Question

shift 5 steps to the right. 2 Marks c) Suppose the following RTT samples (in ms) are collected by TCP (in the given order): 90, 100, 95, 110, 90. Assuming the initial RTT estimate was 100ms, what will the Time_Out Interval used by TCP after the 3'd RTT sample. (Hint: assume a -0.125 and B-0.25). 4 Marks Question 4 (10 Marks) a) Given an IP address 192.168.10.20/24: 2 Marks i) What is the network to which the IP address belongs to? ii) Divide the network in (a.i) into 8 equal subnets, and indicate which one of subnets the address (192.168.10.20) belongs to. Clearly show your working. 6 Marks b) One of the arguments against NAT is that it violates the principle that network layer devices (routers) should only process up to layer 3. How does NAT violate this principle? 2 Marks

Answer 1

C> G iven, RT Somples are : j0, loo, 95, uo, 9o K = 0-125 9 0 Estimated RTT x(RT)+ (t-)(i) Dev RTT = Tninal DevRTr, = ( b, 25 ) 5.93 Timeout Interval = 98·15 + H ( 5.93) terndSamtple : (too) timated RTT (0.8朽)(98.75) (0125Xtoo)+ 98.906 Dev RTT く1100-98.906))+(0-15)(S93) (0-25) = 98.906 + H/hね) Time aut Inte nal Scanned with CamScanner

ample: C5) imated RTT-(o-125) (9s) / o-j)(7 906) 92-41 月ev RTT (0,25) = 39s 395) Time Out Toterval9841 59 mi :. ter'd Sample , Tine-Out. Interval ,1599チMS 0> Given, 9p address: R2.168.10. 30 124 The netuoork is 192-16&too divide he netoork nto 8 ubs 8 Subnets3 bits ) Mask 124. 192.168. o.20 belonqs o 2 168.lo D Subnet Broadcast Sddress: 142 168.lo.3] Scanned with CamScanner CS