Answer:
It is not possible.
Explanation:
In this example, we need to accommodate 473 computers for six clients that are 473 IP addresses.
For this request just we have /22 IPv4 address blocks, this mean
22 red bits 11111111111111111111110000000000 <--- 10 host bits
We must increase red bits to 25, we need these 3 bits to create 6 sub red, in this case, 2^3 = 8 sub red.
Why did we ask 3 bits? Because if we ask only 2, 2^2 = 4, and we need 6 sub red.
25 red bits 11111111111111111111111110000000 7 host bits
In this case, we need more than 260 computers, but just we have 7 bits, this means.
2^7 = 128 and just one customer needs 260, for that is impossible.
Answer:
Please kindly check explainations for the code.
Explanation:
lw $t1, Num1
lw $t2, Num2
lw $t3, Num3
blt $t1, $t2, if
beq $t1, $t2, elseif
else:
add $t0, $t3, 5
sw $t0, Result
endif:
#.....other statements after if-elseif-else
if:
sw $t1, Result
b endif
elseif:
ble $t2, $t3, if2
or $t0, $t1, $t3
sw $t0, Result
b endif
if2:
and $t0, $t2, $t3
sw $t0, Result
b endif
Go to attachment for the onscreen code.
Answer:
Check the explanation
Explanation:
223.1.17/24 indicates that out of 32-bits of IP address 24 bits have been assigned as subnet part and 8 bits for host id.
The binary representation of 223.1.17 is 11011111 00000001 00010001 00000000
Given that, subnet 1 has 63 interfaces. To represent 63 interfaces, we need 6 bits (64 = 26)
So its addresses can be from 223.1.17.0/26 to 223.1.17.62/26
Subnet 2 has 95 interfaces. 95 interfaces can be accommodated using 7 bits up to 127 host addresses can represented using 7 bits (127 = 27)
and hence, the addresses may be from 223.1.17.63/25 to 223.1.17.157/25
Subnet 3 has 16 interfaces. 4 bits are needed for 16 interfaces (16 = 24)
So the network addresses may range from 223.1.17.158/28 to 223.1.17.173/28
<span>Encapsulation is defined as the process of adding a header in front of data supplied by a higher layer (and possibly adding a trailer as well).</span>
