What does a broadcast actually cost?

If C1 requests X while only C0 may hold it, a peer broadcast and an exact holder lookup both target C0. In this two-core case there is no saving for that request. The value of filtering appears when the peer is known absent, or in a larger system.

Use independent stable snapshots derived from the course's X = 0 start, not successive transactions. A row describes what a later request would need to investigate. The ordinary H directory still determines legal permission and data handling; a filter does not invent a new grant rule.

Count probes and responses

Each probe needs a response (Inv-Ack, or Data from the holder), so count both. With N private caches, a broadcast sends N-1 probes and collects N-1 responses. A filter with k candidates, 0 <= k <= N-1, sends k and collects k, plus one filter lookup at H. Why N-1 tag checks per miss scale badly is covered in Choosing Snoop or Directory. These are message counts, not bytes or router-link traversals: a multicast tree may share links, and a slow required response can still dominate latency even when k is small.

1
2
3
4
5
6
7
8
9
10
11
12

def counts(n, requester, candidates):
    peers = set(range(n)) - {requester}
    assert n >= 2 and requester in range(n)
    assert candidates <= peers
    return 2 * len(peers), 2 * len(candidates)
assert counts(2, 1, {0}) == (2, 2)
assert counts(2, 1, set()) == (2, 0)
assert counts(8, 0, {2, 5}) == (14, 4)
assert counts(8, 0, set(range(1, 8))) == (14, 14)
broadcast, filtered = counts(8, 0, {2, 5})
assert broadcast - filtered == 10
print("PASS: probes plus responses, 14 for broadcast and 4 for k = 2 at N = 8")

Related topics

• Shared course model

• Hybrids and Snoop Filters

• A required dirty data source