My Favorite Things About Financial Math!!!

 

 

1). Simple Interest – When money is borrowed or lent, the borrower usually pays the lender for the service. The amount charged is usually called interest.

Formula: I=PRT/100

·        I= interest

·        P= principal amount

·       T= time

·       R= rate per annum

Practice Problem: If $600 is invested in an account and earns $136.50 simple interest over a 3.5 year period, find the annual interest rate.

·        Unknown = R

·        I = PRT/100 changes to: R =100xI/PT

·        R = 100 x 136.50/600 x 3.5 = 6.5%

 

2). Compound Interest – It is common for investors to add (compound) the interest to the principle so that the principle grows during the term of the investment.

·        A= amount of the account

·        P= principal amount

·        T= time

·        R= rate

·        N= # of compounding periods

Practice Problem: $1200 is placed in a savings account that pays 8% annual interest compounded annually. Find the amount in the account and the interest paid after 10 years.

·        P = 1200, R = 8, n = 10

·        Formula: A = 1200(1 + 8/100)^10 = 2590.71

·        Account will contain $2590.71 after 10 years.

o       The interest is $2590.71 - $1200 = 1390.71Interest rate is annual yet it is calculated monthly! Express the interest rate as % per month by dividing it by 12 to get R = 0.5%. The number of interest periods must also be expressed in months (n = 60 months)

·        A = 1500(1 +  0.5/100)^60 = 2023.2752 or 2023 marks

·        The interest paid is:

o       2023 – 1500 = 523 marks

 

3). Currency Conversions

Practice Problem 1:

US	Aus	Sterling	Yen	Mark
1	1.633	0.610	139.490	1.782

·        Convert:

o       $50 US to Japanese Yen

§         $1 US = 139.490 Yen

§         $50 US x 139.490 = 6974.5 Yen

o       15 Sterling to $US

§         $1 US = 0.61 Sterling

§         1 Sterling = 1/0.61 = 1.6393443 dollars

§         15 Sterling =x 1.6393443 = 24.59016 (or $24.59)

·        Alternative: 15/0.61 = 24.59016 (or $24.59)

o       $1500 Australian to Marks

§         $1.633 Australian = 1.782 Marks

§         $1 = 1.782/1.633 = 1.0912431

§         1500 = 1.0912431 = 1636.86 Marks

·        Alternative: 1.782/1.633 x 1500 = 1636.86

o       125 Yen to Sterling

§         1 Yen in Sterling = 0.610/139.490 = 0.00437307

§         125 Yen = 0.00437307 x 125 = 0.5466

§         Alternative: 0.610/139.490 x 125 = 0.5466 (roughly 0.56 Sterling)

 

Practice Problem 2:

	GBP	USD	AUD
GBP	1	1.63	2.5
USD	P	1	Q
AUD	1/2.5	.61	1

-750 USD x AUD: (without commission)

-750 x 1/.61 = $1,229.5 AUD

 

How many AUD do you have if you take 2% commission?

-750 x .02 = 15 (750-15=735)

-735 x (1/.61) = $1,204.5 AUD

 

4). Construction and Use of Tables- Buy and Sell

Practice Problem:

A.) 500 AUD- Simple Interest= 7% for 5 years

B.) 500 AUD to GBP

-Sell at 105 & Buy at 100

 

Convert 7.5 compound quarterly for 5 years

A.) I=PRT/100 --> I=(500)(7)(5)/100 --> I=175

B.) 500 x 105 =52,500(1+((7.5/4)/100)^(4x5)

          =76122/100 GBP --> =761 in AUD

Therefore…

A.) 500+175= 675 AUD

B.) 761 AUD

 

Thus...Option "B" is the best option!

 

5). Linear Programming – Commonly used method in business management in making decisions on operating procedures.

·        Stage 1- Define the variables

·        Stage 2- Write the constraints of the problem as inequalities

·        Stage 3- Find the objective function

·        Stage 4- Plot graphs of the inequalities and identify the feasible region for the problem

·        Stage 5- Use the objective function to find the optimal solution to the problem

Practice Problem: 1 metre of power cable costs $2 and 1 metre of coaxial cable costs $3. Menara has $60 to spend. Write the information as inequalities and show the feasible region as a graph. What is the largest amount of cable the Menara can afford to buy?

·        Stage 1- What are you asked to find? The amount of each type of cable that Menara can afford to buy.

o       P = Number of metres of power cable that Menara buys

o       C = Number of metres of coaxial cable

·        Stage 2- Menara cannot buy a negative amount of either type of cable

o       P is (greater than or equal to) 0, and C is (greater than or equal to) 0

o       2p + 3c is (less than or equal to) 60

·        Stage 3- What is the objective function? Menara wants to buy as much cable of either type as possible

o       Mathematically: Menara wants to maximize A = p + c (the total length of cable bought it p+c)

·        Stage 4- (Due to web page construction constraints, the feasible region graph cannot be drawn)

·        Stage 5- Find the solution that gives the largest value

o       Vertices: p=0, c=0, A=0 (origin)/ p=30, c=0, A=30/ p=0, c=20, A=20

·        Largest value of the objective = 30…thus, the full solution is: Menara should buy 30 metres of power cable, which will cost him $60.

 

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Created by: Caitlin Doyle, Sarah Johnson, and Hedieh Fakhriyazdi (P.1- Pandza)