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 c

Question

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.

in progress 0
Amity 3 years 2021-08-27T07:57:25+00:00 1 Answers 5 views 0

Answers ( )

  1. Dau
    0
    2021-08-27T07:58:55+00:00
    Reply

    Answer:

    Step-by-step explanation:

    From the given information;

    The present value of series A:=\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]

    = 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]

    = 1000 \Big[ 0.917 + 0.922 + 0.926 + 0.930 + 0.934\Big]

    = 1000 \times 4.629

    = \$4629

    Thus, the present value of series A is = $4629

    Present value of series A = Present value of series B

    The \ value \ of\ X = \dfrac{Present \ value \ of \ series \ B }{\Big [\dfrac{1-(1+r)^{-n}}{r} \Big ]}

    The \ value \ of\ X = \dfrac{4629 }{\Big [\dfrac{1-(1+0.09)^{-5}}{0.09} \Big ]}

    The \ value \ of\ X =\dfrac{4629 \times 0.09}{1-0.6499}

    The \ value \ of\ X =\dfrac{416.61}{0.3501}

    The \ value \ of\ X =1189.97

    Thus, the value of X = $1189.97

    2.

    The present value of series A:

    =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]

    =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 ]

    = 1000\Big [ 0.9259 + 0.9839+0.952 + 0.965+0.979\Big ]

    = 1000 \times 4.76059

    \simeq \$4761

    Thus, the present value of series A is = $4761

    Present value of series B =Value \ of \ X \times \Big [ \dfrac{1 - (1+r)^{-n} }{r}\Big ]

    = 1189.97 \times \Big [ \dfrac{1 - (1+0.08)^{-5} }{0.08}\Big ]

    = \dfrac{1189.97}{0.08} \times \Big ( 1 -0.6806\Big )

    = 14874.625 \times 0.3194

    = \$4750

    Thus, the present value of series B = $4750

    3.

    The present value of series A:

    =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]

    =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 ]

    = 1000\Big [ 0.9524 + 0.9932+1.0357 + 1.08+1.127\Big ]

    = 1000 \times 5.1883

    \simeq \$5,188

    Thus, the present value of series A = $5188

    Present value of series B: =Value \ of \ X \times \Big [ \dfrac{1 - (1+r)^{-n} }{r}\Big ]

    = 1189.97 \times \Big [ \dfrac{1 - (1+0.05)^{-5} }{0.05}\Big ]

    = \dfrac{1189.97}{0.05} \times( 0.2165)

    = \$5152.57

    Thus, the present value of series B = $5153

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )