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Notes from Chapter 1
38. Based on past experience, a chemicals firm estimates that the cost of new capacity additions
obeys the law:
f(y) = .0205y^.58
y = is measured in tons per year
f(y) = in millions of dollars.
a = .58
D=Demand is 3,000 tons per year
r = rate of 12 percent per year
(1). Determine the optimal timing of plant additions and the optimal size of each addition.
Find u from a in f(u)=u/((e^u)1) table
Then x = u/r = .75/.12=6.25 years = optimal timing
Y=optimal size=xD=6.25*3000=18,750tons
(2). What is the cost of each addition?
f(y)=.0205(18750)^.58= $6.167million
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 85 percent. Furthermore, a plant of size 6,000 barrels per day costs
$30 million.
a) Find the value of
assuming a relationship of the form
Since doubling capacity increases cost by 85%,
So,
a.
(3 points) Find the value of
assuming a relationship of the form
Assume that
is in units of
barrels per day.
So,
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View Full Document Chapter 2
Moving average is average of n things usually for months and is onestep ahead.
Two step ahead is two forecasts down instead of one. Eg. The twostep ahead forecast for July is
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This note was uploaded on 03/06/2011 for the course ISEN 315 taught by Professor Hiram during the Spring '08 term at Texas A&M.
 Spring '08
 HIRAM

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