Ask question

Ask question

All categories

3 years ago

12

Assume that the arrival of airplanes at a one-runway airport is a Poisson distribution with a mean rate of 8 planes per hour. Th

e landing time (service time, not service rate which would be planes/min) is an exponential distribution with a mean of 5 minutes per plane. What is the mean number of planes in a holding (planes circling around waiting to land) pattern (compute exact number)? a. 1 b. 1.33 c. 2 d. 1.67

1 answer:

Delvig [45]3 years ago

3 0

Answer:

Option b that is 1.33 is the right choice.

Step-by-step explanation:

Given:

Mean rate of arrival $(\lambda)$ = 8 planes/hr

Service time = $5$ minute/plane

Mean service rate $(\mu)$ = $\frac{60}{5}$ = $12$ planes/hr

Applying the concept Poisson-distributed arrival and service rates (exponential inter-arrival and service times)(M/M/1) process:

We have to find mean number of planes waiting to land that is mean number of customers in the queue .

Mean number of customers in queue $(L_q)$ .

⇒ $L_q=\frac{\rho \lambda}{\mu-\lambda}$

Considering, $\rho=\frac{\lambda}{\mu}$ , $\rho$ is also mean number of customers in service.

⇒ $L_q=\frac{ \lambda^2}{\mu(\mu-\lambda)}$

⇒ Plugging the values.

⇒ $L_q=\frac{8^2}{12(12-8)}$

⇒ $L_q=\frac{64}{48}$

⇒ $L_q=1.33$

So,

Mean number of planes in holding and waiting to land = 1.33

You might be interested in

1. Which of the following equations would NOT contain the point (4,12)?

vodka [1.7K]

Answer:

the answer is b hope this helps :)

6 0

3 years ago

Solve for n, please i would really appreciate it :)

Lelu [443]

Answer:

2

Step-by-step explanation:

${ {a}^{b} }^{c} = {a}^{bc } \\ in \: this \: case \\ { ({8}^{n}) }^{3} = {8}^{3n} \\ \\ {8}^{3n} = {8}^{6 } \\ 3n = 6 \\ n = \frac{6}{3} = 2$

6 0

3 years ago

Ranjita types 1080 words in one hour. What is her gross words a minute rate? (a) 15 (b) 16 (c) 18 (d) none of these

Sauron [17]

Answer:

C 18

Step-by-step explanation:

1080/60 = 18

5 0

4 years ago

Find the value of x in each figure <br><br>

Ronch [10]

Answer:

X is 50

Step-by-step explanation:

3 0

3 years ago

Using Laplace transforms, solve x" + 4x' + 6x = 1- e^t with the following initial conditions: x(0) = x'(0) = 1.

professor190 [17]

Answer:

The solution to the differential equation is

$X(s)=\cfrac 1{6} -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)$

Step-by-step explanation:

Applying Laplace Transform will help us solve differential equations in Algebraic ways to find particular solutions, thus after applying Laplace transform and evaluating at the initial conditions we need to solve and apply Inverse Laplace transform to find the final answer.

Applying Laplace Transform

We can start applying Laplace at the given ODE

$x''(t)+4x'(t)+6x(t)=1-e^t$

So we will get

$s^2 X(s)-sx(0)-x'(0)+4(sX(s)-x(0))+6X(s)=\cfrac 1s -\cfrac1{s-1}$

Applying initial conditions and solving for X(s).

If we apply the initial conditions we get

$s^2 X(s)-s-1+4(sX(s)-1)+6X(s)=\cfrac 1s -\cfrac1{s-1}$

Simplifying

$s^2 X(s)-s-1+4sX(s)-4+6X(s)=\cfrac 1s -\cfrac1{s-1}$

$s^2 X(s)-s-5+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}$

Moving all terms that do not have X(s) to the other side

$s^2 X(s)+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}+s+5$

Factoring X(s) and moving the rest to the other side.

$X(s)(s^2 +4s+6)=\cfrac 1s -\cfrac1{s-1}+s+5$

$X(s)=\cfrac 1{s(s^2 +4s+6)} -\cfrac1{(s-1)(s^2 +4s+6)}+\cfrac {s+5}{s^2 +4s+6}$

Partial fraction decomposition method.

In order to apply Inverse Laplace Transform, we need to separate the fractions into the simplest form, so we can apply partial fraction decomposition to the first 2 fractions. For the first one we have

$\cfrac 1{s(s^2 +4s+6)}=\cfrac As + \cfrac {Bs+C}{s^2+4s+6}$

So if we multiply both sides by the entire denominator we get

$1=A(s^2+4s+6) + (Bs+C)s$

At this point we can find the value of A fast if we plug s = 0, so we get

$1=A(6)+0$

So the value of A is

$A = \cfrac 16$

We can replace that on the previous equation and multiply all terms by 6

$1=\cfrac 16(s^2+4s+6) + (Bs+C)s$

$6=s^2+4s+6 + 6Bs^2+6Cs$

We can simplify a bit

$-s^2-4s= 6Bs^2+6Cs$

And by comparing coefficients we can tell the values of B and C

$-1= 6B\\B=-1/6\\-4=6C\\C=-4/6$

So the separated fraction will be

$\cfrac 1{s(s^2 +4s+6)}=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6}$

We can repeat the process for the second fraction.

$\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac A{s-1} + \cfrac {Bs+C}{s^2+4s+6}$

Multiplying by the entire denominator give us

$1=A(s^2+4s+6) + (Bs+C)(s-1)$

We can plug the value of s = 1 to find A fast.

$1=A(11) + 0$

So we get

$A = \cfrac1{11}$

We can replace that on the previous equation and multiply all terms by 11

$1=\cfrac 1{11}(s^2+4s+6) + (Bs+C)(s-1)$

$11=s^2+4s+6 + 11Bs^2+11Cs-11Bs-11C$

Simplifying

$-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C$

And by comparing coefficients we can tell the values of B and C.

$-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C\\-1=11B\\B=-\cfrac{1}{11}\\5=-11C\\C=-\cfrac{5}{11}$

So the separated fraction will be

$\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac {1/11}{s-1} + \cfrac {-s/11-5/11}{s^2+4s+6}$

So far replacing both expanded fractions on the solution

$X(s)=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6} -\cfrac {1/11}{s-1} -\cfrac {-s/11-5/11}{s^2+4s+6}+\cfrac {s+5}{s^2 +4s+6}$

We can combine the fractions with the same denominator

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {-s/6-4/6+s/11+5/11+s+5}{s^2 +4s+6}$

Simplifying give us

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{s^2 +4s+6}$

Completing the square

One last step before applying the Inverse Laplace transform is to factor the denominators using completing the square procedure for this case, so we will have

$s^2+4s+6 = s^2 +4s+4-4+6$

We are adding half of the middle term but squared, so the first 3 terms become the perfect square, that is

$=(s+2)^2+2$

So we get

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{(s+2)^2 +(\sqrt 2)^2}$

Notice that the denominator has (s+2) inside a square we need to match that on the numerator so we can add and subtract 2

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61(s+2-2)/66+316 /66}{(s+2)^2 +(\sqrt 2)^2}\\X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66+194 /66}{(s+2)^2 +(\sqrt 2)^2}$

Lastly we can split the fraction one more

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66}{(s+2)^2 +(\sqrt 2)^2}+\cfrac {194 /66}{(s+2)^2 +(\sqrt 2)^2}$

Applying Inverse Laplace Transform.

Since all terms are ready we can apply Inverse Laplace transform directly to each term and we will get

$\boxed{X(s)=\cfrac 1{6} -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)}$

6 0

4 years ago