Based on the caching mechanism described in the case study, calculate the resulting summary key vector and summary value vector.

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A moving average-based cache compresses a window of recent key-value pairs into a single summary pair. As illustrated, the four most recent key vectors ($\mathbf{k}_{i-3}$ to $\mathbf{k}_{i}$) and value vectors ($\mathbf{v}_{i-3}$ to $\mathbf{v}_{i}$) are averaged independently. This process creates a memory component consisting of one summary key and one summary value, effectively reducing the memory size from four pairs to one (a size of 1x2). The specific calculations are:
$$ \text{Memory Key} = \frac{\mathbf{k}_{i-3}+\mathbf{k}_{i-2}+\mathbf{k}_{i-1}+\mathbf{k}_{i}}{4} $$
$$ \text{Memory Value} = \frac{\mathbf{v}_{i-3}+\mathbf{v}_{i-2}+\mathbf{v}_{i-1}+\mathbf{v}_{i}}{4} $$

Example of a Moving Average-based Cache

Calculating a Compressed Memory State

A system's memory component compresses the four most recent key-value pairs into a single summary pair by independently averaging their respective vectors. What is the most significant trade-off inherent in this compression technique?

A system uses a moving average-based cache that summarizes the four most recent key-value pairs into a single summary pair. The current summary is calculated from pairs at time steps `t-4`, `t-3`, `t-2`, and `t-1`. When a new key-value pair arrives at time step `t`, how is the cache updated?

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