Probability in terms of slots wasted during network contention

Question

Given 3 nodes accessing LAN, with time divided into slots and each node attempting to transmit at the beginning of a slot with probability of transmission $ p1,p2,p3 $ corresponding to each node. I am trying to find the portion of slots wasted due to multiple simultaneous attempts.

This is my solution :

1) Portion of slots wasted equals to the probability that there are 2 or more attempts in a given slot, so $P=1−P(0)−P(1)=1−(1−p)^3−3∗p(1−p)^2$, since the number of nodes is 3.

2) Since each station has its own transmission probability, I calculated slots wasted by each node as : $ SlotsWasted_{node1}=1−3∗p_1(1−p_1)^2−(1−p_1)^3 $

$ SlotsWasted_{node2}=1−3∗p_2(1−p_2)^2−(1−p_2)^3 $

$ SlotsWasted_{node1}=1−3∗p_3(1−p_3)^2−(1−p_3)^3 $

3) Then the sum of those 3 portions is the portion of the overall slots wasted for these 3 nodes.

Is this the right way to do this?

There are general formulas when the probability is just p, but not sure if its correct when each station has a different value for p.

Thank you!

Answer 1

This is not the way to do this. Let's talk about all eight things that can happen when a time-slot starts:

• no node transmits. Probability: $(1 - p_1)(1 - p_2)(1 - p_3)$.
• only node 1 transmits. Probability: $p_1(1 - p_2)(1 - p_3)$.
• only node 2 transmits. Probability: $(1 - p_1)p_2(1 - p_3)$.
• only node 3 transmits. Probability: $(1 - p_1)(1 - p_2)p_3$.
• nodes 1 and 2 transmit. Probability: $p_1 p_2(1 - p_3)$.
• nodes 1 and 3 transmit. Probability: $p_1(1 - p_2)p_3$.
• nodes 2 and 3 transmit. Probability: $(1 - p_1)p_2 p_3$.
• all three nodes transmit. Probability: $p_1 p_2 p_3$.

So the probability that two or three nodes transmit at once is the sum of the last four of these, $$ p_1 p_2(1 - p_3) + p_1(1 - p_2)p_3 + (1 - p_1)p_2 p_3 + p_1 p_2 p_3 \text{.} $$

This doesn't substantially simplify.