

(given)

ET := 1.5

ER := 1.2

(Table 6.7)

1 f HV := 1 + PT⋅ ET − 1 + PR ⋅ ER − 1

(

V := 1300

)

(

PHF := 0.85


)

f p := 0.95 vp = 832.322

f HV = 0.967

(Eq. 6.5)

N := 2

(given)

pc/mi/ln

(Eq. 6.3)

FFS := 45

(given)

LOS C, from Figure 6.2

Problem 6.12 Determine the number o f vehicles added. FFS := 45

(given)

vp := 1900

(from Table 6.11)

PHF := 0.85

f p := 0.95

PT := 0.06

PR := 0.02

ET := 1.5

ER := 1.2

N := 2

(given)

1 f HV := 1 + PT⋅ ET − 1 + PR ⋅ ER − 1

(

)

(

)

f HV := 0.967 (from Problem 6.11)

Vprev := 1300 V := vp ⋅ PHF ⋅ N ⋅ f HV ⋅ f p

V = 2967

V − Vprev = 1667.239

1667 vehicles can be added to the traffic flow

(Eq. 6.3)


Problem 6.13 Determine Determine the level of service. Estimate FFS BFFS := 55 + 5

(posted speed limit + 5 mi/h)

f LW := 1.9

(Table 6.3)

f LC := 0.4

(Table 6.13)

f M := 0

(Table 6.14)

f := 3.75 A


FFS := BFFS − fLW − f LC − f M − A f

FFS = 53.95

(Eq. 6.7)

calculate analysis flow rate PT := 0.08

PR := 0.02

(given)

ET := 3.0

(Table 6.8)

ER := 3.0

(Table 6.9)

1 f HV := 1 + PT⋅ ET − 1 + PR ⋅ ER − 1

(

V := 4000

)

veh/h

(

PHF := 0.90

V vp := PHF ⋅ N ⋅ f HV ⋅ f p S := 51.5 D :=

f HV = 0.833

) N := 3

vp = 1871.345

f p := 0.95 pc/h/ln

(Eq. 6.5)

(given) (Eq. 6.3)

(Figure 6.4)

vp

(Eq. 6.6)

S

D = 36.34 LOS E

veh/mi/ln (Table 6.11 or Figure 6.4)


Probl em 6.14 6.14 Determin Determin e LOS before before and after the developm ent. Before: PT := 0.10

PR := 0.03

(given)

ET := 2.5

ER := 2.0

(Table 6.7)

1 f HV := 1 + P T⋅ ET − 1 + P R ⋅ ER − 1

f HV = 0.847

(Eq. 6.5)

PHF := 0.95

V := 2300

(given)

(

)

(

N := 2

V vp := N ⋅ PHF ⋅ f HV ⋅ f p BFFS := 50 + 5

)

f p := 0.90 vp = 1587.135

(Eq. 6.3)

BFFS = 55

(given)

f LW := 6.6

(Table 6.3)

TLC := 6 + 3

TLC = 9

(Eq. 6.8)

f LC := 0.65


f M := 0.0

(Table 6.14)

f := 1.0 A

(Table 6.15)


FFS = 46.75

S := 46 D :=

vp S

(Eq. 6.7) (from Figure 6.4)

D = 34.5

which is LOS D

(Eq. 6.6)

After: (Table 6.15)

f := 3.0 A FFS := BFFS − fLW − f LC − f M − A f

FFS = 44.75

(Eq. 6.7)

Vnew := 2700 Vnew vp := N ⋅ PHF ⋅ f HV ⋅ f p

vp = 1863.158

(from Figure 6.4)

S := 42 D :=

vp S

(Eq. 6.3)

D = 44.36

which is LOS E

(Eq. 6.6)


Problem 6.15 Determine Determine the level of servic e before and after after tru cks are allowed. Before: PB := 0.02

(given)

EB := 3.5

(Table 6.8)

1 f HVB := 1 + PB ⋅ EB − 1

(

V := 1900

f HVB = 0.952

)

PHF := 0.80

V vp := PHF ⋅ N ⋅ f HVB ⋅ f p1 LOS LOS C

N := 2

(Eq. 6.5)

f p1 := 1.0

vp = 1246.875

pc/h/ln

(given)

(Eq. 6.3)

(fro (from m Tab Table le 6.1 6.11, 1, wit with h FFS FFS = 55 55 mi/h mi/h))

After: Vnew := V + 150 Buses := PB ⋅ V PBnew :=

PT :=

Buses = 38

Buses Vnew

150

PBnew = 0.019

PT = 0.073

Vnew

PTB := PBnew + PT

PTB = 0.092

ETB := 2.5

(Table 6.8)

1 f HV := 1 + PTB⋅ ETB − 1.

(

)

f HV = 0.879

f p2 := 0.97

(given)

Vnew vpnew := PHF ⋅ N ⋅ f HV ⋅ f p2 LOS LOS D

(Eq. 6.5)

vpnew = 1502.577

(Eq. 6.3)

(fro (from m Tab Table le 6.11 6.11,, wit with h FFS FFS = 55 55 mi/ mi/h) h)


Problem 6.16 Determine Determine the directional hourly v olume. calculate FFS BFFS := 55 + 5

BFFS = 60

(posted speed limit + 5 mi/h)

f LW := 1.9 f LC := 0.4

(Table 6.3) (4 ft right shoulder + 6 ft left (undivided))

(Table 6.13)

f M := 1.6

(Table 6.14)

f := 2.5 A

(Table 6.15)


FFS = 53.6

mi/h

(Eq. 6.7 )

calculate heavy vehicle adjustment PT := 0.08

(given)

ET := 1.5

(Table 6.7)

1 f HV := 1 + PT ⋅ E T − 1

(

f HV = 0.962

)

(Eq. 6.5)

calculate directional hourly volume f p := 0.90

PHF := 0.80

Roadway at capacity, so V := vp ⋅ PHF ⋅ N ⋅ f HV ⋅ f p

N := 2

vp := 2062 veh/h V = 2855

(given) (Table 6.11, by interpolation 2000 - 2100)

veh/h


Problem 6.17 Determine Determine the level of service. calculate the heavy vehicle factor V := 1990 PT :=

trucks := 140

buses := 40

trucks + buses

PR :=

rvs := 10

(given)

PT = 0.09

V rvs

PR = 0.005

V

ET := 1.5

(Table 6.10)

ER := 1.2

(Table 6.7)

1 f HV := 1 + PT⋅ ET − 1 + PR ⋅ ER − 1

(

)

(

)

f HV = 0.956

(Eq. 6.5)

determine the level of service N := 2

PHF := 0.85

(given)

f p := 1.0

(Assumed)

V vp := N ⋅ PHF ⋅ f HV ⋅ f p BFFS := 5 + 50

vp = 1225

pc/h/ln

(Eq. 6.3)

BFFS = 55

(given)

f LW := 0.0

(Table 6.3)

f LC := 0.0

(Table 6.13, 6 ft shoulders, divided

f M := 0.0

(Table 6.14)

f := 1 A

(Table 6.15)

FFS := BFFS − fM − f LW − f LC − A f

FFS = 54

(Eq. 6.7)

S := FFS D :=

vp S

D = 22.68

pc/mi/ln

therefore LOS is C

(Eq. 6.6)


Problem 6.18 Determine Determine the number o f vehicles th at can be added. find initial volume FFS := 50 PT := 0.06

PR := 0.02

(given)

ET := 4.5

(Table 6.8)

ER := 4.0

(Table 6.9)

1 f HV := 1 + PT⋅ ET − 1 + PR ⋅ ER − 1

(

N := 3

)

PHF1 := 0.90

vp1 := 1300

(

f HV = 0.787

)

f p := 0.92

(Eq. 6.5)

(given)

(from Table 6.11, for FFS = 50 mi/h, LOS C)

V1 := vp1 ⋅ PHF1 ⋅ N ⋅ f HV ⋅ f p

V1 = 2543

(Eq. 6.3)

find final volume vp2 := 2000

(capacity conditions)

PHF2 := 0.95

(given)

other variables are same as above V2 := vp2 ⋅ PHF2 ⋅ N ⋅ f HV ⋅ f p Vadded := V2 − V1

V2 = 4129 Vadded = 1586

(Eq. 6.3)

vehicles


Problem 6.19 Determine how many access points mu st be blocked. calculate analysis flow rate V := 2500 + 200 PT :=

trucks := 200

trucks

(given)

PT = 0.074

V

(Table 6.8) (Table 6.8)

ET := 2.0 1 f HV := 1 + PT⋅ ET − 1

(

)

f HV = 0.931

V15 := 720 PHF :=

(given)

V

PHF = 0.938

V15 ⋅ 4

f p := 1.0

(Eq. 6.4)

N := 2


(given)

vp = 1547

(Eq. 6.3)

find BFFS FFS := 55

(given)

f LW := 1.9

(Table 6.3)

f LC := 0.2

(Table 6.13)

f M := 1.6

(Table 6.14)

for A f , first convert the 15 access points in 0.62 mile to the number of access points per mile 15 0.62 f := 6.25 A

= 24.2 (using a rounded value of 25 access points/mile)

BFFS := FFS + fLW + f LC + f M + A f

BFFS = 64.95 mi/h

(Table 6.15) (Eq. 6.7)

find density and LOS


S := 54.5

D :=

vp S

(interpolated from Fig. 6.4, for FFS = 55) (Eq. 6.6)

D = 28.4

LOS D

(Table 6.11)

Determine the number of access points that need to be blocked D := 26

S :=

vp D

(to achieve LOS C)

S = 59.5

(Eq. 6.6)

FFS := 60

(interpolated from Fig 6.4)

f := BFFS − FFS − f LW − f LC − f M A

f = 1.25 A

(Eq. 6.7)

for A f = 1.25 mi/h (reduction in FFS), access points per mile = 5 (from Table 6.15) f ⋅ 4) 0.62 = 3.1 ( A 15 − 3 = 12

access points over 0.62 mile (4 to 1 relationship between f A and access points/mile) access points need to be blocked


Problem 6.20 Determine the level of service. ATS: V := 180 V PHF

PHF := 0.90

(given)

= 200

PT := 0.15

(given)

ET := 1.7

(Table 6.18)

1 f HV := 1 + PT⋅ ET − 1

(

f HV = 0.905

)

(Eq. 6.5)

f G := 1.00

(Table 6.15)

V vp := PHF ⋅ fG ⋅ f HV

vp = 221

(Eq. 6.11)

FFS := 65

(given)

f np := 1.9


ATS := FFS − 0.00776⋅ vp − f np

(Eq. 6.12)

ATS = 61.385

(Table 6.21)

mi/h

LOS is A

PTSF: PT := 0.15

(given)

ET := 1.1

(Table 6.18)

f HV :=

1

(

f HV = 0.985

)

1 + PT⋅ ET − 1

V := 180

(Eq. 6.5)

PHF := 0.90

(given)

f G := 1.00

(Table 6.17)


vp = 203

− 0.000879 ⋅ vp BPTSF := 100⋅ 1 − e

f dnp := 22.0 PTSF := BPTSF + fdnp

(Eq. 6.11)

BPTSF = 16.342

(Eq. 6.14) (Table 6.20, by interpolation)

PTSF = 38.342 %

LOS is B (Table 6.21)

(Eq. 6.13)

lower LOS (PTSF) governs, LOS is B


Problem 6.21 Determine th e level of servic e. ATS: V := 540 V15 :=

PHF := 0.87 V

(given)

V15 = 620.69

PHF

PT := 0.05

PR := 0.10

(given)

ET := 1.9

ER := 1.1

(Table 6.18)

1 f HV := 1 + P T⋅ ET − 1 + PR ⋅ ER − 1

(

)

(

f HV = 0.948

)

f G := 0.93

(Eq. 6.5) (Table 6.17)

V vp := PHF ⋅ fG ⋅ f HV Vf := 275

vp = 704.116

(Eq. 6.11)

SFM := 57

FFS := SFM + 0.00776 ⋅

(given)

Vf

FFS = 59.251

f HV

(Eq. 6.9)

f np := 3.0

(Table 6.19)

ATS := FFS − 0.00776⋅ vp − f np

ATS = 50.787

LOS B

(Eq. 6.12)

PTSF: PT := 0.05

PR := 0.10

ET := 1.5

ER := 1.0

(given) (Table 6.18)

1 f HV := 1 + P T⋅ ET − 1 + PR ⋅ ER − 1

(

)

(

f HV = 0.976

)

(Table 6.17)

f G := 0.94 V vp := PHF ⋅ fG ⋅ f HV

vp = 676.816

− 0.000879 ⋅ vp BPTSF := 100⋅ 1 − e

(Eq. 6.11)

BPTSF = 44.839

f dnp := 17.5

PTSF := BPTSF + fdnp

(Eq. 6.5)

(Eq. 6.14) (Table 6.20)

PTSF = 62.339

LOS C

(Eq. 6.13)

lower LOS (PTSF) governs, LOS is C


Problem 6.22 Determine the level of service. ATS: PT := 0.04

PB := 0.03

PR := 0.01

(given)

PT := PT + PB ET := 1.2

ER := 1.0

(Table 6.18)

1 f HV := 1 + PT⋅ ET − 1 + PR ⋅ ER − 1

(

VNB := 440 V PHF

)

(

VSB := 360

f HV = 0.986

)

V := VNB + VSB

V = 800

(Eq. 6.5)

PHF := 0.87

= 919.54

f G := 1.00 vp :=

(given)

(Table 6.17) V

vp = 932.414

PHF ⋅ fG ⋅ f HV

(Eq. 6.11)

BFFS := 60

(given)

f LS := 1.7

(Table 6.16)

f := 4 A

(Table 6.15)

FFS := BFFS − fLS − A f

FFS = 54.3

(Eq. 6.10)

f np := 0

(Table 6.19)

ATS := FFS − 0.00776⋅ vp − f np

ATS = 47.064

mi

LOS C

(Eq. 6.12)

h

PTSF: ET := 1.1

ER := 1.0

(Table 6.18)

1 f HV := 1 + PT⋅ ET − 1 + PR ⋅ ER − 1

(

f G := 1.00

)

(

)

f HV = 0.993

(Eq. 6.5) (Table 6.17)



vp = 925.977

− 0.000879 ⋅ vp BPTSF := 100⋅ 1 − e

VNB V

= 0.55

(Eq. 6.11)

BPTSF = 55.689

so a 55/45 split

f dnp := 0.0

(Table 6.20)

PTSF := BPTSF − fdnp PTSF = 55.689

(Eq. 6.14)

(Eq. 6.13) LOS C, from both PTSF and ATS


Problem 6.23 Determine the percentage of trucks . V15 := 720

(given)

V := V15 ⋅ 4

V = 2880

vp := 3200 Using ATS adjustment values f G := 0.99

(Table 6.17)

PHF := 1 V f HV := PHF ⋅ fG ⋅ vp

f HV = 0.909

(Eq. 6.11)

ET := 1.5

f HV

(Table 6.18)

1

(

(Eq. 6.5)

)

1 + PT⋅ ET − 1

PT = 0.2

PT⋅ 100 = 20

%


Problem 6.24

Determine the maximum percentage of no-passing zones. Determine the flow rate for PTSF PT := 0.08

PB := 0.02

PR := 0.05

(given)

PTB := PT + PB ET := 1.8

ER := 1.0

(Table 6.18)

1 f HV := 1 + PTB ⋅ ET − 1 + PR ⋅ ER − 1

(

V := 500

)

(

)

f HV = 0.926

(Eq. 6.5)

PHF := 0.85

(given)

f G := 0.77

(Table 6.17)


(Eq. 6.11)

vp = 825.057 must find new values, since 825 > 600 f G := 0.94

(Table 6.17)

ET := 1.5

ER := 1.0

(Table 6.18)

1 f HV := 1 + PTB⋅ ET − 1 + PR ⋅ ER − 1

(

)

(

)

f HV = 0.952

(Eq. 6.5)

V vp := PHF ⋅ fG ⋅ f HV vp = 657.071

(Eq. 6.11)

is between 600 and 1200 pc/h, so ok

Determine the percent no-passing zones − 0.000879 ⋅ vp BPTSF := 100⋅ 1 − e

BPTSF = 43.874

(Eq. 6.14)

to maintain maximum passing zones and still LOS B, from Table 6.22, PTSF = 55% PTSF := 55 f dnp := PTSF − BPTSF percentnp := 24.32

f dnp = 11.126 %

(Eq. 6.13, rearranged to solve for f d/np ) (from Table 6.20, with triple interpolation)



Problem 6.26

Determine the freeway's AADT. PT := 0.08

(given)

ET := 2.5

(Table 6.7)

1 f HV := 1 + PT⋅ ET − 1

(

)

PHF := 0.85

f HV = 0.893

(Eq. 6.5)

N := 2

(given)

BFFS := 70

(assumed)

f LW := 1.9

(Table 6.3)

f LC := 1.2

(Table 6.4)

f N := 4.5

(Table 6.5)

f ID := 0.0

(Table 6.6)

FFS := BFFS − fLW − f LC − f N − f ID S := 51.5

( interpolated)

D := 45.0

(for capacity)

D

vp

FFS = 62.4

(Eq. 6.2)

(Eq. 6.6)

S

vp = 2317.5

vp

V PHF ⋅ N ⋅ f HV ⋅ 1.0

(Eq. 6.3)

V = 3517.634 DDHV := V K := 0.12

D := 0.6

AADT⋅ K ⋅ D

DDHV

AADT = 48856

veh day


Alternate Solution: vp := 2325

vp

(capacity at FFS = 62.4, interpolated from Table 6.1 or 6.2)

V PHF ⋅ N ⋅ f HV ⋅ 1.0

V = 3529.018 AADT :=

V K ⋅D

AADT = 49014

veh day


Problem 6.27 Determine the total directional tr affic volum e. f p := 1.0 PT := 0.10

N := 3

vp := 1430

PR := 0.05

(Table 6.1)

PHF := 0.94

(given)

K := 0.20 ET := 2.5

ER := 2.0

(Table 6.7)

1 f HV := 1 + PT⋅ E T − 1 + P R ⋅ E R − 1

(

)

V := vp ⋅ PHF ⋅ N ⋅ f p ⋅ f HV VD :=

V K

(

)

V = 3360

VD = 16802

f HV = 0.833

(Eq. 6.5)

(Eq. 6.3)

veh/day


Problem 6.28

What is the density o f the traffic stream?

V := 2000 veh

V15 := 600

BFFS := 70

f lw := 6.6

PT := 0.12

mi

f lc := 0.8

h

ET := 2.5

mi h

f N := 3.0

)

(

FFS = 52.1

)

f HV = 0.806

v p = 992

PHF⋅ N⋅ f HV⋅ f p

S

h

PHF = 0.833

600⋅ 4

V

v p

f id := 7.5

mi h

(Tables: 6.3, 6.4, 6.5, 6.6, respectively)

(Eq 6.2)

(Table 6.7)

V

for vp = 992,

D :=

mi

ER := 2.0

(

(given)

(assumed)

1 f HV := 1 + PT⋅ ET − 1 + PR ⋅ ER − 1

v p :=

N := 3

f p := 1.0

FFS := BFFS − flw − f lc − f N − f id

PHF :=

PR := 0.06

(Eq 6.5)

(Eq 6.4)

(Eq 6.3)

S := FFS

D = 19.04

pc/mi/ln

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answers 1) using BFFS instead of FFS (with S = BFFS) S := BFFS v p

D :=

D = 14.171

S

2) not adjusting for PHF PHF := 1.0 V v p := PHF⋅ N⋅ f HV⋅ f p

v p = 826.667

v p

D :=

D = 15.867

52.1

3) use level terrain PCE values from Table 6.7 instead of rolling terrain ET := 1.5 f HV :=

1

(

V 600⋅ 4

(

)

f HV = 0.933

PHF = 0.833

V PHF⋅ N⋅ f HV⋅ f p

S := FFS

D :=

)

1 + PT⋅ ET − 1 + PR ⋅ ER − 1

PHF :=

v p :=

ER := 1.2

v p S

v p = 857.6 S = 52.1

D = 16.461


Determine is th e level of service. PHF := 0.84

BFFS := 70

V := 2500 veh

mi

Problem 6.29 PR := 0.04

N := 3 lanes

f p := 1

h

(given)

(assumed)

f lw := 6.6

(Table 6.3)

f n := 3.0 (Table 6.5)

f id := 2.5

(Table 6.6)

f lc := 0.4 (Table 6.4)

ER := 4.0

(Table 6.7)

Determine free flow speed FFS := BFFS − flw − f n − f id − f lc

FFS = 57.5

mi

(Eq 6.2)

h

Determine heavy vehicle factor f HV :=

1

(

)

1 + PR ⋅ ER − 1

f HV = 0.89

(Eq 6.5)

v p = 1111.11 pc/h/ln

(Eq 6.3)

Determine the flow rate V v p := PHF⋅ N⋅ f HV⋅ f p

Interpolation of Table 6.1 Interpolation of Table 6.1 with a vp of 1111.11 pc/h/ln yields a level of service of C

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

Alternative Answers: 1) LOS D 2) LOS A 3) LOS B


Determine the p eak-hour volume. PT := .06 + .04

Problem 6.30

PR := .03

PHF := 0.81

f p := 0.94

N := 2 lanes

v p := 1250 pc/h/ln

BFFS := 60.0

ER := 2.0

mi h

ET := 2.5

(Table 6.7)

f lc := 1.3

(Table 6.13, with TLC = 4 + 2 = 6 from Eq 6.8)

(Table 6.7)

(given)

f lw := 1.9

(Table 6.3) f M := 0.0

(Table 6.14)

f A := 1.25 (Table 6.15, by interpolation)

Determine free flow speed FFS := BFFS − flc − f lw − f M − f A

FFS = 56

mi

(Eq 6.7)

h

Determine heavy-vehicle factor f HV :=

1

(

)

(

)

1 + PT⋅ ET − 1 + PR ⋅ ER − 1

f HV = 0.847

(Eq 6.5)

Determine peak-hour volume V := v p ⋅ PHF⋅ N ⋅ f HV⋅ f p

V = 1613.14

veh

(Eq 6.3, rearranged)

h

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answers: 1) Reverse V and vp V := 1250

veh

v p :=

h

V PHF⋅ N⋅ f HV⋅ f p

v p = 969

veh

V = 1329.79

veh

(different units, supposed to be pc/h/ln)

h

2) Use f p of 1.0 f p := 1.0

V := v p⋅ PHF⋅ N ⋅ f HV⋅ f p

h

3) switch bus/truck and recreational vehicle percentages PR := 0.10 f HV :=

PT := 0.03

f p := 0.94

1

(

)

(

)

1 + PT⋅ ET − 1 + PR ⋅ ER − 1

V := v p⋅ PHF⋅ N ⋅ f HV⋅ f p

f HV = 1

V = 1288.21

veh h


Estimate the free flow speed (in mi/h).

N := 3 lanes

PostedSpeed := 50

BFFS := PostedSpeed + 5

Problem 6.31

mi

(given)

h

BFFS = 55

look up adjustments TLC := 6 + 2

(LCL = 6 ft since undivided highway)

mi f LW := 0.0 h f M := 1.6

mi

(Table 6.3)

(Table 6.14)

h

f LC := 0.9

mi

(Eq 6.8)

(Table 6.13)

h

f A := 0.5 (Table 6.15, with interpolation)

calculate free-flow speed FFS := BFFS − fLW − f LC − f M − f A

FFS = 52

mi

(Eq 6.7)

h

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


A six-lane undivided multilane highway (three lanes in each direction) has 12-ft lanes with 2-ft shoulders on the right side. Estimate the free flow speed (in mi/h).

N := 3 lanes

PostedSpeed := 50

BFFS := PostedSpeed + 5

mi

(given)

h

BFFS = 55

look up adjustments TLC := 6 + 2 f LW := 0.0 f M := 1.6

mi

(Table 6.3)

h mi

(Eq 6.8)

(LCL = 6 ft since undivided highway)

(Table 6.14)

h

f LC := 0.9

mi

(Table 6.13)

h

f A := 0.5 (Table 6.15, with interpolation)

calculate free-flow speed FFS := BFFS − fLW − f LC − f M − f A

FFS = 52

mi

(Eq 6.7)

h

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers 1) Use freeway free-flow speed equation f LC := 1.6

mi h

f N := 3.0

mi h


f ID := 7.5

mi h

FFS = 43

mi

(Eq 6.2)

h

2) Use TLC = 2-ft f LC := 2.8

mi h

FFS := BFFS − fLW − f LC − f M − f A

FFS = 50

3) Misinterpolation of Table 6.15 mi f A := 2.5 h FFS := BFFS − fLW − f LC − f M − f A

FFS = 48

mi h

mi h


Determine the analysis flow rate, based on average travel speed. PHF := 0.92

V := 550

mi

PT := .07 + .03

h

PR := .07


determine heavy vehicle adjustment factor f G := 0.93

(Table 6.17; >600-1200)

ET := 1.9

(Table 6.18; >600-1200)

ER := 1.1

(Table 6.18; >600-1200)

1 f HV := 1 + PT⋅ ET − 1 + PR ⋅ ER − 1

(

)

(

)

f HV = 0.912

(Eq 6.5)

determine 15-min passenger car equivalent flow rate v p :=

V PHF⋅ f G⋅ f HV

v p = 705

pc/h

(Eq 6.11)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answers 1) Misuse Tables 6.17 and 6.18 f G := 0.71

(Table 6.17; 0-600)

ET := 2.5

(Table 6.18; 0-600)

ER := 1.1

(Table 6.18; 0-600)

f HV :=

1

(

)

(

f HV = 0.864

)

1 + PT⋅ ET − 1 + PR ⋅ ER − 1

V v p := PHF⋅ f G⋅ f HV

v p = 974

pc/h/ln

2) Don't include percentage of buses PT := 0.07 f G := 0.93

(Table 6.17; >600-1200)

ET := 1.9

(Table 6.18; >600-1200)

ER := 1.1

(Table 6.18; >600-1200)

1 f HV := 1 + PT⋅ ET − 1 + PR ⋅ ER − 1

(

)

V v p := PHF⋅ f G⋅ f HV

(

f HV = 0.935

)

v p = 688

pc/h

3) Use Equation 6.3 for flow rate N := 2

f p := 1

v p :=

V PHF⋅ N ⋅ f HV⋅ f p

f HV := 0.912 v p = 328

pc/h


Determine the number of lanes required to pr ovide at least LOS D. PHF := 0.88

D := 0.65

f p := 1.0

f HV := 1.0

FFS := 70

mi h

AADT := 75000


(assumed) (Figure 6.7)

K 60 := 0.11

Calculate directional design-hour volume DDHV:= K60 ⋅ D⋅ AADT

(Eq 6.16)

From Table 6.1, the maximum service flow rate for LOS D at 70 mi/h is 2150 pc/h/ln.

Assume a four-lane freeway. Use Eq 6.3 N := 3 lanes v p :=

DDHV PHF⋅ N⋅ f p ⋅ f HV

v p = 2031.25

pc/h/ln

(Eq 6.3)

This value of vp is less than 2150 pc/h/ln, so six lanes (both directions) is sufficient.

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 4 lanes, 8 lanes, 10 lanes




Chapter 7 Traffic Control and Analysis at Signalized Intersections U.S. Customary Units



•


•


•




www.mathcad.com


Problems 7.1 Determine the maximum length of the effective red.

λ := ρ :=

800

λ = 0.222

3600

λ μ

ρ = 0.533

C := 60

μ :=

1500 3600

μ = 0.417

(given) (Eq. 5.27)

(given)

at queue clearing r + t0

+

10

C

t0

C − r − 10

t0

ρ ⋅ r (1 − ρ )

C − r − 10

r

= 23.3

so

(Eq. 7.7)

ρ ⋅ r (1 − ρ ) sec



Problem 7.3 Determine the average vehicle delay. C := 60

g := 25

(given)

r := C − g

r = 35

(Eq. 7.5)

s :=

λ :=

1400

s

3600 500

λ = 0.139

3600

μ := s ρ :=

= 0.389

(given)

μ = 0.389

λ μ

ρ = 0.357

(Eq. 5.27)

2

d :=

r

2 ⋅ C⋅ (1

d

− ρ)

= 15.88

s/veh

(Eq. 7.12)

Problem 7.4 Determine queue dissipation time. r := 40

λ := μ :=

Q r 1440 3600

ρ :=

λ μ

tc :=

ρ ⋅ r (1 − ρ )

Q := 8

(Given)

λ = 0.2

given vehicles over a given time period

μ = 0.4

(Given)

ρ = 0.5

tc

= 40

(Eq. 5.27)

sec

(Eq. 7.7)


Problem 7.5

Determine the arrival rate. calculate the effective red time

μ :=

1500

μ = 0.416667

3600

C := 60

(Given)

D := 5.78 ⋅ 60

D = 346.8

for delay, a graph of the queue gives the area between the arrival and departure curves as C⋅ (9

Area1

− 4 + λ ⋅ g)

triangle above rectangle for full cycle

2 rectangle for full cycle

C⋅ 4

Area2

g⋅(λ ⋅C + 4

Area3

− 2)

triangle under departure function for green time

2

Area1 + Area2 + Area3

D

the λs will cancel out, giving C⋅ ( 9

D

−4+

g)

2

+ C⋅ 4 −

g ⋅ ( C + 4 − 2) 2

plug in variables to solve for g

g

= 43.2

r := C − g

r

= 16.8

sec

(Eq 7.5)

calculate the arrival rate by inspection of the queue diagram

λ ⋅ r

9

−4

λ = 0.298

veh/s

λ ⋅ 3600 = 1071.4

veh/h




Problem 7.7 Determine the arrival rate.

μ :=

3600

(given)

μ=1

3600

C := 60 at queue dissipation

λ ⋅ ( C − 8) r

μ ⋅ ( C − 8 − r )

13

λ

substituting and solving for λ

⎛ C − 8 − 13 ⎞ λ ⎠ ⎝

λ ⋅ ( C − 8)

μ⋅

λ = 0.5

λ ⋅ 3600 = 1800

r :=

13

λ

g := C − r

veh/h

r = 26

g = 34


Problem 7.8 Determine the t otal vehicl e delay. find the arrival rate

μ :=

1800

μ = 0.5

3600

C := 80

veh/s (given)

s

arrival rate during eff. green = 2λ (arrival rate during effective red)

Δ q := 7.9 − 2

difference between queue at beginning and end of effective red

Δ q = 5.9 by inspection of a queuing diagram, we find that equating vertical distances gives 2

+ λ ⋅ ( C − g) + 2λ ⋅ g μg

(1)

(note that r = C - g )

and

λ ⋅ ( C − g)

Δq

or g

C−

Δq λ

(2)

with two equations and two unknowns (λ, g), substituting (2) into (1) gives 2

+ λ⋅

⎡ ⎛ Δ q ⎞⎤ ⎛ Δ q ⎞ ⎛ Δ q ⎞ C− C− + 2 λ ⋅ ⋅ C− μ⋅ C − ⎣ ⎝ λ ⎠⎦ ⎝ λ ⎠ ⎝ λ ⎠

solving for λ yields two solutions

λ1 = 0.118

veh/s

λ2 = 0.157

veh/s

both λ's are feasible, so we must find two delays


determine total delay based on first arrival rate g1 := C −

Δq

g1

λ1

= 29.8

r 1 := C − g1 r 1

(Eq 7.5)

= 50.2

D1 :=

(2

+

7.9) ⋅ r 1 2

+

7.9 ⋅ g1 2

D1

= 366.2

veh-sec

determine total delay based on second arrival rate

g2 := C −

Δq

g2

λ2

= 42.4

r 2 := C − g2 r 2

(Eq 7.5)

= 37.6

D2 :=

(2

+

7.9) ⋅ r 2 2

+

7.9 ⋅ g2 2

D2

= 353.6

veh-sec


Problem 7.8 Alternate Solution with 8.0 vehicles in the queue. Determine the t otal vehicl e delay. find the arrival rate

μ :=

1800

μ = 0.5

3600

C := 80

veh/s (given)

s

arrival rate during eff. green = 2λ (arrival rate during effective red)

Δ q := 8 − 2

difference between queue at beginning and end of effective red

Δq = 6 by inspection of a queuing diagram, we find that equating vertical distances gives 2

+ λ ⋅ ( C − g) + 2λ ⋅ g μg

(1)

(note that r = C - g )

and

λ ⋅ ( C − g)

Δq

or g

C−

Δq λ

(2)

with two equations and two unknowns (λ, g), substituting (2) into (1) gives

⎡ ⎛ Δ q ⎞⎤ ⎛ Δ q ⎞ ⎛ Δ q ⎞ + 2 λ ⋅ ⋅ C− μ⋅ C − 2 + λ⋅ C − C − ⎣ ⎝ λ ⎠⎦ ⎝ λ ⎠ ⎝ λ ⎠ solving for λ yields two solutions

λ1 = 0.125

veh/s

λ2 = 0.15

veh/s

both λ's are feasible, so we must find two delays


determine total delay based on first arrival rate g1 := C −

Δq

g1

λ1

= 32

r 1 := C − g1 r 1

(Eq 7.5)

= 48

D1 :=

(2

+

8) ⋅ r 1 2

+

8 ⋅ g1

D1

2

= 368

veh-sec

determine total delay based on second arrival rate

g2 := C −

Δq

g2

λ2

= 40

r 2 := C − g2 r 2

(Eq 7.5)

= 40

D2 :=

(2

+

8) ⋅ r 2 2

+

8 ⋅ g2 2

D2

= 360

veh-sec


Problem 7.9 Determine the t otal vehicl e delay. s

s := 1800

μ :=

C := 60

r := 25

g := C − r 1 Dt := ⋅ [ 6 2

g

+

(6

3600

= 35

μ = 0.5

λ :=

900 3600

λ = 0.25

(given)

(Eq. 7.5)

sec

+ λ ⋅ 60) ] ⋅ 60 −

1 2

⋅ ( g ⋅ μ ) ⋅ g

Dt

= 503.75

veh-sec


Problem 7.10 Determine the total delay. 1064

λ :=

λ = 0.296

3600

s := 2640

veh/s

μ :=

veh/h

s

μ = 0.733

3600

veh/s

(given)

solving for t 3

+ λ ⋅t

t :=

10

7

t

λ

= 23.7

sec

solving for effective red r := t

+

r = 31.7

8

sec

g := 15

(given)

C := r + g

C = 46.7

(Eq. 7.5) b2 := g

b1 := C h1 := [ 3

+

(3

+ λ ⋅ C) ]

1 Dt := ⋅ b1 ⋅ h1 2

1 Dt := ⋅ [ 3 2

+

−

(3

1 2

h2 := g ⋅ μ

⋅ b2 ⋅ h2

+ λ ⋅ C) ] ⋅ C −

Dt

1 2

= 379.6

⋅ ( g ⋅ μ ) ⋅ g

veh-sec

Dt

= 379.6


Problem 7.11 Determine the t otal vehicl e delay. find the arrival rate 1800

μ :=

μ = 0.5

3600

C := 80

veh/s

(given)

s

6

λ

r C − r

g

(Eq. 7.5)

substituting and solving for r ( 10 +

λ ⋅ g) − μ ⋅ g

[ 10 +

λ ⋅ ( C − r ) ] − μ ⋅ ( C − r )

2

⎡ 10 + 6 ⋅ ( C − r )⎤ − μ ⋅ (C − r ) r ⎣ ⎦

8

+

480

480 r

− 6 − 40 +

+

0.5 ⋅ r 38

0.5 ⋅ r

2

− 38⋅ r +

r = 16

λ :=

r

2

2

0.5 ⋅ r 0

480

0

6

λ = 0.375

r

find total delay 1 Dt := ⋅ ( 4 2 Dt

+

10) ⋅ r +

= 496

1 2

⋅ [ 10 +

[ 4 + ( 0.375⋅ C) ] ] ⋅ ( C − r )

−

1 2

⋅ [ μ ⋅ ( C − r ) ] ⋅ ( C − r )

veh − sec

Alternative Solution

⌠ C 4⋅ C + ⎮ ⌡0

⌠ C−r 0.375⋅ t dt − ⎮ 0.5 ⋅ t dt = 496.006 ⌡0

veh − sec


Problem 7.12 Determine effective green and red ti mes and the t otal vehicle delay. C := 60

s (given)

λ ( t) := 0.22 + Arrivals( t) :=

Arrivals( C)

.012 ⋅ t

⌠ ⎮ λ ( t) dt → .220 ⋅ t + ⎮ ⌡

= 34.8

g :=

⋅t

veh

λC= μg

to clear at the end of the cycle

μ :=

-3 2

6.00⋅ 10

3600

μ=1

3600 Arrivals( C)

μ

r := C − g

veh/s

(given)

g

= 34.8

sec

r

= 25.2

sec

(Eq. 7.5)

determine total delay

Delay :=

⎛ ⌠ 60 ⎮ ⌡0

⎝

⎞ Arrivals( t) dt

⎠

−

1 2

g ⋅ ( μ ⋅ g)

Delay = 222.5

veh − sec


Problem 7.13 Determine the t otal vehicl e delay.

λ1 := λ2 := μ :=

800 3600 500 3600

1200 3600

λ1 = 0.222

λ1 ⋅ 40 = 8.89

λ2 = 0.139

λ2 ⋅ ( 120 − 40) = 11.11

μ = 0.333

μ ⋅ 20 = 6.667

(given)

By inspection of the queuing diagram, (with 8.89 vehicles arriving and 6.67 departing at the end of the first cycle and 14.45 and 13.34 arriving and departing at the end of the second cycle) we find that the first cycle delay is (using triangles)

D1 :=

⎛ 40⋅ 8.89 − ⎝ 2

20⋅ 6.67 ⎞ 2

D1

⎠

= 111.1

veh − sec

similarly, second cycle delay is:

D2 := 40⋅ ( 2.23)

+

( 14.45 − 8.89) ⋅ 40 2

−

( 13.34 − 6.67) ⋅ 20 2

D2 = 133.7

veh − sec

Therefore, total delay 2 cycles later is 111.1 + 133.7 = 244.8

veh − sec




Problem 7.16 Determine the sum of crit ical flow ratios. NBL := SBL :=

330 365

SBL

1750 1125

NBTR :=

3400 1075

SBTR :=

EBL :=

NBL = 0.194

1700

3300

110

WBL :=

80

EBTR :=

WBTR :=

250 1750 285 1800

Σvs := SBL +

phase 2 critical

SBTR = 0.326

= 0.169

WBL

600

phase 1 critical

NBTR = 0.331

EBL

650

= 0.209

phase 3 critical

= 0.133

EBTR = 0.143

WBTR = 0.158

NBTR + EBL

Σvs = 0.709


Problem 7.18 Determine X c. C := 95

(given)

L := 5 ⋅ 4

L

Σvs := 0.235 +

= 20 0.250 + 0.170 + 0.125

(given)

Σvs = 0.78

C

Xc

L ⋅ Xc Xc

− Σvs

(Eq. 7.20)

= 0.988



Problem 7.20 Determine the minim um cy cle length and p hase effective green times. calculate cycle length flow ratios - 0.225, 0.175, 0.200, 0.150 L := 5 ⋅ 4

lost time per phase - 5 sec

(given) L

= 20

sec

Xc := 0.85

(given)

Σvs := 0.225 +

Cmin :=

0.175 + 0.200 + 0.150

L ⋅ Xc Xc

Cmin

− Σvs

= 170

Σvs = 0.75

(Eq. 7.20)

sec

calculate green times for each phase

g1 := ( 0.225) ⋅

g2 := ( 0.175) ⋅

g3 := ( 0.200) ⋅

g4 := ( 0.150) ⋅ g1

+

⎛ Cmin ⎞ Xc

⎝ ⎠ ⎛ Cmin ⎞ Xc

⎝ ⎠ ⎛ Cmin ⎞ Xc

⎝ ⎠ ⎛ Cmin ⎞ ⎝

Xc

⎠

g2 + g3 + g4 + L

= 170

g1

= 45

g2

= 35

g3

= 40

sec

(Eq. 7.22)

g4

= 30

sec

(Eq. 7.22)

sec

sec

sec

(Eq. 7.22)

(Eq. 7.22)

checks


Problem 7.21 Determine the minimum cycle length and effective green times. calculate the minimum cycle length SBL :=

365

NBTR := EBL :=

SBL

1750 1125

= 0.209

NBTR = 0.331

3400

110

EBL

650

= 0.169

Σvs := SBL +

NBTR + EBL

L := 4 ⋅ 3

= 12

L

(given)

Σvs = 0.709 (given)

L ⋅ 0.9 Cmin := 0.9 − Σvs Cmin

= 56.451

C := 60

(Eq. 7.20)

rounds to 60 sec

sec

calculate the effective green times

Σvs⋅ C Xc := C−L

Xc

= 0.886

⎛ C ⎞ g1 := SBL ⋅ ⎝ Xc ⎠

g1

= 14.1

sec

⎛ C ⎞ g2 := NBTR⋅ ⎝ Xc ⎠

g2

= 22.4

sec

⎛ C ⎞ g3 := EBL ⋅ ⎝ Xc ⎠

g3

= 11.5

sec

g1

+

g2 + g3 + L

= 60

sec

(Eq. 7.20)

checks


Problem 7.22 Determine the northbo und approach delay and level of service. calculate northbound left delay C := 60

gL := 14.127 NBL

XL :=

sL ⋅

s L := 1700

XL = 0.824

⎛ gL ⎞

(given)

(Eq. 7.22)

⎝ C ⎠

⎛ gL ⎞ 0.5 ⋅ C⋅ 1 − ⎝ C ⎠ d1L := ⎛ gL ⎞ 1 − XL ⋅ ⎝ C ⎠ T := 0.25

c L := sL ⋅

NBL := 330

2

k := 0.5 gL

(Eq. 7.15)

I := 1.0

(assumed)

c L = 400.265

C

d2L := 900⋅ T⋅

PF := 1.0

d1L = 21.76

⎡ ⎣

(Eq. 7.6)

(XL − 1) + (XL − 1)

2

+

8 ⋅ k ⋅ I⋅ XL ⎤ cL ⋅ T

⎦

d3L := 0

dL := d1L ⋅ PF

+

d2L = 17.322

(Eq. 7.16)

(assumed)

d2L + d3L

= 39.082

dL

(Eq. 7.14)

calculate northbound through delay gTR := 22.411 XTR :=

NBTR := 1125

NBTR

⎛ gTR ⎞ s TR⋅ ⎝ C ⎠

⎛ gTR ⎞ 0.5 ⋅ C⋅ 1 − C ⎠ ⎝ d1TR := ⎛ gTR ⎞ 1 − XTR⋅ C ⎝

s TR := 3400

(given)

XTR = 0.886

2

d1TR = 17.597


c TR := s TR⋅

gTR

T := 0.25

k := 0.5

d2TR := 900⋅ T⋅ PF := 1.0

cTR = 1269.957

C

⎡ ⎣

I := 1.0

(XTR − 1) + (XTR − 1)

2

+

8 ⋅ k ⋅ I⋅ XTR ⎤ c TR⋅ T

⎦

d2TR = 9.312

d3TR := 0

dTR := d1TR⋅ PF

+

d2TR + d3TR

dTR = 26.909

calculate northbound total delay

d :=

NBL ⋅ dL

+

NBTR⋅ dTR

NBL + NBTR

d = 29.7

sec/veh

LOS is C

(Table 7.4)


Problem 7.23 Determine the southbo und approach delay and level of service. calculate southbound left delay C := 60

gL := 14.127 SBL

XL :=

sL ⋅

XL

⎛ gL ⎞

s L := 1750

(given)

= 0.886

(Eq. 7.22)

⎝ C ⎠

⎛ gL ⎞ 0.5 ⋅ C⋅ 1 − ⎝ C ⎠ d1L := ⎛ gL ⎞ 1 − XL ⋅ ⎝ C ⎠ c L := sL ⋅

SBL := 365

gL

T := 0.25

cL

C

d1L = 22.158

⎡ ⎣

(Eq. 7.15)

= 412.038

k := 0.5

d2L := 900⋅ T⋅ PF := 1.0

2

(Eq. 7.6)

I := 1.0

(assumed)

(XL − 1) + (XL − 1)

2

+

8 ⋅ k ⋅ I⋅ XL ⎤ cL ⋅ T

⎦

d3L := 0

dL := d1L ⋅ PF

+

d2L = 23.316

(Eq. 7.16)

(assumed)

d2L + d3L

dL

= 45.474

(Eq. 7.14)

calculate southbound through and right delay gTR := 22.411 XTR :=

SBTR := 1075

s TR := 3300

(given)

SBTR XTR = 0.872

⎛ gTR ⎞ s TR⋅ ⎝ C ⎠


2

d1TR = 17.463


c TR := s TR⋅

gTR

T := 0.25

k := 0.5

d2TR := 900⋅ T⋅ PF := 1.0

cTR = 1232.605

C

⎡ ⎣

I := 1.0

(XTR − 1) + (XTR − 1)

2

+

8 ⋅ k ⋅ I⋅ XTR ⎤ c TR⋅ T

⎦

d2TR = 8.658

d3TR := 0

dTR := d1TR⋅ PF

+

d2TR + d3TR

dTR = 26.121

calculate southbound total approach delay

d :=

SBL ⋅ dL SBL

+ +

SBTR⋅ dTR SBTR

d

= 31

sec/veh

LOS = C

(Table 7.4)


Problem 7.24 Determine the westbound approach delay and level of service. calculate westbound left delay C := 60

XL :=

gL := 11.462

WBL := 80

WBL

XL

⎛ gL ⎞ sL ⋅ ⎝ C ⎠

⎛ gL ⎞ 0.5 ⋅ C⋅ 1 − ⎝ C ⎠ d1L := ⎛ gL ⎞ 1 − XL ⋅ ⎝ C ⎠ c L := s L ⋅

gL

T := 0.25

⎡

(

⎣

PF := 1.0

= 0.698

(Eq. 7.22)

d1L = 22.653

(Eq. 7.6)

I := 1.0

(assumed)

− 1) +

(

XL

2

− 1) +

8 ⋅ k ⋅ I⋅ XL⎤ cL ⋅ T

⎦

d3L := 0

dL := d1L ⋅ PF

+

(Eq. 7.15)

= 114.62

k := 0.5

d2L := 900⋅ T⋅ XL

(given)

2

cL

C

sL := 600

d2L = 29.769

(Eq. 7.16)

(assumed)

d2L + d3L

dL

= 52.422

(Eq. 7.14)

calculate westbound through and right delay gTR := 11.462 XTR :=

WBTR := 285

sTR := 1800

(given)

WBTR XTR = 0.829

⎛ gTR ⎞ s TR⋅ ⎝ C ⎠

⎛ gTR ⎞ 0.5 ⋅ C⋅ 1 − C ⎠ ⎝ d1TR := ⎛ gTR ⎞ 1 − XTR⋅ C ⎠ ⎝

2

d1TR = 23.326


c TR := s TR⋅ T := 0.25

gTR

c TR = 343.86

C k := 0.5

⎡

(

I := 1.0

) + (XTR − 1)

d2TR := 900⋅ T⋅ XTR − 1

⎣

PF := 1.0

2

+

8 ⋅ k ⋅ I⋅ XTR⎤ c TR⋅ T

⎦

d2TR = 20.101

d3TR := 0

dTR := d1TR⋅ PF

+

d2TR + d3TR

dTR = 43.427

calculate westbound total approach delay

d :=

WBL ⋅ dL WBL

+ +

WBTR ⋅ dTR WBTR

d

= 45.4

sec/veh

LOS is D

(Table 7.4)


Problem 7.25 Determine the eastbound approach delay and level of service. calculate eastbound left delay C := 60

gL := 11.462 EBL

XL :=

sL ⋅

XL

⎛ gL ⎞

sL := 650

(given)

= 0.886

(Eq. 7.22)

⎝ C ⎠


EBL := 110

gL

T := 0.25

cL

C

(

d2L := 900⋅ T⋅ XL

⎣

d1L = 23.632

(Eq. 7.15)

= 124.172

k := 0.5

⎡

PF := 1.0

2

− 1) +

(Eq. 7.6)

I := 1.0

(assumed)

(XL − 1)

2

+

8 ⋅ k ⋅ I⋅ XL⎤ cL ⋅ T

⎦

d3L := 0

dL := d1L ⋅ PF

+

d2L = 54.559

(Eq. 7.16)

(assumed)

d2L + d3L

dL

= 78.191

(Eq. 7.14)

calculate eastbound through and right delay gTR := 11.462 XTR :=

EBTR := 250

sTR := 1750

(given)

EBTR XTR = 0.748

⎛ gTR ⎞ s TR⋅ ⎝ C ⎠


2

d1TR = 22.905


c TR := s TR⋅

gTR

T := 0.25

cTR = 334.308

C k := 0.5

⎡

(

I := 1.0

) + (XTR − 1)

d2TR := 900⋅ T⋅ XTR − 1

⎣

PF := 1.0

2

+

8 ⋅ k ⋅ I⋅ XTR⎤ c TR⋅ T

⎦

d2TR = 14.191

d3TR := 0

dTR := d1TR⋅ PF

+

d2TR + d3TR

dTR = 37.096

calculate eastbound total approach delay

d :=

EBL ⋅ dL EBL

+ +

EBTR⋅ dTR EBTR

d

= 49.65

sec/veh

LOS is D

(Table 7.4)


Problem 7.26 Determine the overall i ntersection d elay and LOS. Phase 1

Phase 2

Phase 3

NBL := 330

NBTR := 1125

EBL := 110

SBL := 365

SBTR := 1075

WBL := 80 EBTR := 250 WBTR := 285

dNB := 29.7

dSB := 31.0

dWB

:= 45.4

dEB := 49.65

(from previous questions)

Using Eq. 7.28:

dtotal :=

dtotal

( NBL + NBTR) ⋅ dNB

= 33.8

+

( SBL

( NBL + NBTR) sec/veh

+

+

SBTR) ⋅ dSB

( SBL

+

LOS is C

SBTR)

+

( WBL

+

WBTR) ⋅ dWB

+

( WBL

+

WBTR)

+

+

( EBL

( EBL

+

+

EBTR) ⋅ dEB

EBTR)

(Table 7.4)



Problem 7.28 Determine the minim um cy cle length and effective green times fo r each phase. find minimum cycle length EBL :=

245

WBTR :=

SBL :=

NBL :=

EBL

1750 1030

= 0.14

WBTR = 0.303

3400

255

SBL

1725 225

= 0.148

NBL = 0.132

1700

Σvs := EBL +

WBTR + SBL

L := 4 ⋅ 4

L

Cmin :=

Cmin

= 16

+

Σvs = 0.723

NBL

Xc := 0.95

L ⋅ Xc Xc

(Eq. 7.20)

− Σvs

= 66.996

rounds to 70 seconds

C := 70

sec

find effective green times

Σvs⋅ C Xc := C−L

Xc

C g1 := EBL ⋅ Xc

g1 = 10.5

sec

C g2 := WBTR⋅ Xc

g2 = 22.6

sec

C g3 := SBL ⋅ Xc

g3 = 11

sec

C g4 := NBL ⋅ Xc

g4 = 9.9

sec

g1

+

g2 + g3 + g4 + L

= 0.937

= 70

checks


Problem 7.29 Determine the northbo und approach delay and level of service. calculate northbound left delay C := 70

gL := 9.884 NBL

XL :=

sL ⋅

NBL := 225

(given)

XL = 0.937

⎛ gL ⎞

(Eq. 7.22)

⎝ C ⎠


s L := 1700

gL

cL

C

T := 0.25

2

d1L = 29.752

= 240.04

k := 0.5

d2L := 900⋅ T⋅ PF := 1.0

⎡ ⎣

(Eq. 7.15)

(Eq. 7.6)

I := 1.0

(given)

(XL − 1) + (XL − 1)

2

+

8 ⋅ k ⋅ I⋅ XL ⎤ cL ⋅ T

⎦

d3L := 0

dL := d1L ⋅ PF

+

d2L = 43.883

(Eq. 7.16)

(given)

d2L + d3L

dL = 73.634

calculate northbound through and right delay gTR := 9.884 XTR :=

NBTR := 215

s TR := 1750

(given)

NBTR XTR = 0.87

⎛ gTR ⎞ s TR⋅ ⎝ C ⎠

⎛ gTR ⎞ 0.5 ⋅ C⋅ 1 − C ⎠ ⎝ d1TR := ⎛ gTR ⎞ 1 − XTR⋅ C ⎠ ⎝ c TR := s TR⋅

gTR C

2

d1TR = 29.429

c TR = 247.1


T := 0.25

k := 0.5

d2TR := 900⋅ T⋅

PF := 1.0

⎡ ⎣

(

I := 1.0

)+

XTR − 1

(

)

XTR − 1

2

+

8 ⋅ k ⋅ I⋅ XTR ⎤ c TR⋅ T

⎦

d2TR = 31.652

d3TR := 0

dTR := d1TR⋅ PF

+

d2TR + d3TR

dTR = 61.082

calculate northbound total approach delay

d :=

NBL ⋅ dL

+

NBTR⋅ dTR

NBL + NBTR

d

= 67.5

sec/veh

LOS is E

(Table 7.4)


Problem 7.30 Determine the southbo und approach delay and level of service. calculate eastbound left delay C := 70

gL := 11.0

SBL := 255

SBL

XL :=

sL ⋅

XL

⎛ gL ⎞

= 0.9

(given)

(Eq. 7.22)

⎝ C ⎠


s L := 1725

gL

cL

C

T := 0.25

2

(

d2L := 900⋅ T⋅ XL

⎣

PF := 1.0

(Eq. 7.15)

= 271.1

k := 0.5

⎡

d1L = 29.2

− 1) +

(Eq. 7.6)

I := 1.0

(assumed)

(XL − 1)

2

+

8 ⋅ k ⋅ I⋅ XL⎤ cL ⋅ T

⎦

d3L := 0

dL := d1L ⋅ PF

+

d2L = 41.3

(Eq. 7.16)

(assumed)

d2L + d3L

dL

= 70.5

(Eq. 7.14)

calculate southbound through/right delay gTR := 11.0

XTR :=

SBTR := 235

SBTR

⎛ gTR ⎞ s TR⋅ ⎝ C ⎠


s TR := 1750

XTR = 0.9

(given)

(Eq. 7.22)

2

d1TR = 28.7

(Eq. 7.15)


c TR := s TR⋅

gTR

T := 0.25

C k := 0.5

⎡

(

(assumed)

I := 1.0

) + (XTR − 1)

d2TR := 900⋅ T⋅ XTR − 1

⎣

PF := 1.0

(Eq. 7.6)

cTR = 275

2

+

8 ⋅ k ⋅ I⋅ XTR⎤ c TR⋅ T

⎦

d2TR = 27.2

(assumed)

d3TR := 0

dTR := d1TR⋅ PF

+

(Eq. 7.16)

d2TR + d3TR

dTR = 55.9

(Eq. 7.14)

calculate southbound total approach delay

d :=

SBL ⋅ dL SBL

+ +

SBTR⋅ dTR SBTR

d

= 63.5

sec/veh

LOS is E

(Table 7.4)


Problem 7.31 Determine the westbound approach delay and level of service. calculate westbound left delay C := 70 XL :=

gL := 10.455

WBL := 230

s L := 1725

(given)

WBL XL = 0.893

⎛ gL ⎞ sL ⋅ ⎝ C ⎠


gL

T := 0.25

2

d1L = 29.222

(Eq. 7.15)

c L = 257.641

C k := 0.5

d2L := 900⋅ T⋅ PF := 1.0

(Eq. 7.22)

⎡ ⎣

(

XL

− 1) +

(Eq. 7.6)

I := 1.0

(

XL

(assumed) 2

− 1) +

8 ⋅ k ⋅ I⋅ XL ⎤ cL ⋅ T

⎦

d3L := 0

dL := d1L ⋅ PF

+

d2L = 34.079

(Eq. 7.16)

(assumed)

d2L + d3L

dL

= 63.301

(Eq. 7.14)

calculate westbound through and right delay gTR := 22.623 XTR :=

WBTR := 1030

WBTR

(given)

XTR = 0.937

⎛ gTR ⎞ s TR⋅ ⎝ C ⎠


s TR := 3400

2

d1TR = 23.001


c TR := s TR⋅ T := 0.25

gTR

k := 0.5

d2TR := 900⋅ T⋅ PF := 1.0

c TR = 1098.831

C

⎡ ⎣

I := 1.0

(XTR − 1) + (XTR − 1)

2

+

8 ⋅ k ⋅ I⋅ XTR ⎤ c TR⋅ T

⎦

d2TR = 15.732

d3TR := 0

dTR := d1TR⋅ PF

+

d2TR + d3TR

dTR = 38.733

calculate westbound total approach delay

d :=

WBL ⋅ dL WBL

+ +

WBTR ⋅ dTR WBTR

d

= 43.2

sec/veh

LOS is D

(Table 7.4)




Problem 7.33 Determine the overall intersection d elay and level of service. calculate overall intersection delay by a weighted average of approach delays EBL := 245

WBL := 230

SBL := 255

NBL := 225

EBTR := 975

WBTR := 1030

SBTR := 235

NBTR := 215

dNB := 67.50

dSB := 62.86

dWB

:= 43.22

dEB := 41.85

Using Eq. 7.28:

dtotal :=

dtotal

( NBL + NBTR) ⋅ dNB

= 48.7

+

( SBL

( NBL + NBTR) sec/veh

+

+

SBTR) ⋅ dSB

( SBL

LOS is D

+

SBTR)

+

( WBL

+

WBTR) ⋅ dWB

+

( WBL

+

WBTR)

+

+

( EBL

( EBL

+

+

EBTR) ⋅ dEB

EBTR) (Table 7.4)


Problem 7.34

Determine the optimal cycle length and correspond ing effective green times.

EBL :=

245

1030

WBTR :=

SBL :=

NBL :=

EBL

1750

WBTR = 0.303

3400

255

SBL

1725 225

(from Prob. 7.17)

= 0.148

NBL = 0.132

1700

Σvs := EBL +

= 0.14

WBTR + SBL

+

Σvs = 0.723

NBL

calculate total lost time L := 4 ⋅ 4

L = 16

calculate optimal cycle length 1.5 ⋅ L + 5 Copt := 1.0 − Σ vs

Copt

= 104.7

(Eq. 7.21)

sec

round up to

Copt := 105

sec

calculate phase effective green times

g1 := EBL ⋅

⎛ Copt − L ⎞ ⎝ Σvs ⎠

g1

= 17.2

sec

(Eq. 7.22)

⎛ Copt − L ⎞ g2 := WBTR ⋅ ⎝ Σvs ⎠

g2

= 37.3

sec

(Eq. 7.22)

⎛ Copt − L ⎞ g3 := SBL ⋅ ⎝ Σvs ⎠

g3

= 18.2

sec

(Eq. 7.22)

⎛ Copt − L ⎞ ⎝ Σvs ⎠

g4

= 16.3

sec

(Eq. 7.22)

g4 := NBL ⋅ g1

+

g2

+

g3

+

g4

+

L

= 105

checks




Problem 7.36 Determine the new level of service for the nor thbound approach. Volumes increase 10% from previous values, calculate new volumes NBL := 90⋅ 1.10

NBL = 99

NBT := 340⋅ 1.10

NBT = 374

NBR := 50 ⋅ 1.10

NBR = 55

calculate northbound left delay C := 65

gL := 15.8 NBL

XL :=

sL ⋅

s L := 475

(previously calculated)

XL = 0.857

⎛ gL ⎞

(Eq. 7.22)

⎝ C ⎠

⎛ gL ⎞ 0.5 ⋅ C⋅ 1 − ⎝ C ⎠ d1L := ⎛ gL ⎞ 1 − XL ⋅ ⎝ C ⎠ T := 0.25

2

k := 0.5

d1L = 23.523

(Eq. 7.15)

I := 1.0

(assumed)

c L := 337


d2L := 900⋅ T⋅

⎡ ⎣

⎡

(XL − 1) + (XL − 1)

PF := 1.0

⎣

2

+

8 ⋅ k ⋅ I⋅ XL⎤ ⎤ cL ⋅ T

⎦⎦

d2L = 23.508

d3L := 0

dL := d1L ⋅ PF

+

d2L + d3L

(Eq. 7.16)

(assumed)

dL

= 47.031

LOS for left turn is D

(Eq. 7.14)

calculate northbound through and right delay C := 65

XTR :=

gTR := 15.8 NBT + NBR

⎛ gTR ⎞ s TR⋅ ⎝ C

sTR := 1800


XTR = 0.98


⎛ gTR ⎞ 0.5 ⋅ C⋅ 1 − C ⎠ ⎝ d1TR := ⎛ gTR ⎞ 1 − XTR⋅ C ⎠ ⎝ T := 0.25

k := 0.5

2

d1TR = 24.447

(Eq. 7.15)

I := 1.0

(assumed)

c TR := 1292

d2TR := 900⋅ T⋅


⎡ ⎣

⎡

(XTR − 1) + (XTR − 1) ⎣

2

+

8 ⋅ k ⋅ I⋅ XTR⎤ ⎤ cTR⋅ T

(Eq. 7.16)

⎦⎦

d2TR = 20.788 PF := 1.0

(assumed)

d3TR := 0

dTR := d1TR⋅ PF

+

d2TR + d3TR

dTR = 45.235

(Eq. 7.14)

LOS for through/right turn is D calculate northbound total approach delay

d

:=

NBL ⋅ dL + ( NBT + NBR) ⋅ dTR NBL + NBR + NBT

d

= 45.6

sec/veh

LOS is D

(Table 7.4)


Problem 7.37 Determine the yellow and all-red times. V

:= 30⋅

5280

V

3600

tr := 1.0

= 44

a := 10.0

g := 32.2

(assumed)

w := 60

l := 20

(given)

G := 0.08

calculate yellow time Y := tr +

V 2⋅a

Y := 3.0

+ 2⋅ g⋅G

Y

= 2.749

rounds to 3.0 seconds

(Eq. 7.23)

sec

calculate all-red time AR :=

w+ l

AR

V

AR := 2.0

= 1.818

rounds to 2.0 seconds

(Eq. 7.24)

sec





d1

⎛ g ⎞ 0.5 ⋅ C⋅ 1 − ⎝ C ⎠ :=

c := s ⋅

c

C

= 18.481

(Eq. 7.15)

C

= 1292

⎡ ⎣

d2 := 900⋅ T⋅ ( X − 1) PF := 1

d1

g

1 − X⋅ g

2

⎡( X − 1) 2 + 8 ⋅ k ⋅ I⋅ X⎤ ⎤ c ⋅T ⎦⎦ ⎣

+

d2 = 7.265

d3 := 0

dEBTR := d1 ⋅ PF

+

(Eq. 7.16)

(assumed) d2

+

d3

dEBTR = 25.746

Calculate volume-weighted delay for eastbound approach VEBL := 300

VEBTR := 1100

VEBL ⋅ dEBL

dEB :=

+

VEBL

VEBTR⋅ dEBTR

+

VEBTR

dEB

= 31.628

(Eq. 7.27)

Calculate delay for westbound left-turn lane group C := 65

g := 12.5

X := WBL ⋅

d1

C

s := 1750

X = 0.744

g

⎛ g ⎞ 0.5 ⋅ C⋅ 1 − ⎝ C ⎠ := 1

c := s ⋅

g C

− X⋅ c

d1

g

= 24.74

(Eq. 7.15)

C

= 336.538

⎡ ⎣

d2 := 900⋅ T⋅ ( X − 1) PF := 1

2

+

⎡( X − 1) 2 + 8 ⋅ k ⋅ I⋅ X⎤ ⎤ c ⋅T ⎦⎦ ⎣

d3 := 0

dWBL := d1 ⋅ PF

+

d2 = 13.849

(Eq. 7.16)

(assumed) d2

+

d3

dWBL

= 38.589


Calculate delay for westbound through/right-turn lane group g := 24.7

s := 3400

X := WBTR⋅

d1

C

X = 0.889

g

⎛ g ⎞ 0.5 ⋅ C⋅ 1 − ⎝ C ⎠ := 1

c := s ⋅

− X⋅

d1

g

= 18.872

(Eq. 7.15)

C

g

c

C

⎡ ⎣

d2 := 900⋅ T⋅ ( X − 1) PF := 1

2

= 1292

⎡(X − 1)2 + 8 ⋅ k ⋅ I⋅ X⎤ ⎤ c ⋅T ⎦⎦ ⎣

+

d2

= 9.426

d3 := 0

dWBTR := d1 ⋅ PF

+

(Eq. 7.16) (assumed)

d2

+

dWBTR = 28.297

d3

Calculate volume-weighted delay for westbound approach VWBL := 250

dWB

:=

VWBTR := 1150

VWBL ⋅ dWBL VWBL

+ +

VWBTR ⋅ dWBTR VWBTR

dWB

= 30.135

(Eq. 7.27)

Calculate delay for northbound left-turn lane group g := C 65 := 15.8 X := NBL ⋅

d1

s := 575 C

X = 0.644

g

⎛ g ⎞ 0.5 ⋅ C⋅ 1 − ⎝ C ⎠ := 1

c := s ⋅

g C

− X⋅

2

g

d1

= 22.076

(Eq. 7.15)

C

c

= 139.769


⎡ ⎣

d2 := 900⋅ T⋅ ( X − 1) PF := 1

⎡( X − 1) 2 + 8 ⋅ k ⋅ I⋅ X⎤ ⎤ c ⋅T ⎦⎦ ⎣

+

d2

= 20.632

d3 := 0

dNBL := d1 ⋅ PF

+

(Eq. 7.16) (assumed)

d2

+

dNBL = 42.708

d3

Calculate delay for northbound through/right-turn lane group s := 1800 g 15.8 C

X := NBTR⋅

d1

X = 0.891

g

⎛ g ⎞ 0.5 ⋅ C⋅ 1 − ⎝ C ⎠ := 1 − X⋅

c := s ⋅

g

c

C

d1

g

(Eq. 7.15)

= 23.771

C

= 437.538

⎡ ⎣

d2 := 900⋅ T⋅ ( X − 1) PF := 1

2

⎡( X − 1) 2 + 8 ⋅ k ⋅ I⋅ X⎤ ⎤ c ⋅T ⎦⎦ ⎣

+

d2

= 22.964

d3 := 0

dNBTR := d1 ⋅ PF

(assumed)

+

d2

+

d3

dNBTR = 46.735

Calculate volume-weighted delay for northbound approach VNBL := 90

dNB :=

(Eq. 7.16)

(Eq. 7.27)

VNBTR := 390

VNBL ⋅ dNBL + VNBTR⋅ dNBTR VNBL + VNBTR

dNB

= 45.98

Calculate delay for southbound left-turn lane group C := 65

X := SBL ⋅

g := 15.8 C g

s := 550

X = 0.524


d1

⎛ g ⎞ 0.5 ⋅ C⋅ 1 − ⎝ C ⎠ := 1 − X⋅

c := s ⋅

g

(Eq. 7.15)

C

= 133.692

⎡ ⎣

d2 := 900⋅ T⋅ ( X − 1) PF := 1.0

d1 = 21.336

g

c

C

2

⎡( X − 1) 2 + 8 ⋅ k ⋅ I⋅ X⎤ ⎤ c ⋅T ⎦⎦ ⎣

+

d2

= 13.896

d3 := 0

dSBL := d1 ⋅ PF

+

d2

(Eq. 7.16)

(assumed)

+

d3

dSBL

= 35.232

Calculate delay for southbound through/right-turn lane group g := 15.8

s := 1800

X := SBTR⋅

d1

C

X = 0.846

g

⎛ g ⎞ 0.5 ⋅ C⋅ 1 − ⎝ C ⎠ := 1 − X⋅

c := s ⋅

g

c

C

d1

g

= 23.438

(Eq. 7.15)

C

= 437.538

⎡ ⎣

d2 := 900⋅ T⋅ ( X − 1) PF := 1.0

2

⎡( X − 1) 2 + 8 ⋅ k ⋅ I⋅ X⎤ ⎤ c ⋅T ⎦⎦ ⎣

+

d2

= 17.916

d3 := 0

dSBTR := d1 ⋅ PF

+

(Eq. 7.16)

(assumed)

d2

+

d3

dSBTR = 41.355

Calculate volume-weighted delay for southbound approach VSBL := 70

dSB :=

VSBTR := 370

VSBL ⋅ dSBL VSBL

+ +

VSBTR⋅ dSBTR VSBTR

dSB

= 40.381

(Eq. 7.27)


Calculate volume-weighted delay for entire intersection

( VEBL + VEBTR) ⋅ dEB + ( VWBL + VWBTR) ⋅ dWB + ( VNBL + VNBTR) ⋅dNB + ( VSBL + VSBTR) ⋅ dS d := ( VEBL + VEBTR) + ( VWBL + VWBTR) + ( VNBL + VNBTR) + ( VSBL + VSBTR) d

= 34

LOS of the intersection is C

(Eq. 7.28) (Table 7.4)


Problem 7.40 Determine the new mini mum pedestrian gr een time. Nped := 20

W E := 6

L := 60

(given)

SP := 4.0 GP := 3.2

(assumed)

+

L SP

+

0.27 ⋅ Nped

GP

= 23.6

sec

(Eq. 7.25)



⎛ second ⎞ ⎜⎜ ⎟⎟ term ⎝ ⎠

second term (continuation of equation above)



⎛ second ⎞ ⎜⎜ ⎟⎟ term ⎝ ⎠

second term (continuation of equation above)


What is the g/C ratio fo r this approach? PF := 0.641


The arrival type is AT-5 for a highly favorable progression quality.

f p := 1.00

(Table 7.6 for AT-5)

R p := 1.667

(Table 7.7 for AT-5)

Solve for g/C using Equations 7.31 and 7.32 PF = [1 - R p*(g/C)]*(f p) / [1 - (g/C)]

1− gC := R p

−

PF f p PF

gC

= 0.35

f p


Problem 7.44

Determine the arrival type for thi s approach.

g := 35 s arrivalgreen

C := 100

:= 66

veh

(given)

arrivaltotalcycle

:= arrivalgreen +

105

(given)

Calculate g/C ratio gC :=

g

gC

C

= 0.35

Calculate PVG

PVG:=

arrivalgreen

PVG = 0.386

arrivaltotalcycle

Solve for R p usi ng Eq. 7.32

R p :=

PVG gC

R p

= 1.103

(Eq. 7.32)

Use Table 7.7 to determine arrival type

An Rp of 1.103 falls in the range for arrival type 3 (AT-3).


Problem 7.45

Determine the progression adjustm ent factor.

g := 35 s arrivalgreen

C := 100 s

:= 66

veh

arrivalcycle := arrivalgreen

(given)

+

105

veh

(given)

Calculate g/C ratio

gC :=

g

gC = 0.35

C

Calculate PVG PVG:=

arrivalgreen arrivalcycle

PVG = 0.386

Determine R p R p :=

PVG

R p

gC

= 1.103

(Eq 7.32)

Determine f p For an Rp of 1.103, the arrival rate is AT-3 by use of Table 7.7 f p := 1.00

(Table 7.6 using AT-4)

Calculate PF

PF :=

( 1 − PVG) ⋅ f p 1−

⎛ g ⎞

PF = 0.945

(Eq 7.31)

⎝ C ⎠


Determine the displayed green time for the traffic movement. r := 37 s

C := 70 s

Y := 5 s

AR

:= 2

s

t L := 4 s


Determine the effective green time for a traffic movement g := C − r

g

= 33

s

(Eq 7.5)

Calculate the displayed green time for a traffic movement G := g

− Y − AR +

G = 30 s

tL

(Eq 7.3)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Solve for effective green time g

= 33

s

2) Use all-red time in Eq 7.5 r := 2 G := g

g := C − r

− Y − AR +

tL

g

= 68

s

G = 65 s

3) Add AR time to effective red, and then subtract from cycle length r := 37 s

G := C − ( AR

AR

+

r)

:= 2 G = 31


Problem 7.47

Determine the average delay per vehicle.

μ :=

1900

veh

3600

s

R := 58 s

λ :=

tL := 4 s

550

veh

3600

s

g := 28 s

Y := 3 s

(given) AR

:= 2

s

Determine effective red time and cycle length r := R

+

tL

r

C := r + g

= 62.0

(Eq 7.4)

s

C = 90.0 s

(Eq 7.5)

Calculate traffic intensity

ρ :=

λ μ

(Eq 5.27)

ρ = 0.3

Determine average delay per vehicle 2

λ ⋅ r 1 d avg := ⋅ 2⋅ ( 1 − ρ) λ ⋅ C

d avg

= 30.1

s

(Eq 7.12)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answer: 1) Use displayed red time in Eq 7.12 2

λ ⋅ R 1 d avg := ⋅ 2⋅ ( 1 − ρ) λ ⋅ C

d avg

= 26.3

s

2) Solve for effective green time using 28 seconds as displayed green time G := 28 s

g := G + Y + AR − tL

C := r + g

g

= 29.0

(Eq 7.3)

s

C = 91.0 s 2

d avg

:=

λ ⋅ r 1 ⋅ 2⋅ ( 1 − ρ) λ ⋅ C

d avg

= 29.7

s

3) Solve for total vehicle delay g := 28 s

C := r

+

g

C = 90.0 s

2

λ ⋅ r Dt := 2⋅ ( 1 − ρ)

Dt

= 413.3

s

(Eq 7.11)


Determine the average approach delay (in seconds). v := 750

veh

s

h

:= 3200

Problem 7.48

veh h

(given) C := 100 s

g := 32 s

T := 0.25

d 3 := 0 s

PF := 1.0

k := 0.5 (pretimed control)

(assumed)

I := 1.0 (isolated mode)

Calculate v/c (X) ratio c := s ⋅ X :=

g

= 1024.000

c

C

v

X

c

veh

(Eq 7.6)

h

= 0.732

Calculate uniform delay

⎛ g ⎞ 0.5⋅ C⋅ 1 − ⎝ C ⎠ d 1 := ⎛ g ⎞ 1 − X⋅ ⎝ C ⎠

2

d 1 = 30.198 s

(Eq 7.15)

Calculate random delay

⎡ − 1) + ⎣

d 2 := 900⋅ T⋅ ( X

(X

2

− 1) +

8⋅ k ⋅ I⋅ X ⎤ c⋅ T

⎦

= 34.8

s

d 2 = 4.633 s

(Eq 7.16)

Calculate the total delay d := d 1⋅ PF + d 2 + d 3

d

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answers: 1) Use d1 as final answer.

d 1 = 30.198 s

2) Use d2 as final answer

d 2 = 4.633 s

3) Use T of 15 minutes rather than 0.25 hrs T := 15

⎡ − 1) + ⎣

d 2 := 900⋅ T⋅ ( X

d := d 1⋅ PF + d 2 + d 3

(X d

2

− 1) +

= 35.0

8⋅ k ⋅ I⋅ X⎤ c⋅ T

⎦

d 2 = 4.808

s


Problem 7.49

Calculate the sum of flow ratios for th e critical lane groups. Calculate flow ratios Phase 1 EBL :=

250 1800

WBL :=

300 1800

Phase 2

EBL = 0.139

WBL = 0.167

EBTR :=

1200 3600

WBTR :=

1350 3600

EBTR = 0.333

WBTR = 0.375

Phase 3 SBL :=

75 500

100 NBL := 525 SBTR := NBTR :=

420 1950 425 1950

SBL = 0.15 NBL

= 0.19

SBTR = 0.215 NBTR = 0.218

Calculate sum of flow ratios for critical lane groups Yc := WBL + WBTR + NBTR

Yc

= 0.76

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Use incorrect Phase 1 value Yc := EBL + EBTR + NBTR

Yc

= 0.69

2) Use split phasing for NB & SB Yc := WBL + WBTR + NBL

+ NBTR

Yc

= 0.95

3) Choose the smallest flow ratio for each phase. Yc := EBL + EBTR + SBL

Yc

= 0.622


Calculate the minimum cy cle length.

L := 12 s

Xc

:= 0.9

Problem 7.50

SumCritFlowRatios:= 0.72

(given)

Calculate minimum cycle length

Cmin :=

L⋅ Xc Xc

Cmin = 60.0 s

− SumCritFlowRatios

(Eq 7.20)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Solve for optimum cycle length Copt

:=

1.5⋅ L + 5

Copt

1.0 − SumCritFlowRatios

= 82.1

s

(Eq 7.21)

2) Switch sum and Xc in Eq 7.21 SumCritFlowRatios:= 0.9

Cmin :=

L⋅ Xc Xc


Xc

:= 0.72

Cmin = −48.0s

use absolute value

3) Use Xc = 1.0 Xc

:= 1.0

Cmin :=

SumCritFlowRatios:= 0.72 L⋅ Xc

Xc


Cmin = 42.9 s


Determine the sum of th e yellow and all-red times.

G :=

4

w := 60 ft

100

V := 40⋅

⎛ 5280 ⎞ ⎝ 3600 ⎠

tr := 1.0

l := 16 ft

ft

g := 32.2

s

(given)

ft s

a := 10.0

s

Problem 7.51

2

(assumed)

ft s

2

Determine the yellow time V

Y := tr +

2⋅ a

Y = 3.599s

+ 2⋅ g⋅ G

(Eq 7.23)

Determine the all red time AR :=

+

w

l

AR

V

= 1.295

s

Sum Y & AR AR

+

Y = 4.894 s

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Use assumed value of 20.0 ft for vehicle length l := 20.0 ft AR

+

AR :=

+

w

l

AR

V

= 1.364

s

Y = 4.962 s Y = 3.599s

2) Use Y value only

3) The percent grade is not divided by 100 G := 4

l := 16.0 ft

Y := tr + AR := AR

+

w

V 2⋅ a

+

+ 2⋅ g⋅ G

l

V

AR

Y = 1.211

= 1.295

s

Y = 2.507 s




Chapter 8 Travel Demand and Traffic Forecasting U.S. Customary Units



•


•


•




www.mathcad.com


Problem 8.1 Determine the number o f peak hour trips generated. income := 20

household_size := 2

neighborhood_employment:= 1.0

nonworking := 2

(given)

From Example 1: trips ex1 := 0.12 + 0.09 ⋅ ( household_size) trips ex1

= 0.37

+ 0.011⋅ ( income) − 0.15⋅ ( neighborhood_employment)

trips per household

trips ex1 ⋅ 1700 = 629

total shopping trips

From Example 2: trips ex2 := 0.04 + 0.018⋅ ( household_size) trips ex2

= 0.576

trips per household

trips ex2 ⋅ 1700 = 979

(

total := 1700⋅ tripsex1 total

+ 0.009⋅ ( income) + 0.16⋅ ( nonworking)

total social/recreational trips

+ tripsex2)

= 1608.2 trips

629 + 979 = 1608 total shopping and social/recreational trips


Problem 8.2 Determine the amount o f addition al retail employment needed. income := 20 trips := trips

size := 2

retail

retail := 100

(given)

trips per household x 1700 = 100

1700

= 0.059

trips

0.12 + 0.09⋅ size

0.520 − 0.059 0.15

+ 0.011⋅ income − 0.15⋅ employment

= 3.073

= 3.075

employment

employment ⋅ 100 − retail

= 207.451

so,

208 more retail employees

Problem 8.3 Determine the change in the number of peak hour social/recreational tri ps. income := 15

size := 5.2

nonworking := size

− working

workingnew := 1.2 ⋅ working nonworkingnew := size

working := 1.2 nonworking = 4 workingnew = 1.44

− workingnew

incomenew := 1.1income trips := 0.04 + 0.018⋅ size

(given)

nonworkingnew = 3.76

incomenew = 16.5

+ 0.009⋅ income + 0.16 ⋅ nonworking

trips new := 0.04 + 0.018⋅ size

trips

+ 0.009⋅ incomenew + 0.16 ⋅ nonworkingnew

trips new⋅ 1500 − trips ⋅ 1500 = −37.35

= 0.909 tripsnew = 0.884

trips


Problem 8.4 Determine the number of peak-hour sh opping t rips and the probabili ty that the household wil l make more than 1 peak-hour shop ping tri p. calculate the number of peak hour trips house := 4

income := 150

retail := 3

(given)

BZi := −0.35 + 0.03⋅ house + 0.004⋅ income − 0.10⋅ retail BZi

λi := e

λi = 1.07

BZi

= 0.07

(Example 8.4)

vehicle trips

(Eq. 8.3)

find the probability of making more than 1 trip

− λi

( ) :=

e

P Ti

Ti

⋅ λi

(Eq. 8.2)

Ti!

= 0.342

P ( 0)

P ( 1)

= 0.367

Pmore := 1

− P ( 0) − P ( 1)

Pmore = 0.291

Problem 8.5 Determine the number of peak-hour social/recreational tr ips and t he probability that the household wil l not make a peak-hour social/rec trip. calculate the number of peak-hour trips coeff size

:= 0.025

size := 5

coeff income := 0.008

coeff nonwork := 0.10

(given)

income := 100

nonwork := 5 − 3

(given)

BZi := −0.75 + coeff size ⋅ size BZi

+ coeff income ⋅ income + coeff nonwork⋅ nonwork

(given)

= 0.375 BZi

λ i := e

λ i = 1.455

vehicle-trips

(Eq. 8.3)

calculate the probability of zero peak-hour trips Ti := 0

(given)

− λi

( ) :=

P Ti

P ( 0)

e

Ti

⋅λ i

Ti!

(Eq. 8.2)

= 0.233


Problem 8.6 Determine the number of w ork tri p vehicles that leave during th e peak hour. usersB := 750

userseach_B := 15

usersB

buses :=

buses

userseach_B

cars DL := 2380

= 50

cars SR := 870 carsSR

cars := carsDL + cars

(given)

cars

2

(given)

= 2815

+ buses = 2865 total trip vehicles

Problem 8.7 Determine how m any workers w ill t ake each mode. cost DL := 8.00

cost SR := 4.00

cost B := 0.50

traveltimeDL := 20

traveltimeSR := 20

traveltimeB := 25

(given)

From Example 8.5: UDL := 2.2

− 0.2 ⋅ cost DL − 0.03 ⋅ traveltimeDL

UDL = 0

USR := 0.8

− 0.2 ⋅ cost SR − 0.03 ⋅ traveltimeSR

USR = −0.6

UB := −0.2 ⋅ cost B

− 0.01 ⋅ traveltimeB

UB

= −0.35

Using Eq. 8.7: UDL

e

PDL :=

UDL

e

USR

+e

UB

PDL = 0.444

PDL⋅ 4000 = 1775

drive alone

PSR = 0.244

PSR ⋅ 4000 = 974

shared ride

+e

USR

e

PSR :=

UDL

e

USR

+e

UB

+e

UB

PB :=

e UDL

e

USR

+e

UB

PB

= 0.313

PB ⋅ 4000 = 1251

bus

+e


Problem 8.8 Determine how many shopp ing trip s will b e made to each of the four shopping centers. coeff dist := −0.455

coeff space := 0.0172

U A1 := coeff dist ⋅ ( 2.5)

+ coeff space ⋅ ( 200)

U A1

= 2.303



U A2

= 0.078



U A3

= 2.885



U A4

= 6.362

P A1

= 0.016

P A2

= 0.002

P A3

= 0.029

P A4

= 0.952

(given)

Using Eq. 8.7: U A1

e

P A1 := U A1 U A2 U A3 U A4 +e +e +e e U A2

e

P A2 := U A1 U A2 U A3 U A4 e +e +e +e U A3

e


e

P A4 := U A1 U A2 U A3 U A4 e +e +e +e trips := 4000 Trips to each shopping center: trips 1 := trips ⋅ P A1

trips1

= 66

trips 2 := trips ⋅ P A2

trips2

=7


trips3

= 118


trips4

= 3809


Problem 8.9 Determine the new distributi on of the 4000 shopping tr ips. coeff dist := −0.455

coeff space := 0.0172



U A1

= 2.303



U A2

= 6.098



U A4

= 6.362

(given)

Using Eq. 8.7: U A1

e

P A1 := U A1 U A2 U A4 +e +e e

P A1

= 0.01

P A2

= 0.43

P A4

= 0.56

U A2

e

P A2 := U A1 U A2 U A4 e +e +e U A4

e

P A4 := U A1 U A2 U A4 e +e +e trips := 4000 Trips to each shopping center: trips 1 := trips ⋅ P A1

trips1

= 39


trips2

= 1721


trips4

= 2241


Problem 8.10 Determine how muc h comm ercial floor sp ace is needed. coeff dist := −0.455 trips := trips 4000

4000 3

coeff space := 0.0172 trips

(given)

= 1333

= 0.333

therefore P A := 0.333 as before U A4 := 6.362 which means that U A := 6.362 so for shopping center 1

U A

coeff dist ⋅ ( 2.5)

+ coeff space ⋅ ( space1)

space1 = 436.017

space1 ⋅ 1000 = 436017

2

ft

and for shopping center 2

U A

coeff dist ⋅ ( 5.5)

space2 = 515.378

+ coeff space ⋅ ( space2) space2 ⋅ 1000 = 515378

2

ft


Problem 8.11 Determine the new distri bution of s hoppin g trips by destin ation and mode. time2auto := 15 ⋅ 0.8

time2auto = 12

time2bus := 22 ⋅ 0.8

time2bus

(given)

= 17.6

coeff := 0.012 U A1 := 0.6

− 0.3 ⋅ 8 + coeff ⋅ 250

U A1

= 1.2

UB1 := −0.3 ⋅ 14 + coeff ⋅ 250

UB1

= −1.2

U A2 := 0.6

U A2

= 1.8

UB2

= −0.48

P A1

= 0.323

PB1

= 0.029

P A2

= 0.588

PB2

= 0.06

− 0.3 ⋅ time2auto + coeff ⋅ 400

UB2 := −0.3 ⋅ time2bus

+ coeff ⋅ 400

Using Eq. 8.7: U A1

e

P A1 := U A1 UB1 e +e

U A2

+e

UB2

+e

UB1

PB1 :=

e U A1

e

UB1

U A2

+e

+e

UB2

+e

U A2

e

P A2 := U A1 UB1 e +e

U A2

+e

UB2

+e

UB2

PB2 :=

e U A1

e

UB1

+e

U A2

+e

UB2

+e

total := 900 trips A1 := total ⋅ P A1

trips A1

= 290

trips B1 := total ⋅ PB1

tripsB1

= 26

trips A2 := total ⋅ P A2

trips A2

= 529

trips B2 := total ⋅ PB2

tripsB2

= 54


Problem 8.12 Determine how much c ommercial floor s pace must be added to shoppi ng center 2. total := 900

trips A2

P A2 :=

total tripsB2

PB2 :=

P1 :=

trips A2 := 355

total

total

tripsB2 := 23

P A2

= 0.394

PB2

= 0.026

− ( trips A2 + trips B2)

P1

total

U A1 := 1.2

(given)

= 0.58

UB1 := −1.2

(given)

let x = e UA2 + e UB2

U A1

e

P1

U A1

e x

UB1

+e

UB1

+e

(Eq. 8.7)

+x

= 2.622

time2auto := 15 + 4

time2bus := 22 + 4

time2auto = 19

time2bus

x

0.6−0.3⋅ time2 auto + 0.012 ⋅ space

e

space

= 26

− 0.3⋅ time2 bus + 0.012 ⋅ space

+e

= 499.918

added_space := space − 400 added_space = 99.918

added_space⋅ 1000 = 99918

2

ft


Problem 8.13 Determine the distribution of trips among possible destinations. coeff pop := 0.2

coeff dist := −0.24

U A1 := coeff pop⋅ ( 15.5)

coeff space := 0.09

(given)

+ coeff dist ⋅ ( 7.5) + coeff space ⋅ ( 5)

U A1

= 1.75

U A2 := coeff pop⋅ ( 6.0)


U A2

= 0.9



U A3

= 0.4



U A4

= 0.67

Using Eq. 8.7: U A1

e

P A1 := U A1 U A2 U A3 U A4 e +e +e +e

P A1

= 0.494

P A2

= 0.211

P A3

= 0.128

P A4

= 0.168

U A2

e


e

P A3 := U A1 U A2 U A3 U A4 +e +e +e e

U A4

e

P A4 := U A1 U A2 U A3 U A4 e +e +e +e

trips := 700 trips 1 := trips ⋅ P A1

trips1

= 345


trips2

= 148


trips3

= 90


trips4

= 117


Problems 8.14 Determine the distr ibutio n of the 700 peak-hour s ocial/recreational tri ps. coeff pop := 0.2


U A1 := coeff pop⋅ ( 15.5)

coeff space := 0.09

(given)


U A1

= 1.75



U A2

= 0.9


+ coeff dist ⋅ ( 2.0) + coeff space ⋅ ( 8 + 15)

U A3

= 1.75



U A4

= 0.67

Using Eq. 8.7: U A1

e

P A1 := U A1 U A2 U A3 U A4 e +e +e +e

P A1

= 0.361

P A2

= 0.154

P A3

= 0.361

P A4

= 0.123

U A2

e


e

P A3 := U A1 U A2 U A3 U A4 +e +e +e e U A4

e

P A4 := U A1 U A2 U A3 U A4 +e +e +e e trips := 700 trips 1 := trips ⋅ P A1

trips1

= 253


trips2

= 108


trips3

= 253


trips4

= 86


Problem 8.15 Determine how muc h additional amusement floor space must be added. trips1total := 500⋅ 0.40

trips1total

= 200

P1 := 0.40 P2and3 := 1.0 − P1 U A2 := 2.124

P2and3 = 0.6

UB2 := 0.564

U A3 := 0.178

UB3 := −2.042

(From Ex. 8.8)

let x = e UA1 + e UB1

U A2

P2and3

e

U A2

e x

x

UB2

+e

UB2

+e

U A3

+e

U A3

+e

UB3

+e

UB3

+e

+x

= 7.631 ( 0.9− 0.22 ⋅ 14 + 0.16⋅ 12.4+ .11⋅ space )

e

space

( − 0.22 ⋅ 17 + 0.16⋅ 12.4+ .11⋅ space )

+e

= 18.5

added_space := space added_space = 5.5

− 13

added_space⋅ 1000 = 5523

2

ft


Problem 8.16 Determine how muc h additional amusement floor space must be added. trips := 700

trips2 := 250

trips 2 P A2 := trips

P A2

U A1 := 1.75

= 0.357

U A3 := 0.4

U A4 := 0.67


U A2

P A2

e U A1

e U A2

U A2

+e

+e

U A4

+e

= 1.631

coeff pop := 0.2

U A2

(Eq. 8.7) U A3


coeff pop⋅ ( 6.0)

coeff space := 0.09

(given)

+ coeff dist ⋅ ( 5.0) + coeff space ⋅ ( space2)

space2 = 18.128 space2 − 10.0 = 8.128

8.128⋅ 1000 = 8128

2

ft



Problem 8.18 Determine the user equilibrium and system opti mal route travel times, total travel time, and rout e flows. User equilibrium: t1 x1

8

t2

1

+ 2 ⋅ x2

q := 4

(given)

+ x2 q

t1 8

+ x1

t2

+ x1 1 + 2 ⋅ ( q − x1)

Route flows: x1

= 0.333

x 1 ⋅ 1000 = 333

x 2 := q − x 1

veh/h

x 2 = 3.667

x 2 ⋅ 1000 = 3667

veh/h

Route travel times: t1 := 8

+ x1

t2 := 1

+ 2 ⋅ x2

t1

= 8.33 t2

= 8.33

min min

x 1 ⋅ t 1 ⋅ 1000 + x 2 ⋅ t2 ⋅ 1000 = 33333

veh − min

total travel time

System Optimal: S (x)

S (x)

dx 1

+ x 1) + x 2 ⋅ (1 + 2 ⋅ x2)

q − x1

x2

d

(

x1 ⋅ 8

(

x1 ⋅ 8

S ( x)

+ x 1) + (q − x 1) ⋅ 1 + 2 ⋅ ( q − x1)

−9 + 6 ⋅ x1

setting the derivative = 0


Route flows: x1

= 1.5

x 2 := q

− x1

x1 ⋅ 1000 = 1500 x2

= 2.5

veh/h x 2 ⋅ 1000 = 2500

veh/h

Route travel times: t1 := 8 + x 1 t2 := 1 + 2 ⋅ x 2

t1 = 9.5 t2

=6

min min

x 1 ⋅ t1 ⋅ 1000 + x2 ⋅ t2 ⋅ 1000 = 29250

veh − min

total travel time


Problem 8.19 Determine how many vehicl e hours wil l be saved. hourstotal := 875.97

veh − h

(from Example 8.12)

For System Optimal:

(4 + 1.136⋅ x1) ⋅ x1 + (3 + 3.182⋅ x2) ⋅ x2

z x1

6 − x2

z

4 + 1.136⋅ 6 − x2

(

2

4.318⋅ x2

z d

z

dx x 2 :=

) ⋅ (6 − x2) + (3 + 3.182⋅ x2) ⋅ x2

− 14.632⋅ x2 + 51.264

8.636⋅ x2

− 14.632 0

14.632 8.636

x 2 = 1.694

x 1 := 6 − x 2

x 1 = 4.306

t1 :=

t2 :=

(4 + 1.136⋅x 1) ⋅ x1 ⋅ 1000 60

(3 + 3.182⋅ x2) ⋅ x2 ⋅ 1000

hourstotal

60

− t1 − t2 = 0.962

t1

= 638.052

veh − h

t2

= 236.956

veh − h

veh − h


Problem 8.20 Determine the reduction in peak-hour traffic demand needed. Before reconstruction c 1 := 2.5

c 2 := 4

q := 3.5 S (x )

t1

+ 3⋅

c1

⎝ ⎠

t2

4

+ 2⋅

⎛ x2 ⎞

(given)

c2

⎝ ⎠

thousand vehicles

( 2 + 1.2 ⋅ x1) ⋅ x1 + (4 + 0.5⋅ x2) ⋅ x2 q − x1

x2 S (x )

( 2 + 1.2 ⋅ x1) ⋅ x1 +

S (x)

1.7x1

d

2

⎛ x1 ⎞

2

S ( x)

x 1 :=

+ 0.5 ⋅ ( 3.5 − x1) ⋅ ( 3.5 − x1)

− 5.5 ⋅ x1 + 20.125

3.4x1

dx 1

4

− 5.5

5.5 3.4

x 2 := q − x 1

0

x1

= 1.618

veh/hr

x2

= 1.882

veh/hr

( ) tot_trav2 := ( 4 + 0.5 ⋅ x2) ⋅ x2 ⋅ 1000 tot_trav1 := 2 + 1.2 ⋅ x1 ⋅ x1 ⋅ 1000

tot_trav1

+ tot_trav2 = 15676.5 veh-min

tot_trav1

= 6375.433 veh-min

tot_trav2

= 9301.038 veh-min

total vehicle travel time (before construction)

During reconstruction minimize

S

( 2 + 1.2 ⋅ x1) ⋅ x1 + (4 + 1 ⋅ x2) ⋅ x2 with ( 2 + 1.2 ⋅ x1) ⋅ x1 + 4 + 1( q − x1) (q − x1)

S

2.2 ⋅ x1

S

d dx

2

S

q

− x1

2

− 2 ⋅ x1 + q − 4 ⋅ q − 2 ⋅ q ⋅ x 1

4.4 ⋅ x1

q := 2.2 ⋅ x1

x2

− 2 − 2q 0

−1


Also, from travel times: 15.676

( 2 + 1.2⋅x 1) ⋅ x1 + (4 + 1⋅ x2) ⋅ x2

15.676

2 ⋅ x1

+ 1.2 ⋅ x1 + 4 ⋅ ( q − x1) + (q − x 1)

15.676

2 ⋅ x1

+ 1.2 ⋅ x1 + 4 ⋅ q − 4x1 + q − 2 ⋅ q ⋅ x 1 + x1

15.676

2.2x1

2 2

2

x2

q

− x1

2

2

2

2

− 2 ⋅ x1 + 4q + q − 2 ⋅ q ⋅ x 1

2.2 ⋅ x1 − 1

Substituting q 15.676

with

2

2.2 ⋅ x1

2

− 2 ⋅ x 1 + 4 ⋅ ( 2.2 ⋅ x1 − 1) + (2.2 ⋅ x1 − 1) − 2 ⋅ ( 2.2 ⋅ x1 − 1) ⋅ x 1

Solving for x1 yields: x1

= 1.954

x1 ⋅ 1000 = 1954 veh/h

q := 2.2 ⋅ x1 − 1 x 2 := q

− x1

Δq := 3.5 − q

q = 3.299 x2

= 1.345

thousand vehicles x 2 ⋅ 1000 = 1345

Δq ⋅ 1000 = 201

veh

veh/h

reduction in peak-hour traffic demand needed



Problem 8.22 Determine the user equilibrium t raffic flows .

( ) := 8 + 0.5 ⋅ x1

q := 3

t1 x1

( ) := 1 + 2 ⋅ x2

( ) := 3 + 0.75⋅x 3

t 2 x2

t1 ( q)

= 9.5

t1 ( 0)

=8

t2 ( 0)

=1

t2 ( q)

=7

t3 ( 0)

=3

t3 x 3 t3 ( q)

(given)

= 5.25

So, Route 1 will never be used since t1 (0) > t 2 (q) > t 3 (q) therefore, t2 1

+ 2 ⋅ x2 3 + 0.75 ⋅x 3 x2 + x3

q 1

t3

+ 2 ⋅ x2 3 + 0.75 ⋅ ( q − x2)

x2

= 1.545

x 2 ⋅ 1000 = 1545

veh/h

x 3 ⋅ 1000 = 1455

veh/h

x 3 := q − x 2 x3

= 1.455


Problem 8.23 Determine equilibrium f lows and travel times before and after reconstruct ion begins. Let subscript 11 denote route 1 before construction Let subscript 12 denote route 1 during construction Let subscript 21 denote route 2 before construction Let subscript 22 denote route 2 during construction c 11 := 4

c 21 := 2

c 12 := 3

t12 - t 11 = 35.28 s or 0.588 min

c 22 := 2 (given)

Also, since for UE, t 1 = t 2 ; t 22 - t 21 = 0.588 min (1) t 22

10 + 1.5 ⋅x 22

t 21

10 + 1.5 ⋅x 21

Substituting the above two equations into (1):

(10 + 1.5 ⋅x 22) − (10 + 1.5 ⋅x 21)

0.588

and solving for x21 in terms of x 22 yields: x 21

x 22 − 0.392

( 2)

Also, for the traffic increase: x 22

1.685⋅ x21

or x 21

0.5935⋅ x22

Substituting into (2):

0.5935⋅ x22

x 22 = 0.964

x 22 − 0.392

x 22 ⋅ 1000 = 964

t22 := 10 + 1.5 ⋅x 22

t22 = 11.45

veh/h min

flow on route 2 after construction travel time after construction


t12 := t22

t12

6

⎛ x12 ⎞

+8

⎝ c12 ⎠

x 12 = 2.042

x 12 ⋅ 1000 = 2042

veh/h

flow on route 1 after construction

from User Equilibrium: t11 := t22 − 0.588

t11 = 10.86

min

travel time before construction

t21 := t11

t11

6

+ 8⋅

⎛ x11 ⎞ ⎝ c11 ⎠

x 11 = 2.429

t21

x 11 ⋅ 1000 = 2429

veh/h

flow on route 1 before construction

⎛ x21 ⎞

10 + 3

x 21 = 0.572

⎝ c21 ⎠ x 21 ⋅ 1000 = 572

veh/h

flow on route 2 before construction


Problem 8.24 Determine the user equilib rium flo ws. Determine for q = 10000

( ) := 2 + 0.5 ⋅ x1

q := 10

t1 x 1

t1 ( q)

=7

t1 ( 0)

=2

t2 ( 0)

=1

t2 ( q)

= 11

t3 ( 0)

=4

( ) := 1 + x2

t2 x 2

t3 ( q)

( ) := 4 + 0.2 ⋅ x3

t3 x 3

(given)

=6

so all routes might be used t1

t2

t3

x1 + x2 + x3

q x2

q − x1

t1

t2

2

+ 0.5 ⋅ x1 1 + x 2 1 + 0.5 ⋅ x1

x2 also 2

− x3

t1

t3

+ 0.5 ⋅ x1 4 + 0.2 ⋅ x3

(−2 + 0.5 ⋅ x1)

x3

0.2

−10 + 2.5 ⋅ x1

x3 so

(

) + (−10 + 2.5 ⋅ x1)

x1

= 4.75

x 1 ⋅ 1000 = 4750

t1

= 4.375

q

x 1 + 1 + 0.5 ⋅ x1

veh/h

t2 := t1 x 2 := t2

−1

x 2 = 3.375

x 2 ⋅ 1000 = 3375

veh/h


t3 := t1 t3

x 3 :=

−4 x3

0.2

= 1.875

x3 ⋅ 1000 = 1875

veh/h

Determine for q = 5000

( ) := 2 + 0.5 ⋅ x1

q := 5

t1 x 1

t1 ( q)

= 4.5

t1 ( 0)

=2

t2 ( 0)

=1

t2 ( q)

=6

t3 ( 0)

=4

( ) := 1 + x2

t2 x 2

t3 ( q)

( ) := 4 + 0.2 ⋅ x3

t3 x3

(given

=5

so all routes might be used Routes 1 and 2 have the lowest free-flow travel time, so assume route 3 is not used t1

t2

+ 0.5 ⋅ x1 1 + x2 x 2 1 + 0.5 ⋅ x1 2

x2 1

q

− x1

+ 0.5 ⋅ x1 q − x 1

x1

= 2.667

x 2 := q

x 1 ⋅ 1000 = 2667

− x1

x2

= 2.333

veh/h x2 ⋅ 1000 = 2333

veh/h

Check to see if route 3 is used:

( ) = 3.333

t1 x 1

3.333 < 4

minutes (route 3 free-flow time)

Therefore the assumption that route 3 is not used is valid.


Problem 8.25 Determine the minimum demand so that all routes are used. Flow on route 3 will begin when t1 = t 2 = 4 (which is route 3's free-flow time). so t1

2

x 1 :=

+ 0.5 ⋅ x1 4

4−2

x1

=4

x 2 := 4 − 1

x2

=3

q := x 1 + x 2

q

t2

1

0.5

+ x2 4

=7

thousand vehicles

If q > 7, flow on all three routes If q

≤ 7 flow on routes 1 and 2 only.


Problem 8.26 Determine the total person hours of travel for each situation. SOV veh := 2500

HOV 2veh := 500

HOV3veh := 300

Buses := 20

(given)

vehtotal := SOV veh + HOV2veh + HOV 3veh + Buses vehtotal = 3320 peopletotal := SOV veh⋅ 1 + HOV 2veh⋅ 2 + HOV 3veh⋅ 3 + Buses ⋅ 50 peopletotal = 5400 t 0 := 15

(given)

c 3 := 3600 a) open to all

⎡ t := t 0 ⋅ 1 + 1.15 ⋅

⎣

personhours :=

⎛ vehtotal ⎞ ⎝

c3

6.87

⎠

⎤ t = 24.89

⎦

t ⋅ people total 60

personhours = 2240

person − hours

b) 2+ lane c 1 := 1200

c 2 := 2400

⎡ t SOV := t 0 ⋅ 1 + 1.15 ⋅

⎛ SOV veh ⎞ c2

⎤

⎝

⎡

⎛ HOV2veh + HOV3veh + Buses ⎞

⎣

⎝

⎦

t SOV = 37.834

⎣

t HOV := t 0 ⋅ 1 + 1.15 ⋅

⎠

6.87

c1

6.87

⎤ ⎦

t HOV = 16.261


personhours :=

(

tSOV ⋅ SOVveh + tHOV ⋅ 2 ⋅ HOV2veh + 3 ⋅ HOV3veh + 50⋅ Buses

)

60

personhours = 2362

person − hours

c) 3+ lane

⎡ t2 := t0 ⋅ 1

⎣

+ 1.15 ⋅

⎡ t3 := t0 ⋅ 1

⎣

+ 1.15 ⋅

personhours :=

⎛ SOVveh + HOV2veh ⎞ ⎝

c2

⎠

⎛ HOV3veh + Buses ⎞ ⎝

(

c1

6.87

⎦

6.87

⎠

⎤ t2

= 94.903

t3

= 15.002

⎤ ⎦

) + t3 ⋅ (3 ⋅ HOV3veh + 50⋅ Buses)

t2 ⋅ SOVveh + 2 ⋅ HOV2veh

personhours = 6011

60

person − hours


Problem 8.27 Determine the total person-hours and the minimum m ode shift. t0 := 15

c2 := 2400

SOV

⎡

⎛ SOV ⎞ tSOV := t0 ⋅ 1 + 1.15 ⋅ ⎣ ⎝ c2 ⎠

:= 2500 − 500

⎤

6.87

⎦

= 19.93

tSOV

c 1 := 1200

HOV2 := 500

⎡ tHOV := t0 ⋅ 1

⎣

tHOV

+ 1.15 ⋅

HOV3 := 300

⎛ HOV2 + HOV3 + Buses ⎞ c1

⎝

⎠

Buses := 20 + 10

⎤

6.87

⎦

= 16.371

personhours :=

tSOV ⋅ SOV

+ tHOV ⋅ ( 2 ⋅ HOV2 + 3 ⋅ HOV3 + 50⋅ Buses ) 60

personhours = 1592

person − hours

determine mode shift for all 3 lanes open to all traffic

personhours := 2240.127

(From Problem 8.26)

personmin := personhours ⋅ 60 personmin SOV

= 134407.62

:= 2500

HOV := 500 + 300 + 20

HOVp := 500⋅ 2 + 300⋅ 3 + 20 ⋅ 50 c 1 := 1200

(given)

HOVp = 2900

c2 := 2400

⎡ ⎢

⎛ HOV + 6.87⎤ ⎡ ⎛ SOV − x ⎞ ⋅ ( SOV − x) + t0 ⋅ 1 + 1.15 ⋅ ⎜ t0 ⋅ 1 + 1.15 ⋅ ⎣ ⎝ c2 ⎠ ⎦ ⎣ ⎝ c1 x

= 43.826

⎞ 50 ⎟ ⎠ x

⎤ ⎥

6.87

⎦

⋅ ( HOVp + x )

personmin

so, 100 people must shift from SOV to Bus


Problem 8.28 Determine the user equilib rium and syst em optimal route flows and total travel times. t1

5

+ 3 ⋅ x1

t2

7

+ x2

q := 7

(given)

At System Optimal:

(

x1 ⋅ 5

S (x)

+ 3 ⋅ x1) + x 2 ⋅ ( 7 + x2)

q − x1

x2

(

x1 ⋅ 5

S (x) d

S ( x)

dx 1

+ 3 ⋅ x1) + ( q − x 1) ⋅ 7 + ( q − x1)

8 ⋅ x1

− 16 0

Route flows:

x1

=2

x 1 ⋅ 1000 = 2000

x 2 := q − x 1

x2

=5

veh/h

x 2 ⋅ 1000 = 5000

veh/h

Route travel times:

t1 := 5 t1

+ 3 ⋅ x1

t2 := 7

= 11

t2

(

TravelTimeSO := t1 ⋅ x 1 TravelTimeSO

= 82

+ x2

= 12

+ t2 ⋅ x 2) veh − min

At User Equilibrium: t1

5

t2

+ 3 ⋅ x1 7 + ( q − x1)


Route flows:

x1

= 2.25

x 2 := q

x1 ⋅ 1000 = 2250

− x1

x2

= 4.75

veh/h

x 2 ⋅ 1000 = 4750

veh/h

Route travel times:

t1 := 5 + 3 ⋅ x 1 t1

= 11.75

TravelTimeUE := t1 ⋅ q TravelTimeUE

= 82.25

veh − min


Problem 8.29 Determine the value of the derivative of the user equilibri um math program evaluated at the system optimal solut ion wit h respect to x 1 . q := 7 S (x)

S (x)

2

5 ⋅ x 1 + 1.5 ⋅ x1

S ( x)

dx 1 at d

⌠ ⎮ ( 7 + 3 ⋅ w) dw ⎮ ⌡

q − x1

x2

d

⌠ ⎮ ( 5 + 3 ⋅ w) dw + ⎮ ⌡

4 ⋅ x1

2

+ 49 − 7 ⋅ x1 + 24.5 − 7 ⋅ x 1 + 0.5 ⋅ x1

−9

x 1 := 2 (the SO solution) z

−1

dx 1


Problem 8.30 Determine the user equilibrium f lows and total hourly o rigindestination demand after the improvement.

⎛ x ⎞ t1 ( x , c ) := 3 + 1.5 ⋅ ⎝ c ⎠ c 1 := 2

2

t2 ( x , c ) := 5

c 2 := 1.5

+ 4⋅

⎛ x ⎞ ⎝ c ⎠

(given)

q := 6

(

) = 16.5

t1 0 , c 1

(

)=5

t2 q , c 2

t1 q , c 1 t2 0 , c 2

(

)=3

(

) = 21

so both routes may be used t1

t2

x2

q

− x1

Existing Condtions

3

+ 1.5 ⋅

x1

⎛ x1 ⎞

2

⎝ c1 ⎠

5

+ 4⋅

⎛ q − x1 ⎞ ⎝

c2

⎠

= 4.232

(

)

t1_exist := t1 x 1 , c 1

t1_exist

= 9.715

t2_exist := t1_exist x 2 := q − x 1 x2

5

+

4 ⋅ x2 2.5

5

+ 1.6 ⋅ x2

= 1.768

After capacity expansion, c 2_new := 2.5 t2_new

5

+ 4⋅

x 2_new c 2_new


− (t2_new)

q_new

q + 0.5 ⋅ t2_exist

q_new

q + 0.5 ⋅ t2_exist

q_new

6 + 0.5 ⋅ 9.715 −

q_new

8.3595 − 0.8⋅ x2_new

⎡ ⎣ ⎡ ⎣

⎛

x 2_new ⎞⎤

⎝

c 2_new

− 5 + 4⋅

⎠⎦

x 2_new ⎞⎤ ⎛ 5 + 4⋅ 2.5 ⎠⎦ ⎝

also, q_new

x1_new + x 2_new 8.3595 − x1_new

x 2_new t1

3

1.8

t2

+ 1.5 ⋅

⎛ x1_new ⎞ ⎝

c1

⎛ 8.3595 −x1_new ⎞ 1.8 ⎜ ⎟ 5 + 4⋅ ⎝ c2_new ⎠

2

⎠

x 1_new = 3.968

t1_new := 3 + 1.5 ⋅

x1_new⋅ 1000 = 3968

⎛ x1_new ⎞ ⎝

c1

veh/h

2

⎠

t1_new = 8.9

t2_new := t1_new x 2_new :=

t2_new − 5 4

⋅ c 2_new

q := x1_new + x 2_new

x2_new = 2.44

q

= 6.41

veh/h

x2_new⋅ 1000 = 2440

veh/h

total hourly origin-destination demand


Problem 8.31 Determine if the user equilibrium and sy stem optimal solu tions can be equal. t1

5

+ 4 ⋅ x1

t2

7

+ 2 ⋅ x2

(given)

At System Optimal:

(

x1 ⋅ 5

S (x)

q − x1

x2

2

5 ⋅ x1 + 4 ⋅ x1

S (x)

d

S

dx 1 x1

+ 4 ⋅ x1) + x 2 ⋅ ( 7 + 2 ⋅ x2)

+ 7 ⋅ ( q − x1) + 2 ⋅ ( q − x1)

2

−7 − 4 ⋅ q + 4 ⋅ x 1 + 5 + 8 ⋅ x 1 0

0.333q + 0.1667

At User Equilibrium: t1 5

t2

+ 4 ⋅ x1 7 + 2 ⋅ ( q − x1)

6x1

− 2q − 2 0

substituting, 2q + 1 − 2q − 2

0

This statement is false, therefore the solution is not possible


Problem 8.32 Determine the user equilib rium flo ws. t1 ( x ) := 5

+ 1.5 ⋅ x

t2 ( x) := 12 + 3 ⋅ x

t 3 ( x)

:= 2 + 0.2x

2

(given)

q := 4 t1 ( 4)

= 11

t1 ( 0)

=5

t2 ( 0)

= 12

t2 ( 4)

= 24

t3 ( 0)

=2

t3 ( 4)

= 5.2 Determine Which Routes Are Used

so only routes 1 and 3 will be used t1

5

t3 and x 1

+ 1.5( 4 − x3) 2

0.2 ⋅ x3 x3

4 − x3

2

2

+ 0.2 ⋅ x3

+ 1.5 ⋅ x3 − 9 0

= 3.935

x 3 ⋅ 1000 = 3935

x 1 := q − x 3

x1

( ) = 5.1

min

( ) = 5.1

min

t1 x 1 t3 x 3

= 0.065

veh/h x1 ⋅ 1000 = 65

veh/h


Problem 8.33 Determine user equilibriu m route flo ws and total vehicl e travel time. qo := 4

t1

6

+ 4x1

t2

2

2

+ 0.5x2

(given)

For each minute of travel time greater than 2 minutes, 100 fewer vehicles depart qo

− 0.1( t2 − 2)

x1

+ x2

substitute for t 2 qo

2

− 0.1 2 + 0.5 ⋅x 2 − 2

x1

+ x2

solving for x1 x1

4 − x2

2

− 0.05x2

user equilibrium route flows means t1 = t 2 t1

6

2

+ 4 ⋅ x1 t2 2 + 0.5 ⋅ x2

plugging in x1 2

2

+ 0.5 ⋅ x2 2

0.7x 2

x2

6

2

+ 4 4 − x2 − 0.05 ⋅ x2

+ 4 ⋅ x2 − 20 0

= 3.204

t2 := 2

x 2 ⋅ 1000 = 3204

+ 0.5 ⋅ ( x2)

qnew := qo

2

t2

− 0.1( t2 − 2)

x 1 := qnew − x 2

veh/h

= 7.132 qnew = 3.487

x 1 = 0.283

Total travel time = q new⋅ t2 = 24.868

x 1 ⋅ 1000 = 283

veh/h

or 24,868 vehicle-min


Problem 8.34 Determine the distribut ion of tr affic between restricted and unrestric ted lanes so that total person-hours are minimized. (subscripting: r = restricted, u2 = 2 person using unrestricted, u1 = 1 person using unrestricted)

( xr ⋅ tr ) ⋅ 2 + (xu2⋅ tu) ⋅ 2 + (xu1⋅ tu) ⋅ 1

z ( x) where

x u2 + x u1 and

xu

x u1

3

rewriting

(

)

2 ⋅ x r ⋅ 12 + xr

z ( x)

and x + x r u2

+ 2 ⋅ x u2 ⋅ 12 + 0.5 ⋅ ( 3.0 + xu2) + 3 12 + 0.5 ⋅ ( 3.0 + xu2)

4.0 2

2

z ( x)

24 ⋅ x r + 2 ⋅ xr

z ( x)

24 ⋅ 4 − xu2

z ( x)

96 − 24xu2 + 32 − 16xu2 + 2 ⋅ xu2

z ( x)

3 ⋅ xu2

d

+ 24⋅ xu2 + 3xu2 + x u2 + 36 + 4.5 + 1.5xu2

) + 2 ⋅ (4 − xu2) 2 + 24⋅ xu2 + 3xu2 + xu22 + 36 + 4.5 + 1.5xu2

(

2

2

− 11.5 ⋅ xu2 + 168.5

6 ⋅ xu2 − 11.5

z

dxu2

2

+ 24xu2 + 3xu2 + x u2 + 36 + 4.5 + 1.5xu2

0

x u2 := 1.916

so,

x r := 4.0 − 1.916

x r = 2.084

x r ⋅ 1000 = 2084

veh/h

x u := 3

xu

= 4.916

x u ⋅ 1000 = 4916

veh/h

+ xu2

Total person hours: tr := 12 + 2.083

t r = 14.083

min

tu := 12 + 0.5 ⋅ 4.916

t u = 14.458

min

2 ⋅ 2083⋅

14.083 60

+ 2 ⋅ 1916⋅

2624.1⋅ 60 = 157446

14.458 60

+ 3000⋅

14.458 60

= 2624.1

person − hr

person − min


Problem 8.35 Determine what percentage would take rout e 1 and ho w muc h travel time would b e saved. At User Equilibrium: t1

t2

x1

⎛ 3x2 ⎞ 5+ ⎝ 2 ⎠ x2

2

⎛ x1 ⎞ t1 ( x1) := 5 + ⎝ 2 ⎠

3x2

⎛ x2 ⎞ 7+ ⎝ 4 ⎠

2

⎛ x2 ⎞ t2 ( x2) := 7 + ⎝ 4 ⎠

2

(given)

2

= 0.956

x 1 := 3 ⋅ x 2

x 1 = 2.869

q := x 1 + x 2

q

( )⋅

TravelTime:= t1 x1

= 3.825 q ⋅ 1000 60

TravelTime= 449.861

At System Optimal:

⎡

S (x)

2 ⎡ ⎛ x ⎞ 2⎤ ⎛ x1 ⎞ ⎤ 2 x1 ⋅ 5 + + x2 ⋅ 7 + ⎣ ⎝ 2 ⎠ ⎦ ⎣ ⎝ 4 ⎠ ⎦

q − x2

x1

⎡

S (x)

S (x)

d

2 ⎡ ⎛ x ⎞ 2⎤ ⎛ q − x2 ⎞ ⎤ 2 ( q − x 2) ⋅ ⎣ 5 + ⎝ 2 ⎠ ⎦ + x2 ⋅ ⎣ 7 + ⎝ 4 ⎠ ⎦

S ( x)

dx 2

x2

3

2

−0.1875⋅ x 2 + 2.86875⋅ x2 − 8.973⋅ x2 + 33.1155 2

−0.5625⋅ x 2 + 5.7375⋅ x2 − 8.973 0

= 1.929

x 1 := q − x 2

x1

= 1.896


percentage :=

x1 q

⋅ 100

percentage = 49.6

%

Total travel time:

( )

t1_SO = 5.899

( )

t2_SO = 7.232

t1_SO := t1 x 1 t2_SO := t2 x 2

TravelTimeSO :=

x 1 ⋅ t1_SO + x 2 ⋅ t2_SO 60

⋅ 1000

TravelTimeSO

= 418.892

Savings := TravelTime− TravelTimeSO Savings = 30.97

veh-min


Problem 8.36 Determine the dif ference in total vehicle travel times between user equilibrium and system optimal solutions. At User Equilibrium: t1

5

+ 3.5 ⋅ x1 q−x2

S ( x)

5q − 5x2

dx 2

2

+ 0.5 ⋅ x2

x1

q − x2

(given)

x

S (x)

S ( x)

1

⌠ 2 2 ( 5 + 3.5 ⋅ w) dw + ⎮ ( 1 + 0.5w ) dw ⌡0

⌠ ⎮ ⌡0

d

t2

2

2

3

+ 1.75q − 3.5q ⋅ x2 + 1.75x2 + x 2 + 0.167x 2 2

−5 − 3.5q + 3.5x2 + 1 + 0.5x2

At System Optimal:

S (x)

5x1

2

5 ⋅ q − 5 ⋅ x 2 + 3.5 ⋅ q

S (x)

S ( x)

dx 2

d dx 2

2

+ x2 + 0.5x2

q − x2

x1

d

2

+ 3.5x1

2

1.5x2

SSO ( x )

2

1.5x 2

−

d dx 2

2

3

− 7 ⋅ q ⋅ x 2 + 3.5 ⋅ x2 + x 2 + 0.5 ⋅ x2

+ 7x2 − 4 − 7q 0

SUE ( x)

7

2

+ 7x2 − 4 − 7q − 0.5x2 + 3.5x2 − 4 − 3.5q

7

q := 3.57


Finding total travel time at user equilibrium: 2

0.5x2

+ 3.5x2 − 4 − 3.5( 3.57)

0

x 2u := 3.226 by quadratic, so

gives

Total Travel Time

tt u := [ 5

x 1u := 3.57 − 3.226

+ 3.5( 0.344) ] ⋅ 3.57⋅ 1000

x1u = 0.344

tt u

22314

veh-min

Finding total travel time at system optimal: 2

1.5x2

+ 7 ⋅ x 2 − 4 − 3.5( 3.57)

gives

x 2s := 2.64 by quadratic, so

0

tt 1s := [ 5

+ 3.5( 0.927) ] ⋅ 0.927⋅ 1000

tt 2s := 1

+ 0.5 ( 2.64)

Total := tt 1s

+ tt 2s

2

⋅ 2.64 ⋅ 1000 Total = 19483

x 1s := 3.57 − 2.64 tt 1s

= 7642.651

tt 2s

= 11839.872

x1s

= 0.93

veh-min veh-min

veh − min

Difference := 22314 − 19482 Difference = 2832

veh-min


Determine the probability of t he household making th ree or more peak-hour trip s?

Problem 8.37

Solve Poisson regression BZi := −0.30 + 0.04⋅ ( 4)

e

BZ i

+ 0.005⋅ ( 85) − 0.12⋅ ( 2)

BZi = 0.045

= 1.046 vehicle trips

Calculate probability

P0 :=

e

P1 :=

e

P2 :=

e

⋅ (1.046 ) 0!

− 1.046

0

P0 = 0.351

⋅ (1.046 ) 1!

P1 = 0.368

⋅ (1.046 ) 2!

P2 = 0.192

− 1.046

(

(Eq 8.2)

1

− 1.046

(Eq 8.2)

2

(Eq 8.2)

) = 0.089

1 − P0 + P1 + P2

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Find probability of exatly three trips

P3 :=

e

⋅ (1.046 ) 3!

− 1.046

3

P3 = 0.067

2) Do not subtract probabilities from 1 P0 + P1 + P2 = 0.911

3) Find probability of greater than 3 trips

(

) = 0.022

1 − P0 + P1 + P2 + P3


Problem 8.38

How many workers wi ll use the shared-ride mode?

Cost DL

Cost DL := 5.50

Cost SR :=

TravTimeDL := 21

TravTimeSR := 21

3

Cost B := 1.25 TravTimeB := 27

Calculate utility expressions

(

) − 0.02⋅ ( TravTimeDL)

(

) − 0.04⋅ ( TravTimeSR )

UDL := 2.6 − 0.3 Cost DL

USR := 0.7 − 0.3⋅ Cost SR

(

) − 0.01( TravTimeB)

UB := −0.3⋅ Cost B

UDL = 0.530 USR = −0.690 UB = −0.645

Calculate probability of using carpool mode e

PSR := e

UDL

USR

+e

PSR ⋅ 5000 = 920

USR

+e

UB

PSR = 0.184

(Eq 8.7)

individuals ride in carpools

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−


Alternative Answers: 1) Solve for individuals riding in drive-alone autos

e

PDL := e

UDL

UDL

+e

USR

+e

UB

PDL = 0.623

PDL⋅ 5000 = 3117 individuals

2) Solve for individuals riding in buses e

PB := e

UDL

UB

+e

USR

+e

UB

PB = 0.193

PB⋅ 5000 = 963

individuals

3) Do not divide cost by three

(

) − 0.04⋅ ( TravTimeSR )

USR := 0.7 − 0.3⋅ Cost DL

e

PSR := e

UDL

USR = −1.790

USR

+e

USR

+e

UB

PSR = 0.070

PSR ⋅ 5000 = 349 individuals


Problem 8.39

Determine the number of bus tri ps to sh opping p laza 2. UA1 := 0.25 − ( 0.4⋅ 15) UB1 := 0.0 − ( 0.5⋅ 18)

e

UA1

+e

UB2 := 0.0 − ( 0.5⋅ 19)

+ ( 0.013325 ⋅ )

+ ( 0.013325 ⋅ )

UB2

UB1

PB2⋅ 1100 = 21

UA2 := 0.25 − ( 0.4⋅ 16)

+ ( 0.013275 ⋅ )

e

PB2 :=

+ ( 0.013275 ⋅ )

+e

UA2

+e

PB2 = 0.019

UB2

trips

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternative Answers: 1) Solve for bus trips to plaza 1 e

PB1 := e

UA1

+e

UB1

UB1

PB1⋅ 1100 = 18

+e

UA2

+e

PB1 = 0.016

UB2

trips

2) Mix up commercial floor spacing UA1 := 0.25 − ( 0.4⋅ 15) UB1 := 0.0 − ( 0.5⋅ 18)

e

UA1

+e

PB2⋅ 1100 = 10

UA2 := 0.25 − ( 0.4⋅ 16)

+ ( 0.013325 ⋅ )

e

PB2 :=

+ ( 0.013325 ⋅ )

UB2 := 0.0 − ( 0.5⋅ 19)

+ ( 0.013275 ⋅ )

+ ( 0.013275 ⋅ )

UB2

UB1

+e

UA2

+e

PB2 = 0.009

UB2

trips

3) Solve for auto trips to plaza 2 UA1 := 0.25 − ( 0.4⋅ 15)

+ ( 0.013275 ⋅ )

UB1 := 0.0 − ( 0.5⋅ 18)

+ ( 0.013275 ⋅ )

PA2 :=

e e

UA1

+e

PA2⋅ 1100 = 597

UA2 := 0.25 − ( 0.4⋅ 16) UB2 := 0.0 − ( 0.5⋅ 19)

+ ( 0.013325 ⋅ )

+ ( 0.013325 ⋅ )

UA2

UB1

+e

UA2

+e

UB2

PA2

= 0.542

trips


How many additional vehicle-hours of travel time will be added to the system assuming user-equilibrium c onditions hol d?

Problem 8.40

Check to see if both routes are used t1(0) = 5 min

t 2(0) = 4 min

t1(4) = 9.57 min

t 2(4) = 8.35 min

Both routes are used since t 2(4) > t 1(0) and t1(4) > t 2 (0). Apply user equilibrium and flow conservation to solve for x1 and x 2 5 + 4/3.5(x 1) = 4 + 5/4.6(x 2) Flow conservation: x2 = 4 - x 1 5 + 1.143(x 1) = 4 +1.087(4 - x 1)

x1 := 1.501

x2 := 4 − x1

x2 = 2.5

Find total travel time in hours

⎡ ⎛ x1 ⎞⎤ ⎡ ⎛ x2 ⎞⎤ ⋅ 1501 + 4 + 5⋅ ⋅ 2499 5 + 4⋅ ⎣ ⎝ 3.5 ⎠⎦ ⎣ ⎝ 4.6 ⎠⎦ = 447.7

veh-h

60

Check route usage for reduced-capacity case t1(0) = 5 min

t 2(0) = 4 min

t1(4) = 9.57 min

t 2(4) = 12.70 min

Both routes are used since t 2(6) > t 1(0) and t1(6) > t 2 (0) Apply user equilibrium and flow conservation to solve for x1 and x 2 5 + 4/3.5(x 1) = 4 + 5/2.5(x 2) Flow conservation: x2 = 4 - x 1 5 + 1.143(x 1) = 4 +2(4 - x 1)

x1 := 2.227

x2 := 4 − x1

x2 = 1.8


⎡ ⎛ x1 ⎞⎤ ⎡ ⎛ x2 ⎞⎤ ⋅ 2227 + 4 + 5⋅ ⋅ 1773 5 + 4⋅ ⎣ ⎝ 3.5 ⎠⎦ ⎣ ⎝ 2.5 ⎠⎦ = 503.0

veh-h

60

Find additional vehicle hours 503.0 − 447.7 = 55.3

veh-h

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski. Copyright © 2008, by John Wiley & Sons, Inc. All rights reserved.

Alternative Answers: 1) Solve only for reduced-capacity case Additional vehicle hours = 503.0

veh-h

2) Use 2100 veh-h for route 2 capacity 5 + 4/3.5(x 1 ) = 4 + 5/2.1(x 2 ) Flow conservation: x2 = 4 - x 1 5 + 1.143(x1 ) = 4 +2.381(4 - x 1 )

x1 := 2.145

x2 := 4 − x1

x2 = 1.9


⎡ ⎛ x1 ⎞⎤ ⎡ ⎛ x2 ⎞⎤ 5 + 4⋅ ⋅ 2227 + 4 + 5⋅ ⋅ 1773 ⎣ ⎝ 3.5 ⎠⎦ ⎣ ⎝ 2.1 ⎠⎦ = 525.3

veh-h

60

Find additional vehicle hours 525.3 − 447.7 = 77.6

veh-h

3) Solve only for reduced-capacity case using 2100 veh-h route 2 capacity Additional vehicle hours = 525.3

veh-h


Determine the sys tem-optimal total t ravel time (in veh-h).

7

Route 1:

65 4

Route 2:

50

Problem 8.41

⋅ 60 = 6.46 min ⋅ 60 = 4.80 min

Performance functions: t1

6.46 + 4⋅ x1

t2

4.80 + x2

2

The basic flow conservation identity is: q = x 1 + x 2 = 5.5 Substitute the performance functions for routes 1 and 2 into Eq 8.9

(

) + x2⋅

S( x)

x1⋅ 6.46 + 4⋅ x1

S( x)

6.46⋅ x1 + 4x1

2

2

4.80 + x2 2

+ 4⋅ x2 + x2

From flow conseration, x1 = 5.5 - x 2 ; therefore,

) + 4⋅ ( 5.5 − x2) 2 + 4⋅ x2 + x23

(

S( x)

6.46⋅ 5.5 − x2

S( x)

x2

3

2

+ 4⋅ x2 − 46.46x2 + 156.33

Set the first derivative to zero to find minimum dS( x) dx2

2

3⋅ x2

+ 8⋅ x2 − 46.46 0

which gives

x2 := 2.822

and x := 5.5 − x 1 2

Find system-optimal travel times and system-optimal total travel time

( )

t1 := 6.46 + 4 x1

min

2

t2 := 4.80 + x2 t1 = 17.17

min t2 = 12.76

( 2680t⋅ 1) + (2822t⋅ 2) 60

x1 = 2.68

x2 = 2.82

= 1367.33 veh-h

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Alternative Answers: 1) Simply add up individual travel times t1 and t2 t1 + t2 = 29.94 veh-h

2) Solve for total user-equilibrium travel time using t1 5500t ⋅1 60

= 1574.10

veh-h

3) Solve for total user-equilibirum travel time using t2 5500t ⋅2 60

= 1170.00

veh-h


How many vehicle-hours could be saved?

Problem 8.42

Solve for total number of vehicle hours using given distributions x1 := 2500

x2 := 2000

t1 := 12 + 2.5

t1 = 14.50 x1⋅ t1

Route 1:

t2 = 11.00

= 604.17

60 x2⋅ t2

Route 2:

t2 := 7 + 2⋅ 2.0

= 366.67

60

total = 970.84 vehicle-hours Solve with system-optimal traffic distribution S( x)

:= ( 12 + x1) ⋅ x1 + (7 + 2⋅ x2) ⋅ x2

With flow conservation, x1 = 4.5 - x 2 so that S( x)

2

:= 5⋅ x2 − 14⋅ x2 + 63

Set the first derivative to zero 10x2 − 14

x2 :=

0

t1 := 12 + x1

Route 1:

10

t1 = 15.10

1400t ⋅1 60

Route 2:

14

t2 := 7 + 2⋅ x2

t2 = 9.80

= 352.33

3100t ⋅2 60

x1 := 4.5 − x2

= 506.33

total = 858.66 vehicle-hours Hours saved = 970.84 - 858.66 = 112.18 vehicle-hours saved

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Principle of Highway Engineering and Traffic Analysis, 4 Edition, Solution Manual

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