Answer:
a) Speedup gain is 1.428 times.
b) Speedup gain is 1.81 times.
Explanation:
in order to calculate the speedup again of an application that has a 60 percent parallel component using Anklahls Law is speedup which state that:
Where S is the portion of the application that must be performed serially, and N is the number of processing cores.
(a) For N = 2 processing cores, and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.428 times.
(b) For N = 4 processing cores and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.81 times.
B.) is what most employers are looking for
Answer:
To ensure that the trainees are well acquainted with how to differentiate 568A cables and 568B cables, you must ensure they know appropriate color scheme pattern for the both cables. Also, they are expected to know the correct terminology when using both cables, as 568A is known as a crossover cable and 568B is known as a straight-through cable.
Explanation:
Answer:
if and and not and or ans all
Answer: Site Link
Explanation: Site link is the type connection that is created for the different sites.It helps in the connectivity in the multiple site surrounding known as inter-site and transferring the traffic created by the replication activity.
Other options are incorrect because bridgehead servers is a device to control the domain,subnet is the part of the IP network and domain is used for the identification of address .Thus, the correct option is site link.
