Economics

Economics

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Question 1 (Bad news principle). In the context of the two-period example we studied in class, let the direct cost of investment be I the interest rate be r, and the initial price of a widget at t = 0 be P0. At t = 1, the price becomes

P(

(1

+ u)P0

with probability q

 

q

(1

 

d)P0

with probability 1

 

 

 

 

 

 

value of waiting and seeing? (c) Show that if it`s optimalwasto invest.aftercoma price drop then it`s optimal to invest at t = 0 by showing that, in this case, the di erence

(a) What is the net present value of investing today? (b) Wh t is the net present

$200 now and will always remainresourceth . Howeve , the direct cost of investment I is uncertain.

between (b) and (a) is positive. (d) Derive the initial price Psuch that the rm is just indi erent between investing now and waiting.

Question 2 (Uncertainty over cost). In the context of the two-period example we

studied in class, suppose that there is no uncertainty ov the price of widgets: it is

(a) Suppose that investing today (i.e., at t = 0) costs $1,600, call this I0. Next

year, the investment cost I1

will either increase to $2,400 or decrease to $800, each

 

 

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with probability 0.5. The interest r te is 10% per year. What is the rm`s optimal

investment strategy?

 

 

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(b) Now suppose instead that, to build a factory, one must initially spend $1,000.

rm`s optimal investment strategy?

AfterThist sharedinitial inv stm nt, with probability 0.5 the factory will be complete, but with probability 0.5 an additional $3,000 will be required to complete it. What is the

Question 3 (Uncertainty over interest rates). (a) Consider an investment that yields $1 per year forever. (i) What is the present value of the investment when the interest rate r = 10%? (ii) What is the present value if r can wither equal 5% forever or 15% forever, each with equal probability?

 

(b) In the context of the two-period example we studied in class, x the widget price at $200 forever, and the cost of building the factory at $2,000. Today (at t = 0) the interest rate is r= 10%, but next year (t = 1) it will either increase to 15% or decrease to 5%, each with equal probability. The interest rate will remain at this new level thereafter. (i) What is the rm`s optimal investment strategy? (ii) How does your answer change when r is held always constant at 10%?

Question 4 (Scale versus exibility). An electric utility company is currently at

capacity, so its demand equals its supply, and faces constant demand growth of 100

megawatts (MW) per year, so the utility must add capacity. It has two choices: either

build a 200 MW coal- red power plant at a capital cost of $180 million, or build a 100 MW oiled- red plant at a capital cost of $100 million.

per year for each 100 MW of power. Suppose that thewasprice of coal.remainscomxed forever, so the operating cost of the coal- red plant is always $19 milli n. However,

At current fuel prices, the operating cost of the coal- red plant is $19 million per year

for each 100 MW of power, and the operating cost of the oil- red plant is $20 million

of the coal- red plant is the mresourceindi erent between investing in the coal- red plant and investing in an oil- red plant today?

assume that the oil price will uctuate next year, translating into operating costs

for the oil- red plant as follows. With probability 0.5, the oil- r d plant`s operating

costs will be $30 million per year, and also with probability 0.5, the operating costs will be $10 million. Operating costs will remain xed th after.

(a) What is the rm`s optimal capacity expansion plan? (b) At what capital cost

Question 5 (Three perio ). Let`s extend the two period model from class to three.

 

 

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At t = 0 the price of widgets is P0. At t = 1 it will either increase or decrease by

50% (i.e., P= 1:5Por P= 0:5P0), each with probability 0.5. Then, at t = 2, it

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will aga n e ther increase or decrease by 50% with equal probability. Hence, there are

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three poss ble values for P2: 2:25P0

, 0:75P0, and 0:25P0. The price then remains at

this level for all t 2. Fix I = $1; 600.

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