Computer Information Systems

Question

Manu · Accepted Answer

Step 1: Given Data: Sender transmits data at a total rate of 150 Mbps (100 Mbps + 50 Mbps). The network consists of five routers labeled A, B, C, D, and E. Connections and data rates between the routers are labeled as 𝑥 1 , 𝑥 2 , 𝑥 3 , 𝑥 4 , x 1 ​ , x 2 ​ , x 3 ​ , x 4 ​ , and 𝑥 5 x 5 ​ . Connections and their corresponding data rates: Sender to A: 100 Mbps Sender to C: 50 Mbps A to B: 𝑥 1 x 1 ​ A to E: 𝑥 2 x 2 ​ C to E: 𝑥 3 x 3 ​ C to D: 𝑥 5 x 5 ​ B to E: 𝑥 4 x 4 ​ D to B: 𝑥 5 x 5 ​ D to Receiver: 120 Mbps Equation for Router A: Incoming data: 100 Mbps (from Sender) + 𝑥 2 x 2 ​ (from C) Outgoing data: 𝑥 1 x 1 ​ (to B) + 𝑥 2 x 2 ​ (to E) 100 + 𝑥 2 = 𝑥 1 + 𝑥 2 100 + x 2 ​ = x 1 ​ + x 2 ​ Simplifying, we get: 100 = 𝑥 1 100 = x 1 ​ Equation for Router B: Incoming data: 𝑥 1 x 1 ​ (from A) + 𝑥 5 x 5 ​ (from D) Outgoing data: 𝑥 2 x 2 ​ (to Receiver) + 𝑥 4 x 4 ​ (to E) 𝑥 1 + 𝑥 5 = 𝑥 2 + 𝑥 4 x 1 ​ + x 5 ​ = x 2 ​ + x 4 ​ Equation for Router C: Incoming data: 50 Mbps (from Sender) Outgoing data: 𝑥 3 x 3 ​ (to E) + 𝑥 5 x 5 ​ (to D) 50 = 𝑥 3 + 𝑥 5 50 = x 3 ​ + x 5 ​ Equation for Router D: Incoming data: 𝑥 5 x 5 ​ (from C) + 𝑥 4 x 4 ​ (from E) Outgoing data: 120 Mbps (to Receiver) 𝑥 5 + 𝑥 4 = 120 x 5 ​ + x 4 ​ = 120 Equation for Router E: Incoming data: 𝑥 2 x 2 ​ (from A) + 𝑥 3 x 3 ​ (from C) Outgoing data: 𝑥 4 x 4 ​ (to D) 𝑥 2 + 𝑥 3 = 𝑥 4 x 2 ​ + x 3 ​ = x 4 ​ Now, we substitute 𝑥 1 = 100 x 1 ​ = 100 into the second equation: 100 + 𝑥 5 = 𝑥 2 + 𝑥 4 100 + x 5 ​ = x 2 ​ + x 4 ​ We already have: 50 = 𝑥 3 + 𝑥 5 50 = x 3 ​ + x 5 ​ Solving these equations step by step: From 𝑥 5 + 𝑥 4 = 120 x 5 ​ + x 4 ​ = 120 : 𝑥 4 = 120 − 𝑥 5 x 4 ​ = 120 − x 5 ​ Substitute 𝑥 4 = 120 − 𝑥 5 x 4 ​ = 120 − x 5 ​ into 100 + 𝑥 5 = 𝑥 2 + 𝑥 4 100 + x 5 ​ = x 2 ​ + x 4 ​ : 100 + 𝑥 5 = 𝑥 2 + ( 120 − 𝑥 5 ) 100 + x 5 ​ = x 2 ​ + ( 120 − x 5 ​ ) 100 + 𝑥 5 = 𝑥 2 + 120 − 𝑥 5 100 + x 5 ​ = x 2 ​ + 120 − x 5 ​ 100 + 𝑥 5 + 𝑥 5 = 𝑥 2 + 120 100 + x 5 ​ + x 5 ​ = x 2 ​ + 120 100 + 2 𝑥 5 = 𝑥 2 + 120 100 + 2 x 5 ​ = x 2 ​ + 120 2 𝑥 5 = 𝑥 2 + 20 2 x 5 ​ = x 2 ​ + 20 𝑥 2 = 2 𝑥 5 − 20 x 2 ​ = 2 x 5 ​ − 20 From 50 = 𝑥 3 + 𝑥 5 50 = x 3 ​ + x 5 ​ : 𝑥 3 = 50 − 𝑥 5 x 3 ​ = 50 − x 5 ​ Substitute 𝑥 3 = 50 − 𝑥 5 x 3 ​ = 50 − x 5 ​ into 𝑥 2 + 𝑥 3 = 𝑥 4 x 2 ​ + x 3 ​ = x 4 ​ : 𝑥 2 + ( 50 − 𝑥 5 ) = 𝑥 4 x 2 ​ + ( 50 − x 5 ​ ) = x 4 ​ ( 2 𝑥 5 − 20 ) + 50 − 𝑥 5 = 120 − 𝑥 5 ( 2 x 5 ​ − 20 ) + 50 − x 5 ​ = 120 − x 5 ​ 𝑥 5 + 30 = 120 − 𝑥 5 x 5 ​ + 30 = 120 − x 5 ​ 2 𝑥 5 = 90 2 x 5 ​ = 90 𝑥 5 = 45 x 5 ​ = 45 Now substitute 𝑥 5 = 45 x 5 ​ = 45 back into the equations: 𝑥 4 = 120 − 𝑥 5 = 120 − 45 = 75 x 4 ​ = 120 − x 5 ​ = 120 − 45 = 75 𝑥 2 = 2 𝑥 5 − 20 = 2 ⋅ 45 − 20 = 90 − 20 = 70 x 2 ​ = 2 x 5 ​ − 20 = 2 ⋅ 45 − 20 = 90 − 20 = 70 𝑥 3 = 50 − 𝑥 5 = 50 − 45 = 5 x 3 ​ = 50 − x 5 ​ = 50 − 45 = 5 Final Answer:: { 𝑥 1 = 100 𝑥 2 = 70 𝑥 3 = 5 𝑥 4 = 75 𝑥 5 = 45

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