An electrical engineer is configuring 5 distinct smart meters to be installed at different substations arranged in a line. Each meter is assigned one of two power modes: high-efficiency (H) or standard (S). How many assignments are there such that no two adjacent meters use high-efficiency mode?

An electrical engineer is configuring 5 distinct smart meters to be installed at different substations arranged in a line. Each meter is assigned one of two power modes: high-efficiency (H) or standard (S). How many assignments are there such that no two adjacent meters use high-efficiency mode?

["Title: Valid Configurations for 5 Smart Meters with No Adjacent High-Efficiency Settings", "When deploying 5 smart meters across linearly arranged substations, electrical engineers face critical decisions in configuring each meter’s power mode. Each meter must be set to either High-Efficiency (H) or Standard (S), but a key constraint applies: no two adjacent meters may operate in High-Efficiency mode. This restriction prevents localized power surges and improves grid stability.", "We model this problem as counting the number of valid binary sequences of length 5 (with H = 1 and S = 0) such that no two consecutive 1s appear.", "Let ( a_n ) denote the number of valid configurations for ( n ) meters under the adjacency rule. This is a classic combinatorial problem solved via recurrence relations.", "Recurrence logic:\nConsider the last meter’s mode:", "- If the ( n )-th meter is in Standard (S), the first ( n-1 ) meters can form any valid configuration of length ( n-1 ): ( a_{n-1} ) ways.\n- If the ( n )-th meter is in High-Efficiency (H), the ( (n-1) )-th meter must be in S to avoid two adjacent Hs. Then the first ( n-2 ) meters form any valid configuration: ( a_{n-2} ) ways.", "Thus, the recurrence is:\n[\na_n = a_{n-1} + a_{n-2}\n]", "This is the Fibonacci recurrence. We now establish initial conditions:", "- ( a_1 ): One meter can be H or S → 2 valid configurations → ( a_1 = 2 )\n- ( a_2 ): Valid pairs: SS, SH, HS (HH invalid) → 3 valid → ( a_2 = 3 )", "Using the recurrence:", "[\n\begin{align}\na_3 &= a_2 + a_1 = 3 + 2 = 5 \\na_4 &= a_3 + a_2 = 5 + 3 = 8 \\na_5 &= a_4 + a_3 = 8 + 5 = 13 \\n\end{align}\n]", "Therefore, the total number of valid configurations for 5 smart meters, ensuring no two adjacent high-efficiency modes, is ( \boxed{13} ).", "This result helps engineers plan reliable deployments, leveraging combinatorial logic to avoid power conflicts across substations."]

Related Articles

Trending Articles