91Ó°ÊÓ

Seventeen years ago, the Archer Corporation borrowed \(6,500,000. Since then, cumulative inflation has been 65 percent (a compound rate of approximately 3 percent per year).

a. When the firm repays the original \)6,500,000 loan this year, what will be the effective purchasing power of the \(6,500,000? (Hint: Divide the loan amount by one plus cumulative inflation.)

b. To maintain the original \)6,500,000 purchasing power, how much should the lender be repaid? (Hint: Multiply the loan amount by one plus cumulative inflation.)

c. If the lender knows he will receive only $6,500,000 in payment after 17 years, how might he be compensated for the loss in purchasing power? A descriptive answer is acceptable

Step by step solution

01

:Computation of effective purchasing power

$\begin{array}{rcl}\mathbf{EffectivePurchasingPower}& \mathbf{=}& \frac{\mathbf{Loan}}{\left(1+\mathbf{CumulativeInflation}\right)}\\ & \mathbf{=}& \frac{\mathbf{6}\mathbf{,}\mathbf{500}\mathbf{,}\mathbf{000}}{\mathbf{1}+\mathbf{65}\%}\\ & \mathbf{=}& \mathbf{\$}\mathbf{3}\mathbf{,}\mathbf{939}\mathbf{,}\mathbf{393}\mathbf{.}\mathbf{94}\\ & & \end{array}$

02

:Computation of amount lender be repaid

$\begin{array}{rcl}\mathbf{Amount}\mathbf{toberepaid}& \mathbf{=}& \mathbf{Loan}\mathbf{×}\left(1+\mathbf{CumulativeInflation}\right)\\ & \mathbf{=}& \mathbf{6}\mathbf{,}\mathbf{500}\mathbf{,}\mathbf{000}\mathbf{×}\left(\mathbf{1}+\mathbf{65}\%\right)\\ & \mathbf{=}& \mathbf{\$}\mathbf{10}\mathbf{,}\mathbf{725}\mathbf{,}\mathbf{000}\\ & & \end{array}$

03

:Compensation for the loss in purchasing power

The lender can be compensated with the amount calculated below. the amount of compensation is calculated as the average of the loan amount. This amount will be enough to compensate for the loss in purchasing power.

$\begin{array}{rcl}\mathbf{Compensatedamount}& \mathbf{=}& \frac{\mathbf{Loanamount}}{\mathbf{Period}}\\ & \mathbf{=}& \frac{\mathbf{6}\mathbf{,}\mathbf{500}\mathbf{,}\mathbf{000}}{\mathbf{17}}\\ & \mathbf{=}& \mathbf{\$}\mathbf{382}\mathbf{,}\mathbf{352}\mathbf{.}\mathbf{94}\\ & & \end{array}$

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