file:/Users/ivan/Homes/master/Documents/Webpages/It%20is%20all%20about%20me%20April%202007/It%20is%20all%20about%20me/ivan.unixdaemons.com/blog/index5d38.html?p=87 if I was to give you a feedback channel such that the receiver can tell the sender which bits arrived correctly and which didn’t this will not imporve the capacity! Here is why: Lets use the intuitive protocol. STEP 1: I send n bits. You receive (1-p)n bits and tell me which bits you didn’t receive. STEP 2: I send the broken bits from STEP 1, there are pn of them. You receive (1-p)pn of them. and you inform me again of the ones that you missed. STEP N: I send the bits that you didn’t receive in step N-1. After a while you will have received all the bits. So lets check the rate of our new protocol_with_feedback: EQN missing…i file:/Users/ivan/Homes/master/Documents/Webpages/It%20is%20all%20about%20me%20April%202007/It%20is%20all%20about%20me/ivan.unixdaemons.com/blog/latexrender/mimetexa3dc.gif?4$Rate_{new}%20=%20\frac{total.inf.received}{total.bits.sent.down.the.line}=

file:/Users/ivan/Homes/master/Documents/Webpages/It%20is%20all%20about%20me%20April%202007/It%20is%20all%20about%20me/ivan.unixdaemons.com/blog/latexrender/mimetex2721.gif?4$\qquad\qquad=%20\frac{%20(1-p)n%20+%20(1-p)pn%20+%20(1-p)p^2n%20+\ldots%20}{n%20+%20pn%20+%20p^2n%20+%20\ldots}%20=%201-p = 1-p Crazy!