hw4_solution

# hw4_solution - Transmission time per frame = x = 1000/R R =...

This preview shows pages 1–2. Sign up to view the full content.

6.12 7.3 a = Propagation Delay L R = 270 × 10 - 3 10 3 10 6 = 270 a. U = 1/(1 + 2a) = 1/541 = 0.002 b. Using Equation 7.6: U = W/(1 + 2a) = 7/541 = 0.013 c. U = 127/541 = 0.23 d. U = 255/541 = 0.47 7.4 A B: Propagation time = 4000 × 5 μ sec = 20 msec Transmission time per frame = 1000 100 × 10 3 = 10 msec The effective data rate: r1 = (3 frames)/(10 msec + 2x20 msec) = 60 kbps B C: Propagation time = 1000 × 5 μ sec = 5 msec

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Transmission time per frame = x = 1000/R R = data rate between B and C (unknown) The effective data rate: r2= (1 frame)/(1000/R + 2x5 msec) To avoid buffer overflow at B, there must be r2 >= r1, thus, R>=150 kbps. So, the minimum data rate between B and C is 150 kbps. 7.9 a. … 0 1 2 3 4 5 6 7 0 … b. … 0 1 2 3 4 5 6 7 0 … c. … 0 1 2 3 4 5 6 7 0 …...
View Full Document

## This note was uploaded on 11/02/2010 for the course EE 136 taught by Professor Kang xi during the Spring '10 term at NYU Poly.

### Page1 / 2

hw4_solution - Transmission time per frame = x = 1000/R R =...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online