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- Data Communication & Computer Network Suppose that Jon's and Jenny's computers are connected by an 8 Kbps link (note, 1 Kbps = 103 bps) that uses circuit switching and that Jon sent some number of bits to Jenny over the link. Suppose that circuit switching on the link connecting Jon’s and Jenny’s computers is implemented using Time Division Multiplexing (TDM), that a frame is 1 sec in duration, and that a frame consists of 16 slots. Jon is assigned one slot in every frame. Considering that it took 12 minutes for Jon to send all bits to Jenny, find how many bits were sent by Jon to Jenny.Shielded twisted-pair (STP) cabling was first made popular by IBM when it introduced Type classification for data cabling. Though to purchase and install than UTP, STP offers some distinct advantages. Shielded twisted-pair (STP) cable can up to 600MHZ. STP has an more currently support applications requiring shield that consists of braided . The shield surrounds the 150-ohm, 22 AWG, two-pair conductors. Each is insulated with a to install than UTP. STP cabling solely material and then each twisted pair is individually shielded. STP cabling is because it provides better EMI protection and potential bandwidth, installers consider using fiber-optic cable instead. conductor copper material bandwidth expensive dielectric inner glass Cheap lower Interference expensive WAN higher outerIn GSM, there are 8 separate time slots that makeup what is known as a “TDMA frame’’. Each GSM time slot lasts for 577 µs (microseconds) and contains more than simply a sample of each individual signal. Each time slot comprises a total of 148 bits. Of these 148 bits, only 114 represent voice or other data. The rest are used for a variety of control purposes. How many bits of data are in one TDMA frame?
- A signal in a network has a propagation velocity of 1.77x10" m/s and a delay of 4.3 us introduced by each repeater. If the round trip delay for two computers at either end of a link that includes three repeaters is 52.92 us, calculate the length of this link? Express your answer in kilometers.A frame size 10 million bytes is being sent on a link with with bandwidth of 8 Mbps. The total transmission time is (a.) 1 sec: (b.) 2 secs: (c.) 5 secs: (d.) 10 secs: (e) 20 secs: (f) 50 secs: (g) 100 secs ChooseCalculate the utilzation of a 100 m long 100 Mbps coaxial CSMACD system that transmits 10 Kb packets. (Assume speed of coil signal propagtioe 2x10 misec.) O 36.2% O 97.6% O 99.9% O 1896
- Coaxial cable has been around since local area networking was in its invented. The original designers of Ethernet picked coaxial cable as their "ether" because the coaxial cable is well it has high bandwidth capabilities and low and is easy to install. Coaxial cables are identified by their RG standard. Coaxial cable can have a or stranded core and impedance of 50, 75, or 92 ohms. The Coaxial has wire that carries the signal surrounded by a layer of and another concentric ; both the shield and the inner conductor run along the same axis. The shield also serves as a ground and should be grounded to be effective. Coaxial cable is still widely used for video applications; in fact, its use is increasing due to the greater demand for However, it is not recommended for data installations and is not recognized by the Standard for such. Noise outer two local area networks CCTV wire solid cladding expensive inner insulation one shielded attenuation Telephone shieldIn GSM, a "TDMA frame" is composed of eight distinct time slots. Each GSM time slot is 577 s (microseconds) in duration and includes far more than a snapshot of each individual signal.Each time slot contains 148 bits. Just 114 of these 148 bits reflect speech or other info. The remainder are used for a number of different types of power. How many data bits are included in a single TDMA frame?In GSM, a "TDMA frame" is composed of eight distinct time slots. Each GSM time slot is 577 s (microseconds) in duration and includes far more than a snapshot of each individual signal.. Each time slot contains 148 bits. Just 114 of these 148 bits reflect speech or other info. The remainder are used for a number of different types of power. How many data bits are included in a single TDMA frame?
- Assume you are working as a Network Engineer for a Telecommunications service provider and your task is to estimate the feasibility of a 20 km link between two buildings with one access point and one client radio. The access point is connected to an antenna with 12 dBi gain, with a transmitting power of 500 mW and a receive sensitivity of -84 dBm. The client is connected to an antenna with 10 dBi gain, with a transmitting power of 16 dBm and a receive sensitivity of -82 dBm. The cables in both systems are short, with a loss of 3 dB at each side at the 900 MHz frequency of operation. What is the link margin? By examining the link margin, comment on the reliability of the link.Your answer includes number only, thousands should be separated by a : For example: 1,000 or 1,000,000 • Involved networking devices: • A geostationary statellite • Base 10Mbps a microwave link • Every minute a photo is taken and sends to the base. •S = 2.4 e8 m/sec • Distance (statellite, earth) = 36,000 km Size of the photo x for the microwave link continuously transmitting ? bitsDear Writer could oyu please kindly explin why i get this soltuion and it is different that your solution? The transmission time of a frame can be calculated as follows: Propagation delay (Tp) = distance/speed = 5km / 200000 km/s = 0.025ms Transmission time (Tt) = frame size/bandwidth = 10000 * 8bits / (1Mbps * 10 ^ 6) = 0.008sec = 8ms Total time for transmission = 2 * Tp + Tt(since the frame has to travel from A to D and then D to A) = 2 * 0.025ms + 8ms = 0.05ms + 8ms = 8.05ms The transmission time of the frame from A to D is 8.05 ms. b) To calculate the efficiency of the CSMA/CD protocol in this LAN, we can use the formula: Efficiency = (1 / (1 + 5tprop / tframe)) tprop is the propagation delay tframe is the time it takes to transmit a frame. To calculate tprop, we need to determine the time it takes for a signal to propagate through the 5 Km distance between Station A and Station D. The speed of propagation is 200,000 Km/sec The propagation delay is: tprop = distance /…