Host A Host B Seg-92, 20 bytes data ACK Seg=100, 20 bytes data loss ACK Seg= S4 .20 bytes data time Figure: Cumulative ACKS In Figure: Cumulative ACKS, the value of 54 is timeout-
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- Based on the given block data below, answer the following questions. Hint: First do Google search for ‘packet dump decode’. Then cut and paste the block of data below into the input data section. Then press decode. You will be able to find the answer from the decoding result. Hint Q6: For the payload of the application layer, look at the last few bytes in ASCII. You can use the ASCII character table to manually convert them. If you understanding the layer 4 destination port #, you understand what it is asking for. There is only one kind of information for the port #. Data block to be decoded: 00 11 bc b5 49 40 b8 8d 12 00 0d fa 08 00 45 00 00 37 99 09 00 00 ff 11 ea 9f 0a 07 03 1a 42 aa e8 41 c7 19 00 35 00 23 a3 9e 02 90 01 00 00 01 00 00 00 00 00 00 05 6d 69 74 72 65 03 6f 72 67 00 00 01 00 01 Please answer the following questions after decoded the above data block: Q1: (Layer 4) Protocol is: Q2: Source MAC address: Q3: Destination MAC address: Q4: Source IP address:…The following HDLC frame was received by a host. Does it have an error or not? If so, explain the error. 01111110 11010111110111011101111001110111110110 011111110please typed not hand written Exercise 1: We are sending a 7.4 KB (care this is kilo bytes!) file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 8 Kbps. Assume that the propagation speed is 2 * 10^8 meters/sec, and the distance between source and destination is 40,000 km. Reply to the following questions. Question:How much is the end-to-end delay (transmission delay plus propagation delay)
- 89. A ________ bridge can forward and filter frames and automatically build its forwarding table. a. simple b. dual c. transparent d. none of the abovehello can you convert from this coding below to mbed c++ #define BLYNK_DEVICE_NAME "NodeMCU"#define BLYNK_AUTH_TOKEN "oLt-Y2bvgoy22dT9JP2ZIuLR1MDr9n4h"#define BLYNK_PRINT Serial #define BLYNK_MAX_SENDBYTES 256#include <ESP8266WiFi.h>#include <BlynkSimpleEsp8266.h>#include "dht.h"#define dht_apin D5 // Analog Pin sensor is connected to#define sensorPin A0 // You should get Auth Token in the Blynk App.// Go to the Project Settings (nut icon).char auth[] = "oLt-Y2bvgoy22dT9JP2ZIuLR1MDr9n4h"; char ssid[] = "nizammustafa";char pass[] = "nizamkillerz17"; dht DHT; int sensorValue = 0;int safetyValue;int humidity;int val = 0; const int trigPin = D3;const int echoPin = D2; const int ledPin1 = D8;const int ledPin2 = D7;const int ledPin3 = D6; // defines variableslong duration;int distance;int safetyDistance; void setup() {Serial.begin(9600); // Starts the serial communication;Blynk.begin(auth, ssid, pass);pinMode(trigPin, OUTPUT); // Sets the trigPin as an OutputpinMode(echoPin,…Q5. Imagine you are using a mobile application that needs to fetch data from a specific webserver. Assume the mobile application knows the IP address of the web server. The app requests a data payload from the URL, which consists solely of a small HTML file. With RTT0 representing the RTT between your device and the web server, and considering negligible transmission time for the HTML file, calculate the total time elapsed from when the request is made by the app to when the data is received? Q6. Consider Q5, now assume the HTML file includes references to six additional small images hosted on the same server. Ignoring the actual data transmission times and assuming that there is a direct link between your mobile and the webserver, estimate the total time needed to download the full webpage when using: Non-persistent HTTP Persistent HTTP without using pipelining Persistent HTTP with pipelining Non-persistent HTTP configured to allow three parallel TCP connections. Note that for N…
- C/N Imagine that you are writing the data linksoftware for a line used to send data to you, but not from you. The other end uses HDLC, with a 3-bit sequence number and window size of seven frames. You would like to buffer as many out of sequence frames as possible to enhance efficiency, but you are not allowed to modify the software on the sending side. Is it possible to have a receiver window greater than one and still guarantee that the protocol will never fail? If so, what is the largest window that can be safely used?Which of the OSI layers is responsible for each of the following: a. allows the tasks to be synchronised b. Error detection and correction. c. Allows for flow management. d. Separate data into packets.Which of the following statements is/are correct?I. The fragmentation is applicable for data in the datagram but not for header.II. Reassembly of the fragments must be performed at the destination because,theIntermediate networks may have different maximum transmission unit (MTU) sizes.III. In the IP header, the time-to-live (TTL) field is invariable. (A) I and II(B) I and III(c) II and III(D) All are correct
- A 4000 octet user data is to be transmitted over a network which supports a maximum user data size of 536 octets. Assuming the header in each IP data gram requires 20 octets, derive the number of datagrams (fragments) required and the contents of the following fields: Identification Total length Fragment offset More Fragments flagQuestion 7IN UDP, the checksum is used to detect and correct the corrupted segments. True FalseWhich of the following statements is/are correct?I. The fragmentation is applicable for data in the datagram but not for header.II. Reassembly of the fragments must be performed at the destination because, the intermediatenetworks may have different maximum transmission unit (MTU) sizes.III. In the IP header, the time-to-live (TTL) field is invariable.(A) I and II(B) I and III(C) II and III(D) All are correct