Chapter 03, Problem 007 Consider the following three cash flow series End of Yea

Question

Chapter 03, Problem 007 Consider the following three cash flow series End of Year Cash Flow Series A Cash Flow Series B Cash Flow Series C -$2,610 $2,710 $2,310 $1,910 $1,510 $1,110 $1,000 1.5X 2.0X 2.5X 3.0X 2 4 2Y 5 Determine the values of X and Y so that all three cash flows are equivalent at an interest rate of 16% per year compounded yearly Y: $ carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is ±5. Click here to access the TVM Factor Table Calculato

Explanation / Answer

Present worth (PW) of Series B ($) = - 2,610 + 2,710 x P/F(16%, 1) + 2,310 x P/F(16%, 2) + 1,910 x P/F(16%, 3) + 1,510 x P/F(16%, 3) + 1,110 x P/F(16%, 3)

= - 2,610 + 2,710 x 0.8621** + 2,310 x 0.7432 + 1,910 x 0.6407** + 1,510 x 0.5523** + 1,110 x 0.4761**

= - 2,610 + 2,336.29 + 1,716.79 + 1,223.74 + 833.97 + 528.47

= 4,029.26

(1) For equivalence,

- 1,000 + X x P/F(16%, 1) + 1.5X x P/F(16%, 2) + 2X x P/F(16%, 3) + 2.5X x P/F(16%, 3) + 3X x P/F(16%, 3) = 4,029.26

X x 0.8621** + 1.5X x 0.7432 + 2X x 0.6407** + 2.5X x 0.5523** + 3X x 0.4761** = 5,029.26

X x (0.8621 + 1.1148 + 1.2814 + 1.38075 + 1.4283 = 5,029.26

X x 6.06735 = 5,029.26

X = 829

(2) For equivalence,

Y + Y x P/F(16%, 1) + Y x P/F(16%, 2) + 2Y x P/F(16%, 3) + 2Y x P/F(16%, 3) + 2Y x P/F(16%, 3) = 4,029.26

Y + Y x 0.8621** + Y x 0.7432 + 2Y x 0.6407** + 2Y x 0.5523** + 2Y x 0.4761** = 4,029.26

Y x (1 + 0.8621 + 0.7432 + 1.2814 + 1.1046 + 0.9522) = 4,029.26

Y x 5.9435 = 4,029.26

Y = 678

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