Sentence examples for states busy from inspiring English sources

From Fig. 1b, the channel may be in one of the two states: busy (B) or free (F).

Therefore, for simplicity of analysis, we consider only Markov chain in this paper)with states (busy: the channel is occupied by primary users and cannot be used by secondary users) and (idle: there is no primary user over this channel).

Recall that the channel status of contention period (initial ranging or bandwidth request) can be generally divided as two states: busy and idle states as shown in Figure 3.

A corrected index appears in this week's finance section.Dirty air in UtahSIR – Utah is a great state ("Busy bees", August 31st) though our economic growth comes partly at the expense of residents' health.

λ1, λ2, λ3: Mean arrival rate of the units when the server is in idle state, busy state, and under repair server, respectively.

Based on 802.11 protocols, a WLAN station applies the clear channel assessment (CCA) mechanism to detect the channel state (busy or idle).

The first is to develop the explicit expressions for steady-state availability, steady-state busy period due to failure units in subsystems and steady-state busy period due to failure of supporting unit and profit function for the three systems under study.

The notations used in the formulation of the model are as follows: λ1, λ2, λ3: Mean arrival rate of the units when the server is in idle state, busy state, and under repair server, respectively.

The stationary probabilities of state idle and state busy are thus given by P I = Pr { Θ [ t ] = 0 } = β α + β, P B = Pr { Θ [ t ] = 1 } = α α + β. (2).

The customers arrive in batches in the system following the Poisson distribution with state-dependent arrival rate λ I depending on server's status; 'I' takes value 1, 2, 3, 4 and 5 when the server is in retrial state, busy state, setup state, repair state and in vacation state, respectively.

The explicit expressions for the steady-state availability, state busy period of repairman due to failure of units A k and B k, busy period of repairman due to failure of supporting are as follows: A V 1 1 = P 0 + P 1 + P 2 + P 3 = b 1 b 2 (4) B V 1 1 = P 1 + P 2 + P 3 + P 5 + P 6 = b 3 b 2 (5) B V 1 1 ∗ = P 4 = b 4 b 2. (6).