*/
static void __bfq_bfqq_recalc_budget(struct bfq_data *bfqd,
				     struct bfq_queue *bfqq,
				     enum bfqq_expiration reason)
{
	struct request *next_rq;
	int budget, min_budget;

	min_budget = bfq_min_budget(bfqd);

	if (bfqq->wr_coeff == 1)
		budget = bfqq->max_budget;
	else /*
	      * Use a constant, low budget for weight-raised queues,
	      * to help achieve a low latency. Keep it slightly higher
	      * than the minimum possible budget, to cause a little
	      * bit fewer expirations.
	      */
		budget = 2 * min_budget;

	bfq_log_bfqq(bfqd, bfqq, "recalc_budg: last budg %d, budg left %d",
		bfqq->entity.budget, bfq_bfqq_budget_left(bfqq));
	bfq_log_bfqq(bfqd, bfqq, "recalc_budg: last max_budg %d, min budg %d",
		budget, bfq_min_budget(bfqd));
	bfq_log_bfqq(bfqd, bfqq, "recalc_budg: sync %d, seeky %d",
		bfq_bfqq_sync(bfqq), BFQQ_SEEKY(bfqd->in_service_queue));

	if (bfq_bfqq_sync(bfqq) && bfqq->wr_coeff == 1) {
		switch (reason) {
		/*
		 * Caveat: in all the following cases we trade latency
		 * for throughput.
		 */
		case BFQQE_TOO_IDLE:
			/*
			 * This is the only case where we may reduce
			 * the budget: if there is no request of the
			 * process still waiting for completion, then
			 * we assume (tentatively) that the timer has
			 * expired because the batch of requests of
			 * the process could have been served with a
			 * smaller budget.  Hence, betting that
			 * process will behave in the same way when it
			 * becomes backlogged again, we reduce its
			 * next budget.  As long as we guess right,
			 * this budget cut reduces the latency
			 * experienced by the process.
			 *
			 * However, if there are still outstanding
			 * requests, then the process may have not yet
			 * issued its next request just because it is
			 * still waiting for the completion of some of
			 * the still outstanding ones.  So in this
			 * subcase we do not reduce its budget, on the
			 * contrary we increase it to possibly boost
			 * the throughput, as discussed in the
			 * comments to the BUDGET_TIMEOUT case.
			 */
			if (bfqq->dispatched > 0) /* still outstanding reqs */
				budget = min(budget * 2, bfqd->bfq_max_budget);
			else {
				if (budget > 5 * min_budget)
					budget -= 4 * min_budget;
				else
					budget = min_budget;
			}
			break;
		case BFQQE_BUDGET_TIMEOUT:
			/*
			 * We double the budget here because it gives
			 * the chance to boost the throughput if this
			 * is not a seeky process (and has bumped into
			 * this timeout because of, e.g., ZBR).
			 */
			budget = min(budget * 2, bfqd->bfq_max_budget);
			break;
		case BFQQE_BUDGET_EXHAUSTED:
			/*
			 * The process still has backlog, and did not
			 * let either the budget timeout or the disk
			 * idling timeout expire. Hence it is not
			 * seeky, has a short thinktime and may be
			 * happy with a higher budget too. So
			 * definitely increase the budget of this good
			 * candidate to boost the disk throughput.
			 */
			budget = min(budget * 4, bfqd->bfq_max_budget);
			break;
		case BFQQE_NO_MORE_REQUESTS:
			/*
			 * For queues that expire for this reason, it
			 * is particularly important to keep the
			 * budget close to the actual service they
			 * need. Doing so reduces the timestamp
			 * misalignment problem described in the
			 * comments in the body of
			 * __bfq_activate_entity. In fact, suppose
			 * that a queue systematically expires for
			 * BFQQE_NO_MORE_REQUESTS and presents a
			 * new request in time to enjoy timestamp
			 * back-shifting. The larger the budget of the
			 * queue is with respect to the service the
			 * queue actually requests in each service
			 * slot, the more times the queue can be
			 * reactivated with the same virtual finish
			 * time. It follows that, even if this finish
			 * time is pushed to the system virtual time
			 * to reduce the consequent timestamp
			 * misalignment, the queue unjustly enjoys for
			 * many re-activations a lower finish time
			 * than all newly activated queues.
			 *
			 * The service needed by bfqq is measured
			 * quite precisely by bfqq->entity.service.
			 * Since bfqq does not enjoy device idling,
			 * bfqq->entity.service is equal to the number
			 * of sectors that the process associated with
			 * bfqq requested to read/write before waiting
			 * for request completions, or blocking for
			 * other reasons.
			 */
			budget = max_t(int, bfqq->entity.service, min_budget);
			break;
		default:
			return;
		}
	} else if (!bfq_bfqq_sync(bfqq)) {
		/*
		 * Async queues get always the maximum possible
		 * budget, as for them we do not care about latency
		 * (in addition, their ability to dispatch is limited
		 * by the charging factor).
		 */
		budget = bfqd->bfq_max_budget;
	}

	bfqq->max_budget = budget;

	if (bfqd->budgets_assigned >= bfq_stats_min_budgets &&
	    !bfqd->bfq_user_max_budget)
		bfqq->max_budget = min(bfqq->max_budget, bfqd->bfq_max_budget);

	/*
	 * If there is still backlog, then assign a new budget, making
	 * sure that it is large enough for the next request.  Since
	 * the finish time of bfqq must be kept in sync with the
	 * budget, be sure to call __bfq_bfqq_expire() *after* this
	 * update.
	 *
	 * If there is no backlog, then no need to update the budget;
	 * it will be updated on the arrival of a new request.
	 */
	next_rq = bfqq->next_rq;
	if (next_rq)
		bfqq->entity.budget = max_t(unsigned long, bfqq->max_budget,
					    bfq_serv_to_charge(next_rq, bfqq));

	bfq_log_bfqq(bfqd, bfqq, "head sect: %u, new budget %d",
			next_rq ? blk_rq_sectors(next_rq) : 0,
			bfqq->entity.budget);