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2 years ago

Which load balancing method is supported in equal cost multipath (ECMP) load balancing when SD-WAN is enabled

Engineering

1 answer:

Dominik [7]2 years ago

8 0

Volume load balancing method is supported in equal cost multipath (ECMP) load balancing when SD-WAN is enabled.

<h3>What is load balancing method?</h3>

Load balancing means are regarded as a form of an algorithms or method that is used to rightly share an incoming server request or traffic in the midst or among servers that is from the server pool.

Note that Volume load balancing method is supported in equal cost multipath (ECMP) load balancing when SD-WAN is enabled as that is its role.

Learn more about Volume load balancing from

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The nuclear reactions resulting from thermal neutron absorption in boron and cadmium are 10B5 + 1 n0 ï 7Li3 + 4He2 113Cd48 + 1

kirill115 [55]

Solution :

The nuclear reaction for boron is given as :

$^{10}\textrm{B}_5 + ^{1}\textrm{n}_0 \rightarrow ^{7}\textrm{Li}_3 + ^{4}\textrm{He}_2$

And the reaction for Cadmium is :

$^{113}\textrm{Cd}_48 + ^{1}\textrm{n}_0 \rightarrow ^{114}\textrm{Cd}_48 + \gamma [5 \ \textrm{MeV}]$

We know that it is easier that to shield or stop an alpha particle (i.e. He nucli) as they can be stopped or obstructed by only a few centimetres of the material. However, the gamma rays ( γ ) can penetrate through the material to a greater distance. Therefore, we can choose the first one.

6 0

3 years ago

A good visual lead is .... seconds from the front of the vehicle, focusing in the center of the path of travel. Searching 20 to

Tomtit [17]

A good visual lead is 20-30 seconds from the front of the vehicle, focusing in the center of the path of travel. Searching 20 to 30 seconds ahead, gives you time to assess within the next ..12-15 seconds, actions you may need to take to control an approaching risk.

Discussion:

Keeping eye focus centered in a path of travel at an interval of 20 to 30 seconds away from the vehicle is critical to gaining enough info. as possible in the driving scene. Good targeting sets up good sight lines for referencing and good peripheral fields for observing changes.

It is important to look ahead 12-15 seconds into your target area as one drives. Compromise. space by giving as much space to the greater of two hazards.

Read more on driving visual leads:

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3 0

3 years ago

Consider a single bacterial cell as a discrete particle with a diameter of 1x10-6m and a specific gravity of 1.01. Assuming lami

Alinara [238K]

Answer:

Sedimentation is not a good method.

Explanation:

We need to apply Stoke laws and assume that is valid here.

So,

$V_s= 418(G-1)d^2(\frac{3T+70}{100})$

Replacing the values,

$V_s= (418)(1.01-1)(10^-3)^3*(\frac{3*20+70}{100})\\V_s=5.434*10^-6mm/s$

Here then we calculate the time,

$t_{req}=\frac{x}{v}$

Where x= Distance, v= velocity

$t_{req}=\frac{1foot}{5.434*10^{-6}} = \frac{304.8}{5.434*10^{-6}}\\t_{req}=649.2days$

To calculate the surface required we need first to calculate the volume through the volume,

So,

$Q=4MGD=4*10^6*0.135 (ft^3/day)$

Then,

$V_{req} = Q*t\\V_{req}=4*10^6*0.134*649.2\\V_{req}=34.79*10^7ft^3$

Here we can calculate the surface

$S_{req}= \frac{Volume}{Distance}=\frac{34.79*10^7}{1}\\S_{req}=7987.8 Acres$

<em>So, the requeriment of Area of tank and settlement time is huge, it's not a practical method.</em>

6 0

3 years ago

Four eight-ohm speakers are connected in parallel to an audio power amplifier. The amplifier can supply a maximum driver output

Vera_Pavlovna [14]

Answer:

112.5 watts

Explanation:

The output voltage (V) = 15 volts

The equivalent resistance for the speakers connected in parallel ( $R_T$ ) is gotten by using the formula:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\\\\But\ R_1=R_2=R_3=R_4=8\ ohm\\\\\frac{1}{R_T} =\frac{1}{8} +\frac{1}{8} +\frac{1}{8} +\frac{1}{8} \\\\\frac{1}{R_T} =\frac{1+1+1+1}{8} \\\\\frac{1}{R_T} =\frac{4}{8} \\\\R_T=\frac{8}{4} \\\\R_T=2\ ohm$

The current flowing through the amplifier (I) is:

I = V / $R_T$ = 15 / 2 = 7.5 A

The audio power (P) when outputting this maximum voltage is given by:

P = V² / $R_T$ = I² $R_T$ . Therefore:

P = V² / $R_T$ = 15² / $R_T$ = 112.5 watts

6 0

3 years ago

Which of the following is an example of an iterative process?

Jlenok [28]

Answer:

-testing the solution and redesigning the prototype

3 0

4 years ago