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Section 6.7 – Financial Models. Simple Interest Formula. Compound Interest Formula. Continuous Compounding Interest Formula. Section 6.7 – Financial Models. Example - Simple Interest. What is the future value of a $34,100 principle invested at 4% for 3 years. Examples - Compound Interest. - PowerPoint PPT Presentation
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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
Section 6.7 – Financial Models
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.
Section 6.7 – Financial Models
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.
Section 6.7 – Financial Models
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
Section 6.7 – Financial Models
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
Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models
𝐴(𝑡)=𝐴0𝑒𝑘𝑡
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
Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models
𝐴(𝑡)=𝐴0𝑒𝑘𝑡
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
Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models
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
Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models
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
Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models
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
Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models
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