Bluetooth: Minor optimization for multiple tx connections

When assigning tx quotas for multiple connections, bump fractional amounts to the next higher integer. This can reduce the number of iterations through the tx scheduler. For example, if 5 avail acl buffers are to be used by 3 transmitting connections, the current scheduler will produce quotas of 1, 1, 1, 1, 1. This optimization will instead produce quotas of 2, 2, 1 (ie., 2 less iterations of the tx scheduler).

Signed-off-by Peter Hurley <[email protected]>

---
 net/bluetooth/hci_core.c |    5 ++---

 1 files changed, 2 insertions(+), 3 deletions(-)

net/bluetooth/hci_core.c +1 -1

index 47a31fa..0defa83 100644
@@ -1826,7 +1826,7 @@ static inline struct hci_conn *hci_low_sent(struct hci_dev *hdev, __u8 type, int
        }
 
        if (conn) {
-               int cnt, q;
+               int cnt;
 
                switch (conn->type) {

               case ACL_LINK:

net/bluetooth/hci_core.c +1 -3

@@ -1844,8 +1844,7 @@ static inline struct hci_conn *hci_low_sent(struct hci_dev *hdev, __u8 type, int
                        BT_ERR("Unknown link type");
                }
 
-               q = cnt / num;
-               *quote = q ? q : 1;
+               *quote = (cnt + (num - 1)) / num;
        } else
                *quote = 0;
 
--

1.7.4.1

Thanks. Yes, in the current scheduler a connection can be scheduled multiple times per tasklet run.

For example, supposed that acl_cnt == 3 and 2 connections have packets in their queue. Let's assume for the sake of simplicity that no connections currently have unacked packets (c->sent == 0). Then (with the current scheduler),

 1st iteration: conn list walked => conn A found (equal unacked) quote = 3/2 => 1 for conn A
                conn A now has 1 unacked packet (after send)

 2nd iteration: conn list walked => conn B found (lowest unacked) quote = 2/2 => 1 for conn B
                conn B now has 1 unacked packet (after send)

 3rd iteration: conn list walked => conn A found (equal unacked) quote = 1/2 => 1 for conn A
                conn A now has 2 unacked packets (after send)

With the proposed change, only 2 iterations occur:

 1st iteration: conn list walked => conn A found (equal unacked) quote = 3/2 => 2 for conn A
                conn A now has 2 unacked packets (after send)

 2nd iteration: conn list walked => conn B found (lowest unacked) quote = 1/2 => 1 for conn B
                conn B now has 1 unacked packet (after send)

However, even with this change it may be that connections are scheduled multiple times per tasklet. This can happen if a connection still has equal or fewer unacked packets even after spending it's quota.

But I didn't see in your proposal where you assume that an HCI channel can only be scheduled once (or at least where it breaks down if true).