Arithmetic Gradient Geometric Gradient Mortgages

This Week

•

_{Arithmetic Gradient}

•

Geometric Gradient

Compound Interest Factors

COMPOUND AMOUNT FACTOR

(F/P, i, N) = (1+ i)N

PRESENT WORTH FACTOR

(P/F, i, N) =1/ (1+ i)N

UNIFORM SERIES COMPOUND AMOUNT

FACTOR

(F/A, i, N) = [(1+ i)N –1]/i

SINKING FUND FACTOR

(A/F, i, N) = i/[(1+ i)N –1]

SERIES PRESENT WORTH FACTOR

(P/A, i, N) = [(1+ i)N –1]

i . (1+ i)N

CAPITAL RECOVERY

FACTOR

(A/P, i, N) = i . (1+ i)N

Arithmetic Gradient Series

•

A series of receipts or disbursements that

may start at zero, or may have a base value at

the end of the first period, but then increase

by a constant amount [G] from period to

period.

•

Composed of two components

–

The gradient component [G]

ARITHMETIC GRADIENT SERIES

. .

-1 0 1 2 3 4 _{(N-1) N (N+1)}

G 2G

3G 4G

(N-1)G

(N-2)G

0

RECEIPTS

ARITHMETIC GRADIENT SERIES

G 2G

3G 4G

(N-1)G (N-2)G

0

RECEIPTS

P

ARITHMETIC GRADIANT TO ANNUITY CONVERSION FACTOR

(A/G, i, N)= 1

-

N

i [(1+ i)

N

–1]

(A/G, i, N)= (P/G, i, N) x (A/P, i, N)

A

ARITHMETIC GRADIENT SERIES

ARITHMETIC GRADIENT TO ANNUITY CONVERSION FACTOR

(A/G, i, N)= 1 - N

i [(1+ i)

N

–1]

Arithmetic Gradient with a base

0 1 2 3 4 5 6 7

$500 $700

$900

$1100

$1300

$1500

Arithmetic Gradient with a base

0 1 2 3 4 5 6 7

$200

$400

$600

$800

$1000

$1200

Declining Linear Gradient

$1200

$1000

$800

$600

0 1 2 3 4 5

$1200

$1000

$800

$600

0 1 2 3 4 5

$400

$1200

0 1 2 3 4 5

$200 $400

$600

0 1 2 3 4 5

Arithmetic Gradient to Annuity Conversion

Factor

Your company has just opened and must dispose of its biomedical

waste. The base cost that was quoted to you was $6000 but as

business increases you expect the cost to grow by $100 per month.

If the interest rate is 10% compounded monthly what is the future

cost of the disposal at the end of the first year?

Solution: A’ = $6000, G = $100, i = (10/12)%, N=1 x 12

A_tot = $6540.11 F = $82,180.20

%C_XSXa ! % + % + M 2

M ! V Q

+

% + ! M 2 M B2 * ! % * % M 2

! * %

Shifted Gradient Example: i = 10%

•

Consider the following Cash Flow

“Shifted” negative gradient, on a negative cash flow.

The PW point in time is at t = 3 (not t = 0)

Arithmetic Gradient Series to Annuity

Conversion Factor

CASES

MEANING

A>0 and G>0

A>0 and G<0

A<0 and G>0

Winning the Lottery

4 1MPPMSR

O O

O O

O

%C_XSXa ! % + !

!

8MQI TIVMSH ! 4 %

Option 2: P_Opt2 =$3,818,363

Winning the Lottery

Option 1: P = $3.44 million

Atot = 189,000 + 7000(A/G, 0.045,25) = 189,000 + 7000(9.7560955) = 257,292.67

Now move this to time 1 A' = 189,000

Time 1: [175,000 + 257 292.67 (P/A, 0.045,25)] Today: [ ] * (P/F, 4.5%, 1)

= $3 818 363

GEOMETRIC GRADIENT SERIES

“A set of receipts or disbursements that change

by a constant proportion from one period to the

next in a sequence of periods”

RECEIPTS

A(1+g)(N-2)

A

A(1+g)(N-1)

A(1+g)

A(1+g)A(1+g)2

g is positive

g is negative

GEOMETRIC GRADIENT SERIES

A(1+g)

A(1+g)(N-2)

A

RECEIPTS

A(1+g)2

A(1+g)3

A(1+g)(N-1)

GEOMETRIC GRADIENT SERIES TO PRESENT WORTH

CONVERSION FACTOR

Where,

1+

i

0

= [1+ i]

[1+g]

i0 – growth adjusted interest rate

(P/A, g, i, N)= (P/A, i

0

, N) /(1+g)

GEOMETRIC GRADIENT SERIES TO PRESENT WORTH

CASES

MEANING

PROCEDURE

i>g>0

i

Use tables or

formula

g>i>0

i

Use formula

ONLY

g=i>0

i

Special Case:

P = N [A/(1+g)]

GEOMETRIC GRADIENT SERIES TO PRESENT WORTH

CONVERSION FACTOR: Application

Time to Retire

• Your retirement benefits are worth $50,000 per year which should cover your cost of living immediately upon retirement

• However cost of living is expected to increase by 5%/yr

• Your investments can make 7%/yr

• How much more should you invest when you retire to cover you for 25 years

0 1 2 3 4 5 6 23 24 25

$50,000

-$50,000

-$50,000(1.05)

Time to Retire

Total Benefits:

•

P

= $582,679

= 50 000(P/A, 7%,25) <--positive

Cost of Living:



_{P= $940,696}

=

Required Additional Savings:

When N Approaches Infinity

•

Long-lived projects may be modelled as

Geometric Gradient to Present

Worth Conversion Factor

Your company has just opened and must dispose of its biomedical waste. The base cost that was quoted to you was $6000 but as business increases you expect the cost to grow by 1% per month. If the interest rate is 10% compounded monthly what is the future cost of the disposal at the end of the first year?

Solution: A = $6000, g = 1%, i = (10/12)%=0.8333%, N=1 x 12, since g>i, use formula

P=$72057.70

F = $79,603.08

Since 1% growth, the g is positive

Mortgages….

•

Would be one of the financial situation most of us

would face

•

Amortization

is the number of years it would take

to repay a mortgage loan in full for a given interest

rate and payment schedule.

•

The

term

of the mortgage refers to the number of

Mortgages

Bobby has just bought a house for $200 000. He paid $50 000 down, and the rest of the cost has been obtained from a mortgage. The mortgage has a nominal interest rate of 10%, compounded monthly with a 10-year

amortization period. The term of the mortgage is 5 years. What are

Bobby’s current monthly payments? How much does he owe in 5 years?

Solution:

A = $1982.26

P = 200 000 - 50 000 = $150 000 N = 10 * 12 = 120

A = P(A/P, i,N)

=150 000 (A/P, 0.0083, 120)

i = r/m = 10%/12 = 0.00833

F = P (F/P, i, N) - A (F/A, i, N)

= 150 000 (F/P, 0.0083, 60) - 1 982.26 (F/A, 0.0083, 60) = 150 000 (1.645309) - 1 982.26 (77.437072)

Mortgages

How much does he owe in 5 years?

Solution:

After 5 years, Bobby still owes $93 295.94.

F = $93 295.94

Payment Breakdown

N

A

I

P

P

Mortgages (Cash Flow)



_{After 5 years, Bobby will need to renegotiate his}

mortgage, where Bobby can choose a new TERM.



_{Usually, the annuity payment amount will not}

change, however, by changing the interest rate, the

actual Amortization Period would have changed.

Buying Into Perfection

If the purchase price is $1,900,000 and the

interest rate is 6% compounded monthly,

how long will it take you to pay off the

mortgage if you make monthly payments of

3

bedrooms

2

bathrooms

Car Payments

• Determine the monthly payments required if you want to borrow $30,000 from a bank to buy a car, at 6%, interest (compounded monthly) for 6 years.

• For the previous car loan, when you hand the bank your cheque for the 24th monthly payment, how much of that cheque is

interest, and how much is going toward principal reduction?

• How much total interest will you end up paying for the car loan?

i = r/m = 6%/12 = 0.005

N =72

A = P(A/P,i,N)

Car Payments

• Determine the monthly payments required if you want to borrow $30,000 from a bank to buy a car, at 6% interest (compounded monthly) for 6 years.

Car Payments

• For the 24th monthly payment, how much of that cheque is interest, and how much is going toward principal reduction?

• The interest then is = 107.80, and the rest is principal reduction, 497.19-107.8 = 389.39

30 000 = 497.19(P/A, 0.005, 23) + F(P/F, 0.005, 23)

F_23 = $21 560 OR

F_23 = $21 560

30 000 = 497.15(P/A, 0.005, 24) + F_24(P/F, 0.005, 24) F_24 = $21 169.92

F_23 - F_24 = $389.08

F_23 --> F_24 = 21 560(F/P, 0.005, 1) = 21 560 (1 + 0.005)^1 = 107.80

Car Payments

•

Total Interest = 5797.68

How much total interest will you end up paying fo the car loan?

= (A * N) - P