| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem, we need to break it down into the different types of frames that will be exchanged and calculate the total number of bytes for each. 1. **Poll Frames**: The primary station will send a poll frame to each of the four secondary stations. Each poll frame is 32 bytes. - Total poll frames = 4 (one for each station) - Total bytes for poll frames = 4 * 32 = 128 bytes 2. **Data Frames**: Each station can send only one frame in response to a poll. Since each station has 5 frames to send, but can only send one, we consider only one data frame per station. Each data frame is 1000 bytes. - Total data frames = 4 (one from each station) - Total bytes for data frames = 4 * 1000 = 4000 bytes 3. **ACK Frames**: After each data frame, the primary station will send an ACK frame. Each ACK frame is 32 bytes. - Total ACK frames = 4 (one for each data frame received) - Total bytes for ACK frames = 4 * 32 = 128 bytes 4. **NAK Frames**: Since the problem does not specify any errors, we can assume there will be no NAK frames sent. 5. **Total Bytes Exchanged**: To find the total number of bytes exchanged, we add the bytes for poll frames, data frames, and ACK frames. - Total bytes = Total bytes for poll frames + Total bytes for data frames + Total bytes for ACK frames - Total bytes = 128 + 4000 + 128 = 4256 bytes Therefore, the total number of bytes exchanged in the network is **4256 bytes**. |
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| deepseekr1 | To determine the total bytes exchanged in the polling network: 1. **Sequence per Data Frame**: - **Poll Frame** (Primary to Secondary): 32 bytes - **Data Frame** (Secondary to Primary): 1000 bytes - **ACK Frame** (Primary to Secondary): 32 bytes **Total per data frame**: \(32 + 1000 + 32 = 1064\) bytes. 2. **Frames per Station**: - Each secondary sends **5 data frames**, requiring 5 cycles of Poll → Data → ACK. **Bytes per secondary**: \(5 \times 1064 = 5320\) bytes. 3. **All Four Secondaries**: - Total for 4 stations: \(4 \times 5320 = 21,\!280\) bytes. **Final Answer**: \boxed{21280} bytes. |
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| gpto1 | To solve this problem, we need to calculate the total number of bytes exchanged during the communication between the primary station and the four secondary stations using polling. **Given:** - Number of secondary stations = 4 - Each station has 5 frames to send - Size of each data frame = 1000 bytes - Size of each poll frame = 32 bytes - Size of each ACK (Acknowledgment) frame = 32 bytes - Each station can send only one frame in response to a poll **Steps:** 1. **Calculate the total number of data frames:** - Each secondary station has 5 frames to send. - Total data frames = Number of stations × Number of frames per station - Total data frames = 4 stations × 5 frames = **20 data frames** 2. **Calculate the total number of polls:** - Since each station can send only one frame per poll, the primary station must poll each station once for each frame it needs to send. - Total polls per station = Number of frames per station = 5 polls - Total polls = Number of stations × Polls per station - Total polls = 4 stations × 5 polls = **20 poll frames** 3. **Calculate the total number of ACK frames:** - After each data frame is sent by a secondary station, the primary station sends an ACK. - Total ACK frames = Total data frames = **20 ACK frames** 4. **Calculate the total bytes exchanged:** **a. Bytes for poll frames:** - Total poll bytes = Number of poll frames × Size of each poll frame - Total poll bytes = 20 polls × 32 bytes = **640 bytes** **b. Bytes for data frames:** - Total data bytes = Number of data frames × Size of each data frame - Total data bytes = 20 data frames × 1000 bytes = **20,000 bytes** **c. Bytes for ACK frames:** - Total ACK bytes = Number of ACK frames × Size of each ACK frame - Total ACK bytes = 20 ACKs × 32 bytes = **640 bytes** **d. Total bytes exchanged:** - Total bytes = Poll bytes + Data bytes + ACK bytes - Total bytes = 640 bytes + 20,000 bytes + 640 bytes = **21,280 bytes** **Answer:** **21,280 bytes are exchanged in total.** |
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| GPT-4o | To solve this problem, we need to calculate the total number of bytes exchanged in the network, considering the given constraints and the polling mechanism. ### Given Data: - Number of stations: 1 primary and 4 secondary - Size of a data frame: 1000 bytes - Size of a poll frame: 32 bytes - Size of an ACK frame: 32 bytes - Size of a NAK frame: 32 bytes - Each station has 5 frames to send - Each station can send only one frame in response to a poll ### Steps to Calculate Total Bytes Exchanged: 1. **Polling Overhead**: - Each secondary station is polled once per data frame. - Total polls for one secondary station to send all 5 frames: \( 5 \) - Total polls for all 4 secondary stations to send all their frames: \[ 4 \text{ stations} \times 5 \text{ polls per station} = 20 \text{ polls} \] - Bytes for polling: \[ 20 \text{ polls} \times 32 \text{ bytes per poll} = 640 \text{ bytes} \] 2. **Data Frames**: - Each station sends 5 data frames. - Total data frames sent by all stations: \[ 4 \text{ stations} \times 5 \text{ frames per station} = 20 \text{ frames} \] - Bytes for data frames: \[ 20 \text{ frames} \times 1000 \text{ bytes per frame} = 20000 \text{ bytes} \] 3. **Acknowledgement Frames**: - Each data frame sent requires an ACK frame. - Total ACK frames: \[ 20 \text{ data frames} \times 1 \text{ ACK per data frame} = 20 \text{ ACKs} \] - Bytes for ACK frames: \[ 20 \text{ ACKs} \times 32 \text{ bytes per ACK} = 640 \text{ bytes} \] ### Total Bytes Exchanged: Adding up all the bytes for polls, data frames, and ACKs: \[ \text{Total bytes} = 640 \text{ bytes (polls)} + 20000 \text{ bytes (data frames)} + 640 \text{ bytes (ACKs)} \] \[ \text{Total bytes} = 21280 \text{ bytes} \] Thus, the total number of bytes exchanged in the network is **21,280 bytes**. |
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| llama3.1-70B-Instruct | A classic problem in computer networking! Let's break down the problem step by step: 1. The primary station sends a poll frame to each of the 4 secondary stations. Since each poll frame is 32 bytes, the total number of bytes sent by the primary station is: 4 (secondary stations) × 32 (bytes per poll frame) = 128 bytes 2. Each secondary station responds with a data frame (1000 bytes) or an ACK/NAK frame (32 bytes). Since each station has 5 frames to send, and each station can send only one frame in response to a poll, each secondary station will send: 5 (frames to send) × 1000 (bytes per data frame) = 5000 bytes (if all frames are data frames) or 5 (frames to send) × 32 (bytes per ACK/NAK frame) = 160 bytes (if all frames are ACK/NAK frames) Since we don't know the exact mix of data and ACK/NAK frames, let's assume the worst-case scenario: all frames are data frames. So, each secondary station sends 5000 bytes. 3. The primary station receives the responses from each secondary station and sends an ACK/NAK frame (32 bytes) for each response. Since each secondary station sends 5 frames, the primary station sends: 4 (secondary stations) × 5 (frames per station) × 32 (bytes per ACK/NAK frame) = 640 bytes 4. Now, let's calculate the total number of bytes exchanged: Primary station (poll frames): 128 bytes Secondary stations (data frames): 4 (stations) × 5000 (bytes per station) = 20,000 bytes Primary station (ACK/NAK frames): 640 bytes Total bytes exchanged: 128 + 20,000 + 640 = 20,768 bytes Therefore, the total number of bytes exchanged is 20,768 bytes. |
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| llama3.1-405B-Instruct | To solve this problem, we need to calculate the total number of bytes exchanged between the primary station and the secondary stations. 1. Each secondary station has 5 frames to send, and each frame is 1000 bytes. So, the total number of bytes sent by each secondary station is 5 x 1000 = 5000 bytes. 2. Since there are 4 secondary stations, the total number of bytes sent by all secondary stations is 4 x 5000 = 20,000 bytes. 3. For each frame sent by a secondary station, the primary station sends a poll frame (32 bytes), an ACK frame (32 bytes), or a NAK frame (32 bytes). Since each station can send only one frame in response to a poll, the primary station sends a total of 4 x 5 = 20 poll frames, 20 ACK frames, and 0 NAK frames (assuming all frames are received correctly). 4. The total number of bytes sent by the primary station is 20 x 32 (poll frames) + 20 x 32 (ACK frames) = 1280 bytes. 5. Therefore, the total number of bytes exchanged is the sum of the bytes sent by the secondary stations and the primary station: 20,000 + 1280 = 21,280 bytes. The final answer is 21,280. |
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| DeepSeek-Coder-V2 | To solve the problem, we need to calculate the total number of bytes exchanged in the network using polling, given the constraints and frame sizes. 1. **Identify the number of frames each station needs to send:** - Each station has 5 frames to send. 2. **Determine the number of polls required:** - Since each station can send only one frame in response to a poll, each station will require 5 polls to send all 5 frames. 3. **Calculate the total number of polls for all stations:** - There are 4 secondary stations. - Each secondary station requires 5 polls. - Total polls for secondary stations = 4 stations * 5 polls/station = 20 polls. 4. **Include the primary station's polls:** - The primary station polls each secondary station. - Total polls including the primary station = 20 polls. 5. **Calculate the total number of frames sent:** - Each station sends 5 frames. - Total frames sent by secondary stations = 4 stations * 5 frames/station = 20 frames. 6. **Calculate the total number of control frames (polls, ACKs, and NAKs):** - Each poll frame is 32 bytes. - Each ACK frame is 32 bytes. - Each NAK frame is 32 bytes. - Total control frames = 20 polls + 20 ACKs + 20 NAKs = 60 control frames. 7. **Calculate the total number of bytes for control frames:** - Each control frame is 32 bytes. - Total bytes for control frames = 60 control frames * 32 bytes/frame = 1920 bytes. 8. **Calculate the total number of bytes for data frames:** - Each data frame is 1000 bytes. - Total bytes for data frames = 20 data frames * 1000 bytes/frame = 20000 bytes. 9. **Calculate the total number of bytes exchanged:** - Total bytes exchanged = Total bytes for control frames + Total bytes for data frames. - Total bytes exchanged = 1920 bytes + 20000 bytes = 21920 bytes. Therefore, the total number of bytes exchanged in the network is: \[ \boxed{21920} \] |
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| Qwen2-72B-Instruct | To solve this problem, we need to break it down into the different types of frames that will be exchanged and calculate the total number of bytes for each. 1. **Poll Frames**: The primary station will send a poll frame to each of the four secondary stations. Each poll frame is 32 bytes. - Total poll frames = 4 (one for each station) - Total bytes for poll frames = 4 * 32 = 128 bytes 2. **Data Frames**: Each station can send only one frame in response to a poll. Since each station has 5 frames to send, but can only send one, we consider only one data frame per station. Each data frame is 1000 bytes. - Total data frames = 4 (one from each station) - Total bytes for data frames = 4 * 1000 = 4000 bytes 3. **ACK Frames**: After each data frame, the primary station will send an ACK frame. Each ACK frame is 32 bytes. - Total ACK frames = 4 (one for each data frame received) - Total bytes for ACK frames = 4 * 32 = 128 bytes 4. **NAK Frames**: Since the problem does not specify any errors, we can assume there will be no NAK frames sent. 5. **Total Bytes Exchanged**: To find the total number of bytes exchanged, we add the bytes for poll frames, data frames, and ACK frames. - Total bytes = Total bytes for poll frames + Total bytes for data frames + Total bytes for ACK frames - Total bytes = 128 + 4000 + 128 = 4256 bytes Therefore, the total number of bytes exchanged in the network is **4256 bytes**. |
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| baichuan4 | To solve this problem, we need to calculate the total number of bytes exchanged during the polling process. We have one primary station and four secondary stations. Each secondary station has 5 frames to send, but according to the problem, each station can send only one frame in response to a poll. Here's the breakdown of the communication process: 1. The primary station sends a poll to the first secondary station. 2. The secondary station responds with one data frame (1000 bytes). 3. The primary station acknowledges the received frame (ACK - 32 bytes). 4. The primary station sends a poll to the next secondary station. 5. This process repeats until all secondary stations have sent their one frame each. Since each secondary station can send only one frame per poll, it will take five rounds of polling for each secondary station to send all its frames. Now, let's calculate the total bytes exchanged: - There are 4 secondary stations, and each will send 5 frames (1000 bytes each) over 5 polling rounds. So, the total data bytes sent by secondary stations is 4 * 5 * 1000 = 20,000 bytes. - For each frame sent by a secondary station, the primary station will send an ACK (32 bytes). So, the total ACK bytes sent by the primary station is 4 * 5 * 32 = 640 bytes. - The primary station will send a poll to each secondary station before receiving a frame. Since there are 4 secondary stations, and the polling process happens 5 times, the total poll bytes sent by the primary station is 4 * 5 * 32 = 640 bytes. Adding all these together, we get the total bytes exchanged: Total Bytes Exchanged = Data Bytes Sent by Secondary Stations + ACK Bytes Sent by Primary Station + Poll Bytes Sent by Primary Station Total Bytes Exchanged = 20,000 + 640 + 640 = 21,280 bytes Therefore, the total number of bytes exchanged in this polling process is 21,280 bytes. |
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| doubao | The primary station will send a poll frame of 32 bytes to each of the 4 secondary stations, so that's 4 * 32 = 128 bytes for the poll frames. Each secondary station will send a data frame of 1000 bytes and an ACK frame of 32 bytes when polled, so for each secondary station that's 1000 + 32 = 1032 bytes. Since there are 4 secondary stations, that's 4 * 1032 = 4128 bytes for the data and ACK frames from the secondary stations. Adding the bytes from the poll frames to the bytes from the data and ACK frames, the total number of bytes exchanged is 128 + 4128 = 4256 bytes. |
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| minimax | To solve this problem, we need to calculate the total number of bytes exchanged during the polling process. We'll consider the following steps: 1. The primary station sends a poll to each secondary station. 2. Each secondary station sends a data frame in response to the poll. 3. The primary station sends an ACK or NAK for each data frame received. Given: - Data frame size = 1000 bytes - Poll, ACK, and NAK frame size = 32 bytes each - Each station has 5 frames to send, but they can only send one frame in response to a poll. Since each station can send only one frame in response to a poll, we'll consider that each station sends one frame after being polled. Let's calculate the total bytes exchanged: 1. Polling phase: - The primary station sends a poll to each of the four secondary stations. - Total bytes for polling = 4 polls * 32 bytes per poll = 128 bytes 2. Data transmission phase: - Each secondary station sends one data frame in response to the poll. - Total bytes for data transmission = 4 stations * 1000 bytes per data frame = 4000 bytes 3. Acknowledgment phase: - The primary station sends an ACK for each data frame received. - Total bytes for ACKs = 4 ACKs * 32 bytes per ACK = 128 bytes There is no need to consider NAKs in this scenario since we are not given any information about errors in transmission, and we assume that each frame is successfully received. Now, let's add up all the bytes exchanged: Total bytes exchanged = Bytes for polling + Bytes for data transmission + Bytes for ACKs Total bytes exchanged = 128 bytes (polls) + 4000 bytes (data frames) + 128 bytes (ACKs) Total bytes exchanged = 4256 bytes Therefore, the total number of bytes exchanged during the polling process is 4256 bytes. |
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| kimi | To solve this problem, we need to consider the following: 1. Each station has 5 frames to send, and each frame is 1000 bytes. 2. The size of a poll, ACK, and NAK frame is 32 bytes. 3. Each station can send only one frame in response to a poll. Since there are four secondary stations and one primary station, the primary station will poll each secondary station four times to get all the frames. For each secondary station: - The primary station sends a poll frame (32 bytes). - The secondary station sends back a frame (1000 bytes) and an ACK frame (32 bytes) to confirm the reception of the poll. So, for one secondary station, the total bytes exchanged are: - Poll frame: 32 bytes - Data frame: 1000 bytes - ACK frame: 32 bytes Total for one secondary station = 32 + 1000 + 32 = 1064 bytes Since there are four secondary stations, the total bytes exchanged for all secondary stations are: Total for all secondary stations = 1064 bytes/station * 4 stations = 4256 bytes However, we must also consider the initial poll from the primary station to each secondary station. This happens once for each station, so for four stations, we have: Initial poll frames = 32 bytes/station * 4 stations = 128 bytes Adding this to the total for all secondary stations gives us: Total bytes exchanged = 4256 bytes + 128 bytes = 4384 bytes This is the total number of bytes exchanged for the secondary stations. We must also consider the primary station sending its own frames. Since the primary station also has 5 frames to send, we need to add the bytes for these frames as well. For the primary station: - Data frames: 5 frames * 1000 bytes/frame = 5000 bytes Adding this to the total bytes exchanged gives us: Total bytes exchanged = 4384 bytes + 5000 bytes = 9384 bytes So, the total number of bytes exchanged in the network is 9384 bytes. |
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| wenxinyiyan | To solve this problem, we need to calculate the total number of bytes exchanged during the polling process. Let's break down the communication into steps and calculate the bytes exchanged in each step. 1. **Polling Phase**: * The primary station sends a poll frame to each secondary station. Since there are 4 secondary stations, 4 poll frames will be sent. * Size of each poll frame = 32 bytes * Total bytes sent during polling phase = 4 * 32 = 128 bytes 2. **Response Phase**: * After receiving the poll frame, each secondary station sends one data frame back to the primary station. * Size of each data frame = 1000 bytes * Since each station has 5 frames to send, but can only send one frame in response to a poll, the process will repeat 5 times for each station. * Total bytes sent by each secondary station = 5 * 1000 = 5000 bytes * Total bytes sent by all secondary stations = 4 * 5000 = 20000 bytes 3. **Acknowledgment Phase**: * After receiving each data frame, the primary station sends an ACK or NAK frame back to the secondary station. * Size of each ACK/NAK frame = 32 bytes * Since there are 5 data frames sent by each secondary station, 5 ACK/NAK frames will be sent to each station. * Total bytes sent to each secondary station = 5 * 32 = 160 bytes * Total bytes sent to all secondary stations = 4 * 160 = 640 bytes Now, let's calculate the total number of bytes exchanged: Total bytes exchanged = Polling Phase + Response Phase + Acknowledgment Phase Total bytes exchanged = 128 bytes + 20000 bytes + 640 bytes Total bytes exchanged = 20768 bytes Therefore, the total number of bytes exchanged during the polling process is 20768 bytes. |
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| zhipu | To determine the total number of bytes exchanged in the network using the polling protocol, we need to consider the bytes exchanged for each poll, the data frames sent by each station, and the ACK or NAK frames sent in response to each data frame. ### Step-by-Step Calculation: 1. **Poll Frame:** - Each poll frame is 32 bytes. 2. **Data Frame:** - Each data frame is 1000 bytes. 3. **ACK/NAK Frame:** - Each ACK or NAK frame is 32 bytes. ### Sequence of Events for Each Poll: 1. **Primary Station sends a Poll to a Secondary Station:** - 32 bytes (Poll frame) 2. **Secondary Station sends a Data Frame:** - 1000 bytes (Data frame) 3. **Primary Station sends an ACK/NAK:** - 32 bytes (ACK or NAK frame) ### Total Bytes Exchanged per Poll and Response: - For each poll and the corresponding data frame and ACK/NAK, the total bytes exchanged are: \[ 32 \text{ (Poll)} + 1000 \text{ (Data)} + 32 \text{ (ACK/NAK)} = 1064 \text{ bytes} \] ### Total Exchanges: - There are 4 secondary stations, and each station has 5 frames to send. - Each station can send only one frame in response to a poll, so there will be 5 polls for each station. ### Total Bytes for All Polls and Responses: - For each station: \[ 5 \text{ polls} \times 1064 \text{ bytes per poll} = 5320 \text{ bytes} \] - For all 4 stations: \[ 4 \text{ stations} \times 5320 \text{ bytes per station} = 21280 \text{ bytes} \] ### Final Calculation: - Therefore, the total number of bytes exchanged in the network is 21280 bytes. ### Solution: \[ \boxed{21280 \text{ bytes}} \] |
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| gpto1 | To solve this problem, we need to calculate the total number of bytes exchanged during the communication between the primary station and the four secondary stations using polling. **Given:** - Number of secondary stations = 4 - Each station has 5 frames to send - Size of each data frame = 1000 bytes - Size of each poll frame = 32 bytes - Size of each ACK (Acknowledgment) frame = 32 bytes - Each station can send only one frame in response to a poll **Steps:** 1. **Calculate the total number of data frames:** - Each secondary station has 5 frames to send. - Total data frames = Number of stations × Number of frames per station - Total data frames = 4 stations × 5 frames = **20 data frames** 2. **Calculate the total number of polls:** - Since each station can send only one frame per poll, the primary station must poll each station once for each frame it needs to send. - Total polls per station = Number of frames per station = 5 polls - Total polls = Number of stations × Polls per station - Total polls = 4 stations × 5 polls = **20 poll frames** 3. **Calculate the total number of ACK frames:** - After each data frame is sent by a secondary station, the primary station sends an ACK. - Total ACK frames = Total data frames = **20 ACK frames** 4. **Calculate the total bytes exchanged:** **a. Bytes for poll frames:** - Total poll bytes = Number of poll frames × Size of each poll frame - Total poll bytes = 20 polls × 32 bytes = **640 bytes** **b. Bytes for data frames:** - Total data bytes = Number of data frames × Size of each data frame - Total data bytes = 20 data frames × 1000 bytes = **20,000 bytes** **c. Bytes for ACK frames:** - Total ACK bytes = Number of ACK frames × Size of each ACK frame - Total ACK bytes = 20 ACKs × 32 bytes = **640 bytes** **d. Total bytes exchanged:** - Total bytes = Poll bytes + Data bytes + ACK bytes - Total bytes = 640 bytes + 20,000 bytes + 640 bytes = **21,280 bytes** **Answer:** **21,280 bytes are exchanged in total.** |
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| deepseekv3 | To calculate the total bytes exchanged in the polling-based network, let's break down the process step by step. ### Given: - **Primary station**: 1 - **Secondary stations**: 4 - **Data frame size**: 1000 bytes - **Poll frame size**: 32 bytes - **ACK frame size**: 32 bytes - **NAK frame size**: 32 bytes - **Each station has 5 frames to send** - **Each station can send only one frame in response to a poll** ### Process: 1. **Polling Cycle**: - The primary station sends a poll frame to each secondary station. - Each secondary station responds with a data frame (if it has data to send) or a NAK frame (if it has no data). - The primary station then sends an ACK frame in response to the data frame. 2. **Bytes Exchanged per Polling Cycle**: - **Primary to Secondary**: Poll frame = 32 bytes - **Secondary to Primary**: Data frame = 1000 bytes or NAK frame = 32 bytes - **Primary to Secondary**: ACK frame = 32 bytes (if a data frame was received) 3. **Total Polling Cycles**: - Each station has 5 frames to send, and it can send only one frame per poll. - Therefore, each station will be polled 5 times. - Total polling cycles = Number of stations × Number of polls per station = 4 × 5 = 20 4. **Total Bytes Exchanged**: - For each polling cycle where a data frame is sent: - Poll frame: 32 bytes - Data frame: 1000 bytes - ACK frame: 32 bytes - Total per cycle: 32 + 1000 + 32 = 1064 bytes - For each polling cycle where a NAK frame is sent: - Poll frame: 32 bytes - NAK frame: 32 bytes - Total per cycle: 32 + 32 = 64 bytes - Since each station has 5 frames to send, all polling cycles will involve data frames. - Total bytes exchanged = Total polling cycles × Bytes per cycle = 20 × 1064 = 21280 bytes ### Final Answer: **Total bytes exchanged: 21,280 bytes** |
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