Find Average Number Of Slot Times A Station Waits

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Average number of bits = sum (p_i)log2(1/p_i) for i = 2 through 12. Using the probabilities given in the figure above the average number of bits of information provided by the sum of two dice is 3.2744. So if we had the perfect encoding, the expected length of the transmission would be 3274.4 bits. Solution (a) rtt = 2 x (1 x 103/100000 x 103+ 8000/10x109) =21.6 x 10-6 number of frames that can be sent during rtt = 21.6 x 10-6/0.8 x 10-6 = 27 (b) 5 since 25 = 32 > 27 (c) max-wind-size -> 25 -> window-max = 25 -1 = 31 example 6: the following data fragment occurs in the middle of a data stream for which the byte-stuffing algorithm described.

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All the developed models are based on calculating station's delay time at the transmission slot (s) plus the average time that station defers at backoff slots before successful transmission. The mathematical models are simple, computationally fast and can be used to build admission control algorithms. 18 POISSON PROCESS 197 Nn has independent increments for any n and so the same holds in the limit. We should note that the Heads probability does not need to be exactly λ n, instead it suﬃces that this probability converges to λ when multiplied by n. Similarly, we do not need all integer multiplies of 1 n, it is enough that their number in [0,t], divided by n, converges to t in probability. Ten thousand airline reservation stations are competing for the use of a single slotted ALOHA channbel. The average station makes 18 requests/hour. A slot is 125 μsec. What is the approximate total channel load ? Answer: Each terminal makes one request every 200 sec, for a total load of 50 requests/sec. Hence.

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Waiting time for B = 0 * T slot = 0. Therefore, both stations will transmit at the same time and hence collision occurs. When A chooses K = 0 and B chooses K = 1 -> Waiting time for A = 0 * T slot = 0 Waiting time for B = 1 * T slot = T slot. Therefore, A transmits the packet and B waits for time T slot for transmitting and hence A wins. When.

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819:2 = 56:875 stations The maximum number of stations is 56 2 (10 pts) Two stations (C and D) are on the same broadcast domain approximately 500 meters apart. There are 3 repeaters inline between the two stations, each with a delay of 5 s. Suppose C and D transmit 1500 bytes at the same exact time and their frames collide.

Train wait problem, probability - Mathematics Stack Exchange.

Since demand varied during the day, the number of tellers provided would have to vary to meet the customer-service and teller- utilization goals. 70 60 Day 4/20/91: Friday Time Period: 11:45-12:45 Total Arrivals: 431 customers Average Time Between Arrivals: 8.4 Ser 50 40 Number of Observations 30 20 10 0 1 3 5 7 9 11 21 23 25 27 29 31 13 15 17.

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Stop‐and‐Wait sends a packet and waits for its ACK before sending the next packet Total # packets = 80000/(1000 - 40) = 84 [There is a 40 byte header] Hence total time taken = 84 * RTT [RTT is used since there is a wait for the ACK] = 84 * 10 = 840 ms or 0.84 s b. The model aims to control the delay times of express and normal vehicle classes such that the ratio of their average delay times tracks a target relative delay rate in real time. Each station can.

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Throughput of slotted aloha = Efficiency x Bandwidth = 0.368 x 100 Kbps = 36.8 Kbps Total Number Of Stations- Throughput of slotted aloha = Total number of stations x Throughput of each station Substituting the values, we get- 36.8 Kbps = N x 100 bits/sec ∴ N = 368 Thus, required value of N = 368. To gain better understanding about Aloha,. Eytan Modiano Slide 8 Example • Suppose a train arrives at a station according to a Poisson process with average inter-arrival time of 20 minutes • When a customer arrives at the station the average amount of time until the next arrival is 20 minutes - Regardless of when the previous train arrived • The average amount of time since the last departure is 20 minutes!.

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E (w) = λ/ (µ (µ-λ)) Let the random variable v denote the total time that a customer has to spend in the system including the service. E (v) = 1/ (µ -λ) This explanation of queuing theory and the associated math and formulas for calculating the average wait time within queue systems has been published with input from Prof. Dr. Rajalakshmi.

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Ensure you request for assistant if you can’t find the section. When you are done the system will automatically calculate for you the amount you are expected to pay for your order depending on the details you give such as subject area, number of pages, urgency, and academic level. After filling out the order form, you fill in the sign up details. If station chooses a number k, then- Back off time = k x Time slot where value of one time slot = 1 RTT Example- Consider the following scenario where stations A and D start transmitting their data simultaneously- For simplicity, We consider the value of time slot = 1 unit. Thus, back off time = K units. Enter the email address you signed up with and we'll email you a reset link.

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After the first collision, each stations waits either 0 or 1 slot times before trying again. If say, two stations collide, and each one picks up the same number, they will collide again. After second collision, each one picks up either 0,1,2,3. If a third collision occurs, then the next time the number of slots to wait is chosen from 0 to 23-1.

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The maximum throughput for the slotted ALOHA occurs at G = 1.0 ( Equation 6.21) and it is equal to 1/ e or about 0.368. This implies that at the maximum throughput, 36.8% of the time slots carry successfully transmitted packets. The best channel utilization with the slotted ALOHA protocol is 36.8% — twice the pure ALOHA protocol. Christmas music comprises a variety of genres of music regularly performed or heard around the Christmas season.Music associated with Christmas may be purely instrumental, or, in the case of carols or songs, may employ lyrics whose subject matter ranges from the nativity of Jesus Christ, to gift-giving and merrymaking, to cultural figures such as Santa Claus, among other topics.

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A second order polynomial function was developed to predict truck turn times, with an adjusted R 2 of 0.7381, using the average truck number served by a crane as the predictor variable.