28.a) Sixteen-bit messages are transmitted using a Hamming code. How many check bits are needed to ensure that the receiver can detect and correct single-bit errors? (b) Show the bit pattern transmitted for the message 1101001100110101. Assume that even parity is used in the Hamming code.
a) Parity bits are needed at positions 1, 2, 4, 8, and 16 for messages that do not extend beyond 31 bits (including the parity bits). Our message has 16 data bits and 5 parity bits, for a total of 21 bits.
b. We can write the code-word ab1c101d0011001e10101 —> , the bit pattern transmitted is 011110110011001110101.

29. What is the remainder obtained by dividing x 7 + x 5 + 1 by the generator polynomial x 3 + 1? You must show the computation.
The remainder is x 2 + x + 1

30. A bit stream 10011101 is transmitted using the standard CRC method described in the text. The generator polynomial is x 3 + 1. Show the actual bit string transmitted. Suppose that the third bit from the left is inverted during transmission. Show that this error is detected at the receiver’s end.
The frame is 10011101.
The generator is 1001.
The message after appending three zeros is 10011101000.
The remainder on dividing 10011101000 by 1001 is 100
actual bit string transmitted is 10011101100. The received bit stream with an error in the third bit from the left is 10111101100. Dividing this by 1001 produces a remainder of 100, which is different from 0. Thus, the receiver detects the error and can ask for a retransmission.

31. In the Stop-and-Wait protocol with window =1, consider a 50kbps satellite channel with a 500-msec round-trip propagation delay. Assume, we are sending 1000-bit frames via satellite. Say, at t = 0 the sender starts sending the first frame. At t = 20 msec (check how?), the frame has been completely sent. Not untiil, t = 270 msec, th…..
(a) The sender was blocked for 500 msec out of 520 msec, that is, the sender was blocked for 96.15% of the time.
(b) Bandwidth efficiency is therefore, only about (100-96.1)% or, 3.85%.

32. Give at least one reason why PPP uses byte stuffing instead of bit stuffing to prevent accidental flag bytes within the payload from causing confusion.
The PPP was clearly designed to be implemented in software, not in hardware as bit-stuffing protocols. With a software implementation, working entirely with bytes is much simpler than working with individual bits. In addition, PPP was designed to be used with modems, and modems accept and transmit data in units of 1 byte, not 1 bit.

33. Given the PPP format (Figure below), what is the minimum overhead to send an IP packet using PPP? Count only the overhead introduced by PPP itself, not the IP header overhead. What is the maximum overhead?
At its smallest, each frame has 2 flag bytes, 1 protocol byte, and 2 checksum bytes, for a total of 5 overhead bytes per frame. For maximum overhead, 2 flag bytes, 1 byte each for address and control, 2 bytes for protocol and 4 bytes for checksum. This totals to 10 overhead bytes.

34. The cost of a fast microprocessor has dropped to the point where it is now possible to put one in each modem. How does that affect the handling of telephone line errors? Does it negate the need for error checking/correction in layer 2 (link or, Data-link layer)?
bits have been sent over the line without any error-correcting scheme in the physical layer (or, layer 1). The presence of a CPU in each modem makes it possible to include an error-correcting code in layer 1 to greatly reduce the effective error rate seen by layer 2.The error handling by the modems can be done totally transparently to layer 2. Many modems now have built-in error correction.errors at layer 2 are still possible. This can happen, for example, because of loss of data as it is transferred from layer 1 to layer 2 due to lack of buffer space.

35. A 100-km-long cable runs at the T1 data rate. The propagation speed in the cable is 2/3 the speed of light in vacuum. How many bits fit in the cable?
The propagation speed in the cable is 200,000 km/sec, or 200 km/msec, so a 100-km cable will be filled in 500 micro-sec. Each T1 frame is 193 bits sent in 125 micro-sec (or, data-rate of T1 is also well known as 1.544 Mbps). This corresponds to four frames, or 772 bits on the cable.

36. A 12-bit Hamming code whose hexadecimal value is 0xE4F arrives at a receiver. What was the original value in hexadecimal? Assume that not more than 1 bit is in error.
Here, we assume that the number the bits from left to right starting at bit 1, the received message was 1110 0100 1111.
p1 (1) => 0001, encoded position ‘1’ when the parity positing 1 is having a ‘1’ message bit.
p2(1) => 0010,     m3(1)=> 0011.       m6(1)=> 0110,      m9(1)=> 1001,      m10(1)=>1010,         m11(1)=>1011,            m12(1)=>1100.
0010, by doing column wise XOR operation.
we find the syndrome to be 0010. So, the parity bit #2, i.e. p2 should have contained ‘0’ instead of ‘1’.  Changing the p2 (see the change in green color), the correct bit pattern becomes 1010 0100 1111, which in hex is 0xA4F.   , removing the parity bits we get the original 8-bit data value of 0xAF.

37. Suppose that A, B, and C are simultaneously transmitting 0 bits, using a CDMA system with the chip sequences of the given figure. What is the resulting chip sequence? Figure (a) Chip sequences and (b) corresponding signals of stations A, B, C.
The result is obtained by negating each of A, B, and C and then adding the three chip sequences. Alternatively, the three can be added and then negated. The result is (+3 +1 +1 -1 -3 -1 -1 +1).