Downloadable Solution Manual for Transportation Infrastructure Engineering a Multim 1

Chapter 2 Transportation Systems Models 2.1 What are some of the basic characteristics of systems?

• • • •

All of the components of a system have to be present and arranged in a specific way for the system to operate as intended. When one element is changed, there will be side effects. Systems tend to have specific purposes within the larger system in which they are embedded. Systems have feedback, which allows for the transmission and return of information.

2.2 List the different components of transportation systems.

• • •

Physical Elements Human Resources Operating Rules

2.3 What types of problems are best addressed using time-space diagrams? problems are addressed using cumulative plots?

• •

Which

Time-Space diagrams are used in cases where many vehicles interact while sharing a common travel way. Cumulative plots deal with problems involving traffic flow through one or more restrictions along a travel way.

2.4 A freight and passenger train shares the same track. The average speed of the freight train is 45 mph, whereas that of the passenger train is 70 mph. The passenger train is scheduled to leave 20 minutes after the freight train departs from the same station. Determine the location where a siding would need to be provided to allow the passenger train to pass the freight train. Also determine the time it takes for the freight train to arrive at the siding.

9

Chapter 2: Transportation Systems Models

(1) X

(

1 3

+

X 70

T p =

1

+

X

Since there is no specified safety distance for the 45 3 70 siding, the siding will occur where the trains would meet, so T p - Tf = 0

Tf = =

)-

X 45

= 0.10 ; X = 29.4 miles.

(2) T f =

X 45

=

29.4 45

= 0.66 hrs = 39.6 minutes.

2.5 Three friends embark on a trip using a tandem bicycle that can carry two of them at a time. To complete the trip, they proceed as follows: first, two friends (friends A & B) ride the bicycle at an average speed of 16 mph for exactly 15 minutes. While this is taking place, the third friend (friend C) walks at an average speed of 4 mph. After 15 minutes, friend A drops friend B and then rides back to meet friend C at an average speed of 18 mph. In the meantime, friend B continues the trip by walking at an average speed of 4 mph. When friend A meets friend C, they ride together at an average speed of 17 mph, until they meet friend B. The cycle just described is then repeated until the trip is completed. Determine the average speed of the friends.

X1 = 15 min = 0.25 h Y1 = 4 mi/h (X1 + X2) X2 ) Y1 = 16 mi/h(X1) – 18(X2) 4 × (.25 + X 2) = 16 × .25 − 18 × X 2 X 2 =

3

X 3 =

3

h = 0.136 h 22 Y2 = 4 mi/h (X1 + X2) + 17 mi/h (X3) Y2 = 16 mi/h (X1) + 4 mi/h (X2 + X3) 3 3 4 × (.25 + ) + 17 × X 3 = 16 × .25 + 4 × ( + X 3) 22 22

13

h = 0.231 h

Avg. Vel. =

Y 2 X 1 + X 2 + X 3

= 8.86 mi / h

10


2.6 A freight train and a passenger train share the same rail track. The freight train leaves station A at 8:00 a.m. The train travels at a speed of 30 mph for the first 10 minutes, and then continues to travel at a speed of 40 mph. At 8:35 a.m., the passenger train leaves station A. The passenger train travels first at a speed of 50 mph for 5 minutes, and then continues to travel at a speed of 70 mph. Determine the location of the siding where the freight train will have to be parked to allow the faster passenger train to pass through. As a safety precaution, it is determined that the time headway between the two trains should not be allowed to fall below 5 minutes.

The time-space diagram for this problem is as shown be low.

11


40 m ph

70 m ph

X

30 mp h

8:00 am

50 m ph

8:10 a m

8:40 am

8:35 am Tf

Tp

From this diagram, Tf = 1/6 + (X – 5)/40 T p = 35/60 + 5/60 + (X – 4.16667)/70 The required time headway is 5 min; therefore, T p – Tf = 5/60 min = 0.083 h 1/6 + X/40 – 1/8 - 7/12 - 1/12 - X/70 + 4.166667/70 = 1/12 Therefore, X = 45 mi

2.7 Travelers arriving at a certain airline counter at a given airport arrive according to the pattern shown below. It is estimated that on average it takes 45 seconds to serve a customer at the counter. For the first 30 minutes (i.e. from 9:00 to 9:30), the airline has only two counters open. At 9:30, however, a third counter opens and remains open until 10:30. Time Period 9:00 – 9:15 9:15 – 9:30 9:30 – 9:45 9:45 – 10:00 10:00 – 10:15 10:15 – 10:30

15 min Count 45 60 55 40 35 55

12

Cumulative Count 45 105 160 200 235 290


(a) Draw a cumulative plot showing the arrival and departure patterns for the travelers at the airline counters; (b) What is the length of the queue at 9:30? (c) What is the maximum length of the queue? (d) What is the time at which no-one remains in line? (e) What is the total wait time for all customers in units of customer/minutes?

If it takes 45 seconds to serve a customer, one counter can accommodate a total of 15*60/45 = 20 customer/15 minutes. Given this, the calculations proceed as follows: Time 9:00 9:15 9:30 9:45 10:00 10:15 10:30

Arrival Departure Queue 0 0 0 45 40 5 105 80 25 160 140 20 200 200 0 235 235 0 290 290 0

(b) Length of queue at 9:30 p.m. is equal to 105 – 80 = 25 travelers (c) Maximum length of the queue = 25 travelers (d) The time at which no one remains in line is 10:00 a.m. (e) Total wait is given by the area between the arrival and departure curves as follows: ½*5*15 + (5+25)/2*15 + (25+20)/2*15 + ½*20*15 = 750 passenger-minutes

2.8 An incident occurs on a freeway that has a capacity in the northbound direction, before the incident, of 4400 veh/hr and a constant flow rate of 3200 veh/hr during the morning commute. At 7:30 a.m., a traffic accident occurs and totally closes the freeway (i.e. reduces its capacity to zero). At 7:50 a.m., the freeway is partially opened with a capacity of 2000 veh/hr. Finally, at 8:10 a.m., the wreckage is removed and the freeway is restored to full capacity (i.e. 4400 veh/hr). (a) Draw cumulative vehicles’ arrival and departure curves for the scenario described; (b) Determine the total magnitude of delay, in units of veh-h, from the moment the accident occurs to the moment the queue formed totally dissipates.

13


(a)

p h v 0 0 6 3

7:30 am

h p v 0 0 4 4

Y

p h v 0 2 0 0 7:50 am 8:10 am X

(b) 2 Y = 3200 × ( + X ) 3 1 Y = 2000 × + 4400 × X 3 2 1 Y = 3200 × ( + X ) = 2000 × + 4400 × X 3 3 X = 1.222 hours Y = 6044.44 vehicles. The total magnitude of delay, in units of veh-h is the area under the curve. ½*(2/3+1.222)*6044.44 – ½*1/3*666.67 – ½*1.222*(6044.44-666.67) –1.222*666.667 = 1496.3 veh-h

14


2.9 A six-lane freeway (i.e. 3 lanes in each direction) has a capacity of 6000 veh/hr under normal conditions. On a certain day, an accident occurs at 4:00 p.m. The accident initially results in blocking two of the three lanes of the freeway, and hence reduces the capacity to only 2000 veh/hr for that direction. At 4:30 p.m., the freeway capacity is partially restored back to a value of 4000 veh/hr. Finally, at 5:00 p.m., the accident is totally cleared, and the full capacity of the freeway is restored. Given that the traffic demand at the accident site is given by the following table, determine: Time Period 4:00 – 4:15 4:15 – 4:30 4:30 – 4:45 4:45 – 5:00 5:00 – 5:15 5:15 – 5:30 5:30 – 5:45 5:45 – 6:00

15-min Volume 700 900 1100 1200 800 700 1100 900

(a) The maximum length of the queue formed at the accident site; (b) The time the queue dissipates; and (c) The total delay.

(a) The calculations can be tabulated as follows: Time 4:00 4:15 4:30 4:45 5:00 5:15 5:30 5:45 6:00

Arrival Departure Queue 0 0 0 700 500 200 1600 1000 600 2700 2000 700 3900 3000 900 4700 4500 200 5400 6500 7400

The maximum length of the queue is therefore 900 vehicles.

(b) The cumulative plot for this problem is shown be low.

15


4:45

As can be seen from the above cumulative plot, the time at which the queue dissipates occurs between 5:15 and 5:30. Between 5:15 and 5:30, the arrival rate is 2800 veh/h, and the departure rate is 6000 veh/h (see Figure below). Therefore, to calculate the time, T, into the 15-min interval during which the queue dissipates, we have: 200 + 2800 * T = 6000 * T 200 = 3200 * T T = 1/16 hours or 3.75 minutes Therefore, the queue dissipates at time 5:18.75 p.m.

16


h v p 0 0 2 8

200 veh

h p v 0 0 0 6

T

(c) The total delay is given by the area between the arrival and departure curves as follows: ½ * 15 * 200 + (200+600)/2*15 + (600+700)/2*15 + (700+900)/2*15 + ½*(900+200)*15 + ½*3.75*200 = 37875 veh-min or 631.25 veh-h

2.10 To evaluate the condition of bike paths, a condition index was developed which rates the surface condition of each bike path segment on a scale from 0 to 100, with 100 referring to a segment in perfect condition. The table below shows the inspection data for several bike path segments in terms of the Condition Index (CI) for each segment, along with its age (i.e. the number of years since its construction): Condition Index (CI) 100 98 96 93 100 93 88 86 100 95 90 83 100 92 85 82 81

Age (years) 0 0.5 1.2 3 0 2 4 7 0 2 4 8 0 3 6 9 10

17


It is postulated that the deterioration of bike path surfaces can be expressed using the following equation: 2 CI = a + b*(AGE) + c*(AGE) Using regression, develop a deterioration prediction curve for bike paths. Plot the resulting curve to show the typical deterioration trend for bike paths.

Linear regression can be performed using Microsoft Excel. follows: 2

CI 100 98 96 93

Age 0 0.5 1.2 3

Age 0 0.25 1.44 9

100 93 88 86 100 95 90 83 100 92 85 82 81

0 2 4 7 0 2 4 8 0 3 6 9 10

0 4 16 49 0 4 16 64 0 9 36 81 100

SUMMARY OUTPUT

Regression Statistics Multiple R 0.992971349 R Square 0.9859921 Adjusted R Square 0.983990971 Standard Error 0.852158585 Observations 17

Coefficients 99.84326958 -3.03900983 0.118281978

Intercept Age Age^2

Therefore, the model can be expressed as: 2

CI = 99.84 – 3.039*AGE + 0.118*AGE This model can be graphed as shown below. Condition Index vs. Age 100 x e d n I n o i t i d n o C

90 80 70 60 0

2

4

The returned output as

6

8

10

Age (years)

18

12


2.11 Develop a relationship between the total number of trips generated by a wholesale tire store and the gross floor area of the store. The data set below the average number of vehicle trips/day to and from several tire stores. How well does your model fit the data? How many trips per day are generated by a store with a gross floor area of 1000 sq. ft? Store ID 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Trip Ends/Day 170 300 250 350 340 200 230 250 100 400 150 380 220 270 280

Gross Floor Area 9 14.5 12 17 18 11 14 16 6 19 8 17.5 11.5 12.5 14

A linear regression analysis of the data can be performed with the resulting relationship:. Y = 21.2197 X − 23.596 The model fit the given data reasonably well as R 2 = 0.905 , which is close to 1. For a store with a gross floor area of 1000 sq. ft., If X = 10 Y = 212.197 – 23.56 = 189 trip ends/day

2.12 It is postulated that the relationship between the average speed of a traffic stream, u, in mph, and the density (which gives the number of vehicles/unit length), k , in veh/mile for a given transportation facility can be expressed as follows: k j u = c ln k where c and k j are parameters. To fit the above equation, average speeds and density were collected from the facility at different times of the day and different

19


usage levels. The data collected are as shown below. Use regression to fit the above equation to the data. What are the values for the two parameters c and k j? Speed, u (mph) 53 40 37 10 25 20 48 45 27 15 13 35 34 37

The given equation, u = c ln

Density, k (veh/mile) 22 44 52 125 70 85 27 36 65 90 97 58 60 51

k j k

, can be made linear, u = c ln k j − c ln k

u

ln k

53

3.091042

40

3.78419

37

3.951244

10

4.828314

25

4.248495

20

4.442651

48

3.295837

45

3.583519

27

4.174387

15

4.49981

13

4.574711

35

4.060443

34

4.094345

37

3.931826

where c ln k is the y-intercept and c ln k j is the slope and independent variable c ln k j = 139.6387

c = 26.802

20


Therefore,

26.802 * ln k j = 139.6387 ln k j = 5.21 k j = 183 veh/mi

2.13 In the context of probability theory, explain what is meant by a random variable.

A random variable is a special type of a probability model that assigns a value to each outcome.

numerical

2.14 Provide examples of random variables that arise within the context of transportation systems problems, and that follow each of the following probability distributions: (1) binomial distribution; (2) geometric distribution; (3) Poisson distribution; and (4) normal distribution.

•

Binomial Distribution: A random variable in a binomial distribution has a value of one with probability p and value of 0 with probability 1- p. An example would be the probability of a certain number of aircrafts of a certain type landing at an airport during a given time period. • Geometric Distribution: The random variable in a geometric distribution represents the probability that the first success will occur on the xth trial. An example would be the probability that the fifth vehicle at an intersection would turn left. • Poisson Distribution: The random variable in a Poisson distribution represents the probability that exactly x units will arrive during time interval t . The Poisson distribution best describes the arrival pattern of vehicles at an intersection or travelers at an airport counter. • Normal Distribution: The random variable is dependent upon a certain mean and

 1  x − µ  2  standard deviation. f ( x) = exp −   . σ 2π  2  σ   1

The distribution of

driving speeds along a road can be described using a Normal distribution.

2.15 Differentiate between the probability distribution function (pdf) and the cumulative distribution function (cdf).

The probability density function associates each value of a discrete random variable to its probability. 0 ≤ p(x) ≤ 1 ∑ p(x) = 1

21


The cumulative distribution function adds probability values for the random variable that are less than or equal to x. Increases at each of the values that the random variable assumes. F(x) = P[ X ≤ x]

2.16 In the context of probability density functions, what does the p-fractile function compute? Illustrate using a simple diagram.

In a normal distribution the p-fractile is the number g such that the probability that X is less than g is equal to p.

p

1-p g

2.17 The following table lists the observed speeds of a number of trains as they pass a certain point midway between two stations. You are asked to determine: (1) the average speed; (2) the variance; and (3) the standard deviation.

22


Train ID 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Speed (mph) 71 69 73 55 62 53 67 45 54 63 67 41 75 48 43 55 58 59 63 67

The mean, variance, and standard deviation can be readily calculated a spreadsheet or 2 statistical analysis software package. The mean is 59.4 mi/h; the variance 95.8 (mi/h) , and the standard deviation 9.79 mi/h.

2.18 Pedestrians arrive at a signalized intersection crossing at the rate of 600 pedestrians/hr. The duration of the red interval for the pedestrians at that intersection is 45 seconds. Assuming that pedestrians’ arrival pattern can be described using a Poisson distribution, what is the probability that there will be more than 10 pedestrians waiting to cross at the end of the pedestrian red interval?

λ = 600/3600 = 0.167 pedestrian/second (average arrival rate) λ t = (0.167 ped/sec) * 45 sec = 7.5 pedestrians (expected number of arrivals) P (X > 10)

= 1 – P(X <= 10) = 1 – POISSON (10, 7.5) = 0.138

23


2.19 Airplanes arrive at an airport at an average rate of 10 aircraft/hr. Assuming that the arrival rate follows a Poisson distribution, calculate the probability that more than 4 aircraft would land during a given 15 minutes.

The probability that more than 4 aircraft would land during this time is the same as 1 – the probability that 4 or less will show. 1aircraft λ = 6 min t = 15min P (X > 4) = 1 – P(X ≤ 4) 1 – POISSON (4, 2.5) = 0.1088

2.20 An approach to an intersection carries an average volume of 1000 veh/hr with 15% of the vehicles desiring to turn left. The cycle length at the intersection is equal to 75 seconds. The city would like to construct a left-turn bay at the intersection in order to minimize the probability of the left-turn vehicles blocking the through lane. You are asked to determine the minimum length of the left-turn bay at that approach so that the probability of an arriving left-turning vehicle not finding enough room on the left-turn bay is less than 10%. Assume the average vehicle length is equal to 20 ft. Assume Poisson arrivals.

λ = 0.15*(1000 veh/h)/(3600 sec/h) = 0.04167 veh/sec λ t = (0.04167 veh/sec)* 75 sec = 3.125 veh What we are seeking here is the minimum number of vehicles, N, that the turning bay should accommodate so that the probability of X > N is less than 10%. We know that P(X > N) is equal to 1.0 – P(X ≤ N). Therefore, P(X > N) less than 10% is equivalent to P(X ≤ N) greater than 0.90. The calculations can be performed using Excel as follows: N 1 2 3 4 5 6

P(X ≤ N) 0.18124 0.395776 0.61925 0.79384 0.902959 0.959791

24


As can be seen, the smallest number of vehicles with a probability of occurrence of less than 0.10 is 5 vehicles. Therefore, the left turn bay should be equal to 100 ft (20 ft for each of 5 vehicles).

2.21 A trucking company has enough capacity to transport 2000 tons of a certain material per week. If the weekly demand for transporting such material is normally distributed with a mean of 1750 tons, and a standard deviation of 300 tons, determine: (a) The probability that within a given week, the company would have to turn down requests for transporting the material; (b) The capacity that the company should maintain so that the probability of it turning down transportation requests is less than 5%.

(a)

Any spreadsheet or statistical analysis software package can be used. Using Excel, P (X > 2000) = 1.0 – P(X ≤ 2000) = 1.0 – NORMDIST(2000, 1750, 300, 1) = 1.0 - 0.797671619 = 0.2023

(b)

Using Excel If probability of P(X ≥ g) = 0.05 then P(X ≤ g) = 0.95 g = NORMINV(0.95,1750,300) g = 2243.456088 ≈ 2300 ton

2.22 A company operates small ferryboats between a small island and the mainland. Each ferryboat can carry a maximum of 6 vehicles. It is estimated that vehicles arrive at the ferryboat dock at the rate of 12 veh/hr. The company is interested in determining the frequency at which the ferryboats should be operated so that the probability of a vehicle being left at the dock because the boat does not have enough capacity does not exceed 10%. Assume that vehicles arrive according to a Poisson distribution.

For this problem, λ = 12 veh/h = 12/60 or 0.2 vehicles/min What we are seeking is the frequency or the time headway between the ferry arrivals that would ensure that the probability of having more than 6 vehicles arriving in that period is less than 10%. Therefore, P (X > 6) ≤ 0.10

25


We proceed by assuming different time headways (i.e. 5 min, 10 min, … etc.), and each time calculating P(X > 6) which is equal to 1.0 – P (X ≤ 6) as shown below. T 5 10 15 20 25

λ t 1 2 3 4 5

P(X ≤ 6) 0.999917 0.995466 0.966491 0.889326 0.762183

P(X > 6) 8.32411E-05 0.004533806 0.033508535 0.110673978 0.237816537

From the table, it can be seen that having a ferry every almost 20 minutes (to be more precise, every 19 minutes) would ensure that the probability of having more than 6 vehicles waiting for a ferry is less than 10%.

2.23 An airport serves four different types of aircraft: Heavy (H); Large (L); Medium (M) and Small (S). During a typical hour, 40 Heavy aircraft, 50 Large, 60 Medium, and 70 Small aircraft land. Determine the probability that: (a) The next aircraft to land is a Small aircraft; (b) In a stream of 20 aircraft, at least 5 Small aircraft would land; (c) The first Medium aircraft would be the fifth aircraft.

(a) 70 40 + 50 + 60 + 70

= 0.318

(b)

This requires use of the binomial distribution. P(X ≥ 5) = 1.0 – P(X ≤ 4) P(X ≤ 4) = BINOMDIST (4 ,20 ,0.318 ,1) = 0.187767 P(X ≥ 5) = 1.0 – P(X ≤ 4) = 1 − 0.187767 = 0.812233 P(X ≥ 5) = 0.812

(c)

This requires use of the geometric distribution. 60 = 0.273 40 + 50 + 60 + 70 x–1

P (X = 5) = (1 – p) p 4 = (1 – 0.273) * 0.273 = 0.076 = 0.076

26


2.24 In the context of queuing theory, what is the difference between the time a customer spends in the queue, and the time she spends in the system?

•

A customer is in the queue when entering the waiting line and until service is provided. • A customer is in the system when entering the waiting line and until service is completed.

2.25 What are examples of queuing systems in transportation systems.

• • • • •

Vehicles waiting at a tollbooth Airplanes waiting to land or take off at a runway Vehicles waiting to pass through a work zone Trucks or ships waiting to be unloaded at a marine terminal People waiting to renew their driver’s licenses.

2.26 On what basis are different types of queues distinguished?

• • •

Customer arrival patterns Customer departure Service patterns and queue disciplines.

2.27 Why do queues form?

Queues form when the arrival rate is greater than the departure rate or as a result of the stochastic fluctuations in arrival patterns.

2.28 At a given airport, aircraft arrive at an average rate of 8 aircraft/hr following a Poisson distribution. The average landing time for an aircraft is 5 minutes. However, this time varies from one aircraft to another. This variation can be assumed to be exponentially distributed. Determine: (a) The average number of aircraft awaiting clearance to land; (b) The average time an aircraft spends in the system; (c) The probability that there will be more than 5 aircraft awaiting clearance to land.

27


This is an M/M/1 problem λ = 8 aircrafts/hour µ = 12 aircrafts/hour ρ = 8/12 = 0.66667 The average number of aircrafts awaiting clearance: (a) The average length of the queue: Q =

ρ 2 (1 − ρ )

0.66667 2

=

(1 − 0.66667) )

= 1.333 aircraft

(b) The average time an aircraft spends in the system: t =

1 ( µ − λ )

1

=

(12 − 8) )

= 0.25 hr/aircraft or 15 minutes/aircraft

(c) The steady-state probability that exactly 5 aircraft are in the system: p n = (1 − ρ ) ρ

n

5

= (1 – 0.66667) * 0.66667 = 0.044

2.29 Travelers arrive at the ticket counter of a particular airport at the rate of 90 customers/hr following a Poisson distribution. The average service time per customer is more or less fixed and is equal to 30 seconds. Determine the average queue length, the average waiting time in the queue and the average time spent in the system. customers 1hour customers λ = 90 × = 1.5 hour 60 min min sec customers customers µ = 60 ×1 =2 1 min 30 sec min λ λ λ p = if <1 = .75 < 1 µ µ µ This is an M/D/1 Queue. Therefore ρ 2 .75 2 Q = = = 1.125travelers 2(1 − ρ ) (1 − .75)

W =

t =

ρ 2 µ (1 − ρ ) 2 − ρ

2 µ (1 − ρ )

=

=

.75 2( 2)(1 − .75) 2 − .75

2( 2)(1 − .75)

= .75

= 1.25

min traveler min

traveler

28


2.30 Travelers arrive at a ticket counter of an Amtrak train station at the rate of 100 travelers/hr. It has been estimated that it takes an average of 30 seconds to serve each customer at the counter. Assuming that arrivals can be described using a Poisson distribution, determine the average wait time in the queue, and the average number of customers waiting in the queue.

Based upon the information given, it is reasonable to assume that this is an M/D/1 Queue. λ = 100/60 = 1.6667 travelers/min µ = 60/30 = 2 travelers/min ρ = 1.66667/2 = 0.83333 Therefore, the average wait time: W =

ρ 2 µ (1 − ρ )

=

.83333 2( 2)(1 − .0.8333)

= 1.25

min traveler

The average number of customers in the queue: ρ 2 .83333 2 Q = = = 2.08 travelers 2(1 − ρ ) 2 * (1 − .83333)

2.31 Travelers at an airport arrive at a given security check point at the rate of 120 travelers/hr. Check-in times for passengers vary according to a negative exponential distribution with an average value of 25 seconds per passenger. Determine: (a) The average number of passengers waiting in line in front of the security check point; (b) The average waiting time for passengers; (c) The average time a passenger spends in the system.

This is an M/D/1 Queue. (a) λ = 120

customers 1hour

×

=2

customers

hour 60 min min sec customers customers µ = 60 ×1 = 2.4 1 min 25 sec min λ λ λ p = if <1 = .8333 < 1 µ µ µ

29


2

Q =

ρ

(1 − ρ )

.8333

=

2

(1 − .8333)

= 4.1655travelers

(b) W =

λ µ ( µ − λ )

=

2 2.4( 2.4 − 2)

= 2.08333

min travelers

(c) t =

1 ( µ − λ )

=

1 ( 2.4 − 2)

= 2.5 min

2.32 In the previous problem, it is desired to limit the probability that there would be more than 7 passengers in line to a value less than 5%. Determine the maximum rate of arrival that should be allowed.

The probability that we have n passengers in the system is given by: p n = (1 − ρ ) ρ n Also, the probability that we have more than 7 passengers is equal to: 1.0 – P(0) – P(1) – P(2) – P(3) – P(4) – P(5) – P(6) – P(7) The value of ρ that would ensure that the above expression is less than 0.05 is the correct answer. This can be found by trial and error using Microsoft Excel as shown below. N 0 1 2 3 4 5 6 7

0.25 0.1875 0.140625 0.105469 0.079102 0.059326 0.044495 0.033371

P(N ≤ 7) P(N > 7)

0.899887 0.100113

po 0.75

As can be seen, the requested value of ρ is 0.75 Therefore, the arrival rate λ is equal to 0.75 * 2.4 = 1.8 travelers/min or 108 travelers/h.

30


2.33 Cite four examples of optimization problems arising within the field of Transportation Infrastructure Engineering.

• • • •

In the planning process the transportation planner is typically seeking the optimal allocation or use of the available funds. During the design stage, the transportation engineer identifies the optimal alignment for a proposed transportation travelway, which would minimize the cost of construction and earthwork. In traffic signal design, the traffic engineer is typically trying to come up with an optimal signal plan that would minimize the total travel time or travel delay. In transportation infrastructure maintenance and management, optimization techniques are commonly used to determine optimal treatment strategies and the optimal timing for implementing that strategy.

2.34 What are the three basic steps in formulating optimization models?

• • •

Identification of the decision variables Formulation of the objective function Formulation of the model’s constraints.

2.35 A transit authority must repair 120 subway cars/month. At the same time, the authority must refurbish 60 subway cars. Each task can be done in the authority’s facility or can be contracted out. Private contracting increases the cost by $1000/car repaired, and by $1500/car refurbished. Car repair and refurbishing take place in three shops, namely the Assembly shop, the Machine shop and the Paint shop. Repairing a single car consumes 2% of the Assembly shop capacity, and 2.5% of the Machine shop capacity. On the other hand, refurbishing a single car takes up 1.5% of the Assembly shop capacity, and 3% of the Paint shop. Formulate the problem of minimizing the monthly expense for private contracting as a linear program, and solve it using Microsoft Excel’s Solver.

Decision Variables: X1 = number of cars repaired in house X2 = number of cars repaired by contracting X3 = number of cars refurbished in house X4 = number of cars refurbished by contracting Objective Function: Minimize Z = 1000 X2 + 1500 X4

31


Constraints: Assembly Shop: 2X1 + 1.5X3 <= 100 Machine Shop: 2.5X1 <= 100 Paint Shop: 3X3 <= 100 Number of cars repaired X1 + X2 = 120 Number of cars refurbished X3+ X4 = 60 The problem can be solved using Excel Solver as follows:

X1 Assembly 2 Shop Machine Shop 2.5 Paint Shop Number 1 repaired Number refurbished Contribution 25 Value

X2

X3

X4

1.5 3 1 1 1000 95

1 1500 33.33333 26.66667

100 62.5 100

<= <= <=

100 100 100

120 60 135000

= =

120 60

Therefore, the transit authority should repair 25 cars and refurbish 33 in house. The rest will have to be contracted out (i.e. 95 repairs and 27 refurbishes).

2.36 You have been asked to prepare the best plan to transport finished products from three production plants to four market places. The production capacities for the three plants are 2000, 3500, and 4000 units, respectively. At the same time, the demand that has to be met at each market place is for 3200 units at market 1, 2800 units at market 2, 2000 at market 3, and 1500 at market 4. The unit shipping costs are given in the table below.

Plant 1 Plant 2 Plant 3

Market 1 4.5 11 5

Market 2 6.5 4 7

32

Market 3 4 12 8

Market 4 7 3 4


This is a typical transportation problem. The solution can be easily formulated and derived using Microsoft Excel Solver as follows: Market 1

Market 2

Market 3

Market 4

Supply

Plant 1

4.5

6.5

4

7

2000

Plant 2

11

4

12

3

3500

Plant 3

5

7

8

4

4000

Demand

3200

2800

2000

1500

Market 1

Market 2

Market 3

Market 4

Plant 1

0

0

2000

0

2000

Plant 2

0

2800

0

700

3500

Plant 3

3200

0

0

800

4000

3200

2800

2000

1500

Transportation Cost

40500

33

Downloadable Solution Manual for Transportation Infrastructure Engineering a Multim 1

Recommend Documents