Systems Architecture
7th Edition
ISBN: 9781305080195
Author: Stephen D. Burd
Publisher: Cengage Learning
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Chapter 3, Problem 3RQ
Explanation of Solution
Excess notation:
- Signed integer can be represented with a fixed number of bits using the format known as excess notation.
- The range of ordinary binary numbers are divided by this format and it uses lower half for negative values and upper half for positive values...
Explanation of Solution
Twos complement notation:
- In base two numbering system, the complement of “1” is “0” and the complement of “0” is ‘1”.
- To find complement of a bit string, “1” is substituted for all values of ȁ...
Explanation of Solution
The purpose of twos complement notation:
Digital electronic circuitry uses twos complement notation for the following reasons;- The sign is represented using the leftmost bit.
- It uses fixed number of bits.
- It requires only two logic circuits to perform single bit addition...
Explanation of Solution
Representation of integer value:
- Positive integers are represented as ordinary binary values but twos complement of negative integers are represented by adding one to the complement of nonnegative integers.
- Example:
- The twos complement notation for the nonnegative integer “
- Twos complement notation for the negative integer “
- The twos complement notation for the nonnegative integer “
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Now unmount the filesystem.
Create a sparse image file, called book.img, of the 2GiB hard drive.
Use a hexdump program to look at book.img and work out the offsets that store the contents of the
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00000000: 0000 0000 0000 0000 0000 0000 0000 0000 .................00000010: 5468 6520 7261
696e 2069 6e20 5370 6169 The rain in Spai00000020: 0000 0000 0000 0000 0000 0000 0000
0000.................00000030: 6e20 6661 6c6c 7320 6d61 696e 6c79 206f n falls mainly 000000040:
6e20 7468 6520 706c 6169 6e2e 0000 0000 n the plain........
Compress the filesystem using gzip by running:
#gzip book.img
This should leave you with a file book.img.gz. You will use scp to transfer this file in a later task.
Enter your ranges in answers2.json as the answer for question1. To…
Task 4: Examine floppy.img
Mount floppy.img at a suitable mountpoint inside your VM.
Look inside the mountpoint directory. You should see a number of files in the top-level directory.
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Chapter 3 Solutions
Systems Architecture
Ch. 3 - Prob. 1VECh. 3 - Prob. 2VECh. 3 - A(n) __________ is an integer stored in double the...Ch. 3 - Prob. 4VECh. 3 - Assembly (machine) language programs for most...Ch. 3 - Prob. 6VECh. 3 - Prob. 7VECh. 3 - Prob. 8VECh. 3 - Prob. 9VECh. 3 - A(n) __________ is an array of characters.
Ch. 3 - Most Intel CPUs use the __________, in which each...Ch. 3 - Prob. 12VECh. 3 - A(n) __________ contains 8 __________.Ch. 3 - Prob. 14VECh. 3 - The result of adding, subtracting, or multiplying...Ch. 3 - Prob. 16VECh. 3 - Prob. 17VECh. 3 - Prob. 18VECh. 3 - Prob. 19VECh. 3 - Prob. 20VECh. 3 - Prob. 21VECh. 3 - Prob. 22VECh. 3 - ___________ occurs when the result of an...Ch. 3 - In a CPU, _______ arithmetic generally is easier...Ch. 3 - In the ________, memory addresses consist of a...Ch. 3 - Prob. 26VECh. 3 - Data represented in ________ is transmitted...Ch. 3 - Prob. 28VECh. 3 - Prob. 29VECh. 3 - A(n) ____________ is one instance or variable of a...Ch. 3 - Prob. 1RQCh. 3 - Why is binary data representation and signaling...Ch. 3 - Prob. 3RQCh. 3 - Prob. 4RQCh. 3 - Prob. 5RQCh. 3 - Prob. 6RQCh. 3 - Prob. 7RQCh. 3 - Why doesnt a CPU evaluate the expression 'A' = 'a'...Ch. 3 - Prob. 9RQCh. 3 - What primitive data types can normally be...Ch. 3 - Prob. 11RQCh. 3 - How is an array stored in main memory? How is a...Ch. 3 - Prob. 14RQCh. 3 - Prob. 1PECh. 3 - Prob. 2PECh. 3 - Prob. 4PECh. 3 - Prob. 5PECh. 3 - Prob. 6PECh. 3 - Prob. 1RPCh. 3 - Prob. 2RPCh. 3 - Prob. 3RP
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- ▼ description: answer: question2: ▼ description: answer: "What are the offset ranges for the book file content within book.img?" "YOUR ANSWER GOES HERE" "What are the offset ranges for the content in unused space within floppy.img?" "YOUR ANSWER GOES HERE"arrow_forwardTask 3: Creating a Sample Hard Drive Image book.img Add a new (virtual) hard drive of capacity 2GiB to your Kali Linux VM. Create a FAT filesystem on the hard drive using mkfs. fat. Be careful to use your new 2GiB drive! It would be a good idea to use 1sblk to verify the block device name before proceeding. Mount the new hard drive at a mount point of your choice. Find a book from Project Gutenberg by an author whose last name begins with the same letter as your last name (look under "Browsing Options" and then "Authors" A-Z). Download the plain text UTF-8 and ensure it is saved into your mounted drive (it should be saved within the mount point directory). For example, using The Royal Guide to Wax Flower Modelling by Emma Peachey (a cracking read, no doubt) we can download the plain text file using:arrow_forward- Is FD A a partial or transitive dependency?- Is FD B a partial or transitive dependency?- Is FD C a partial or transitive dependency?- In what normal form is the relation Parking Tables?- Convert the relation Parking Tables to a set of 3NF relations.arrow_forward
- What happens to MAC addresses as frames travel from one node (device) to another? What happens to MAC addresses as frames travel from one network to another? What happens to IP addresses as packets travel from one node (device) to another? What happens to IP addresses as packets travel from one network to another?arrow_forwardCan you check if my explation is correct: Challenge: Assume that the assigned network addresses are correct. Can you deduce (guess) what the network subnet masks are? Explain while providing subnet mask bits for each subnet mask. [Hint: Look at the addresses in binary and consider the host ids] To assume that the network addresses are correct, we take a look at the network ID and IP addresses. Take 20.0.0.0/8, we have an /8 prefix which is Class A, that provide us with 255.0.0.0. This means that there are 24 host bits while the network bits are 8. When compared to the IP address 20.255.0.7, we can perform an additional operation with the subnet mask giving us 20.0.0.0. The same happened with 20.0.0.89, which we also got 20.0.0.0. Therefore, these two IP addresses fall within the respective /8 subnet. For 200.100.10.0/30, the /30 is Class C with a subnet mask of 255.255.255.252, meaning 30 network bits and only two host bits. When applying the subnet mask and the IP addresses…arrow_forwardIt is possible to sort an array of n values using pipeline of n filter processes.The first process inputs all the values one at a time, keep the minimum, and passes the others on to the next process. Each filter does the same thing; it receives a stream of values from the previous process, keep the smallest, and passes the others to the next process. Assume each process has local storage for only two values--- the next input value and the minimum it has seen so far. (a) Developcode for filter processes. Declare the channels and use asynchronous message passing. Hint:Define an array of channels value[n] (int), and a set of filter processes Filter[i = 0 ton-1]. Each process Filter[i] (where 0 <= i <= n-2) receives a stream of integers through channelvalue[i], keeps the smallest, and sends all other integers to channel value[i+1]. The last processFilter[n-1] receives only one integer through channel value[n-1] and does not need to send anyinteger further.arrow_forward
- It is possible to sort an array of n values using pipeline of n filter processes.The first process inputs all the values one at a time, keep the minimum, and passes the others on to the next process. Each filter does the same thing; it receives a stream of values from the previous process, keep the smallest, and passes the others to the next process. Assume each process has local storage for only two values--- the next input value and the minimum it has seen so far. (a) Developcode for filter processes. Declare the channels and use asynchronous message passing. Hint:Define an array of channels value[n] (int), and a set of filter processes Filter[i = 0 ton-1]. Each process Filter[i] (where 0 <= i <= n-2) receives a stream of integers through channelvalue[i], keeps the smallest, and sends all other integers to channel value[i+1]. The last processFilter[n-1] receives only one integer through channel value[n-1] and does not need to send anyinteger further.arrow_forwardIt is possible to sort an array of n values using pipeline of n filter processes.The first process inputs all the values one at a time, keep the minimum, and passes the others on to the next process. Each filter does the same thing; it receives a stream of values from the previous process, keep the smallest, and passes the others to the next process. Assume each process has local storage for only two values--- the next input value and the minimum it has seen so far. (a) Developcode for filter processes. Declare the channels and use asynchronous message passing. Hint:Define an array of channels value[n] (int), and a set of filter processes Filter[i = 0 ton-1]. Each process Filter[i] (where 0 <= i <= n-2) receives a stream of integers through channelvalue[i], keeps the smallest, and sends all other integers to channel value[i+1]. The last processFilter[n-1] receives only one integer through channel value[n-1] and does not need to send anyinteger further.arrow_forwardIt is possible to sort an array of n values using pipeline of n filter processes.The first process inputs all the values one at a time, keep the minimum, and passes the others on to the next process. Each filter does the same thing; it receives a stream of values from the previous process, keep the smallest, and passes the others to the next process. Assume each process has local storage for only two values--- the next input value and the minimum it has seen so far. (a) Developcode for filter processes. Declare the channels and use asynchronous message passing. Hint:Define an array of channels value[n] (int), and a set of filter processes Filter[i = 0 ton-1]. Each process Filter[i] (where 0 <= i <= n-2) receives a stream of integers through channelvalue[i], keeps the smallest, and sends all other integers to channel value[i+1]. The last processFilter[n-1] receives only one integer through channel value[n-1] and does not need to send anyinteger further.arrow_forward
- It is possible to sort an array of n values using pipeline of n filter processes.The first process inputs all the values one at a time, keep the minimum, and passes the others on to the next process. Each filter does the same thing; it receives a stream of values from the previous process, keep the smallest, and passes the others to the next process. Assume each process has local storage for only two values--- the next input value and the minimum it has seen so far. (a) Developcode for filter processes. Declare the channels and use asynchronous message passing. Hint:Define an array of channels value[n] (int), and a set of filter processes Filter[i = 0 ton-1]. Each process Filter[i] (where 0 <= i <= n-2) receives a stream of integers through channelvalue[i], keeps the smallest, and sends all other integers to channel value[i+1]. The last processFilter[n-1] receives only one integer through channel value[n-1] and does not need to send anyinteger further.arrow_forwardI need help: Challenge: Assume that the assigned network addresses are correct. Can you deduce (guess) what the network subnet masks are? Explain while providing subnet mask bits for each subnet mask. [Hint: Look at the addresses in binary and consider the host ids]arrow_forwardI would like to know if my answer statment is correct? My answer: The main difference is how routes are created and maintained across different networks. Static routing establishes router connections to different networks from the far left and far right. The Dynamic routing focus emphasizes immediate connection within the router while ignoring the other connections from different networks. Furthermore, the static routing uses the subnet mask to define networks such as 25.0.0.0/8, 129.60.0.0/16, and 200.100.10.0/30, which correspond to 255.0.0.0, 255.255.0.0, and 255.255.255.252. On the other hand, dynamic routing uses the wildcard mask to inverse the subnet mask, where network bits become 0 and host bits become 1, giving us 0.0.0.255, 0.0.255.255, and 0.0.0.3. Most importantly, the CLI commands used for Static and Dynamic routing are also different. For static routing, the “IP route” corresponds with the network, subnet mask, and next-hop IP address. In contrast, dynamic routing uses…arrow_forward
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