2.3.2 Bit stuffing to I-frame

At an insert of bits it is necessary to consider, that bits in fields go continuously. There are no five units in succession in the given sequence, that’s why bit stuffing isn’t realized.

2.4 Hdlc procedure support frames construction

Objective: To learn support frames construction algorithm and to acquire skills of frames construction.

Task to the practical seminar:

1. To learn items 1.5.1, 1.5.3, 1.5.4 of this teaching manual.

2. To composite two S-frames for acknowledgement of correct frame with the number of received frame as penultimate figure of number of the credit book; and for request of error frame with the number of received frame as the last figure of number of the credit book. Use address field the same as in item 2.3.1.

3. To composite four U-frames for data transmitting: set asynchronous balance mode (basic or extended), unnumbered acknowledgment, disconnect, disconnect mode.

4. To make bit stuffing in S- and U-frames.

Initial data:

Parameter	Penultimate figure of number of the credit book
	1	2	3	4	5	6	7	8	9	0
Request command	RR	REJ	SREJ	RR	REJ	SREJ	RR	REJ	SREJ	REJ
Parameter	Last figure of number of the credit book
	1	2	3	4	5	6	7	8	9	0
Format of the frames	Extended	Basic	Extended	Basic	Extended	Basic	Extended	Basic	Extended	Basic

2.4.1 S-frames construction

The difference of support frames is in control field. Consider S-frame construction in basic format with the next initial data:

• S-frame function RR;

• Station address 102;

• The number of received frame 9.

Address field is 11001101 (see item 2.3.1).

Consider control field. N_R = 9 modulo 8 = 1 → 001. As the answer on the sent command is required, bit P = 1. RR function definer is 00 and identifier of the frame type is 01. So, control field is 00110001.

The check is calculated by formula (1.4), using address and control fields for coding. For basic format it is the next sequence in polynomial form (multiplied by x¹⁶): x³¹+x³⁰+x²⁷+x²⁶+x²⁴+x²¹+x²⁰+x¹⁶ divided by polynomial P(x) = x¹⁶+x¹²+x⁵+1. The remainder of dividing (16 bits) is check field. R(x) = x¹⁴ + x¹³ + x⁹ + x⁸ + 1 → 01100011 00000001.