指数退避平均碰撞数

Question

我正在研究一种简单的指数退避算法，我想知道我的结果是否正确。以下是假设：

• 有N个站

• 每个电台有1个要发送的数据包

• 最初，所有电台都尝试发送时隙0

• 当两个或多个电台想要在同一时段发送其帧时发生冲突（因此所有电台的帧在第一轮中发生碰撞）

• 当发生冲突时，工作站使用指数退避功能计算其等待时间（即，在c冲突之后，它将等待从0到2 ^ c - 1的随机数量的时隙）

我用Java编码。以下是使用10个站点的运行结果：

Slot 0:  Station 0 Station 1 Station 2 Station 3 Station 4 Station 5 Station 6 Station 7 Station 8 Station 9 
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Collisions in round 1: Station 0 Station 1 Station 2 Station 3 Station 4 Station 5 Station 6 Station 7 Station 8 Station 9 
Slot 0:  
Slot 1:  Station 1 Station 3 Station 5 Station 6 
Slot 2:  Station 0 Station 2 Station 4 Station 7 Station 8 Station 9 
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Collisions in round 2: Station 0 Station 1 Station 2 Station 3 Station 4 Station 5 Station 6 Station 7 Station 8 Station 9 
Slot 0:  
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Slot 2:  Station 3 
Slot 3:  Station 0 Station 2 Station 4 Station 6 
Slot 4:  Station 1 Station 8 Station 9 
Slot 5:  Station 5 Station 7 
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Collisions in round 3: Station 0 Station 1 Station 2 Station 4 Station 5 Station 6 Station 7 Station 8 Station 9 
Slot 0:  
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Slot 2:  Station 3 
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Slot 5:  Station 6 
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Slot 7:  Station 9 
Slot 8:  Station 0 
Slot 9:  Station 1 Station 4 
Slot 10:     Station 8 
Slot 11:     Station 2 
Slot 12:     Station 7 
Slot 13:     Station 5 
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Collisions in round 4: Station 1 Station 4 
Slot 0:  
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Slot 2:  Station 3 
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Slot 5:  Station 6 
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Slot 7:  Station 9 
Slot 8:  Station 0 
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Slot 10:     Station 8 
Slot 11:     Station 2 
Slot 12:     Station 7 
Slot 13:     Station 5 
Slot 14:     Station 1 
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Collisions in round 5: No collisions!
Total number of collisions: 77
Station 0 sent at time slot 8 with 3 collisions.
Station 1 sent at time slot 14 with 4 collisions.
Station 2 sent at time slot 11 with 3 collisions.
Station 3 sent at time slot 2 with 2 collisions.
Station 4 sent at time slot 23 with 4 collisions.
Station 5 sent at time slot 13 with 3 collisions.
Station 6 sent at time slot 5 with 3 collisions.
Station 7 sent at time slot 12 with 3 collisions.
Station 8 sent at time slot 10 with 3 collisions.
Station 9 sent at time slot 7 with 3 collisions.

这看起来不错吗？我似乎无法找到使用此功能的N站的平均碰撞次数，所以我不确定我是否搞砸了。非常感谢任何帮助。

Answer 1

一个电台可以尝试在一个时隙中多次发送，这似乎很奇怪。在我看来，退避值应该是电台跳过的时隙数。因此，如果站A在时间段1发生冲突并且选择跳过0个时隙，则它将被安排在时隙2处再次发送。它不会尝试在同一时隙期间重新发送。

这更有意义。一个电台在一个时段内最多可以尝试一次发送。

因此，如果每个人都试图在时隙0发送，他们都会发生冲突并选择跳过从0到2 ^ i-1（其中i是当前时隙）的随机数量的时隙。因为2 ^ i-1是0，所以他们都会尝试在时隙1发送。

平均，其中一半将尝试在时隙2期间发送，一半将尝试在时隙3期间发送。假设奇数站选择不跳过，并且均匀 - 编号的站都跳过了一个插槽。在时间段1结束时，您的州看起来像这样：

slot    stations
 2      1, 3, 5, 7, 9
 3      2, 4, 6, 8, 10

在时段2期间也会发生冲突。每个站将选择跳过0,1,2或3个时隙。在时间段2之后，您的状态将类似于：

slot    stations
 3      1, 2, 4, 6, 8, 10
 4      3, 7
 5      9
 6      5

由于时隙3中的冲突，这六个站将在时隙4到10之间展开（选择跳过0到7个时隙）。我会让你建立相应的状态。

最终，您将到达当前时段有一个站点的位置，然后该站点可以发送。

如果以这种方式考虑问题，调试代码变得更加容易，因为您可以构建表并在每个时间段结束时显示状态。这将告诉您是否正确选择了退避值，以及是否根据所选的退避值正确安排了工作站。