Answer:
Memory-mapped I/O and port-mapped I/O are two complementary methods of performing input/output between the central processing unit and peripheral devices in a computer. An alternative approach is using dedicated I/O processors, commonly known as channels on mainframe computers, which execute their own instructions.
<span>Data consolidation is the gathering and/or integration of data coming from many and diverse sources and going into a single destination. Meanwhile this process is taking place, different data sources are consolidated, into a single data store.
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Answer:d) Mono-alphabetic substitution rules
Explanation: Poly-alphabetic cipher is the cryptography that is done by using the many alphabets in substitutional way. They are supposed to have a good security because they use cipher text alphabets for the encryption and then decrypts it.
Substitution of the alphabets is done by the rule of the mono-alphabetic substitution rules.Other given options are incorrect because they are not used in the cryptography of the alphabetic cipher. Thus option (d) is the correct answer.
Network layer protocols specify the <u>packet structure</u> and processing used to carry data between hosts.
<h3>What is the Network layer protocols about?</h3>
The header, payload, as well as the trailer of a network packet are its three component pieces. The underlying network architecture or protocol employed determines the format and size of a network packet. A network packet resembles a package in concept.
Therefore, Network layer protocols define the packet format and processing needed to transfer data from one host to another host, in contrast to the Transmission layer (OSI Layer 4), which controls the data transport between the programs operating on each end computer.
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Answer:
Check the explanation
Explanation:
1.
System = 256Byte = 8 bit
Cache = 64B , block size = 16 byte.
A) Direct Mapped Cache:
Block offset = log (Block size) = log 16 = 4bit
Total # of block inside cache = 4.
Therefore index offset = log 4 = 2bit.
Remaining is tag bits.
Therefore tag bits = 8-(4+2) = 8-6 = 2 bits
Tag Index offset Block offset
(2 bits) (2 bits) (4 bits)
Fully associative cache :
In fully associative cache, any of main memory block can be placed anywhere in cache. Therefore index offset =0 bits.
Therefore tag bits = 8-block offset bit= 8-4 =4bits
Tag Block offset
(4 bits) (4 bits)
