Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The following problem arises in molecular biology. The surface of a bacterium consists of several sites at which foreign molecules-some acceptable and some not-become attached. We consider a particular site and assume that molecules arrive at the site according to a Poisson process with parameter \(\lambda\). Among these molecules a proportion \(\alpha\) is acceptable. Unacceptable molecules stay at the site for a length of time that is exponentially distributed with parameter \(\mu_{1}\), whereas an acceptable molecule remains at the site for an exponential time with rate \(\mu_{2}\). An arriving molecule will become attached only if the site is free of other molecules. What percentage of time is the site occupied with an acceptable (unacceptable) molecule?

Short Answer

Expert verified
The percentage of time the site is occupied by an acceptable molecule is given by: Percentage of time occupied by acceptable molecule = \(100 \times \frac{λα\times \frac{1}{μ_2}}{λ \times ( \frac{α}{μ_2} + \frac{1-α}{μ_1})}\) And the percentage of time the site is occupied by an unacceptable molecule is given by: Percentage of time occupied by unacceptable molecule = \(100 \times \frac{λ(1-α)\times \frac{1}{μ_1}}{λ \times ( \frac{α}{μ_2} + \frac{1-α}{μ_1})}\) Use these formulas with given values of λ, α, μ₁, and μ₂ to calculate the percentages.

Step by step solution

01

Identifying the given parameters

We are given the following parameters: 1. Molecules arrive according to a Poisson process with parameter λ. 2. A proportion α of the arriving molecules is acceptable. 3. Unacceptable molecules stay for an exponentially distributed time with parameter μ₁. 4. Acceptable molecules stay for an exponentially distributed time with parameter μ₂.
02

Calculate probabilities related to arrivals

Given the proportion α of acceptable molecules, we can say that the probability of a molecule being acceptable upon arrival is P(A) = α. Therefore, the probability of a molecule being unacceptable is P(U) = 1 - α.
03

Calculate probabilities related to departure

Since the length of time an unacceptable molecule stays attached is exponentially distributed with parameter μ₁, we can say that the departure rate of an unacceptable molecule is λ_U = λ(1 - α)μ₁. Similarly, since the length of time an acceptable molecule stays attached is exponentially distributed with parameter μ₂, we can say that the departure rate of an acceptable molecule is λ_A = λαμ₂.
04

Calculate probabilities of site occupation

Using the provided information, we can now calculate the probabilities of the site being occupied by an acceptable molecule and an unacceptable molecule. In steady state, we can write: P(Occupied by acceptable molecule) = \(\frac{\text{Arrival rate of acceptable molecules} \times \text{average time occupied by acceptable molecule}}{\text{Total Arrival rate} \times (\text{average time occupied by acceptable molecule} + \text{average time occupied by unacceptable molecule})}\) P(Occupied by acceptable molecule) = \(\frac{λα\times \frac{1}{μ_2}}{λ \times ( \frac{α}{μ_2} + \frac{1-α}{μ_1})}\) Similarly, P(Occupied by unacceptable molecule) = \(\frac{λ(1-α)\times \frac{1}{μ_1}}{λ \times ( \frac{α}{μ_2} + \frac{1-α}{μ_1})}\)
05

Calculate the percentages of time the site is occupied

Now that we have probabilities for site occupation by an acceptable and unacceptable molecule, we can calculate the percentage of time the site is occupied by each type: Percentage of time occupied by acceptable molecule = \(100 \times P(\text{Occupied by acceptable molecule})\) Percentage of time occupied by unacceptable molecule = \(100 \times P(\text{Occupied by unacceptable molecule})\) Using values of λ, α, μ₁, and μ₂, these percentages can be calculated to determine the percentage of time the site is occupied by an acceptable and unacceptable molecule.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider two machines that are maintained by a single repairman. Machine \(i\) functions for an exponential time with rate \(\mu_{i}\) before breaking down, \(i=1,2 .\) The repair times (for either machine) are exponential with rate \(\mu .\) Can we analyze this as a birth and death process? If so, what are the parameters? If not, how can we analyze it?

Consider two machines, both of which have an exponential lifetime with mean \(1 / \lambda .\) There is a single repairman that can service machines at an exponential rate \(\mu .\) Set up the Kolmogorov backward equations; you need not solve them.

Let \(O(t)\) be the occupation time for state 0 in the two-state continuous-time Markov chain. Find \(E[O(t) \mid X(0)=1]\).

Consider a system of \(n\) components such that the working times of component \(i, i=1, \ldots, n\), are exponentially distributed with rate \(\lambda_{i} .\) When a component fails, however, the repair rate of component \(i\) depends on how many other components are down. Specifically, suppose that the instantaneous repair rate of component \(i, i=1, \ldots, n\), when there are a total of \(k\) failed components, is \(\alpha^{k} \mu_{i}\) (a) Explain how we can analyze the preceding as a continuous-time Markov chain. Define the states and give the parameters of the chain. (b) Show that, in steady state, the chain is time reversible and compute the limiting probabilities.

Consider a graph with nodes \(1,2, \ldots, n\) and the \(\left(\begin{array}{l}n \\\ 2\end{array}\right) \operatorname{arcs}(t, j), i \neq j, i, j,=1, \ldots, n\) (See Section 3.6.2 for appropriate definitions.) Suppose that a particle moves along this graph as follows: Events occur along the arcs \((i, j)\) according to independent Poisson processes with rates \(\lambda_{i j} .\) An event along arc \((i, j)\) causes that arc to become excited. If the particle is at node \(i\) at the moment that \((i, j)\) becomes excited, it instantaneously moves to node \(j, i, j=1, \ldots, n .\) Let \(P_{j}\) denote the proportion of time that the particle is at node \(j .\) Show that $$ P_{j}=\frac{1}{n} $$ Hint: Use time reversibility.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free