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CE361 Introduction to Transportation Engineering |Posted: Tuesday 4 December 2007 | |

|Homework 10 (not assigned) Solutions | |

Moving Freight and People

1. Trade Between Countries – Timber and Lamb. Repeat FTE Example 12.1, but with the data contained in the table below.

| |Country Q |Country A |

|Population |687,000 |980,000 |

|Timber use |2.96 m3 per person per year |1.40 m3 per person per year |

|Timber price |$426 per m3 |$705 per m3 |

|Lamb consumption |22.25 lb per person per year |14.4 lb per person per year |

|Lamb price |$2.86/lb |$1.94/lb |

It costs $4.36/m3 to ship timber and $7.79/cwt to ship lamb. How much will the consumers of each country save by importing the good with the lower price? Show your calculations and summarize your results using the format of FTE Table 12.2. What tariff could each country apply to the imported product? Explain the reason for the tariff you recommend.

(a) Sample calculations: Total cost to buy domestic lamb in Country Q = 687,000 pop x 22.25 lbs/pop x $2.86/ lbs = $43,717,245. Total cost to buy imported lamb in Country Q excluding transport costs = 687,000 pop x 22.25 lbs /pop x $1.94/ lbs = $29,654,355. Cost to transport lamb from Country A to Country Q = 687,000 pop x 22.25 lbs /pop x $7.79/100 lbs = $1,190,760. Persons in Country Q will save $43,717,245 – ($29,654,355 + $1,190,760) = $12,872,130 by importing lamb from Country A. In a similar fashion, persons in Country A will save $377,000,000 by importing timber from Country Q. The facsimile of FTE Table 12.2 below is produced by a spreadsheet that applied rounding of large values. Timber = Product X and Lamb = Product Y.

[pic]

(b) Import tariff strategies are as follows.

• If the objective is to protect its relatively inefficient domestic industry, a country’s import tariff should be at least as large as the difference in retail price, minus the transportation costs.

• If the tariff has the purpose of raising revenues, the tariff should be no larger than that margin. For timber to Country A, that difference is $377,000,000, or $377,000,000/(980,000*1.4) = $274.78/m3.

Either answer will be acceptable, but you should be aware of the other strategy.

2. Mixed Freight Train. FTE Exercise 12.12, Parts (a) and (b)

(a) Total weight = 70,000 + 60,000 = 130,000 lbs / 2000 lbs per ton = 65 tons

(12.5) Rtt = [pic]+(20*G) = [pic]+(20*.25)

Rtt = (7.415 + 5.0) = 12.415 lbs/ton

(12.8) TEdraw = 12.415 lbs/ton * 120 cars * 65 tons per car = 96,840 lbs

(b) (12.10) = (12.11): 96,840 + (1500*N) = (375*5500*N*0.91)/60; Solve for N=3.25 ( 4 engines

3. Horsepower for 3x4 tow. A 3 x 4 tow (each scow is 35 by 195 feet) is carrying coal and has a draft of 8.5 feet. The desired speed in still water is 5.3 knots. If the water is 13 feet deep, estimate the horsepower to be delivered to the tow if the engine/propeller system is 70 percent efficient.

Follow steps in FTE Example 12.14 so that we can use (12.14).

(12.15) f = 0.0106 L-0.031 = 0.0106 * (4*195)-0.031 = 0.008623.

(12.17) Fr = [pic] = 0.0565. Because Fr ................
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