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(ELEC315)[2010](s)midterm~ee_pax^_10325.pdf
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315.Midterm.Exam.
Note:..(a).Answers.dont.count.unless.explanations.are.provided..
(b).No.partial.credit.will.be.given..So.your.answer.must.be.precise.and.your.explanation.must.be.unambiguous...
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1.
(2.pts).Assume.that.the.unit.of.the.TCP.window.is.byte..The.RTT.between.a.sender.and.a.receiver.of.a.TCP.channel.is.about.100.ms..What.is.the.maximum.bandwidth.achievable.for.this.TCP.channel?...

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Solution:..
The.TCP.window.is.limited.by.the.receivers.buffer.size.and.network.congestion.situation..On.considering.the.maximum.bandwidth.we.assume.the.buffer.size.and.congestion.is.good.enough.and.can.be.ignored..Therefore,.the.maximum.bandwidth.is.limited.by.the.16\bit.receive.window.field.in.a.TCP.segment.header..Since.2^16\1.=.65536\1.=.65535,.the.maximum.bandwidth.achievable.is.(Maximum.window.size)./.RTT.=.655350.Bytes.per.second..
Marking.scheme:.
Note:.if.you.get.655360,.you.get.no.marks..If.you.turn.the.result.into.bps,.if.the.answer.is.right,.you.still.get.full.mark..No.partial.mark.for.this.question..
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2.
(2.pts).John.performed.the.following.command.nslookup..Ctype=a.cs.ust.hk.and.got.the.following.result..Is.it.a.domain.name.or.a.machine.name?.Explain...

Server: dns03.netvigator.com
Address: 203.198.23.208

Name: cs.ust.hk
Address: 143.89.41.167



Solution:..
Generally,.if.no.special.implementation.is.used,.when.you.use.nslookup.Ctype=awith.a.machine.name.(or.host.name),.an.IP.address.will.be.return..If.you.try.with.a.domain.name,.no.IP.will.be.return,.for.example.ece.ust.hk..However,.cs.ust.hk.is.actually.a.domain.name.but.their.department.basically.assigns.an.A.record.for.their.domain.and.this.IP.address.is.their.MX.server..In.this.case,.everyone.will.get.these.two.points..
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Marking.scheme:.
All.students.get.2.points.here.no.matter.what.you.write..

3.
(3.pts).A.filter.type.firewall.blocks.some.applications.based.on.certain.bits.in.the.protocol..Suppose.a.company.uses.a.proxy.server.(such.as.the.one.shown.in.Problem.4).to.block.all.HTTP.connections.originating.from.the.company..What.fields.in.the.IP/TCP.header.and.which.bits.of.these.fields.will.be.examined.by.the.proxy.server?.Assume.that.all.HTTP.servers.use.TCP.port.80..

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Solution:.
Firstly.we.should.check.the.IP.header.to.see.whether.it.carries.TCP.segment..So.we.need.to.check.the.Protocol.field.in.the.IP.header.to.see.whether.it.is.6.(indicating.TCP)..
Then.we.need.to.check.the.TCP.header..We.need.to.check.the.destination.port.field.to.see.whether.it.is.80..Since.we.want.to.block.the.connections.initialed.from.the.company,.we.should.block.it.in.the.first.step.of.three\way.handshake..So.we.need.to.check.whether.SYN.field.is.1.and.ACK.field.is.0..
Marking.scheme:.
Once.you.get.3.of.them,.you.get.full.mark.C.3.points..1.point.for.1.key.point.you.write..If.you.only.say,.like.should.check.11\16.bits,.you.get.no.mark..If.you.only.say.port.or.source.port,.it.is.different.from.destinati