Answer:
It might not work efficiently.
Explanation:
The function of the protocol rtd3.0 is to transfer data to a receiver from a sender.
As soon as the receiver received the packet transferred by the sender, we respond acknowledge (Ack) to the sender so that sender can confirm the receiver has gotten it.
The receiver will not send any acknowledgment if the packet he receives are such that bits contain error or not in order.
After that timeout, the packet will be re-transmitted by the sender.
It is then possible the protocol might seem to be inefficient if a packet is sent many times, because other packets will have to wait to sent until the current packet is sent successfully.
The solution to prevent this kind of issue is to allow for the occurrence of premature timeouts.
Answer:
a database stores a large sum of data
Explanation:
its used to keep track of things like student names bank accounts and other things
Answer: ....
If one load balancer fails, the secondary picks up the failure and becomes active. They have a heartbeat link between them that monitors status. If all load balancers fail (or are accidentally misconfigured), servers down-stream are knocked offline until the problem is resolved, or you manually route around them.
Explanation:
Load balancing is a technique of distributing your requests over a network when your server is maxing out the CPU or disk or database IO rate. The objective of load balancing is optimizing resource use and minimizing response time, thereby avoiding overburden of any one of the resources.
The goal of failover is the ability to continue the work of a particular network component or the whole server, by another, should the first one fail. Failover allows you to perform maintenance of individual servers or nodes, without any interruption of your services.
It is important to note that load balancing and failover systems may not be the same, but they go hand in hand in helping you achieve high availability.
Answer:
Logic for a program
Explanation:
//Here ind = index
//declare the number
number ind
number sum
number avg
number SIZE = 20
number num[SIZE] = {0,0,0,0,0,0,0,0,0,0
,0,0,0,0,0,0,0,0,0,0}
getReady()
while ind < SIZE
getNumbers()
stop
getReady()
ind = 0
sum = 0
return
getNumbers()
cout<< “Enter a number for position ”, ind
input numbers[ind]
sum = sum + numbers[ind]
ind = ind + 1
return
;
finishUp()
avg = sum/SIZE
ind = 0
while ind < SIZE
output numbers[ind], avg – numbers[index]
ind = ind + 1
return
Modify the program in 2a
number index
number sum
number avg
number actualSize
number SIZE = 10
number number[SIZE] = 0
