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7.6: Applications


Learning Objectives

  • Use the compound and continuous interest formulas.
  • Calculate doubling time.
  • Use the exponential growth/decay model.
  • Calculate the rate of decay given half-life.

Compound and Continuous Interest Formulas

Recall that compound interest occurs when interest accumulated for one period is added to the principal investment before calculating interest for the next period. The amount (A) accrued in this manner over time (t) is modeled by the compound interest formula:

(A(t)=Pleft(1+frac{r}{n} ight)^{n t})

Here the initial principal (P) is accumulating compound interest at an annual rate (r) where the value (n) represents the number of times the interest is compounded in a year.

Example (PageIndex{1}):

Susan invested $(500) in an account earning (4 frac{1}{2})% annual interest that is compounded monthly.

a. How much will be in the account after (3) years?

b. How long will it take for the amount to grow to $(750)?

Solution

In this example, the principal (P =) $(500), the interest rate (r = 4 frac{1}{2})% (= 0.045), and because the interest is compounded monthly, (n = 12). The investment can be modeled by the following function:

(A(t)=500left(1+frac{0.045}{12} ight)^{12 t})
(A(t)=500(1.00375)^{12 t})

a. Use this model to calculate the amount in the account after (t=3) years.

(egin{aligned} A(color{Cerulean}{3}color{black}{)} &=500(1.00375)^{12(color{Cerulean}{3}color{black}{)}} &=500(1.00375)^{36} & approx 572.12 end{aligned})

Rounded off to the nearest cent, after (3) years, the amount accumulated will be $(572.12).

b. To calculate the time it takes to accumulate $(750), set (A (t) = 750) and solve for (t).

(egin{array}{l}{A(t)=500(1.00375)^{12 t}} {color{Cerulean}{750}color{black}{=}500(1.00375)^{12 t}}end{array})

This results in an exponential equation that can be solved by first isolating the exponential expression.

(egin{aligned} 750 &=500(1.00375)^{12 t} frac{750}{500} &=(1.00375)^{12 t} 1.5 &=(1.00375)^{12 t} end{aligned})

At this point take the common logarithm of both sides, apply the power rule for logarithms, and then solve for (t).

(egin{aligned} log (1.5) &=log (1.00375)^{12 t} log (1.5) &=12 t log (1.00375)frac{log (1.5)}{color{Cerulean}{12log (1.00375)}}&color{black}{=} frac{cancel{12}tcancel{log (1.00375)}}{cancel{color{Cerulean}{12: log (1.00375)}}} frac{log(1.5)}{12:log(1.00375)}&=tend{aligned})

Using a calculator we can approximate the time it takes.

(t=log (1.5) /(12 * log (1.00375)) approx 9) years

Answer:

a. $(572.12)

b. Approximately (9) years

The period of time it takes a quantity to double is called the doubling time20. We next outline a technique for calculating the time it takes to double an initial investment earning compound interest.

Example (PageIndex{2}):

Mario invested $(1000) in an account earning (6.3)% annual interest, that is compounded semi-annually. How long will it take the investment to double?

Solution

Here the principal (P =) $(1,000), the interest rate (r = 6.3)% (= 0.063), and because the interest is compounded semi-annually (n = 2). This investment can be modeled as follows:

(A(t)=1,000left(1+frac{0.063}{2} ight)^{2 t})
(A(t)=1,000(1.0315)^{2 t})

Since we are looking for the time it takes to double $(1,000), substitute $(2,000) for the resulting amount (A (t)) and then solve for (t).

(egin{aligned} color{Cerulean}{2,000} &color{black}{=}1,000(1.0315)^{2 t} frac{2,000}{1,000} &=(1.0315)^{2 t} 2 &=(1.0315)^{2 t} end{aligned})

At this point we take the common logarithm of both sides.

(egin{aligned} 2 &=(1.0315)^{2 t} log 2 &=log (1.0315)^{2 t} log 2 &=2 t log (1.0315) frac{log 2}{2 log (1.0315)} &=t end{aligned})

Using a calculator we can approximate the time it takes:

(t=log (2) /(2 * log (1.0315)) approx 11.17) years

Answer:

Approximately (11.17) years to double at (6.3)%.

If the investment in the previous example was one million dollars, how long would it take to double? To answer this we would use (P =) $(1,000,000) and (A (t) =) $(2,000,000):

(egin{aligned} A(t) &=1,000(1.0315)^{2 t} color{Cerulean}{2,000,000} &color{black}{=}1,000,000(1.0315)^{2 t} end{aligned})

Dividing both sides by (1,000,000) we obtain the same exponential function as before.

(2=(1.0315)^{2 t})

Hence, the result will be the same, about (11.17) years. In fact, doubling time is independent of the initial investment (P).

Interest is typically compounded semi-annually ((n = 2)), quarterly ((n = 4)), monthly ((n = 12)), or daily ((n = 365)). However if interest is compounded every instant we obtain a formula for continuously compounding interest:

(A(t)=P e^{r t})

Here (P) represents the initial principal amount invested, (r) represents the annual interest rate, and (t) represents the time in years the investment is allowed to accrue continuously compounded interest.

Example (PageIndex{3}):

Mary invested $(200) in an account earning (5 frac{3}{4})% annual interest that is compounded continuously. How long will it take the investment to grow to $(350)?

Solution

Here the principal (P =) $(200) and the interest rate (r = 5 frac{3}{4})% (= 5.75)% (= 0.0575). Since the interest is compounded continuously, use the formula (A (t) = Pe^{rt}). Hence, the investment can be modeled by the following,

(A(t)=200 e^{0.0575 t})

To calculate the time it takes to accumulate to $(350), set (A (t) = 350) and solve for (t).

(egin{array}{r}{A(t)=200 e^{0.0575 t}} {color{Cerulean}{350}color{black}{=}200 e^{0.0575 t}}end{array})

Begin by isolating the exponential expression.

(egin{aligned} frac{350}{200} &=e^{0.0575 t} frac{7}{4} &=e^{0.0575 t} 1.75 &=e^{0.0575 t} end{aligned})

Because this exponential has base (e), we choose to take the natural logarithm of both sides and then solve for (t).

(egin{array}{l}{ln (1.75)=ln e^{0.0575 t}}quadquadcolor{Cerulean}{Apply:the:power:rule:for:logarithms.} {ln (1.75)=0.0575 t ln e} quadcolor{Cerulean}{Recall:that: ln e=1.} {ln (1.75)=0.0575 t cdot 1} {frac{ln (1.75)}{0.0575}=t}end{array})

Using a calculator we can approximate the time it takes:

(t=ln (1.75) / 0.0575 approx 9.73 quad years)

Answer:

It will aprroximately (9.73) years.

When solving applications involving compound interest, look for the keyword “continuous,” or the keywords that indicate the number of annual compoundings. It is these keywords that determine which formula to choose.

Exercise (PageIndex{1})

Mario invested $(1,000) in an account earning (6.3)% annual interest that is compounded continuously. How long will it take the investment to double?

Answer

Approximately (11) years.

www.youtube.com/v/Z_ilJHIQvWQ

Modeling Exponential Growth and Decay

In the sciences, when a quantity is said to grow or decay exponentially, it is specifically meant to be modeled using the exponential growth/decay formula21:

(P(t)=P_{0} e^{k t})

Here (P_{0}), read “(P) naught,” or “(P) zero,” represents the initial amount, k represents the growth rate, and (t) represents the time the initial amount grows or decays exponentially. If (k) is negative, then the function models exponential decay. Notice that the function looks very similar to that of continuously compounding interest formula. We can use this formula to model population growth when conditions are optimal.

Example (PageIndex{4}):

It is estimated that the population of a certain small town is (93,000) people with an annual growth rate of (2.6)%. If the population continues to increase exponentially at this rate:

  1. Estimate the population in (7) years' time.
  2. Estimate the time it will take for the population to reach 120,000 people.

Solution

We begin by constructing a mathematical model based on the given information. Here the initial population (P_{0} = 93,000) people and the growth rate (r = 2.6)% (= 0.026). The following model gives population in terms of time measured in years:

(P(t)=93,000 e^{0.026 t})

a. Use this function to estimate the population in (t = 7) years.

(egin{aligned} P(t) &=93,000 e^{0006(color{Cerulean}{7}color{black}{)}} &=93,000 e^{0.182} & approx 111,564 quad people end{aligned})

b. Use the model to determine the time it takes to reach (P (t) = 120,000) people.

(egin{aligned} P(t) &=93,000 e^{0.026 t} color{Cerulean}{120,000} &color{black}{=}93,000 e^{0.026 t} frac{120,000}{93,000} &=e^{0.026 t} frac{40}{31} &=e^{0.026 t} end{aligned})

Take the natural logarithm of both sides and then solve for (t).

(ln left(frac{40}{31} ight)=ln e^{0.026 t})
(ln left(frac{40}{31} ight)=0.026 t ln e)
(ln left(frac{40}{31} ight)=0.026 t cdot 1)
(frac{ln left(frac{40}{31} ight)}{0.026}=t)

Using a calculator,

(t=ln (40 / 31) / 0.026 approx 9.8quad years)

Answer:

  1. (111,564) people
  2. (9.8) years

Often the growth rate (k) is not given. In this case, we look for some other information so that we can determine it and then construct a mathematical model. The general steps are outlined in the following example.

Example (PageIndex{5}):

Under optimal conditions Escherichia coli (E. coli) bacteria will grow exponentially with a doubling time of (20) minutes. If (1,000) E. coli cells are placed in a Petri dish and maintained under optimal conditions, how many E. coli cells will be present in (2) hours?


Figure (PageIndex{1}): Escherichia coli (E. coli)

Solution

The goal is to use the given information to construct a mathematical model based on the formula (P (t) = P_{0} e^{kt}).

Step 1: Find the growth rate (k). Use the fact that the initial amount, (P_{0} = 1,000) cells, doubles in (20) minutes. That is, (P (t) = 2,000) cells when (t = 20) minutes.

(egin{aligned} P(t) &=P_{0} e^{k t} color{Cerulean}{2,000} &color{black}{=}1,000 e^{k color{Cerulean}{20}} end{aligned})

Solve for the only variable (k).

(egin{aligned} 2,000 &=1,000 e^{k 20} frac{2,000}{1,000} &=e^{k 20} 2 &=e^{k 20} ln (2) &=ln e^{k 20} ln (2) &=k 20 ln e ln (2) &=k 20 cdot 1 frac{ln (2)}{20} &=k end{aligned})

Step 2: Write a mathematical model based on the given information. Here (k ≈ 0.0347), which is about (3.5)% growth rate per minute. However, we will use the exact value for (k) in our model. This will allow us to avoid round-off error in the final result. Use (P_{0} = 1,000) and (k=ln (2) / 20):

(P(t)=1,000 e^{(ln (2) / 20) t})

This equation models the number of E. coli cells in terms of time in minutes.

Step 3: Use the function to answer the questions. In this case, we are asked to find the number of cells present in (2) hours. Because time is measured in minutes, use (t = 120) minutes to calculate the number of E. coli cells.

(egin{aligned} P(color{Cerulean}{120}color{black}{)} &=1,000 e^{(ln (2) / 20)(color{Cerulean}{120}color{black}{)}} &=1,000 e^{ln (2) cdot 6} &=1,000 e^{ln 2^{6}} &=1,000 cdot 2^{6} &=64,000 ext { cells } end{aligned})

Answer:

In two hours (64,000) cells will be present.

When the growth rate is negative the function models exponential decay. We can describe decreasing quantities using a half-life22, or the time it takes to decay to one-half of a given quantity.

Example (PageIndex{6}):

Due to radioactive decay, caesium-137 has a half-life of (30) years. How long will it take a (50)-milligram sample to decay to (10) milligrams?

Solution

Use the half-life information to determine the rate of decay (k). In (t = 30) years the initial amount (P_{0} = 50) milligrams will decay to half (P (30) = 25) milligrams.

(egin{aligned} P(t) &=P_{0} e^{k t} 25 &=50 e^{k 30} end{aligned})

Solve for the only variable, (k).

(egin{aligned}25&=50 e^{130} frac{25}{50}&=e^{30 k} ln left(frac{1}{2} ight)&=ln e^{30 k} ln left(frac{1}{2} ight)&=30 k ln e frac{ln 1-ln 2}{30}&=kquadquadquadcolor{Cerulean}{Recall:that:ln1=0.} -frac{ln 2}{30}&=kend{aligned})

Note that (k=-ln frac{2}{30} approx-0.0231) is negative. However, we will use the exact value to construct a model that gives the amount of cesium-137 with respect to time in years.

(P(t)=50 e^{(-ln 2 / 30) t})

Use this model to find (t) when (P (t) = 10) milligrams.

(egin{aligned}10&=50 e^{(-ln 2 / 30) t} frac{10}{50}&=e^{(-ln 2 / 30) t} ln left(frac{1}{5} ight)&=ln e^{(-ln 2 / 30) t} ln 1-ln 5&=left(-frac{ln 2}{30} ight) t ln e quadcolor{Cerulean} { Recall: that: ln e=1. } -30frac{(ln 1-ln5)}{ln 2}&=t-frac{30(0-ln 5)}{ln 2}&=t frac{30ln 5}{ln 2}&=tend{aligned})

Answer:

Using a calculator, it will take (t ≈ 69.66) years to decay to (10) milligrams.

Radiocarbon dating is a method used to estimate the age of artifacts based on the relative amount of carbon-14 present in it. When an organism dies, it stops absorbing this naturally occurring radioactive isotope, and the carbon-14 begins to decay at a known rate. Therefore, the amount of carbon-14 present in an artifact can be used to estimate the age of the artifact.

Example (PageIndex{7}):

An ancient bone tool is found to contain (25)% of the carbon-14 normally found in bone. Given that carbon-14 has a half-life of (5,730) years, estimate the age of the tool.

Solution

Begin by using the half-life information to find (k). Here the initial amount (P_{0}) of carbon-14 is not given, however, we know that in (t = 5,730) years, this amount decays to half, (frac{1}{2} P_{0}).

(P(t)=P_{0} e^{k t})
(frac{1}{2} P_{0}=P_{0} e^{k 5,730})

Dividing both sides by (P_{0}) leaves us with an exponential equation in terms of (k). This shows that half-life is independent of the initial amount.

(frac{1}{2}=e^{k5,730})

Solve for (k).

(egin{aligned}ln left(frac{1}{2} ight)&=ln e^{k 5,730} ln1-ln2 &= 5,730kln e frac{0-ln 2}{5,730}&=k -frac{ln 2}{5,730}&=kend{aligned})

Therefore we have the model,

(P(t)=P_{0} e^{(-ln 2 / 5,730) t})

Next we wish to the find time it takes the carbon-14 to decay to (25)% of the initial amount, or (P (t) = 0.25P_{0})

.(0.25 P_{0}=P_{0} e^{(-ln 2 / 5,730) t})

Divide both sides by (P_{0}) and solve for (t).

(egin{aligned} 0.25 &=e^{(-ln 2 / 5,730) t} ln (0.25) &=ln e^{(-ln 2 / 5,730) t} ln (0.25) &=left(-frac{ln 2}{5,730} ight)_{t ln e} -frac{5,730 ln (0.25)}{ln 2} &=t11,460&approx t end{aligned})

Answer:

The tool is approximately (11,460) years old.

Exercise (PageIndex{2})

The half-life of strontium-90 is about (28) years. How long will it take a (36) milligram sample of strontium-90 to decay to (30) milligrams?

Answer

(7.4) years

www.youtube.com/v/Oto0ihiyvBc

Key Takeaways

  • When interest is compounded a given number of times per year use the formula (A(t)=Pleft(1+frac{r}{n} ight)^{n t}).
  • When interest is to be compounded continuously use the formula (A(t)=P e^{r t}).
  • Doubling time is the period of time it takes a given amount to double. Doubling time is independent of the principal.
  • When amounts are said to be increasing or decaying exponentially, use the formula (P(t)=P_{0} e^{k t}).
  • Half-life is the period of time it takes a given amount to decrease to one-half. Half-life is independent of the initial amount.
  • To model data using the exponential growth/decay formula, use the given information to determine the growth/decay rate (k). Once (k) is determined, a formula can be written to model the problem. Use the formula to answer the questions.

Exercise (PageIndex{3})

  1. Jill invested $(1,450) in an account earning (4 frac{5}{8})% annual interest that is compounded monthly.
    1. How much will be in the account after (6) years?
    2. How long will it take the account to grow to $(2,200)?
  2. James invested $(825) in an account earning (5 frac{2}{5})% annual interest that is compounded monthly.
    1. How much will be in the account after (4) years?
    2. How long will it take the account to grow to $(1,500)?
  3. Raul invested $(8,500) in an online money market fund earning (4.8)% annual interest that is compounded continuously.
    1. How much will be in the account after (2) years?
    2. How long will it take the account to grow to $(10,000)?
  4. Ian deposited $(500) in an account earning (3.9)% annual interest that is compounded continuously.
    1. How much will be in the account after (3) years?
    2. How long will it take the account to grow to $(1,500)?
  5. Bill wants to grow his $(75,000) inheritance to $(100,000) before spending any of it. How long will this take if the bank is offering (5.2)% annual interest compounded quarterly?
  6. Mary needs $(25,000) for a down payment on a new home. If she invests her savings of $(21,350) in an account earning (4.6)% annual interest that is compounded semi-annually, how long will it take to grow to the amount that she needs?
  7. Joe invested his $(8,700) savings in an account earning (6 frac{3}{4})% annual interest that is compounded continuously. How long will it take to earn $(300) in interest?
  8. Miriam invested $(12,800) in an account earning (5 frac{1}{4})% annual interest that is compounded monthly. How long will it take to earn $(1,200) in interest?
  9. Given that the bank is offering (4.2)% annual interest compounded monthly, what principal is needed to earn $(25,000) in interest for one year?
  10. Given that the bank is offering (3.5)% annual interest compounded continuously, what principal is needed to earn $(12,000) in interest for one year?
  11. Jose invested his $(3,500) bonus in an account earning (5 frac{1}{2})% annual interest that is compounded quarterly. How long will it take to double his investment?
  12. Maria invested her $(4,200) savings in an account earning (6 frac{3}{4})% annual interest that is compounded semi-annually. How long will it take to double her savings?
  13. If money is invested in an account earning (3.85)% annual interest that is compounded continuously, how long will it take the amount to double?
  14. If money is invested in an account earning (6.82)% annual interest that is compounded continuously, how long will it take the amount to double?
  15. Find the annual interest rate at which an account earning continuously compounding interest has a doubling time of (9) years.
  16. Find the annual interest rate at which an account earning interest that is compounded monthly has a doubling time of (10) years.
  17. Alice invested her savings of $(7,000) in an account earning (4.5)% annual interest that is compounded monthly. How long will it take the account to triple in value?
  18. Mary invested her $(42,000) bonus in an account earning (7.2)% annual interest that is compounded continuously. How long will it take the account to triple in value?
  19. Calculate the doubling time of an investment made at (7)% annual interest that is compounded:
    1. monthly
    2. continuously
  20. Calculate the doubling time of an investment that is earning continuously compounding interest at an annual interest rate of:
    1. (4)%
    2. (6)%
  21. Billy’s grandfather invested in a savings bond that earned (5.5)% annual interest that was compounded annually. Currently, (30) years later, the savings bond is valued at $(10,000). Determine what the initial investment was.
  22. In 1935 Frank opened an account earning (3.8)% annual interest that was compounded quarterly. He rediscovered this account while cleaning out his garage in 2005. If the account is now worth $(11,294.30), how much was his initial deposit in 1935?
Answer

1. (1) $(1,912.73) (2) (9) years

3. (1) $(9,356.45) (2) (3.4) years

5. (5.6) years

7. (frac{1}{2}) year

9. $(583,867)

11. (12.7) years

13. (18) years

15. (7.7)%

17. (24.5) years

19. (1) (9.93) years (2) (9.90) years

21. $(2,006.44)

Exercise (PageIndex{4})

  1. The population of a small town of (24,000) people is expected grow exponentially at a rate of (1.6)% per year. Construct an exponential growth model and use it to:
    1. Estimate the population in (3) years’ time.
    2. Estimate the time it will take for the population to reach (30,000) people.
  2. During the exponential growth phase, certain bacteria can grow at a rate of (4.1)% per hour. If (10,000) cells are initially present in a sample, construct an exponential growth model and use it to:
    1. Estimate the population in (5) hours.
    2. Estimate the time it will take for the population to reach (25,000) cells.
  3. In 2000, the world population was estimated to be (6.115) billion people and in 2010 the estimate was (6.909) billion people. If the world population continues to grow exponentially, estimate the total world population in 2020.
  4. In 2000, the population of the United States was estimated to be (282) million people and in 2010 the estimate was (309) million people. If the population of the United States grows exponentially, estimate the population in 2020.
  5. An automobile was purchased new for $(42,500) and (2) years later it was valued at $(33,400). Estimate the value of the automobile in (5) years if it continues to decrease exponentially.
  6. A new PC was purchased for $(1,200) and in (1.5) years it was worth $(520). Assume the value is decreasing exponentially and estimate the value of the PC four years after it is purchased.
  7. The population of the downtown area of a certain city decreased from (12,500) people to (10,200) people in two years. If the population continues to decrease exponentially at this rate, what would we expect the population to be in two more years?
  8. A new MP3 player was purchased for $(320) and in (1) year it was selling used online for $(210). If the value continues to decrease exponentially at this rate, determine the value of the MP3 player (3) years after it was purchased.
Answer

1. (1) About (25,180) people (2) About (14) years

3. About (7.806) billion people

5. About $(23,269.27)

7. (8,323) people

9. (3,715) years

11. (9.8) days

13. About (3,689) years old

15. (26.6) days

17. (9.9)%

19. (98.8)%

21. (3,648) years

Exercise (PageIndex{5})

Solve for the given variable:

  1. Solve for (t: A = Pe^{rt})
  2. Solve for (t: A = P(1 + r)^{t})
  3. Solve for (I: M=log left(frac{I}{l_{0}} ight))
  4. Solve for (H^{+}: pH = -log left(H^{+} ight))
  5. Solve for (t: P = frac{1}{1+e^{−t}})
  6. Solve for (I: L=10 log left(I / 10^{-12} ight))
  7. The number of cells in a certain bacteria sample is approximated by the logistic growth model (N(t)=frac{1.2 imes 10^{5}}{1+9 e^{-0.32t}}), where (t) represents time in hours. Determine the time it takes the sample to grow to (24,000) cells.
  8. The market share of a product, as a percentage, is approximated by the formula (P(t)=frac{100}{3+e^{-0.44 t}}) where (t) represents the number of months after an aggressive advertising campaign is launched.
    1. What was the initial market share?
    2. How long would we expect to see a (3.5)% increase in market share?
  9. In chemistry, pH is a measure of acidity and is given by the formula (mathrm{pH}=-log left(H^{+} ight)), where (H^{+}) is the hydrogen ion concentration (measured in moles of hydrogen per liter of solution.) What is the hydrogen ion concentration of seawater with a pH of (8)?
  10. Determine the hydrogen ion concentration of milk with a pH of (6.6).
  11. The volume of sound, (L) in decibels (dB), is given by the formula (L=10 log left(I / 10^{-12} ight)) where (I) represents the intensity of the sound in watts per square meter. Determine the sound intensity of a hair dryer that emits (70) dB of sound.
  12. The volume of a chainsaw measures (110) dB. Determine the intensity of this sound.
Answer

1. (t=frac{ln (A)-ln (P)}{r})

3. (I=I_{0} cdot 10^{M})

5. (t=ln left(frac{P}{1-P} ight))

7. Approximately (2.5) hours

9. (10^{-8}) moles per liter

11. (10^{-5}) watts per square meter

Exercise (PageIndex{6})

  1. Which factor affects the doubling time the most, the annual compounding (n) or the interest rate (r)? Explain.
  2. Research and discuss radiocarbon dating. Post something interesting you have learned as well as a link to more information.
  3. Is exponential growth sustainable over an indefinite amount of time? Explain.
  4. Research and discuss the half-life of radioactive materials.
Answer

1. Answer may vary

3. Answer may vary

Footnotes

20The period of time it takes a quantity to double.

21A formula that models exponential growth or decay: (P(t)=P_{0} e^{k t}).

22The period of time it takes a quantity to decay to one-half of the initial amount.


7.6: Applications

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0000313576 00000 n 0000314220 00000 n 0000316631 00000 n trailer > startxref 318068 %%EOF


Other Aspects of YARN Security

AM/RM Token Refresh

The AM/RM token is renewed automatically the AM pushes out a new token to the AM within an allocate message. Consult the AMRMClientImpl class to see the process. Your AM code does not need to worry about this process

Token Renewal on AM Restart

Even if an application is renewing tokens regularly, if an AM fails and is restarted, it gets restarted from that original ApplicationSubmissionContext. The tokens there may have expired, so localization may fail, even before the issue of credentials to talk to other services.

How is this problem addressed? The YARN Resource Manager gets a new token for the node managers, if needed.

  1. The token passed by the RM to the NM for localization is refreshed/updated as needed.
  2. Tokens in the app launch context for use by the application are not refreshed. That is, if it has an out of date HDFS token —that token is not renewed. This also holds for tokens for for Hive, HBase, etc.
  3. Therefore, to survive AM restart after token expiry, your AM has to get the NMs to localize the keytab or make no HDFS accesses until (somehow) a new token has been passed to them from a client.

This is primarily an issue for long-lived services (see below).

Unmanaged Application Masters

Unmanaged application masters are not launched in a container set up by the RM and NM, so cannot automatically pick up an AM/RM token at launch time. The YarnClient.getAMRMToken() API permits an Unmanaged AM to request an AM/RM token. Consult UnmanagedAMLauncher for the specifics.

Identity on an insecure cluster: HADOOP_USER_NAME

In an insecure cluster, the application will run as the identity of the account of the node manager, typically something such as yarn or mapred. By default, the application will access HDFS as that user, with a different home directory, and with a different user identified in audit logs and on file system owner attributes.

This can be avoided by having the client identify the identify of the HDFS/Hadoop user under which the application is expected to run. This does not affect the OS-level user or the application’s access rights to the local machine.

When Kerberos is disabled, the identity of a user is picked up by Hadoop first from the environment variable HADOOP_USER_NAME, then from the OS-level username (e.g. the system property user.name).

YARN applications should propagate the user name of the user launching an application by setting this environment variable.

Note that this environment variable is picked up in all applications which talk to HDFS via the hadoop libraries. That is, if set, it is the identity picked up by HBase and other applications executed within the environment of a YARN container within which this environment variable is set.

Oozie integration and HADOOP_TOKEN_FILE_LOCATION

Apache Oozie can launch an application in a secure cluster either by acquiring all relevant credentials, saving them to a file in the local filesystem, then setting the path to this file in the environment variable HADOOP_TOKEN_FILE_LOCATION. This is of course the same environment variable passed down by YARN in launched containers, as is similar content: a byte array with credentials.

Here, however, the environment variable is set in the environment executing the YARN client. This client must use the token information saved in the named file instead of acquiring any tokens of its own.

Loading in the token file is automatic: UGI does it during user login.

The client is then responsible for passing the same credentials into the AM launch context. This can be done simply by passing down the current credentials.

Timeline Server integration

The Application Timeline Server can be deployed as a secure service —in which case the application will need the relevant token to authenticate with it. This process is handled automatically in YarnClientImpl if ATS is enabled in a secure cluster. Similarly, the AM-side TimelineClient YARN service class manages token renewal automatically via the ATS’s SPNEGO-authenticated REST API.

If you need to prepare a set of delegation tokens for a YARN application launch via Oozie, this can be done via the timeline client API.

Cancelling Tokens

Applications may wish to cancel tokens they hold when terminating their AM. This ensures that the tokens are no-longer valid.

This is not mandatory, and as a clean shutdown of a YARN application cannot be guaranteed, it is not possible to guarantee that the tokens will always be during application termination. However, it does reduce the window of vulnerability to stolen tokens.


The Application list template allows you to customize the body of the application list page in the developer portal.

Default template

Controls

The Product list template may use the following page controls.

Data model

Property Type Description
Paging Paging entity. The paging information for the applications collection.
Applications Collection of Application entities. The applications visible to the current user.
CategoryName string The category of application.

Sample template data


Application under Rule 7.6-1.1 Application Fee

Lawyers or paralegals who plan to retain, occupy office space with, use the services of, partner or associate with, or employ in any capacity having to do with the practice of law or provision of legal services any person who in Ontario or elsewhere, has been disbarred and struck off the Rolls, has had their licence to practise law or to provide legal services revoked, has been suspended, has had their licence to practise law or to provide legal services suspended, has undertaken not to practise law or to provide legal services, or who has been involved in disciplinary action and been permitted to resign or to surrender their licence to practise law or to provide legal services, and has not had their licence restored must formally apply in advance to the Law Society for permission to do so under rule 7.6-1.1 of the Rules of Professional Conduct or subrule 6.01(6) of the Paralegal Rules of Conduct.

The application process is initiated by an Ontario lawyer entitled to practise law or a paralegal entitled to provide legal services in Ontario along with the former or suspended licensee completing an Application Under Rule 7.6-1.1 Applicant Lawyer&rsquos Information form or an Application Under Subrule 6.01(6) Applicant Paralegal&rsquos Information form, along with the Information for a Plan of Supervision Under Rule 7.6-1.1/Subrule 6.01(6) form. The former or suspended licensee must complete the Information for a Rule 7.6-1.1/Subrule 6.01(6) Application from a Former or Suspended Licensee or a Licensee who has Given an Undertaking Not to Practise Law or Provide Legal Services. All forms should be submitted to the Law Society by the applicant.

A non-refundable application fee in the amount of $200 + HST must be submitted with the application.


Front

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Eliminate jams and ensure your production area is always clera of old corrugated containers (OCC). Increase bale density and lower haul-away costs.

Manufacturing plants have numerous forms of waste and BloApCo has a system that can handle whatever your facility might have.

If your company needs to dispose of paper, foil, light metal, wood pallets and skids or plastic, one of our systems can provide the answer. Designed and built in our Germantown, WI, plant, our products fall into two main categories — shredding and trim handling systems. All BloApCo shredders use our patented “Pierce-and-Tear” shredding concept. Ripping teeth, mounted on three shafts, counter-rotate at different speeds to “Pierce-and-Tear” material into irregular-sized pieces which can be conveyed easily into hoppers or deposited into balers, packed into refuse hoppers or into self-contained energy-generating systems.


239 thoughts on &ldquoPublished Applications&rdquo

Im getting an error when i try to add an Local App Access Application as an Published Application.
Error Id: XDDS:BFC354DE

Exception:
Error Source : Citrix Studio
StackTrace: System.InvalidOperationException Sequence contains no elements
at System.Linq.Enumerable.First[TSource](IEnumerable`1 source)
at Citrix.Console.DesktopGroups.UI.Dialogs.CreateLocalAccessAppWizard.GenerateSummary(Object sender, EventArgs e)
at Citrix.Console.CommonControls.ViewModelBase.FireEvent(EventHandler handler)
at Citrix.Console.Shared.UI.Pages.SummaryPageViewModel.LoadPageData()
at Citrix.Console.CommonControls.Wizard.PageBaseViewModel.LoadPageDataInternal()
at Citrix.Console.CommonControls.Wizard.PageBaseViewModel.b__16_20()
at Citrix.Console.CommonControls.ViewModelBase.Invoke(Action action)
at Citrix.Console.CommonControls.ViewModelBase.CallWithErrorHandling(Action action)

For an example Application like Chrome with this parameter:
C:Program Files (x86)GoogleChromeApplicationchrome.exe
WorkingDirectory: C:Program Files (x86)GoogleChromeApplication

Carl, in testing my 2019 vda servers and publishing Windows Explorer i have noticed when i make a change to a file or folder it does not show the change until i refresh the content. this is not happening in my production 2016 vda servers.
I am not sure what to look for. Can you point me in a direction to research this issue ?

If you Google File Explorer not refreshing, you’ll find many possible solutions. Maybe one of them will help you.

carl, Do I use Office Professional Plus 2019 on a windows 2019 VDA server in our CAVD environment ?

You can. You can also install Office 365.

Hi Carl, i have citrix 7.15LTSR and have a SaaS Client Application which downloads all the librarys into %Appdata% i cant change the installation path. can you recommend a citrix publish app solution for this. Do you download the content into a shared folder or do you create somekind of a sandbox app? thanks for your help, kind regards, adrian

Is this a ClickOnce app? Usually I let the app install for each user and then roam the user’s profile. Sometimes I can move the files from AppData to Program Files and publish it from there but it doesn’t always work.

I am looking to configure it in such a way that the application which is part of an app group, launches from ONLY the servers tags were assigned to and not any other, even if an app which is outside of that group is running from some server which has no tags assigned. I followed this article which talks about disabling session sharing: (https://support.citrix.com/article/CTX232362)

However, it only works if I launch the *app1* first (which is part of the app group), followed by the *app2* (not part of app group). The sessions for apps are created on different servers as expected. But if *app2* is launched first, and from a server which has no tags assigned, then *app1*, when launched, launches from the same server app2 has already been running….without taking the tag restriction into account which it should have had (being part of the app group).

Is there any way to achieve that? Or I just HAVE to create a new DG? Why does it work one way but not the other way?

Any help is really appreciated!

& ‘C:Program FilesCitrixReceiver StoreFrontScriptsImportModules.ps1’

The above are only needed to import the legacy -DS StoreFront SDK which is in very old versions of StoreFront. All -STF cmdlets should autoimport AFTER StoreFront is installed on new PS sessions so long as PS is 4.0 or higher.

Sadly a lot of docs are incorrect and suggest doing this.

Hi Carl, I know there is a way in the store advanced settings to make desktop look like a published application, but is there anyway to make a published application look like a desktop in receiver/workspace from the storefront?

I would be happy to explain why I would want to do that if interested.

Is there any way to publish same APP-V application.exe with diferents arguments? I´m using single Admin mode APPV with Studio.

1 – MicrosoftEdgeApplicationmsedge.exe http://citrix.com
2 – MicrosoftEdgeApplicationmsedge.exe http://google.com
3 – MicrosoftEdgeApplicationmsedge.exe http://xxxxx.xxx

I´ve tried many ways, but appv seems get only the first one “exe”.. and ignore anothers. When i add the package in studio, applications TAB show me just one app.

I don’t often get to point out typos, your sites are so good… but I found one!

Thanks again for your great sites. : )

LOL. Thanks for pointing that out. Hopefully it’s fixed now.

Hi Looking for guidance. I have a W2K12 CVAD7.1912 CU1 site. I am looking for a way to restrict access to a published application when the users are connecting remotely. (NO Netscaler Gateway)

1. Users connect remotely using the customers RAS solution
2. Once authenticated the users connect to the internal Storefront servers https://storefront.domain.net and access to their Published apps.

I have
* Specific Delivery Group – contains 2 VDAs – which host 2 x Published apps
* Both Apps have – a specific domain security group to give access

I am looking for a way
* to restrict access to ONE of the applications when they users are connecting remotely via RAS
* Users should have access to both Apps when onsite
* If above not possible to restrict just one of the apps – then a solution that would restrict Both apps

* Users should still have access to other applications presented from different VDAs/Delivery groups

Citrix Storefront currently has a single store

Is there a way of using a Whitelist (easier than a blacklist) to allow access to the one specific “restricted” App once connection comes from a “Subnet” on the “Whitelist” (all the internal device subnets)

Any advice much appreciated

Delivery Groups can be filtered by source IP address – https://support.citrix.com/article/CTX128232. This is the entire Delivery Group.

You can create a separate StoreFront store (probably on separate StoreFront server) and filter apps there. Configure RAS to allow access to the filtered store but not the normal store.


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FAQ: How does XenApp/XenDesktop 7.6 differ from XenApp 6.5?

Q: What are the main Differences between XenApp/XenDesktop 7.6 and XenApp 6.5?

A: If you are familiar with XenApp 6.5 and previous versions of XenApp, it may be helpful to think of XenApp 7.6 and XenDesktop 7.6 in terms of how they differ from those versions. Although they are not exact equivalents, the following table helps map functional elements from XenApp 6.5 and previous versions to XenApp 7.6 and XenDesktop 7.6.

Previous component in XenApp 6.5 : Component equivalent in XenApp/XenDesktop 7.6:
Independent Management Architecture (IMA) FlexCast Management Architecture (FMA)
Farm Site Worker Group machine catalog, Delivery Group
Worker Worker Group machine catalog, Delivery Group
Remote Desktop Services (RDS) or Terminal Services machine Server OS machine, Server OS VDA
Zone and Data Collector Delivery Controller
Delivery Services Console Citrix Studio and Citrix Director
Publishing applications Delivering applications
Data store Database
Load Evaluator Load Management Policy
Administrator Delegated Administrator, Role, Scope

Feature Descriptions:

FlexCast Management Architecture (FMA)
FMA is a service-oriented architecture that allows interoperability and management
modularity across Citrix technologies. FMA provides a platform for application delivery,
mobility, services, flexible provisioning, and cloud management.
FMA replaces the Independent Management Architecture (IMA) used in XenApp 6.5 and
previous versions.

These are the key elements of FMA in terms of how they relate to elements of XenApp 6.5
and previous versions:

Delivery Sites
Farms were the top-level objects in XenApp 6.5 and previous versions. In XenApp 7.6 and
XenDesktop 7.6, the Delivery Site is the highest level item. Sites offer applications and
desktops to groups of users.
FMA requires that you must be in a domain to deploy a site. For example, to install the
servers, your account must have local administrator privileges and be a domain user in
the Active Directory.

Machine catalogs and Delivery Groups
Machines hosting applications in XenApp 6.5 and previous versions belonged to Worker
Groups for efficient management of the applications and server software. Administrators
could manage all machines in a Worker Group as a single unit for their application
management and load-balancing needs. Folders were used to organize applications and
machines.
In XenApp 7.6 and XenDesktop 7.6, you use a combination of machine catalogs and
Delivery Groups to manage machines, load balancing, and hosted applications or
desktops.

Virtual Delivery Agents
In XenApp 6.5 and previous versions, worker machines in Worker Groups ran applications
for the user and communicated with data collectors. In XenApp 7.6 and XenDesktop 7.6,
the VDA communicates with Delivery Controllers that manage the user connections.

Delivery Controllers
In XenApp 6.5 and previous versions there was a zone master responsible for user
connection requests and communication with hypervisors. In XenApp 7.6 and XenDesktop
7.6, Controllers in the Site distribute and handle connection requests.
XenApp 6.5 and previous versions, zones provided a way to aggregate servers and
replicate data across WAN connections. Although zones have no exact equivalent in
XenApp 7.6 and XenDesktop 7.6, you can provide users with applications that cross WANs
and locations. You can design Delivery Sites for a specific geographical location or data
center and then allow your users access to multiple Delivery Sites. App Orchestration
with XenApp 7.6 and XenDesktop 7.6 provides capabilities for managing multiple Sites in
multiple geographies.

Citrix Studio and Citrix Director
Use the Studio console to configure your environments and provide users with access to
applications and desktops. Studio replaces the Delivery Services Console in XenApp 6.5
and previous versions.
Administrators use Director to monitor the environment, shadow user devices, and
troubleshoot IT issues. To shadow users, Microsoft Remote Assistance must be enabled it
is enabled by default when the VDA is installed.

Delivering applications
XenApp 6.5 and previous versions used the Publish Application wizard to prepare
applications and deliver them to users. In XenApp 7.6 and XenDesktop 7.6, you use Studio
to create and add applications to make them available to users who are included in a
Delivery Group. Using Studio, you first configure a Site, create and specify machine
catalogs, and then create Delivery Groups within those machine catalogs. The Delivery
Groups determine which users have access to the applications you deliver.

Database
XenApp 7.6 and XenDesktop 7.6 do not use the IMA data store for configuration
information. They use a Microsoft SQL Server database to store configuration and session
information.

Load Management Policy
In XenApp 6.5 and previous versions, load evaluators use predefined measurements to
determine the load on a machine. User connections can be matched to the machines
with less load.
In XenApp 7.6 and XenDesktop 7.6, use load management policies for balancing loads
across machines.

Delegated Administrators
In XenApp 6.5 and previous versions, you created custom administrators and assigned them permissions based on folders and objects. In XenApp 7.6 and XenDesktop 7.6, custom administrators are based on role and scope pairs. A role represents a job function and has defined permissions associated with it to allow delegation. A scope represents a collection of objects. Built-in administrator roles have specific permissions sets, such as help desk, applications, hosting, and catalog. For example, help desk administrators can work only with individual users on specified sites, while full administrators can monitor the entire deployment and resolve system wide IT issues.


Refusal of application

An application may be refused on the grounds that the FCA is not satisfied that the principal purpose of the publication or service is neither of those mentioned in article 54(1)(a) or (b) of the Regulated Activities Order (see PERG 7.4.5 G ). An application may also be refused on the grounds that the FCA considers that the vehicle through which advice is to be given is not a newspaper, journal, magazine or other periodical publication, a regularly updated news or information service or a service consisting of the broadcast or transmission of television or radio programmes. Where an application is refused, the FCA will issue a notice which will give a statement of the reasons for the refusal in that case. If the application is refused, the applicant, if he is an unauthorised person, will need to consider whether it is appropriate to continue to publish the periodical or provide the service without authorisation or exemption.


Watch the video: Citrix Administrator Training Tutorials for Beginners. Citrix Admin Online Training (October 2021).