section 6.7 – financial models

Section 6.7 – Financial Models

𝐼=𝑃𝑟𝑡𝐼=𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡𝑃=𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒 𝑖𝑛𝑣𝑒𝑠𝑡𝑒𝑑𝑟=𝑎𝑛𝑛𝑢𝑎𝑙 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡𝑟𝑎𝑡𝑒𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑𝑎𝑠𝑎𝑑𝑒𝑐𝑖𝑚𝑎𝑙

Simple Interest Formula

Compound Interest Formula

𝐴=𝑃 ∙(1+𝑟𝑛 )

𝑛 ∙𝑡

Continuous Compounding Interest Formula𝐴=𝑃 𝑒𝑟 ∙ 𝑡

𝐴=𝑟𝑒𝑡𝑢𝑟𝑛𝑜𝑛 h𝑡 𝑒𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒


𝐼=𝑃𝑟𝑡Example - Simple Interest

Examples - Compound Interest

𝐴=𝑃 ∙(1+𝑟𝑛 )

𝑛 ∙𝑡

What is the future value of a $34,100 principle invested at 4% for 3 years

$21,000 is invested at 13.6% compounded quarterly for 4 years. What is the return value?

The amount of $12,700 is invested at 8.8% compounded semiannually for 1 year. What is the future value?

𝐼=(34100 )(.04)(3)𝐼=$ 4092.00

𝐹𝑢𝑡𝑢𝑟𝑒𝑉𝑎𝑙𝑢𝑒=34100+4092.00𝐹𝑢𝑡𝑢𝑟𝑒𝑉𝑎𝑙𝑢𝑒=$ 38,192.00

𝐴=12700 ∙(1+0.088

2 )2∙ 1

𝐴=$13,842.19

𝐴=21000∙ (1+0.136

4 )4 ∙ 4

𝐴=$35,854.85

Section 6.7 – Financial ModelsExamples - Compound Interest

𝐴=𝑃 ∙(1+𝑟𝑛 )

𝑛 ∙𝑡

Example - Continuous Compounding Interest𝐴=𝑃 𝑒𝑟 ∙ 𝑡

If you invest $500 at an annual interest rate of 10% compounded continuously, calculate the final amount you will have in the account after five years.

How much money will you have if you invest $4000 in a bank for sixty years at an annual interest rate of 9%, compounded monthly?

𝐴=4000 ∙(1+0.0912 )

12∙ 60

𝐴=$ 867,959.49

𝐴=500𝑒0.10 ∙ 5 𝐴=$ 824.36

Effective Interest Rate – the actual annual interest rate that takes into account the effects of compounding.


Which is better, to receive 9.5% (annual rate) continuously compounded or 10% (annual rate) compounded 4 times per year?

Compounding n times per year:

Continuous compounding:

Continuous compounding

𝑟𝑒=𝑒0.095−1𝑟𝑒=0.1074=10.74 %

Compounding 4 times per year𝑟𝑒=𝑒𝑟−1

Present Value – the initial principal invested at a specific rate and time that will grow to a predetermined value.


How much money do you have to put in the bank at 12% annual interest for five years (a) compounded 6 times per year and (b) compounded continuously to end up with $2,000?

Compounding n times per year:

Continuous compounding:

𝑃=2000𝑒− 0.12 ∙5P

P 𝑃=$1097.62

Continuous compoundingCompounding 6 times per year

Example


What rate of interest (a) compounded monthly and (b) continuous compounding is required to triple an investment in five years?

𝑟=0.2217=22.17 %

6 0√3=1+𝑟12

𝐴=𝑃 ∙(1+𝑟𝑛 )

𝑛 ∙𝑡

3𝑃=𝑃 ∙(1+𝑟

12 )12∙ 5

3=(1+𝑟12 )

60

12 6 0√3=12+𝑟12 6 0√3−12=𝑟

𝐴=𝑃 𝑒𝑟 ∙ 𝑡

3𝑃=𝑃𝑒𝑟 ∙5

3=𝑒𝑟 ∙ 5

𝑙𝑛3=𝑙𝑛𝑒𝑟 ∙5

𝑙𝑛3=5𝑟𝑟=

𝑙𝑛35

𝑟=0.2197=21.97 %

𝐶𝑜𝑚𝑝𝑜𝑢𝑛𝑑𝑒𝑑 h𝑀𝑜𝑛𝑡 𝑙𝑦 𝐶𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠𝐶𝑜𝑚𝑝𝑜𝑢𝑛𝑑𝑒𝑑

Uninhibited Exponential Growth

Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models

𝐴(𝑡)=𝐴0𝑒𝑘𝑡

𝐴 (𝑡 )=𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟 h𝑤𝑒𝑖𝑔 𝑡 𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒𝐴0=𝑖𝑛𝑖𝑡𝑖𝑎𝑙𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟 h𝑤𝑒𝑖𝑔 𝑡𝑘=𝑟𝑎𝑡𝑒 𝑜𝑓 h𝑔𝑟𝑜𝑤𝑡 ; h𝑡 𝑒𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡(𝑘>0)

Uninhibited Exponential Decay𝐴(𝑡)=𝐴0𝑒𝑘𝑡

𝐴 (𝑡 )=𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟 h𝑤𝑒𝑖𝑔 𝑡 𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒𝐴0=𝑖𝑛𝑖𝑡𝑖𝑎𝑙𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟 h𝑤𝑒𝑖𝑔 𝑡𝑘= h𝑡 𝑒𝑟𝑎𝑡𝑒𝑜𝑓 𝑑𝑒𝑐𝑎𝑦 ; h𝑡 𝑒𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑘<0)

Examples



The population of the United States was approximately 227 million in 1980 and 282 million in 2000. Estimate the population in the years 2010 and 2020.

𝐴 (𝑡 )=227𝑒0.010848(2010−1980 )282=227𝑒𝑘(2000−1980)

282227=𝑒20𝑘

𝑘=𝑙𝑛 282

22720

=0.010848

𝐴 (𝑡 )=314.3𝑚𝑖𝑙𝑙𝑖𝑜𝑛

𝑙𝑛 282227=20𝑘

Find k

𝐹𝑟𝑜𝑚2010𝐶𝑒𝑛𝑠𝑢𝑠 :308.7𝑚𝑖𝑙𝑙𝑖𝑜𝑛2020

2010

𝐴 (𝑡 )=227𝑒0.010848(2020−1980)

𝐴 (𝑡 )=350.3𝑚𝑖𝑙𝑙𝑖𝑜𝑛

Examples



A radioactive material has a half-life of 700 years. If there were ten grams initially, how much would remain after 300 years? When will the material weigh 7.5 grams?

𝐴 (𝑡 )=10𝑒− 0.00099(300)1=2𝑒𝑘 (700)

0.5=𝑒700𝑘

𝑘=𝑙𝑛0.5700

𝐴 (𝑡 )=7.43 𝑔𝑟𝑎𝑚𝑠𝑙𝑛0.5=700𝑘

Find k

or

300 years

𝐴 (𝑡 )=10𝑒𝑙𝑛 0.5700 (300)

𝐴 (𝑡 )=7.43 𝑔𝑟𝑎𝑚𝑠𝑘=−0.000990

7.5=10𝑒−0.00099𝑡

𝑡=290.6 𝑦𝑒𝑎𝑟𝑠or

7.5 grams

7.5=10𝑒𝑙𝑛0.5700 𝑡

𝑡=290.5 𝑦𝑒𝑎𝑟𝑠

0.75=𝑒−0.00099 𝑡

𝑙𝑛0.75=−0.00099 𝑡

0.75=𝑒𝑙𝑛0.5700 𝑡

𝑙𝑛0.75=𝑙𝑛0.5700 𝑡

Newton’s Law of Cooling


𝑢 (𝑡 )=𝑇+(𝑢0−𝑇 )𝑒𝑘𝑡

𝑢 (𝑡 )=𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒𝑜𝑓 𝑎 h𝑒𝑎𝑡𝑒𝑑𝑜𝑏𝑗𝑒𝑐𝑡𝑎𝑡 𝑎𝑛𝑦 𝑔𝑖𝑣𝑒𝑛𝑡𝑖𝑚𝑒𝑇=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒𝑜𝑓 h𝑡 𝑒𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑚𝑒𝑑𝑖𝑢𝑚

𝑘=𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑐𝑜𝑜𝑙𝑖𝑛𝑔𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡(𝑘<0)

𝑢0=𝑖𝑛𝑖𝑡𝑖𝑎𝑙𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒𝑜𝑓 h𝑡 𝑒h𝑒𝑎𝑡𝑒𝑑 𝑜𝑏𝑗𝑒𝑐𝑡

Newton’s Law of Cooling


𝑢 (𝑡 )=𝑇+(𝑢0−𝑇 )𝑒𝑘𝑡

ExampleA pizza pan is removed at 3:00 PM from an oven whose temperature is fixed at 450 F into a room that is a constant 70 F. After 5 minutes, the pizza pan is at 300 F. At what time is the temperature of the pan 135 F?

300=70+(450−70)𝑒𝑘5Find k

230=380𝑒𝑘5

230380=𝑒𝑘5

𝑙𝑛 2338=5𝑘

𝑘=𝑙𝑛 23

385

=−0.100418

t @ 135 F135=70+(450−70)𝑒− 0.100418𝑡

65=380𝑒−0.100418 𝑡

65380=𝑒− 0.100418𝑡

𝑙𝑛 1375=−0.100418 𝑡

𝑡=17.45𝑚𝑖𝑛𝑢𝑡𝑒𝑠𝑎𝑏𝑜𝑢𝑡 3 :17𝑃𝑀

Logistic Growth/Decay


𝑃 (𝑡 )=𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒𝑎 ,𝑏 ,𝑐=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 h h𝑡 𝑟𝑜𝑢𝑔 𝑑𝑎𝑡𝑎𝑎𝑛𝑎𝑙𝑦𝑠𝑖𝑠(𝑎>0𝑎𝑛𝑑𝑐>0)

h𝑔𝑟𝑜𝑤𝑡 𝑚𝑜𝑑𝑒𝑙𝑖𝑓 𝑏>0

𝑃 (𝑡 )= 𝑐1+𝑎𝑒−𝑏𝑡

𝑑𝑒𝑐𝑎𝑦𝑚𝑜𝑑𝑒𝑙𝑖𝑓 𝑏<0

Logistic Growth/Decay


𝑃 (𝑡 )= 𝑐1+𝑎𝑒−𝑏𝑡

ExampleThe logistic growth model relates the proportion of U.S. households that own a cell phone to the year. Let represent 2000, represent 2001, and so on. (a) What proportion of households owned a cell phone in 2000, (b) what proportion of households owned a cell phone in 2005, and (c) when will 85% of the households own a cell phone?

2000

𝑃 (𝑡 )=0.2375=23.75 %

𝑃 (𝑡 )= 0.951+3𝑒− 0.32(0)

𝑃 (𝑡 )=0.951+3

2005

𝑃 (𝑡 )=0.5916=59.16 %

𝑃 (𝑡 )= 0.951+3𝑒− 0.32(5)

𝑃 (𝑡 )= 0.951.60569

𝑡=2000−2000=0 𝑡=2005−2000=5P(t) = 85%

0.85=0.95

1+3𝑒−0.32 𝑡

0.85 (1+3𝑒−0.32𝑡)=0.950.85+2.55𝑒−0.32 𝑡=0.95

2.55𝑒−0.32 𝑡=0.10𝑒−0.32 𝑡=0.039216

−0.32𝑡=𝑙𝑛 0.039216𝑡=10.12 𝑦𝑒𝑎𝑟𝑠→2010 𝑡𝑜2011

section 6.7 – financial models

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