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I am having some difficulty understanding how to interpret the following question, mostly due to the fact that it says to do it without converting interest rate.

Consider a stream of payments over a 30 year period with the following

  1. For the first 10 years, payments of \$100 are made at the beginning of every quarter.

  2. For the second 10 years, payments of \$250 are made semi annually at the end of every half year.

  3. In the final 10 years, payments are made at the end of every 6 months, the first payment is of \$300, and every subsequent payment increases by \$50.

For the first 10 years, $i^{(12)}=0.06$, for the second 10 year, $i=0.08$ and the final 10 years, $i^{(2)}=0.07$.

Find the cost of such stream.

I am mostly confused in regard to what is meant to solve without converting interest rates.

My thoughts

We can consider each of the three 10 year blocks one by one and discount or accumulate to a proper time.

For the first 10 years this is an annuity due with 40 payments of \$100 each, however we are given that $i^{(12)}=0.06$.

I know I could use this to solve for $i^{(4)}$ but it says without converting interest.

The second ten years is similar, except it is an annuity due, here we have an effective interest rate of $0.08$, 20 payments of \$250 but since interest is changing I don't know how I could accumulate this to an appropriate time, say $t=30$.

And for the third 10 years, there are even more steps.

So mostly I am confused about how to solve this question without converting interest.

Suppose we bring everything to time $t=30$.

For $(1)$, $(1+i)^{20}100s_{40}^{\bullet \bullet}$. I would have used $i^{(4)}$ if converting interest hadn't been restricted.

For $(2)$, $(1+i)^{10}250s_{20}$ with $i=0.08$.

For $(3)$, the interest here was given matches what we want I believe so that should eliminate some of the difficulty. The rest would be using geometric series.

So can anyone help clear this up? What would be the correct way to solve this and the correct answer to such problem? Is any of what I have correct so far?

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    Consider (1) and let $i=i_{12}$. If the payments were done monthly, you would consider $100 + 100 (1+i)^{-1} + 100 (1+i)^{-2} + \ldots$ with 120 terms. If payments are done quarterly, update this formula to $100 + 100 (1+i)^{-3} + 100(1+i)^{-6} + \ldots$ with 40 terms. – 2017-02-17
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    @mlc can you elaborate on this at all? – 2017-02-19
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    What do $i^{(12)}$, $i$, and $i^{(2)}$ mean? – 2017-02-21
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    They correspond to interest computed monthly , an effective rate of interest and interest compounded semiannually – 2017-02-21

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The first thing to do is to extract the useful information from the text and put it in a clear overview. The writer of the exercise clearly tries to trick you into making mistakes, so double check.

$$ \begin{array}{} \mathbf{Block} & \mathbf{Payments} & \mathbf{Pay\ at} & \mathbf{Amount} & \mathbf{Interest\ rate} \\ 1 & 40\text{ quarterly} & \text{begin of period} & $100 & 0.06 \text{ compounded monthly} \\ 2 & 20\text{ semi annually} & \text{end of period} & $250 & 0.08 \text{ compounded yearly} \\ 3 & 20\text{ semi annually} & \text{end of period} & $300, $ 350, \cdots, $1250 & 0.07 \text{ compounded semi annually} \end{array} $$

Below the definitions of the different interest rates:

$$ \begin{array}{} \mathbf{Interest\ rate} & \mathbf{Multiply by} \\ 0.06 \text{ compounded monthly} & 1+\frac{0.06}{12} \text{ every month} \\ 0.08 \text{ compounded yearly} & 1+0.08 \text{ every year} \\ 0.07 \text{ compounded semi annually} & 1+\frac{0.07}{2} \text{ every half year} \end{array} $$

So this means that, for example:

  • \$100 will be worth $100 \cdot \left(1+\frac{0.06}{12}\right)^6$ after half a year of interest rate 0.06 compounded monthly
  • \$100 will be worth $100 \cdot \left(1+\frac{0.07}{2}\right)^{1/2}$ after a quarter of interest rate 0.07 compounded semi annually

But we're not going forward in time. We're going backward. We want to know what a future payment will cost now. So instead of multiplying, we divide. Again two examples:

  • Paying \$100 after half a year of interest rate 0.06 compounded monthly is equivalent to paying $100 \cdot \left(1+\frac{0.06}{12}\right)^{-6}$ now
  • Paying \$100 after 18 months of interest rate 0.08 compounded yearly is equivalent to paying $100 \cdot \left(1+0.08\right)^{-3/2}$ now

All 80 payments are made at different times. So we all have to multiply them to different numbers to calculate the present value, and then we add them all together.

But, it's probably easier to calculate the value for each block at the time of the beginning of the block, and then multiply the value of the whole block to get the present value. Using that approach, the formulas become:

$$ \begin{array}{} B_1 = \displaystyle \sum_{i=0}^{39} 100 \cdot \left(1+\frac{0.06}{12}\right)^{-3i} & \text{(value now)} \\ B_2 = \displaystyle \sum_{i=1}^{20} 250 \cdot \left(1+0.08\right)^{-i/2} & \text{(value after 10 years)} \\ B_3 = \displaystyle \sum_{i=1}^{20} (250+50i) \cdot \left(1+\frac{0.07} {2}\right)^{-i} & \text{(value after 20 years)} \\ T = \displaystyle B_1 + \left(1+\frac{0.06}{12}\right)^{-120} \cdot \left( B_2 + (1+0.08)^{-10} B_3\right) & \text{(value now)} \end{array} $$

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    Seems good so far but as you said it seems very easy to make mistakes, so id appreciate seeing what others get as the final answer to compare – 2017-02-23
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    Even if there is a hidden slip (which seems not), (+1) for the logical and systematic steps. – 2017-02-23
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    @Paul for example, in each B , are you using diffirent i? – 2017-02-24
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    @Quality I'm not sure if I understand your question. Are you talking about the variable $i$ I used in the formulas for each $B$, or the interest rate? Also, to clear things up, do you understand everything before the final formulas and do you know how to read the $\sum$ sign? – 2017-02-25