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Short-coming:
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          W(tγ) = γ 
  						   
  	     tγ				 
  	   -----					 
  	    RTT					 
    ==>	  2       =  γ 
  						   
  	    tγ					 
   <==>   ----- = log2(γ)	 
  	   RTT					 
  						   
   <==>	  tγ = RTT × log2(γ)                
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                     RTT sec
      |<-------------------------------->|
     W(t) packets
   ---+----------------------------------+---------           
       \                               / (ACKs)
        \                             /
	W(t) segments en route to destination
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(We don't need to be very precise, we only need a ball park figure)
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           W(t)
   B(t) = ------
           RTT
 TCP switches over to Congestion Avoidance at t = tγ / tγ\ |---| \RTT/ -1 + 2 T(tγ) = ----------- ....... (5) ln(2) From: γ = 2tγ/RTT ...... (3) We have: -1 + γ T(tγ) = ----------- ...... (6) ln(2) Congestion Avoidance phase: ============================== t - tγ W(t) = ------- + γ for t ≥ tγ RTT t - tγ γ B(t) = ------- + ------ for t ≥ tγ RTT2 RTT 0∫t B(t) dt = 0∫tγ B(t) dt + tγ∫t B(t) dt (Use Eq. (6)) = T(tγ) + tγ∫t B(t) dt -1 + γ = --------- + tγ∫t B(t) dt ln(2) -1 + γ = --------- + tγ∫t { [(t - tγ)/RTT2] + γ/RTT } dt ln(2) 
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(This relationship between T(t) and W(t) is similar to the relationship between velocity and distance traveled:)
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        RTT
   ρ = ------    
        RTT0
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where RTT0 is some pre-determined round trip time (e.g., 25 msec).
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where
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           +--
	   |		      RTT
	   |	ρ 2(ρ*t/RTT) = ----- 2(t/RTT0)                  . . . . . . . . (SS)
	   |		      RTT0
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  WH(t) =  |
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	   |	       t - tγ,0           RTT     t - tγ,0
	   |	ρ ( ρ ---------- + γ ) = ----- ( --------- + γ )      . . . . (CA)     
	   |	         RTT		  RTT0      RTT0
           +--
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                   +--
		   |	   1
		   |   = ----- 2(t/RTT0)                  . . . . . (SS)
		   |	  RTT0
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           WH(t)   |
  BH(t) = -----    |
           RTT     |
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		   |       1      t - tγ,0
		   |   = ----- ( --------- + γ )          . . . . (CA)     
		   |      RTT0     RTT0
		   +--
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You can see that the throughput function of Hybla TCP is INDEPENDENT of RTT
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            +--
	    |	WHi + α      . . . . (during SS)    
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   WHi+1 =   |
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	    |	WHi + β      . . . . (during CA)
	    +--
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such that:
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| I.e.: 
            +--
            |   WHi + 2ρ - 1      . . . . (SS)    
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   WHi+1 =   |
	    |	       ρ2
            |   WHi + -----       . . . . (CA)
	    |	       WHi
	    +--
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with WH0 = ρ (initial CWND at slow start)
 In a homework, 
   you can  verify 
	this fact 
 
	(it's a lot easier to verify the correctness of a solution
	than to find a solution)
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