Jonathan'Nguy'603'799'761'Homework'4'1.'These'ACK'packets'do'not'need'sequence'numbers'because'the'rdt3.0'protocol'uses'a'stopEandEwait'procedure,'so'all'the'packets'must'be'in'order'when'they'are'sent'and'ACK’d.'''2.''a)'If'the'receiver'is'expecting'the'sequence'k,'and'the'window'size'is'3,'the'latest'ACK'is'must’ve'sent'could'have'been'k"#1.'Because'the'window'size'is'3,'the'earliest'ACK'that'could'be'sent'is'k"–"3"–"1,'which'is'k"–"4.'With'these'two,'we'know'that'the'possible'ACK’s'being'sent'back'is'between'k"–"4"and'k"–"1."'b)'If'the'receiver'is'expecting'the'sequence'k'next,'that'means'it'must'have'sent'ACK’s'up'until'k"–"1.'If'none'of'those'ACK’s'have'been'received'by'the'sender,'the'sender’s'sequence'numbers'would'be'between'k"–"N"="k"–"3"and'k"–"1.'Next,'if'the'sender'received'all'of'those'ACK’s,'that'means'that'the'next'sequence'numbers'would'be'between'k"and'k"+"N"–"1="k"+"2.''3.''a)'The'minimum'possible'latency'='2'RTT’s'+'size'of'object/transmission'rate.''The'minimum'W'that'achieves'that'is:'!"" + !/!!/!''Plugging'in'28'kbps'to'the'first'equation,'we'get:'2 ∗ .1 +1000003500= 28.77!!"#$%&!!''To'get'the'minimum'W,'we'plug'in'the'values'to'the'second'equation:'. 1 + 536/(280008)536/(280008)= 1.6'So'the'minimum'window'size'is'2.''b)'The'minimum'latency'for'100'kbps'is:'2 ∗ .1 +1000001000008= 2 ∗ .1 + 8 = 8.2!!"#$%&!'To'get'the'minimum'window'size:'. 1 + 536/(1000008)536/(1000008)= 3.33'So'the'minimum'window'is'4.''''c)'The'minimum'latency'for'10'Mbps'is:'2 ∗ .1 +100000100000008= 2 ∗ .1 + .08 = .28!!"#$%&!'To'get'the'minimum'window'size:'. 1 + 536/(100000008)536/(100000008)=
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