Unlike a byte that is always 8 bits, word size can vary from machine to machine. So, to visualize word-addressable memory, we need to know the word size beforehand. Let's say we have a 16-bit system (word size is 2 bytes) and a memory unit that has 10 address lines. How do we visualize such a memory? First, we determine the size of the memory given the number of address lines. Because there are 10 address bits, we can have 210 unique addresses, hence 1024 addresses. If we are working with something word-addressable, that means each word - not each byte-has a unique address, and we can thus store 1024 words. The memory size is 1 kiloword, or equivalently, 2kilobytes because each word is 2 bytes long: Each row is a word; each word has its own address; each word is 2 bytes long Addr 0 1 2 3 1021 1022 1023 Addr (Bin) 00 0000 0000 00 0000 0001 00 0000 0010 00 0000 0011 11 1111 1101 11 1111 1110 11 1111 1111 Contents 1000110101110000 。。。。。。。0 0 0 0 0 1 1 1 0 1 1 110101000000000 00000000000 0 1 1 1 1 1 1 1 0 0 0 0 1 0 0 1 10000 10 0 1 1 1 0 0 0 10 0 0 1 0 1 0 1 0 1 0 01100 1 1 0 0 1 1 0 0 1 1 0 0 1 0 0 1 0 1 1 0 1 1 011000111000110 Address space from 0 to 2"-1 where n is the number of address bits Can store 1024 words. Because each word is 2 bytes, can store 2048 bytes or 2kB The address of the first word is 0, the second word is 1, the third word is 2, and the last word is 1023. Now, try to fill in the blanks for a word-addressable machine with 4 address lines, but the word size is 64 bits instead: • the memory size is 60 bytes • the address of the first word is 0 • the address of the second word is 1 • the address of last word is 15 words, or 64
Unlike a byte that is always 8 bits, word size can vary from machine to machine. So, to visualize word-addressable memory, we need to know the word size beforehand. Let's say we have a 16-bit system (word size is 2 bytes) and a memory unit that has 10 address lines. How do we visualize such a memory? First, we determine the size of the memory given the number of address lines. Because there are 10 address bits, we can have 210 unique addresses, hence 1024 addresses. If we are working with something word-addressable, that means each word - not each byte-has a unique address, and we can thus store 1024 words. The memory size is 1 kiloword, or equivalently, 2kilobytes because each word is 2 bytes long: Each row is a word; each word has its own address; each word is 2 bytes long Addr 0 1 2 3 1021 1022 1023 Addr (Bin) 00 0000 0000 00 0000 0001 00 0000 0010 00 0000 0011 11 1111 1101 11 1111 1110 11 1111 1111 Contents 1000110101110000 。。。。。。。0 0 0 0 0 1 1 1 0 1 1 110101000000000 00000000000 0 1 1 1 1 1 1 1 0 0 0 0 1 0 0 1 10000 10 0 1 1 1 0 0 0 10 0 0 1 0 1 0 1 0 1 0 01100 1 1 0 0 1 1 0 0 1 1 0 0 1 0 0 1 0 1 1 0 1 1 011000111000110 Address space from 0 to 2"-1 where n is the number of address bits Can store 1024 words. Because each word is 2 bytes, can store 2048 bytes or 2kB The address of the first word is 0, the second word is 1, the third word is 2, and the last word is 1023. Now, try to fill in the blanks for a word-addressable machine with 4 address lines, but the word size is 64 bits instead: • the memory size is 60 bytes • the address of the first word is 0 • the address of the second word is 1 • the address of last word is 15 words, or 64
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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please correct me, i tried inputing 60 words and 64 byte but was wrong the other answer is correct except for this two
answer the following with the highlight red
Transcribed Image Text:Unlike a byte that is always 8 bits, word size can vary from machine to machine. So, to visualize word-addressable memory, we need to know the word size beforehand. Let's say we have a 16-bit system (word size is 2 bytes) and a
memory unit that has 10 address lines. How do we visualize such a memory?
First, we determine the size of the memory given the number of address lines. Because there are 10 address bits, we can have 210 unique addresses, hence 1024 addresses. If we are working with something word-addressable, that
means each word - not each byte - has a unique address, and we can thus store 1024 words. The memory size is 1 kiloword, or equivalently, 2kilobytes because each word is 2 bytes long:
Each row is a word; each word has its own address; each word is 2 bytes long
Addr
0
1
2
3
1021
1022
1023
Addr (Bin)
00 0000 0000
00 0000 0001
00 0000 0010
00 0000 0011
11 1111 1101
11 1111 1110
11 1111 1111
1
0
• the address of last word is 15
0
• the address of the second word is
1
0
1
1
0
0 0
0
0
1 1
1
0
1
0 1
0 1
1 1 1 1 1
1
0
0
0 0
10101
0 0 1 1 0
0 1 1 0 0
Contents
0 101 1 1
000000
0
0
1 0 0
1 0 0 1 1
0 1
0 1 1
O
1
01100110010
0 1
1 0
1
1
1
0
0
0
0
0 0
0 0 1 0 0 1
0
0 0 1 0
O
0000
0
1
0
Address space from 0 to 2"-1 where n is the number of address bits
Can store 1024 words. Because each word is 2 bytes, can store 2048 bytes or 2kB
0 1
1
1
The address of the first word is 0, the second word is 1, the third word is 2, and the last word is 1023.
Now, try to fill in the blanks for a word-addressable machine with 4 address lines, but the word size is 64 bits instead:
• the memory size is 60
words, or 64
bytes
• the address of the first word is 0
0 1 1
1
0 0
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