Answer to Question #205604 in Economics of Enterprise for lokesh

Question #205604

F.2. Assume that 362 years ago Peter Minuit, governor of the Dutch West India Company, bought Manhattan Island from the Indians for $24 worth of beads, cloth, and trinkets.

(a) If the Indians had insisted on $24 cash and invested the $24 at 6 percent, what would be compounded amount be worth now?

(b) If the compounding were continuous instead of annual, how much would the $24 investment be worth now?

(c) Assume the actual value of Manhattan Island, land only (no improvements), is estimated to be 28 billion. What is the average percent increase in value with the annual compounding each year?


1
Expert's answer
2021-06-15T12:34:41-0400

Solution:

a.). A = P(1 + i)n

Where: P = Principal amount = $24

            i = interest rate = 6% = 0.06

            n = No. of years = 362 years

"A = 24(1+0.06)^{362} = \\$34,748,372,427"

The compounded amount will now be worth = "\\$34,748,372,427"

 

b.). Continuous compounding formula: "A = P\\times e^{(i\\times t)}"

Where: P = Principal amount = $24

            i = interest rate = 6% = 0.06

            n = No. of years = 362 years

            e = mathematical constant approximated as 2.7183.

 

"A = 24\\times 2.7183^{(0.06\\times 362)} =" A = 24*2.7183(0.06*362) ="\\$65,035,494,477"


The $24 investment will now be worth = "\\$65,035,494,477"


c.). To get the average percent increase in value with the annual compounding each year, we use the Compound annual growth rate formula:

"CAGR = (\\frac{EV}{BV})^{\\frac{1}{n} } - 1"

Where: EV = Ending Value = $28,000,000,000,000

            BV = Beginning value = $24

            n = No. of years = 362 years

"CAGR = (\\frac{28,000,000,000,000}{24})^{\\frac{1}{362} } - 1 = 0.07977 = 7.98\\%"


The average percent increase in value with the annual compounding each year = "7.98\\%"



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS