{'fraction_non_alphanumeric': 0.0311587147030185, 'fraction_numerical': 0.006815968841285297, 'mean_word_length': 4.976744186046512, 'pattern_counts': {'":': 0, '<': 0, '<?xml version=': 0, '>': 0, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 0, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
Component-level redundancy is better than system-level redundancy for channel graphs

A well known engineering principle in introducing redundancy to a system is that component-level redundancy is better than system-level redundancy. However, the mathematical demonstration of this principle is usually for coherent systems constructed by one sequential synthesis and one parallel synthesis (the proof consists of showing that the parallel synthesis should be at the component level). In this article we demonstrate the same principle for systems constructed by parallel-sequential-parallel synthesis, one parallel synthesis at the component level and one at the system level. We show that the parallel synthesis at the component level should have more redundancy. The particular vehicle we use for this demonstration is the comparison of blocking probabilities for a 2-port network known as channel graphs. The crux of the mathematical demonstration is the lemma that [1 − Π (1 − P)]1/a decreases in a, 0 ≤ pi ≤ 1, i = 1,…, n.