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

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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can someone explain this subnet design please
H.W(2) :- Complete the Following Table:- PC-PT PC-PT PC3 PCI 192. 168. 1.0/30 192. 168. 2.0/26 2950-24 Switch2 2950-24 Switch! 1841 1841 Router1 Router2 192. 168.3.0/25 PC-PT PC-PT PC4 PC2 Device Interface IP Address Subnet Mask Gateway Router1 Router2 PC1 PC2 РСЗ PC4
(a) What is the network address for this host? (b) What is the range of host addresses in dotted- decimal and binary? (c) What is the broadcast address? (d) How many host addresses? 192.165.1.0 11000000.10100101.00000001.00000000 255.255.255.224 11111111.11111111.11111111.11100000
Given the following information (address pairs)192.171.247.18 178.221.56.1 10.10.11.1192.173.252.19 178.221.24.1 10.10.12.1What subnet mask will place these addresses on different network segments and provide the greatest number of hosts for heach network segments? How many hosts? What's the first and last host address)What subnet mask will place these addresses on the same address and provide the
# Mask Bits Subnet Mask # Subnets 255.255.255.192 4 4 6 2345 # Hosts/ Subnet 62
You need to subnet a class C network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use? A) 255.255.255.192 B) 255.255.255.224 C) 255.255.255.240 D) 255.255.255.248Please explain the answer
Given the IP address: 78.56.2.6/18 (basis is /16) Determine the following addresses for subnet 2: Network address (Sample Answer: 192.168.12.0/25): Blank 1 Broadcast Address (Sample Answer: 192.168.12.255/25): Blank 2 1st and last Host address (Sample Answer: 192.168.12.1/25- 192.168.12.254/25): Blank 3
Short answer27. A company is assigned to an address block 136.23.12.64/26, now it needs to be further divided into four subnets of the same size. Question: (1) How long is the network prefix of each subnet? (2) How many addresses are there in each subnet? (3) What is the address block of each subnet? (4) What is the minimum address that each subnet can be assigned to a host? (5) What is the maximum address that each subnet can be assigned to a host?
[15] IP ADDRESS DECODING: IP: 139.182.148.50 NM: 255.255.254.0 128 64 32 1 1 1 NA: NETWORK ADDRESS 16 8 1 1 OR BA: BROADCAST ADDRESS AND 255.255.254.0 = 11111111.11111111.11111110.00000000 10001010.10110110.10010100.00000000 139. 182. 148. 0 NA: 4 1 BA: 2 1 1 1 139.182.148.50=10001010.10110110.10010100.00110010 139.182.148.50 = 10001010.10110110.10010100.00110010 255.255.254.000000000.00000000.00000001.11111111 PRACTICE: PLEASE CALCULATE [NA: Network Address] AND [BA: Broadcast Address] IP: 172.16.0.100 ● NM: 255.255.252.0 10001010.10110110.10010101.11111111 139. 182. 149 . 255 (First Address of the Network) (Last Address of the Network)
Find & define the IP addressing problem. 192.168.1.33 Default gateway 192.168.162 1900 FO/27 FO/26 192.168.1.66 Default gateway: 192.168.1.95 FOO 192.168.1.62/27 SO/O FO/2 50/0 2950 192.168.1.65 Default gateway 192.168.1.95 192.168.1.97/27 192.168.1.100/27 F0/3 FO/1 FOO 192.168.1.95/27
40. Determine the first useable address, the last useable address, the network address, the subnet mask and the subnet broadcast address for the 125th useable subnet when subnetting the following: a. 10.0.0.0 when borrowing 14 bits b. 172.20.0.0 when borrowing 7 bits
PC4 192.168.B. 120/24 PC3 192.168.B.125/24 PC1 192.168.A.5/24 PC2 192.168 A 10/24 Switch 1 Questions : D1 192.168.A. 1/24 Switch 2 R1 Network Address Destination E1 192.168.B.1/24 E2 10.200.B.5/28 D2 10.200.B.10/28 Value of A = 29, B = 32, C = 33, D = 34 D3 10.10.B.193/28 E3 10.21.B.5/30 F2 10.21.B.6/30 Next hop F3 10.10.B.200/28 R3 1. Determine the Routing Table for R1, R2, and R3 using the following table F1 192.168.C.1/24 Switch Interface PC6 192.168.C.16/24 PC5 192.168.C.18/24 2. With the Routing Table that you have specified in point d, which path will PC2 take when communicating with PC3 and PC5 and vice versa?