I am trying to calculate the checksum of the following packet:
A TCP packet captured with wire shark
But I never managed to get the correct checksum (0x67ea).
I tried to calculate it like follows (of course with using one's complement sum):
source IP + destination IP + TCP protocol + (TCP header length +
payload length) + payload
c0a8 + ae80 + c0a8 + ae01 + 0006 + 0026 + 6c69 + 646f + 720a = 420df
With the one's complement: 4 + 20df = 20e3
not(20e3) = df1c
Which is defiantly not the current checksum.
I also notice that every time I send the same payload, the checksum is changing so I guess it can't be those same unchanging variables and it must be more (for example timestamp..).
What are the exact parameters and the formula that checksum uses?
and how can I calculate it?
Thanks for your help!
You might've overlooked TCP fixed header and TCP options. These bytes also contribute to the checksum (not just TCP payload). Please note that the TCP checksum field (ddff) is replaced with 0 for correct calculation.
IPv4 SRC + IPv4 DST + IPv4 Protocol + TCP Segment Length +
TCP Fixed Header (with checksum field set to 0) +
TCP Options +
TCP Payload
gives you
c0a8 + ae80 + c0a8 + ae01 + 0006 + 0026 +
115c + dcba + 28d5 + 41da + 64e8 + 6a10 + 8018 + 01fe + 0000 + 0000 +
(csum)
0101 + 080a + 5c86 + c6f8 + bd62 e36f +
6c69 + 646f + 720a
which equals 9980c. Next, 9 + 980c = 9815, and, finally, this will give you 67ea.
Related
I have three equations and three unknowns. When I run it in Maple, I get the error "division by zero". What should I do to fix this error?Do you know the reason for this, please?
This is the Maple code
restart;
r := 32*(6*m^2*xi + m^2 + 1)*(alpha*m^2 - 4*xi*sqrt(m^2/xi) + alpha)^2/(m^2*(alpha*m^2 - 16*xi*sqrt(m^2/xi) + alpha)^2*xi);
ns := -((m^2 + 1)*alpha*(alpha^2*(1/6 + xi)*m^6 + ((2*xi + 1/2)*alpha^2 - 3500*xi^2 - (1750*xi)/3)*m^4 + ((xi + 1/2)*alpha^2 - 8*xi^2 - (1762*xi)/3)*m^2 + alpha^2/6 - 4*xi)*sqrt(m^2/xi) - (2*alpha^2*(1/6 + xi)*m^8)/3 + (((2*xi)/3 + 1/3)*alpha^2 - 16*xi^2 - (8*xi)/3)*m^6 + (((10*xi)/3 + 5/3)*alpha^2 + (8*xi)/3)*m^4 + ((2*xi + 17/9)*alpha^2 + (16*xi)/3)*m^2 + (2*alpha^2)/3)/(2*sqrt(m^2/xi)*m^2*(6*m^2*xi + m^2 + 1)*xi*(alpha*m^2 - 4*xi*sqrt(m^2/xi) + alpha));
nt := -((alpha*m^2 - 4*xi*sqrt(m^2/xi) + alpha)*(alpha*(m^2 + 1)*sqrt(m^2/xi) - 4*m^2))/(12*sqrt(m^2/xi)*xi*m^2);
solve([0.9605 <= ns and ns <= 0.9693, 0 < r and r < 0.056, -0.76 < nt and nt < 0.52], {alpha, m, xi});
I tried to use help maple and use the command it wrote. But either I don't know how to write the command or this command is not correct to use.
here is the link to some pictures of my work:command in help maple
error in program
I'm thinking about how to reduce the total data transfer size of WebSocket.
From my understanding, WebSocket header has only 2 bytes at minimum. So I first thought that if the message payload has 8 bytes for example, sending this message is like [payload] + [websocket header] = 8 + 2 = 10 bytes of data transfer in total.
However, if I think about it, TCP header has minimum length of 20 bytes and so does the IP[V4] header. Also variable(?) ethernet header α bytes depending on hardware and protocol is added.
Does this mean that sending 98 bytes of message once (= [payload] + [websocket header] + [ip header] + [tcp header] + [ethernet header] = 98 + 2 + 20 + 20 + α > 140 bytes) is generally better than sending 8 bytes of data five times (= ([payload] + [websocket header] + [ip header] + [tcp header] + [ethernet header]) * 5 = (8 + 2 + 20 + 20 + α) * 5 > 250 bytes) ?
Protocol for BM62 Bluetooth Module
I just have a simple question about how the Checksum algorithm works for a particular Bluetooth module (BM62). The picture above has the UART protocol explained, and it explains the checksum rule, but I am having trouble understanding how it actually works, and cannot seem to guess the checksum value as it is given in the example in the picture.
The idea seems to be that you need to come up with CHKSUM such that LENH + LENL + OPCODE + PARAM + CHKSUM has 0 in the least significant byte. So, let's do the summation in 8 bits (or modulo 256):
LENH + LENL + OPCODE + PARAM + CHKSUM = 0
CHKSUM = -(LENH + LENL + OPCODE + PARAM)
IOW, CHKSUM = -(0 + 2 + 1 + 0) = -3 = 0xFD. (Remember that all of this was done in 8 bits).
You can verify that CHKSUM satisfies the requirement (you're now doing everything in 16 bits):
0 + 2 + 1 + 0 + 0xFD = 0x100
And that has 0 in the least significant byte. If we did this in 8 bits as well, we'd get 0 instead of 0x100 and that would also be a valid way to check correctness.
What is the last IP address that can be assigned to a class c subnet with a prefix of 192.168.56.128/26?
How did you go about getting this answer (so that I can work them out myself in the future?)
Thanks!
26-bit mask means 32 bits (address length) - 26 bits (mask length) = 6 address bits
2^6 = 64 addresses (0..63)
192.168.56.128 + 0 = 192.168.56.128 subnet
192.168.56.128 + 1 = 192.168.56.129 first address
.
.
192.168.56.128 + 62 = 192.168.56.190 last address
192.168.56.128 + 63 = 192.168.56.191 broadcast address
I have the next infix expression:
(i730 + ssg2LQ) + ((+ G0 /(3064 + 68324.06)) / 28)
and I try to convert it to postfix notation, for this I used some on-line converters and all of them gave me the answer
1730 sg2LQ + G0 3064 68324.06 + / + 28 / +
but someone told me that the answer was
i730 sg2LQ + G0 + 3064 68324.06 + / 28 / +
and now I am confused, who is right? who is wrong? how is this expression properly converted?
The second answer
i730 sg2LQ + G0 + 3064 68324.06 + / 28 / +
is definitely wrong because it will first compute (i730 + ssg2LQ) + G0.
Perhaps this answer was obtained by a tool that was confused by unary +.