Adjusted Value Calculation — adj

This function calculates an adjusted value over a time interval with optional discounting. This is useful for instances when adding cycles may not be desirable, so one can perform "cycle-like" calculations without needing cycles, offering performance speeds. See the vignette on avoiding cycles for an example in a model.

Usage

adj_val(
  curtime,
  nexttime,
  by,
  expression,
  discount = NULL,
  vectorized_f = FALSE
)

Arguments

curtime: Numeric. The current time point.
nexttime: Numeric. The next time point. Must be greater than or equal to curtime.
by: Numeric. The step size for evaluation within the interval.
expression: An expression evaluated at each step. Use .time as the variable within the expression.
discount: Numeric or NULL. The discount rate to apply, or NULL for no discounting.
vectorized_f: boolean, FALSE by default. If TRUE, evaluates the expression once using .time as a vector. If FALSE, it repeatedly evaluates the expression with time as a single value (slower).

Value

Numeric. The calculated adjusted value.

Details

The user can use the .time variable to select the corresponding time of the sequence being evaluated. For example, in curtime = 0, nexttime = 4, by = 1, .time would correspond to 0, 1, 2, 3. If using nexttime = 4.2, 0, 1, 2, 3, 4

Note

In resource-constrained models where event timing is unpredictable (e.g., patients waiting in resource queues), use accum_backwards = TRUE in run_sim() and compute adj_val(prevtime, curtime, ...) at each event reaction. This ensures both time boundaries are known and the age/time adjustment is exact, avoiding prospective errors when a resource event fires earlier than expected.

Examples

# Define a function or vector to evaluate
bs_age <- 1
vec <- 1:8/10

# Calculate adjusted value without discounting
adj_val(0, 4, by = 1, expression = vec[floor(.time + bs_age)])
#> [1] 0.25
adj_val(0, 4, by = 1, expression = .time * 1.1)
#> [1] 1.65
#same result since .time * 1.1 can be vectorized w.r.t time
adj_val(0, 4, by = 1, expression = .time * 1.1, vectorized_f = TRUE)
#> [1] 1.65

# Calculate adjusted value with discounting
adj_val(0, 4, by = 1, expression = vec[floor(.time + bs_age)], discount = 0.03)
#> [1] 0.2463061