Over a TCP connection, suppose host A sends two segments to host B, host B sends an acknowledgement for each segment, the first acknowledgement is lost, but the second acknowledgement arrives before the timer for the first segment expires is True.
True
<u>Explanation:</u>
In network packet loss is considered as connectivity loss. In this scenario host A send two segment to host B and acknowledgement from host B Is awaiting at host A.
Since first acknowledgement is lost it is marked as packet lost. Since in network packet waiting for acknowledgement is keep continues process and waiting or trying to accept acknowledgement for certain period of time, once period limits cross then it is declared as packet loss.
Meanwhile second comes acknowledged is success. For end user assumes second segments comes first before first segment. But any how first segment expires.
Answer and Explanation:
Each subnet can only be power of 2s.
Thus first subnest would have 2^7 = 128 IPs (100 were required)and second subnet would have 2^6 = 64 IPs (42 were required).
Subnet A with 100 Hosts required for 192.168.100.0/25 network:
Subnet mask -> 255:255:255:128
(Mask for 11111111:11111111:11111111:1000000)
Number of usable Host : 2^7 - 1 - 1 = 126
(Since we cannot use IP addresses ending in 0 and 127)
IP Network Address:
192.168.100.0/25
First Valid IP address:
192.168.100.1
Last Valid IP address:
192.168.100.126
Subnet Broadcast IP:
192.168.100.127
Subnet B with 42 Hosts required for 192.168.100.128/26 network:
Subnet mask -> 255:255:255:192
(Mask for 11111111:11111111:11111111:1100000)
Number of usable Host : 2^6 - 1 - 1 = 64-2 = 62
(Since we cannot use IP addresses ending in 0 and 63)
IP Network Address:
192.168.100.128/26
First Valid IP address:
192.168.100.129
Last Valid IP address:
192.168.100.190
Subnet Broadcast IP:
192.168.100.191
Answer:
7 bytes
Explanation:
<u>2 Address Instruction</u>
The 2 address instruction consist 3 components in the format.
One is opcode,other two are addresses of destination and source.
<u>Example-</u>
load b,c | Opcode destination address,source address
add a,d | Opcode destination address,source address
sub c,f | Opcode destination address,source address
Opcode consists of 1 bytes whereas destination address and source address consist of 3 bytes each.
(1+3+3) bytes=7 bytes
A computer chip is a data space stored on your computer. It is there to to keep all of your computer information in it therefor. A computer chip works with your computer to help it work.
Answer:
In Asia
Explanation:
I just guessed and got it right on APEX lol
