Question

A major oil company is considering the optimal timing for the construction of new refineries. From past experience, each doubling of the size of a refinery at a single location results in an increase in the construction costs of about 68 percent. Furthermore, a plant size of 10,000 barrels per day costs $6 million. Assume that the demand for the oil is increasing at a constant rate of two million barrels yearly and the discount rate for future costs is 15 percent.
a. Find the values of k and a assuming a relationship of the form f(y) _ kya.
Assume that y is in units of barrels per day.
b. Determine the optimal timing of plant additions and the optimal size of each plant.
c. Suppose that the largest single refinery that can be built with current technology is 15,000 barrels per day. Determine the optimal timing of plant additions and the optimal size of each plant in this case. (Assume 365 days per year for your calculations.)

Answer 1

Answer:

a, 22276.07

b. $32.9157 million

c.$29.9669million

Explanation:

Find the values of k   and  a assuming a relationship of the form   Assume that  f(y)=ky^a is in units of barrels per day.

$\frac{f(2y)}{f(y)} =1.75=\frac{k(2y)^a}{k(y)^a} =2^{a} =a=\frac{Ln(1.75)}{Ln(2)} =0.8073$

$f(y)=ky^a=k=\frac{f(y)}{y^a} =\frac{25}{6000^0.807} =22276.07$

b. Determine the optimal timing of plant additions and the optimal size and cost of each plant addition.a=0.8073, rx=0.41

optimal timing x=rx/r=2.05yrs

optimal size xD=2.05(1.5)

3.075million barrels/year

$f(y)=ky^a=0.0223(\frac{3.075*10^5}{365} )^0.8073=32.9157million$

$32.9157 million

c. Suppose that the largest single refinery that can be built with current technology is 7,500 barrels per day. Determine the optimal timing of plant additions and the optimal size and cost of each plant in this case

Optimal size xD=min$\frac{307500}{365} ,7500= 7500 barrls/day =2.735million barrels /year$

Optimal timing will be X^*=x*D/D=2.7375/1.5=1.825 year

optimal cost f(y)=ky^a=0.0223(7500)^0.8073=$29,9669 milion

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