feat: shuffle accounts within same sort group to prevent thundering herd
Add post-sort shuffle for accounts with identical (priority, loadRate, lastUsedAt) to break deterministic ordering when concurrent requests read the same scheduler snapshot. Applies to both Antigravity and OpenAI scheduling paths, plus the sortAccountsByPriorityAndLastUsed helper. Keeps upstream CallCount/ModelLoadInfo scheduling intact; shuffle is additive and only randomises within equivalent-rank groups.
This commit is contained in:
@@ -1218,6 +1218,7 @@ func (s *GatewayService) SelectAccountWithLoadAwareness(ctx context.Context, gro
|
||||
return a.account.LastUsedAt.Before(*b.account.LastUsedAt)
|
||||
}
|
||||
})
|
||||
shuffleWithinSortGroups(routingAvailable)
|
||||
|
||||
// 4. 尝试获取槽位
|
||||
for _, item := range routingAvailable {
|
||||
@@ -2027,6 +2028,79 @@ func sortAccountsByPriorityAndLastUsed(accounts []*Account, preferOAuth bool) {
|
||||
return a.LastUsedAt.Before(*b.LastUsedAt)
|
||||
}
|
||||
})
|
||||
shuffleWithinPriorityAndLastUsed(accounts)
|
||||
}
|
||||
|
||||
// shuffleWithinSortGroups 对排序后的 accountWithLoad 切片,按 (Priority, LoadRate, LastUsedAt) 分组后组内随机打乱。
|
||||
// 防止并发请求读取同一快照时,确定性排序导致所有请求命中相同账号。
|
||||
func shuffleWithinSortGroups(accounts []accountWithLoad) {
|
||||
if len(accounts) <= 1 {
|
||||
return
|
||||
}
|
||||
i := 0
|
||||
for i < len(accounts) {
|
||||
j := i + 1
|
||||
for j < len(accounts) && sameAccountWithLoadGroup(accounts[i], accounts[j]) {
|
||||
j++
|
||||
}
|
||||
if j-i > 1 {
|
||||
mathrand.Shuffle(j-i, func(a, b int) {
|
||||
accounts[i+a], accounts[i+b] = accounts[i+b], accounts[i+a]
|
||||
})
|
||||
}
|
||||
i = j
|
||||
}
|
||||
}
|
||||
|
||||
// sameAccountWithLoadGroup 判断两个 accountWithLoad 是否属于同一排序组
|
||||
func sameAccountWithLoadGroup(a, b accountWithLoad) bool {
|
||||
if a.account.Priority != b.account.Priority {
|
||||
return false
|
||||
}
|
||||
if a.loadInfo.LoadRate != b.loadInfo.LoadRate {
|
||||
return false
|
||||
}
|
||||
return sameLastUsedAt(a.account.LastUsedAt, b.account.LastUsedAt)
|
||||
}
|
||||
|
||||
// shuffleWithinPriorityAndLastUsed 对排序后的 []*Account 切片,按 (Priority, LastUsedAt) 分组后组内随机打乱。
|
||||
func shuffleWithinPriorityAndLastUsed(accounts []*Account) {
|
||||
if len(accounts) <= 1 {
|
||||
return
|
||||
}
|
||||
i := 0
|
||||
for i < len(accounts) {
|
||||
j := i + 1
|
||||
for j < len(accounts) && sameAccountGroup(accounts[i], accounts[j]) {
|
||||
j++
|
||||
}
|
||||
if j-i > 1 {
|
||||
mathrand.Shuffle(j-i, func(a, b int) {
|
||||
accounts[i+a], accounts[i+b] = accounts[i+b], accounts[i+a]
|
||||
})
|
||||
}
|
||||
i = j
|
||||
}
|
||||
}
|
||||
|
||||
// sameAccountGroup 判断两个 Account 是否属于同一排序组(Priority + LastUsedAt)
|
||||
func sameAccountGroup(a, b *Account) bool {
|
||||
if a.Priority != b.Priority {
|
||||
return false
|
||||
}
|
||||
return sameLastUsedAt(a.LastUsedAt, b.LastUsedAt)
|
||||
}
|
||||
|
||||
// sameLastUsedAt 判断两个 LastUsedAt 是否相同(精度到秒)
|
||||
func sameLastUsedAt(a, b *time.Time) bool {
|
||||
switch {
|
||||
case a == nil && b == nil:
|
||||
return true
|
||||
case a == nil || b == nil:
|
||||
return false
|
||||
default:
|
||||
return a.Unix() == b.Unix()
|
||||
}
|
||||
}
|
||||
|
||||
// selectByCallCount 从候选账号中选择调用次数最少的账号(Antigravity 专用)
|
||||
|
||||
Reference in New Issue
Block a user