Mrezi 2. Kolokvium zadaci reseni

10.3.84.129/20

Subnet IP:

10.3.0101|0000.00000000
10.3.80.0

-----------------------------
IP: 194.149.172.0/25
Subnet IP: 194.149.172.0|0000000
         = 194.149.172.0
Broadcast IP: 194.149.172.0|1111111
		    = 194.149.172.127
Range: [194.149.172.0 - 194.149.172.127]



Mreza			|	Hosts	| Block		| # of bits		|	SM	|	Subnet
1				|  	18		|	32		|	5			|	/27	| 194.149.172.0
2				|	32		|	64		|	6			|	/26	| 194.149.172.64
3				|	 6		|	 8		|	3			|	/29	| 194.149.172.40
centar			|	12		|	16		|	4			|	/28	| 194.149.172.48
lab				|    4		|	 8		|	3			|	/29	| 194.149.172.32

Kancelarija 3:
194.149.172.43/29
Subnet IP: 194.149.172.00101|011
		 = 194.149.172.00101000
		 = 194.149.172.40

		 zafateni: 0-32, 64-128, 40-48, 48-64, 32-40



------------------------------------------------------------------

	15.0.4.0/25   =  15.0.4.|00000000/24
    15.0.4.128/25 =  15.0.4.|10000000/24
				  =  15.0.4.0/24


	15.0.4.0/24 = 15.0.0000010|0.00000000/23
	15.0.5.0/24 = 15.0.0000010|1.00000000/23
				= 15.0.4.0/23




    15.0.6.0/24 = 15.0.0000011|0.|00000000/23
    15.0.7.0/24 = 15.0.0000011|1.|00000000/23
				= 15.0.6.0/23

	15.0.4.0/23 = 15.0.000001|0|0.00000000/22
	15.0.6.0/23 = 15.0.000001|1|0.00000000/22
				= 15.0.4.0/22


----------------------------------------------------

se sifrira: 101001101

r(1) = 101, r(2) = 110, r(3) = 111

101|001|101
101|110|111
------------xor
000|111|010
------------Rs
001|101|100

=> 001101100

------------------------------------------------------

za prakjanje: 3000B = 2980 + 20B
MTU: 1000B (zadusuvacki)
   = 980 + 20B

fragment 1: 2980 - 980
            preostanati: 2000

		Length: 1000B FF=1

fragment 2: 2000 - 980
			preostanati: 1020

		Length: 1000B FF=1

fragment 3: 1020 - 980
			preostanati: 100
		Length: 1000B FF=1

fragment 4: 100 - 980 < 0

		Length: 100+20 = 120B FF = 0


---------------------------------------------------------

10.3.84.129/20
10.3.0101|0000.00000000 = 10.3.80.0/20