223.23.223.10910.10.250.1|126.123.23.1223.69.230.250Custom Subnet maskNumber of needed subnetsNumber of needed usable hostsNetwork AddressAddress classDefault subnet maskCustom subnet maskTotal number of subnetsTotal number of host addressesNumber of usable addressesNumber of bits borrowed601,000128.77.0.0Number of needed usable hosts Network AddressAddress classTotal number of subnetsTotal number of host addressesNumber of usable addressesNumber of bits borrowedWhat is the 2nd subnet range?What is the subnet number for the 5th subnet?What is the subnet broadcast address for the 7th subnet?What are the assignable addresses for the 10th6.126.0.0.0 129 22 230.19255.255.255.0Section BPart 1 (Please show working)Binary To Decimal Conversion12800011011| 101010106432| 16| 841 2|1Answers011011111111100001010101Decimal To Binary Conversion1235013820010AddressClass192.14.2.0148.17.9.1193.42.1.1126.8.156.0Circle the network portion of these addressesCircle the host portion of these addresses:209.240.80.78198.125.87.177199.155.77.56223.250.200.222117,89,56.4517.45.222.45126.201.54.231191.41.35.112215.45.45.0192.200.15.0209.240.80.78198.125.87.177Using the IP address and subnet mask shown write out the network address192.149.24.191255.255.255.0150.203.23.19255.255.0.010.10.10.10255.0.0.0186.13.23.110255.255.255.0223.69.230.250255.255.0.0Using the IP address and subnet mask shown write out the host address:222.49.49.11128.23.230.19| 255.255.0.030 10.1010.10.10.10255.0.0.0200.113.123.11255.255.255.0223.169.23.20255.255.0.0Write the correct default subnet mask for each of the following addresses:191.249.234.191

Name: 223.23.223.10910.10.250.1|126.123.23.1223.69.230.250Custom Subnet mas
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Description: 223.23.223.10910.10.250.1|126.123.23.1223.69.230.250Custom Subnet maskNumber of needed subnetsNumber of needed usable hostsNetwork AddressAddress classDefault subnet maskCustom subnet maskTotal number of subnetsTotal number of host addressesNumber o

128.77.0.0 Number of usable subnet = 60 Number of subnet bits = ceil(log 60) = 6 bits. Address…

Engineering

Computer Science

223.23.223.109 10.10.250.1 126.123.23.1 223.69.230.250 Custom Subnet mask Number of needed subnets Number of needed usable hosts Network Address Address class Default subnet mask Custom subnet mask Total number of subnets Total number of host addresses Number of usable addresses Number of bits borrowed 60 1,000 128.77.0.0 Number of needed usable hosts Network Address Address class Total number of subnets Total number of host addresses Number of usable addresses Number of bits borrowed What is the 2nd subnet range? What is the subnet number for the 5th subnet? What is the subnet broadcast address for the 7th subnet? What are the assignable addresses for the 10th 126.0.0.0

223.23.223.109 10.10.250.1 126.123.23.1 223.69.230.250 Custom Subnet mask Number of needed subnets Number of needed usable hosts Network Address Address class Default subnet mask Custom subnet mask Total number of subnets Total number of host addresses Number of usable addresses Number of bits borrowed 60 1,000 128.77.0.0 Number of needed usable hosts Network Address Address class Total number of subnets Total number of host addresses Number of usable addresses Number of bits borrowed What is the 2nd subnet range? What is the subnet number for the 5th subnet? What is the subnet broadcast address for the 7th subnet? What are the assignable addresses for the 10th 126.0.0.0

C++ for Engineers and Scientists

4th Edition

ISBN:9781133187844

Author:Bronson, Gary J.

Publisher:Bronson, Gary J.

Chapter2: Problem Solving Using C++using

Section2.3: Data Types

Problem 9E: (Practice) Although the total number of bytes varies from computer to computer, memory sizes of...

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9.18 A certain computer provides its users with a virtual-memory space of 232 bytes. The computer has 218 bytes of physical memory. The virtual memory is implemented by paging, and the page size is 4096 bytes. How many pages are possible in the system? How many frames can it have?
CA_10 Let the virtual address be V bits and the virtual addtess space be byte-addressable, the page size be P KB (and P is a power of 2), and the the main memory size be MM MB(where [MM MB]) is divide into [P KB]). (a) How large is the user's memory spac (b) How large is the main memory page frames? (c) How large is the page table (in number of table entries)? (d)How many of the virtual memory bits need to be translated? (e) How many bits will be produced if the virtual-to-pyysical address translation is "successful" (f) How many bits does a physical address have, and how are each of these bits obtained?
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A set of eight data bytes is stored in memory locations starting from XX70H. Write a program to add two bytes at a time and store the sum in the same memory locations, low order sum replacing the first byte and a carry replacing the second byte. If any pair does not generate a carry, the memory location of the second byte should be cleared. Data(H) F9, 38, A7, 56,98,52, 8F, F2
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A.) What are the contents in hex of the memory location at the following address in binary: 0000 1100 0000 1111? (Enter hex like the following example: 0x2A3F) In all computer models, the contents of a memory location can be interpreted as data. B.) Interpret the contents at address 0xFFFF as a two's complement integer in base 10. C.) Interpret the contents at address 0x0C0C as two ASCII characters. D.) Interpret the contents at the same address as C.) above as an unsigned integer in base 10.
CA_6 We study the properties of cache memory, and for reasons of easier design and efficient circuits, we assume that the cache capacity is 2i Bytes, and cache line size is 2j Bytes, with i and j being natural numbers: (a) How many bits should the tag field have? And can the tag field contain 0 bit (i.e., be empty)? Elaborate (b) Repeat the above for the index field. (c) Repeat the above for the byte-offset field. (d) Finally, depict a figure showing a cache line, indicate what fields it possibly has, state the possible sizes of these fields, and explain the uses of these fields.
Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. Could you hand draw the page table, if possible a) Suppose that the OS uses a two-level page table. Draw the page table. (Assume that frames 7 through 221 are free, so you can allocate space for the page table there.) In addition, suppose that the page-table directory storage comprises a whole number of consecutive full frames. (For examples: if the directory entry is 2 bytes, the entry’s storage comprises 1 frame; if the directory entry is 260 bytes, the entry’s storage comprises 2 consecutive frames.) b) What is the size of the two-level page table
(Practice) Although the total number of bytes varies from computer to computer, memory sizes of millions and billions of bytes are common. In computer language, the letter M representsthe number 1,048,576, which is 2 raised to the 20th power, and G represents 1,073,741,824, which is 2 raised to the 30th power. Therefore, a memory size of 4 MB is really 4 times 1,048,576 (4,194,304 bytes), and a memory size of 2 GB is really 2 times 1,073,741,824 (2,147,483,648 bytes). Using this information, calculate the actual number of bytes in the following: a. A memory containing 512 MB b. A memory consisting of 512 MB words, where each word consists of 2 bytes c. A memory consisting of 512 MB words, where each word consists of 4 bytes d. A thumb drive that specifies 2 GB e. A disk that specifies 4 GB f. A disk that specifies 8 GB
Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. a) Suppose that the OS uses a two-level page table. Draw the page table. (Assume that frames 7 through 221 are free, so you can allocate space for the page table there.) In addition, suppose that the page-table directory storage comprises a whole number of consecutive full frames. (For examples: if the directory entry is 2 bytes, the entry’s storage comprises 1 frame; if the directory entry is 260 bytes, the entry’s storage comprises 2 consecutive frames.) b) What is the size of the two-level page table c) Now, translate the virtual addresses shown in question I(e) to physical addresses for the two-level page table. Show how you obtain your answers. 0x0389 0xDF78 0x0245 0x8012
) Consider the following sequence of virtual memory references (in decimal) generatedby a single program in a pure paging system:100, 110, 1400, 1700, 703, 3090, 1850, 2405, 2460, 4304, 4580, 3640a. Derive the corresponding reference string of pages (i.e. the pages the virtual addressesare located on), assuming a page size of 1024 bytes. (Assume that page numberingstarts at 0)b. For the page sequence derived above, determine the number of page faults for each ofthe following page replacement strategies, assuming that two (2) page frames areavailable to the program.i. LRUii. FIFOiii. OPT (Optimal)
Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. 8 bits are used for offset, 8 bits are used for page # and the max number of pages a process can have is 256. e. Translate the following virtual addresses to physical addresses, and show how you obtain the answers. (Hint: You do not need to convert hexadecimal numbers to decimal ones.) 0x0389 0xDF78 0x0245 0x8012 f) Now, suppose that the OS uses a two-level page table. Draw the page table. (Assume that frames 7 through 221 are free, so you can allocate space for the page table there.) In addition, suppose that the page-table directory storage comprises a whole number of consecutive full frames. (For examples: if the directory entry is 2 bytes, the entry’s storage comprises 1 frame; if the directory entry is 260 bytes, the entry’s storage comprises 2 consecutive frames.) g)What is the size of the two-level page table…