Mathematics, 15.02.2021 20:30 lmoleary7466
Consider the following two cash flow series of payments: Series A is a geometric series increasing at a rate of 9.5% per year. The initial cash payment at the end of year 1 is $1,000. The payments occur annually for 5 years. Series B is a uniform series with payments of value X occurring annually at the end of years 1 through 5. You must make the payments in either Series A or Series B.
a. Determine the value of X for which these two series are equivalent if your TVOM is i = 9%. $
b. If your TVOM is 8%, would you be indifferent between these two series of payments? Enter the PW for each series to support this choice.
c. If your TVOM is 5%, would you be indifferent between these two series of payments? Enter the PW for each series to support this choice.
Answers: 3
![=\Big[1000 \times \dfrac{(1.095)^0}{(1.09)^1}\Big]+\Big[1000 \times \dfrac{(1.095)^1}{(1.09)^2}\Big]+...+\Big[1000 \times \dfrac{(1.095)^4}{(1.09)^5}\Big]](/tpl/images/1120/2737/023e9.png)
![= 1000 \Big [ \dfrac{1}{1.09}+ \dfrac{1.095}{1.1881}+ \dfrac{1.199}{1.295}+\dfrac{1.313}{1.912}+\dfrac{1.438}{1.539}\Big]](/tpl/images/1120/2737/dc1d7.png)
![= 1000 \Big[ 0.917 + 0.922 + 0.926 + 0.930 + 0.934\Big]](/tpl/images/1120/2737/bfaf8.png)
![The \ value \ of\ X = \dfrac{Present \ value \ of \ series \ B }{\Big [\dfrac{1-(1+r)^{-n}}{r} \Big ]}](/tpl/images/1120/2737/e1794.png)
![The \ value \ of\ X = \dfrac{4629 }{\Big [\dfrac{1-(1+0.09)^{-5}}{0.09} \Big ]}](/tpl/images/1120/2737/23a0e.png)
![=1000 \times \Big[\dfrac{(1.095)^0}{(1.08)^1}+ \dfrac{(1.095)^1}{(1.08)^2}+...+\dfrac{(1.095)^4}{(1.08)^5}\Big]](/tpl/images/1120/2737/14065.png)
![=1000 \Big [ \dfrac{1}{1.08}+ \dfrac{1.095}{1.1664}+\dfrac{1.199}{1.2597}+\dfrac{1.313}{1.3605}+\dfrac{1.438}{1.4693}\Big ]](/tpl/images/1120/2737/bc334.png)
![= 1000\Big [ 0.9259 + 0.9839+0.952 + 0.965+0.979\Big ]](/tpl/images/1120/2737/64e89.png)
![Value \ of \ X \times \Big [ \dfrac{1 - (1+r)^{-n} }{r}\Big ]](/tpl/images/1120/2737/96d6c.png)
![= 1189.97 \times \Big [ \dfrac{1 - (1+0.08)^{-5} }{0.08}\Big ]](/tpl/images/1120/2737/60355.png)
![=1000 \times \Big[\dfrac{(1.095)^0}{(1.05)^1}+ \dfrac{(1.095)^1}{(1.05)^2}+...+\dfrac{(1.095)^4}{(1.05)^5}\Big]](/tpl/images/1120/2737/f4de2.png)
![=1000 \Big [ \dfrac{1}{1.05}+ \dfrac{1.095}{1.1025}+\dfrac{1.199}{1.1576}+\dfrac{1.313}{1.2155}+\dfrac{1.438}{1.276}\Big ]](/tpl/images/1120/2737/822b4.png)
![= 1000\Big [ 0.9524 + 0.9932+1.0357 + 1.08+1.127\Big ]](/tpl/images/1120/2737/2aaf1.png)
![= 1189.97 \times \Big [ \dfrac{1 - (1+0.05)^{-5} }{0.05}\Big ]](/tpl/images/1120/2737/2079c.png)
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Consider the following two cash flow series of payments: Series A is a geometric series increasing a...
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