ISE 2014 Chapter 3 Section 3.11 Deferred Annuities

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ISE 2014 Chapter 3

Section 3.11 Deferred Annuities

If we are looking for a present (future) equivalent sum at time other than one period prior to the first cash flow in the series (coincident with the last cash flow in the series) then we are dealing with a deferred annuity. See page 90.

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Example: Deferred Annuities

How much do you need to deposit today into an account that pays 3% per year so that you can make 10 equal annual withdrawals of

$1,000, with the first withdrawal being made seven years from now?

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Example: Deferred Annuities Continued

P₆=$1,000(P|A, 3%, 10) = $8,530.20

P₀=P₆(P|F, 3%, 6) = $8,530.20(0.8375) = $7,144 Or

P₀ =A(P|A, 3%, 16) – A (P|A, 3%, 6)

=$1,000(12.5611) - $1,000(5.4172)

=$12,561.10 - $5,417.20

= $7,143.90

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Section 3.12 Multiple Interest Factors

Deferred annuities can be transformed to an equivalent present or future sum in a single step with the use of multiple interest factors.

Some situations include multiple unrelated sums or series, requiring the problem be broken into

components that can be individually solved and then re-integrated. See page 93.

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Example: Multiple Interest Factors

Given:

Find: P₀, F₆, F₇, A_1-6 P₀ =

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Multiple Interest Factors Example Continued

3.31 3.31 1.331

F6 = 800(F/A, 10%, 3) + 500(F/A, 10%, 3)(F/P, 10%, 3)

1.331

- 1000(F/P, 10%, 3) = $3,520 F7 = ?

A1-6 = P0(A/P, 10%, 6) = 1987(0.2296) = $456

= F6(A/F, 10%, 6) = 3520(0.1296) = $456

=

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Section 3.13. Uniform Gradient Series

What if we have cash flows (revenues or expenses) that are projected to increase or decrease by a uniform amount each period? (e.g. maintenance costs, rental income).

We call this a uniform gradient series (G).

We can have positive or negative gradients if the slope of the cash flows is positive or negative, respectively.

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A. Positive Gradient Example

G = constant by which the cash glows increase or decrease each period. G = $100 in the above example.

Note that G₁=$0. Even though N = 8, their will only be seven actual increases. The first G amount ($100) occurs at the end of year 2.

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A. Positive Gradient Example Continued

Find P for the example positive gradient cash flow

(see Table C-13, p 634)

5.3349 16.029

P₀ = $250 (P|A, 10%, 8) + $100 (P|G, 10%, 8) = $2,937

Find A (annual equivalent) for the example cash flow.

3.0045

A_1-8 = $250 + 100 (A|G, 10%, 8) = $550.45

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B. Negative Gradient Example

- The gradient is negative (e.g., an increasing cost) and the uniform series is positive (e.g., steady stream of revenue).

Given: G = -250, I = 10%/year

and the Cash Flow for Year 1 = $1,250 Year 2 = $1,000 Year 3 = $750 Year 4 = $500 Year 5 = $250 Underlying annuity (A) = ?

Find: Equivalent values for these cash flows at:

P₀ and A_1-5

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B. Negative Gradient Example - Solution

3.7908 6.86

P⁰= 1,250(P | A, 10%, 5) - 250(P | G, 10%, 5) = $3,023.5

1.8101

A^1-5= 1,250 - 250(A | G, 10%, 5) = $797.48

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Section 3.16. Nominal and Effective Interest Rates

Nominal interest (r) = interest compounded more than one interest

period per year but quoted on an annual basis.

* Example: 16%, compounded quarterly

Effective interest (i) = actual interest rate earned or charged for a specific time period.

* Example: 16%/4 = 4% effective interest for each of the four quarters during the year.

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Nominal and Effective Interest Rates (continued)

Relation between nominal interest and effective interest:

where

i = effective annual interest rate r = nominal interest rate per year

M = number of compounding periods per year

M

r

= interest rate per interest period

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Nominal and Effective Interest Rates – Examples

Find the effective interest rate per year at a nominal rate of 18%

compounded (1) quarterly, (2) semiannually, and (3) monthly.

(1) Quarterly compounding

(2) Semiannual compounding

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Nominal and Effective Interest Rates (continued)

(3) Nominal interest rate is 18% compounded monthly.

Interpretation: This is a normal statement of an interest rate where the related time period is one year, and the

subperiod is one month.

R = 18%; M = 12; i_M = 1 ½ %;

What if N=3 and P = $1,000, find F.

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Nominal and Effective Interest Rates – Example

A credit card company advertises an A.P.R. of 16.9% compounded daily on unpaid balances. What is the effective interest rate per

year being charged?

r = 16.9% M = 365

i_eff =

) 1

365 169 .

1 0

(

+ ³⁶⁵ − =0.184 or 18.4% per year

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Section 3.17 and 3.18. Nominal and Effective Interest Rates

Find P₀ when r = 12%, compounded monthly

CFD goes here

We have monthly cash flows so we need to use a monthly interest rate.

i/mo. = r/M = 12% / 12 = 1% per month

P₆ = $500 (P|A, 1%, 6) + $100 (P|G, 1%, 6) = $4,330 P₀ = P₆ (P|F, 1%, 6) = $4,079

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ISE 2014

Nominal and Effective Interest Rates

Two situations we’ll deal with in Chapter 3:

(1) Cash flows are annual. We’re given r per year and M. Procedure: find i/yr. =

r

M

1 1

M

 +  −

 

 

and discount/compound annual cash flows at i/yr.

(2) Cash flows occur M times per year.

We’re given r per year and M. Find

the interest rate that corresponds to

M, which is r/M per time period

(e.g., quarter, month). Then

discount/compound the M cash flows

per year at r/M for the time period

given.

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Interest Problems with Compounding more often than once per Year – Example A

* Example:

If you deposit $1,000 now, $3,000 four years from now followed by five quarterly deposits decreasing by $500 per quarter at an interest rate of 12% per year compounded

quarterly, how much money will you have in you account 10 years from now?

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Example A Solution

r/M = 3% per quarter

Year 3.75 = 15th Quarter P ^{@yr. 3.75}= P ^{qtr. 15}

= 3000(P/A, 3%, 6) - 500(P/G, 3%, 6) = $9713.60

F ^{yr. 10}= F ^{qtr. 40}

= 9713.60(F/P, 3%, 25) + 1000(F/P, 3%, 40) =

= $23,600.34

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Interest Problems with Compounding more often than once per Year – Example B

* Example:

If you deposit $1,000 now, $3,000 four years from now, and $1,500 six years from now at an interest rate of 12%

per year compounded semiannually, how much money will you have in your account 10 years from now?

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Example B – Solution

i per year =

) 1 2

12 . 1 0

(

+ ¹² − ^{= 0.1236}

F = $1,000(F/P, 12.36%, 10) + $3,000(F/P, 12.36%, 6) +

$1,500(F/P, 12.36%, 4) or r/M = 6% per half-year

F = 1000(F/P, 6%, 20) + 3000(F/P, 6%, 12)+ 1500(F/P, 6%, 8)

= $11,634.50

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Nominal and Effective Interest Example

Given: You’ve borrowed $22,000 to buy a new car at 2.9%, compounded monthly. Your loan is for 36 months.

Find: a) Draw a cash flow diagram for this loan.

b) How much will your monthly payment be?

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Problem 3-36 ($100,000 loan)

1) Draw a cash flow diagram to illustrate the situation.

P = present equivalent of the loan payments

A = loan payments on a thirty year repayment plan

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Problem 3-36 ($100,000 loan)

2) Find the value of the loan payments, A

A = P (A | P, 8%, 30) = $100,000(0.0888) = $8,880

3) Set up the equation to solve for F₁₂

P =A(P|A, 8%,8) + 3A(P|A, 8%,4)(P|F, 8%,8) + F₁₂ (P|F, 8%,12) 100,000 = 8,880(5.7466) + 26,640(3.3121)(0.5403) + F₁₂ (0.3971)

100,000 = 51,030 + 47,673 + (0.3971)F₁₂ (0.3971)F ₁₂ = $1,297

F12 = $3,266

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Example: Problem 3-95

What is the value of the following CFD?

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Problem 3-95 Solution

1.15

F₁ = -$1,000(F/P,15%,1) - $1,000 = -$2,150

1.15

F₂ = F 1 (F/P,15%,1) + $3,000 = $527.50

1.10 1.06

F₄ = F 2 (F/P,10%,1)(F/P,6%,1) = $615.07

(add cash flow handouts here from original PDF)