Sentence examples for frequency scenario from inspiring English sources

High demand frequency scenario (f = 1): One demand element is assigned to each day ( t in TW_{j} ).

The detailed assignment of demand elements to due dates within the feasible time window [defined in Step 1 (a)] is accomplished as follows: High demand frequency scenario (f = 1): One demand element is assigned to each day ( t in TW_{j} ).

For this purpose, we propose that all parameters (whatever frequency scenario is considered) be modeled by the same 6-parameter function f that has the following expression: f σ d B = a 1 + a 2 ⋅ 1 - σ d B a 3 1 + σ d B a 3 a 4 1 a 4 ⋅ 1 1 + σ d B a 5 a 6, (21).

We therefore used two different imputation strategies (a low and a high sexual frequency scenario) for partners four and upwards.

36 The low sexual frequency scenario, that very likely underestimates UAS frequency, yielded an R03M of one already at a per act transmission probability of 3.3%.

We imputed for each individual with >3 UAS partners, in the low frequency scenario, one UAS act per partnership four and upwards (the logically smallest possible value) and for the high frequency scenario the smallest reported number of UAS acts among his three partners for which UAS act data were reported.

For a small M, it would be better to employ ICI cancellation schemes for high normalized Doppler frequency scenarios.

(b) The average size of demand elements is set to D f)=Y/n(f) hours depending on f = 1, 3, 7 for the individual demand frequency scenarios.

On the other hand, the improved multidimensional folding (IMDF) scheme, which uses eigenvector instead of eigenvalue for estimating frequencies, can resolve identical frequency scenarios by introducing the randomness on the sample data [18, 19].

Medium and low demand frequency scenarios (f = 3 and f = 7): Demand elements are randomly assigned to periods ( t in TW_{j} ) until the number of demand elements determined in Step 1 (b) is reached.

With f representing the demand frequency factor for distinguishing between low, medium and high demand frequency, the average number of generated demand elements is determined as n(f) = N/f giving a total of n(1) = 3840, n(3) = 1280, and n(7) = 549 demand elements on average generated in the three demand frequency scenarios with f = 1, 3, and 7, respectively.