Assignment #2 (sep2019)-gdb3023-solution

GDB3023 ENGINEERING ECONOMICS & ENTREPRENEURSHIP SEPTEMBER 2019 ASSIGNMENT #2 [CLO1]

2-3

A group of enterprising engineering students has developed a process for extracting combustible methane gas from cow manure (don’t worry, the exhaust is odorless). With a specially adapted internal combustion engine, the students claim that an automobile can be propelled 15 miles per day from the “cow gas” produced by a single cow. Their experimental car can travel 60 miles per day for an estimated cost of $5 (this is the allocated cost of the methane process equipment—the cow manure is essentially free). (a.) (b.)

How many cows would it take to fuel 1,000,000 miles of annual driving by a fleet of cars? What is the annual cost? How does your answer to Part (a) compare to a gasoline-fueled car averaging 30 miles per gallon when the cost of gasoline is $3.00 per gallon?

Solution (a)

# 𝑐𝑜𝑤𝑠 =

𝑚𝑖𝑙𝑒𝑠 𝑦𝑒𝑎𝑟 𝑑𝑎𝑦𝑠 𝑚𝑖𝑙𝑒𝑠 (365 )(15 ) 𝑦𝑒𝑎𝑟 𝑑𝑎𝑦

1,000,000

𝐴𝑛𝑛𝑢𝑎𝑙 𝑐𝑜𝑠𝑡 = (1,000,000

(b)

= 182.6 𝑜𝑟 183 𝑐𝑜𝑤𝑠

𝑚𝑖𝑙𝑒𝑠 $5 )( ) = $83,333 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 𝑦𝑒𝑎𝑟 60 𝑚𝑖𝑙𝑒𝑠

𝐴𝑛𝑛𝑢𝑎𝑙 𝑐𝑜𝑠𝑡 𝑜𝑓 𝑔𝑎𝑠𝑜𝑙𝑖𝑛𝑒 =

𝑚𝑖𝑙𝑒𝑠 𝑦𝑒𝑎𝑟 𝑚𝑖𝑙𝑒𝑠 30 𝑔𝑎𝑙𝑙𝑜𝑛

1,000,000

$3

(𝑔𝑎𝑙𝑙𝑜𝑛) = $100,000 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟

It would cost $16,667 more per year to fuel the fleet of cars with gasoline.

2-4

A municipal solid-waste site for a city must be located at Site A or Site B. After sorting, some of the solid refuse will be transported to an electric power plant where it will be used as fuel. Data for the hauling of refuse from each site to the power plant are shown in Table P2-4. Table P2-4

Average hauling distance Annual rental fee for solid-waste site Hauling cost

Site A 4 miles $5,000 $1.50 / yd3-mile

Site B 3 miles $100,000 $1.50 / yd3-mile

If the power plant will pay $8.00 per cubic yard of sorted solid waste delivered to the plant, where should the solid-waste site be located? Use the city’s viewpoint and assume that 200,000 cubic yards of refuse will be hauled to the plant for one year only. One site must be selected.

Solution Cost Rent Hauling Total

Site A = $5,000 (4)(200,000)($1.50) = $1,200,000 $1,205,000

Site B = $100,000 (3)(200,000)($1.50) = $900,000 $1,000,000

Note that the revenue of $8.00/yd3 is independent of the site selected. Thus, we can maximize profit by minimizing total cost. The solid waste site should be located in Site B.

2-13 A large company in the communication and publishing industry has quantified the relationship between the price of one of its products and the demand for this product as Price = 150 – (0.01 × Demand) for an annual printing of this particular product. The fixed costs per year (i.e., per printing) is $50,000 and the variable cost per unit is $40. What is the maximum profit that can be achieved if the maximum expected demand is 6,000 units per year? What is the unit price at this point of optimal demand?

Solution p = 150 - 0.01D

CF = $50,000

cv = $40/unit

Profit = 150D - 0.01D2 - 50,000 - 40D = 110D - 0.01D2 - 50,000 d(Profit)/dD = 110 - 0.02D = 0 D = 5,500 units per year, which is less than maximum anticipated demand At D = 5,500 units per year, Profit = $252,500 and p = $150 - 0.01(5,500) = $95/unit.

2-15 A company produces and sells a consumer product and is able to control the demand for the product by varying the selling price. The approximate relationship between price and demand is 𝑝 = $38 +

2,700 5,000 − , 𝐷 𝐷2

𝑓𝑜𝑟 𝐷 > 1,

where p is the price per unit in dollars and D is the demand per month. The company is seeking to maximize its profit. The fixed cost is $1,000 per month and the variable cost (CV) is $40 per unit. (a.) (b.)

What is the number of units that should be produced and sold each month to maximize profit? Show that your answer to Part (a) maximizes profit.

Solution (a) 𝑃𝑟𝑜𝑓𝑖𝑡 = [38 +

2700 5000 − 2 ] 𝐷 − 1000 − 40𝐷 𝐷 𝐷

= 38𝐷 + 2700 −

5000 − 1000 − 40𝐷 𝐷

𝑃𝑟𝑜𝑓𝑖𝑡 = −2𝐷 −

5000 + 1700 𝐷

𝑑(𝑃𝑟𝑜𝑓𝑖𝑡) 5000 = −2 + 2 − 0 𝑑𝐷 𝐷 𝑜𝑟, 𝐷2 =

(b)

5000 − 2500 𝑎𝑛𝑑 𝐷∗ = 50 𝑢𝑛𝑖𝑡𝑠 𝑝𝑒𝑟 𝑚𝑜𝑛𝑡ℎ 2

𝑑2 (𝑃𝑟𝑜𝑓𝑖𝑡) −10,000 = < 0 𝑓𝑜𝑟 𝐷 < 1 𝑑𝐷2 𝐷3 Therefore, D* = 50 is a point of maximum profit.

2-29 One component of a system’s life-cycle cost is the cost of system failure. Failure costs can be reduced by designing a more reliable system. A simplified expression for system life-cycle cost, C, can be written as a function of the system’s failure rate: 𝐶𝐼 𝐶 = + 𝐶𝑅 . 𝜆. 𝑡 𝜆

CI CR 𝜆 t

Where;

(a.) (b.) (c.)

= = = =

investment cost ($ per hour per failure) system repair cost system failure rate (failure/operating hour) operating hours

Assume that CI , CR, and t are constants. Derive an expression for 𝜆, say 𝜆∗, that optimizes C. Does the equation in Part (a) correspond to a maximum or minimum value of C? Show all work to support your answer. What trade-off is being made in this problem?

Solution (a)

𝑑𝐶 𝑑𝜆

=−

𝐶𝐼 𝜆2

+ 𝐶𝑅 𝑡 = 0 𝐶

𝜆2 = 𝐶 𝐼𝑡

Or,

𝑅

𝑎𝑛𝑑 𝜆∗ = (𝐶𝐼 /𝐶𝑅 𝑡)1/2

We are only interested in

positive root. (b)

𝑑2 𝐶 𝑑𝜆2

=

2𝐶𝐼 𝜆3

> 0 𝑓𝑜𝑟 𝜆 > 0

Therefore, 𝜆* results in a minimum life-cycle cost value. (c)

Investment cost versus total repair cost