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,network size with 4 connections per node (ants),after 1 hop you can reach 4 nodes,network size with 4 connections per node (ants),after 2 hop you can reach 42 = 16 nodes,network size with 4 connections per node (ants),after 3 hop you can reach 43 = 64 nodes,network size with 4 connections per node (ants),in a network of n nodes the average hops to reach any other node is logc n where c is the mean number of connections per node,network size with 10 or more connections per node (mute),in a network of n nodes the average hops to reach any other node is logc n where c is the mean number of connections per node,network size with 10 or more connections per node (mute),in a network of n nodes the average hops to reach any other node is logc n log10 100 = 2 so with 100 nodes we have still just 2 hops in the mean case,network size with 10 or more connections per node (mute),if the net has less than 100 nodes then the following inequality will hold log10 n 2 iif n 100 so we have an average hops number to any node that is less than 2,ants - mute comparison,we said that ants extablishes less connections than mute in the mean case so it follows that the average hops number to reach any point of the net is higher even when the mute net has more nodes,ants - mute comparison,infact even with 100 nodes mute needs just 2 hops in the worst case to reach any point of the network ants instead with only 25 nodes needs 2,5 hops in the mean case,ants - mute comparison,if ants needs more hops than mute why ants tranfers are faster than mute ones?,ants - mute comparison,the reason have to be found in the differences of the protocols used mute uses a msg ack protocol for any transfer (data and control) so graphically,ants - mute comparison,to complete the sending of a message mute waits for the ask to be received,msg,ants - mute comparison,this produces a low throughput, in particular: throughput = latency = 4 hops (in the example),msg,ants - mute comparison,using this protocol for control message is ok. cause we need to be sure that a message has been delivered instead using this protocol for data transfer is known to be very inefficient!,ack,ants - mute comparison,infact to send a packet we need to wait that the former has been received. so we can send a packet only every 4 hops!,ack,ants - mute comparison,you can easily understand that adding up network congestion and low latency that are tipical of ad-hoc networks to this inefficient protocol is not a good idea!,ants - mute comparison,as we cant do anything to reduce latency we have to act in order to increase throughput and reduce net load (balancing the load over the highest number of connections available),ants - mute comparison,ants grants a higher throughput because it doesnt wait for acks before sending the next message in data tranfers,ants - mute comparison,ants grants an higher throughput because it doesnt wait for acks before sending the next message in data tranfers,file requestor,file owner,ants - mute comparison,first a msg ack exchange is performed to initiate the transfer process,file request,ants - mute comparison,once the owner received the request it sends an ack and begin the file transfer,file request,ants - mute comparison,file parts are sent without waiting for acks the source waits only for an initial ack (to assure that the receiver is ready to receive) this grants an high throughput and this is the main reason why ants is actually faster than mute,ack,ants - mute comparison,the initiate transfer message is the only one the requires the ack!,ack,initiate transfer,ants - mute comparison,the initiate transfer message is the only one the requires the ack!,initiate transfer,ants - mute comparison,the initiate transfer message is the only one the requires the ack!,ack,ants - mute comparison,the initiate transfer message is the only one the requires the ack!,ack,ants - mute comparison,once received the ack, the source can initiate transfer sending groups of parts and stopping only if acks are not received after n parts are sent. i.e. if there are more than n message sent still waiting for their acks. (window of size n),part 1,ants - mute comparison,so if acks are delivered without interruptions we get a constant flux of data!,part 2,part 1,ants - mute comparison,in order to improve the transfer we need not only a constant flux but also a way to distribute load over all the available connections from the source to the destination,part 3,part 2,ack 1,ants - mute comparison,ants let this happend because if more paths are present to our destination they are all used in a way proportional to the load of each one,part 4,part 3,ack 2,ack 1,ants - mute comparison,ants let this happend because if more paths are present to our destination they are all used in a way proportional to the load of each one,part 4,part 3,ack 2,ack 1,stalled,part 5,ants - mute comparison,ants let this happend because if more paths are present to our destination they are all used in a way proportional to the load of each one,part 4,ack 3,ack 2,ack 1,stalled,part 6,part 5,part 7,ants - mute comparison,this way of working applies to a single chunks transfer (each chunk is made up of 10 parts). if you consider that ants has multiple sources download you can understand why, even if a single chunk stalls, the node keeps a constant flux of data.,part 4,ack 3,ack 2,ack 1,stalled,part 6,part 5,part 7,ants - mute comparison,infact even if a source stalls, you can always have another source that is not sta
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