course Mth 271 010. Income Streams
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Given Solution: `aThe total money flow would be $20,000 per year for 6 years, or $120,000. 8% simple interest for 3 years would be 3 * 8% = 24%, and 24% of $120,000 is $28,800, so your first estimate would be $28,800. You therefore expect to end up with something not too far from $148,800. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q002. Money usually doesn't earn simple interest. Interest is almost always compounded. If the interest on $120,000 was compounded annually at 8% for 3 years, what would be the final amount? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8% of $120,000 = $9600 after 1 year we will have $120,000 + $9600 = $129,600. After another year we will have : 8% of $129,600 + $129,600= $139,968. After a third year you will have $139,968 * 1.08 = $151,165. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThere is a more efficient way to calculate this, but we'll see that shortly. 8% of $120,000 is $9600, so after 1 year we will have $120,000 + $9600 = $129,600. Note that this result could have been obtained by multiplying $120,000 by 1.08. We will use this strategy for the next two years. After another year we will have $129,600 * 1.08 = $139,968. If you wish you can take 8% of $129,600 and add it to $129,600; you will still get $139,968. After a third year you will have $139,968 * 1.08 = $151,165. Note that this beats the $148,800 you calculated with simple interest. This is because each year the interest is applied to a greater amount than before; previously the interest was just applied to the starting amount. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q003. Money can earn interest that is compounded annually, quarterly, weekly, daily, hourly, or whatever. For given interest rate the maximum interest will be obtained if the money is compounded continuously. When initial principle P0 is compounded continuously at rate r for t interest periods (e.g., at rate 8% = .08 for t years), the amount at the end of the time is P = P0 * e^(r t). If the $120,000 was compounded continuously at 8% annual interest for 3 years, what would be the amount at the end of that period? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P0 = $120,000 r = .08 t = 3 P = $120,000 * e^(.08 * 3) = $120,000 * e^(.24) = $152,549. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe would have P0 = $120,000, r = .08 and t = 3. So the amount would be P = $120,000 * e^(.08 * 3) = $120,000 * e^(.24) = $152,549. This beats the annual compounding by over $1,000. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `q004. This keeps getting better and better. We end up with more and more money. Now let's see how much money we would really end up with, if the money started compounding continuously as soon as it 'flowed' into the account. As a first step, we note that 6 years is 72 months. About how much would the money flowing into the account during the 6th month be worth at the end of the 72 months? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: monthly flow is: $20,000 / 12 = $1,666.67. The $1,666,67 that flows in during the 6th month will have about 72 - 5.5 = 66.5 months to grow. 66.5 months is 66.5/12 = 5.54 years( appr) So the $1,666,67 would grow at the continuous rate of 8 % for 5.54 years. Resulting in a principle of $1,666,67 * e^(.08 * 5.54) = $2596.14. confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFlowing into the account at a constant rate of $20,000 per year, we see that the monthly flow is $20,000 / 12 = $1,666.67. The $1,666,67 that flows in during the 6th month will have about 72 - 5.5 = 66.5 months to grow. [ Note that we use the 5.5 instead of 6 because the midpoint of the 6th month is 5.5 months after the beginning of the 72-month period (the first month starts at month 0 and ends up at month 1, so its midway point is month .5; the nth month starts at the end of month n-1 and ends at the end of month n so its midway point is (n - 1) + .5). ] 66.5 months is 66.5/12 = 5.54 years, approximately. So the $1,666,67 would grow at the continuous rate of 8 percent for 5.54 years. This would result in a principle of $1,666,67 * e^(.08 * 5.54) = $2596.14. STUDENT COMMENT: I don’t really know why we would use the midpoint of the month. Is that exactly when the money “flows in?” INSTRUCTOR RESPONSE: That's pretty much it, though the word 'exactly' isn't quite accurate. Since money 'flows in' from the beginning of the month to the end, the rate at the midpoint of the month is a better choice than either the rate at the beginning or the rate at the end. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `q005. How much will the money flowing into the account during the 66th month be worth at the end of the 72 months? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ` $1,666,67 would flow into the account during this month. The midpoint of this month = 65.5 months after the start of the 72-month period, therefore the money will have an average of about 6.5 months, or 6.5 / 12 = .54 years( approx.) to grow. the money that flows in during the 66th month will grow to: $1,666,67 * e^(.08 * .54) = $1737, approx.. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAgain $1,666,67 would flow into the account during this month. The midpoint of this month is 65.5 months after the start of the 72-month period, so the money will have an average of about 6.5 months, or 6.5 / 12 = .54 years, approx., to grow. Thus the money that flows in during the 66th month will grow to $1,666,67 * e^(.08 * .54) = $1737, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q006. We could do a month-by-month calculation and add up all the results to get a pretty accurate approximation of the amount at the end of the 72 months. We would end up with $152,511, not quite as much as our last estimate for the entire 72 months. Of course this approximation still isn't completely accurate, because the money that comes in that the beginning of a month earns interest for longer than the money that comes in at the end of the month. We could chop up the 72 months into over 2000 days and calculate the value of the money that comes in each day. But even that wouldn't be completely accurate. We now develop a model that will be completely accurate. We first imagine a short time span near some point in the 72 months, and calculate the value of the income flow during that time span. We are going to use symbols because our calculation asked to apply to any time span at anytime during the 72 months. We will use t for the time since beginning of the 6-year period, to the short time span we are considering; in previous examples t was the midpoint of the specified month: t = 5.5 months in the first calculation and 66.5 months in the second. We will use `dt (remember that this stands for the symbolic expression delta-t) for the duration of the time interval; in the previous examples `dt was 1 month. For a time interval of length `dt, how much money flows? Assume `dt is in years. If this money flow occurs at t years from the beginning of the 6-year period, then how long as it have to grow? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: amount in 1 year is $20,000 amount in `dt years is $20,000 * `dt. Money that flows in near the time t years from the beginning of the 6-year period will have (6-t) years to grow. confidence rating:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe amount in 1 year is $20,000 so the amount in `dt years is $20,000 * `dt. Money that flows in near the time t years from the beginning of the 6-year period will have (6-t) years to grow. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q007. How much will the $20,000 `dt amount received at t years from the start grow to in the remaining (6-t) years? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P0 will grow to P0 e^( r t) in t years in 6-t years amount $20,000 `dt will grow to $20,000 `dt e^(.08 (6-t) ). confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAmount P0 will grow to P0 e^( r t) in t years, so in 6-t years amount $20,000 `dt will grow to $20,000 `dt e^(.08 (6-t) ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `q008. So for a contribution at t years from the beginning of the 6-year period, at what rate would we say that money is being contributed to the final t = 6 year value? Let y stand for the final value of the money, and `dy for the contribution from the interval `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The money that flows in during time interval `dt will grow to `dy = $20,000 `dt * e^(.08 (6-t) ) the rate is about: `dy / `dt = [ $20,000 `dt * e^(.08 (6-t) ) ] / `dt = $20,000 e^(.08 ( 6-t) ) As `dt decreases to 0, this expression approaches the exact rate y ' = dy / dt at which money is being added to the final value. confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe money that flows in during time interval `dt will grow to `dy = $20,000 `dt * e^(.08 (6-t) ), so the rate is about `dy / `dt = [ $20,000 `dt * e^(.08 (6-t) ) ] / `dt = $20,000 e^(.08 ( 6-t) ). As `dt shrinks to 0, this expression approaches the exact rate y ' = dy / dt at which money is being contributed to the final value. STUDENT COMMENT: I do not really understand the last statement in the solution. INSTRUCTOR RESPONSE: This explanation leaves out some technicalities, but the last statement comes down to something very much like the following: The shorter the time interval `dt, the closer everything in the interval is to the midpoint and the less the rate varies over the interval. For both of these reasons, the approximation improves greatly as `dt shrinks. If `dt keeps shrinking, it approaches zero and the approximation approaches an exact value. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating:ok ********************************************* Question: `q009. As we just saw the rate at which money accumulates is y' = dy / dt = 20,000 e^(.08(6-t)). How do we calculate the total quantity accumulated given the rate function and the time interval over which it accumulates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `aThe rate function is the derivative of the quantity function. The antiderivative of a rate-of-change function is a change-in-quantity function. Find the total change in a amount from the rate function by finding the change in this antiderivative. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe rate function is the derivative of the quantity function. An antiderivative of a rate-of-change function is a change-in-quantity function. We therefore calculate the total change in a quantity from the rate function by finding the change in this antiderivative. STUDENT COMMENT: I am not sure what this question is asking. I do know that y’ refers to the derivative, but I know that as the rate of change. Is it that same as the rate at which the money accumulates? And to find the total accumulation, wouldn’t we just use the rate function, which I guess would be the antiderivative of the derivative we just found? INSTRUCTOR COMMENT: Very good. Just a little clarification: y ' is the rate of change of y with respect to t. Since y is the future value of the money flow, y ' is the rate of change of future value with respect to clock time. That would indeed be what we mean in this context by 'the rate at which the money accumulates'. For the 6-year cycle, the rate at which future value is accumulating is y = dy/dt = $20,000 e^(.08 ( 6-t) ). y ' is the derivative of y, so y is an antiderivative of y '. So you are right about the antiderivative. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique Rating:ok ********************************************* Question: `q010. If the money that flows into the account t years along the 6-year cycle adds to the final value of the money at the previously mentioned rate y' = dy / dt = 20,000 e^(.08(6-t)), then what function describes the final value of the money accumulated through time t? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The function must be an antiderivative of the rate function y ' = $20,000 e^(.08 ( 6-t) ). y ' = 20,000 e^(.08 * 6) e^(-.08 t) = 20,000 * 1.616 e^(-.08 t) = 32,321 e^(-.08 t). antiderivative of e^(-.08 t) = -1/.08 e^(-.08 t) + c = -12.5 e^(-.08 t) + c. y ' = $32,321 e^(-.08 t) y = $32,321 * -12.5 e^(-.08 t) = -404,160 e^(-.08 t) + c. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe function must be an antiderivative of the rate function y ' = $20,000 e^(.08 ( 6-t) ). The y ' function can be expressed as y ' = $20,000 e^(.08 * 6) e^(-.08 t) = $20,000 * 1.616 e^(-.08 t) = $32,321 e^(-.08 t). The general antiderivative of e^(-.08 t) is -1/.08 e^(-.08 t) + c = -12.5 e^(-.08 t) + c. The general antiderivative of y ' = $32,321 e^(-.08 t) is y = $32,321 * -12.5 e^(-.08 t) = -404,160 e^(-.08 t) + c. The value of the constant c could be determine if we knew, for example, the amount of money present at t = 0. However as we will see in the next step, the value of c doesn't affect the change in the quantity we are considering in this problem. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `q011. How much money accumulates during the 6 years? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y represents the money accumulated through t years. The result: -404,160 e^(-.08 * 6 ) + c - [ -404,160 e^(-.08 * 0 ) + c ] = $154,072. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `ay represents the money accumulated through t years. We subtract the money assumulated at 0 years from the money accumulated at 6 years to get the amount accumulated during the time interval from t = 0 to t = 6 years. The result is -404,160 e^(-.08 * 6 ) + c - [ -404,160 e^(-.08 * 0 ) + c ] = $154,072.