Theoretical and Applied Informatics
ISSN1896–5334
V ol.18(2006),no.3
pp.213–236
De?ection Routing in an All-Optical Network in2D Grid:Performance
Evaluation
A.B ORRERO M.a,b,F.Q UESSETTE b
a Laboratoire PRiSM Universitéde Versailles
45,Avenue desétats-Unis,78035Versailles Cedex,France
b Laboratorio Escuela Básica,Facultad de Ingeniería Universidad de Los Andes
Núcleo La Hechicera,5101Mérida,Venzuela
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success is based especially on the promise of data transmission rates with several or-ders of magnitude higher than current networks.As it is well known,in an all-optical networks the signals may cross the net from end to end in purely optical form,without conversions to the electronic domain and back,while still in transit.The high speeds (to which we have referred above),are presented by maintaining the signals in optical form,in all the route by the network,(thereby avoiding the prohibitive overhead of con-version to and from the electronic form).Among other characteristics that support this success,are the low latency and the large bandwidth,which seemingly only optics can provide.Another change that has happened and should be considered,is the diminution and even elimination of the use of intermediate storage devices.In the all-optical net-works domain,this type of devices,"buffers",has been implemented with?xed delay lines or recirculation loops.It is indeed for facing these requirements that have arisen new routing paradigms,new routing strategies in all-optical networks communications. The present work has as object the study of some of these strategies,as the De?ection Routing[2,4],also known as"Hot-potato"routing[8].Traditionally,this method has been characterized to avoid the intermediate storage messages(message buffering).It means that,in De?ection routing,there are no buffers to store in packets at the switch nodes.Each node forwards each packet closer to its destination whenever possible(that is,whenever the desired links are not already assigned to other packets).If two or more packets want to use the same link at the same time,only one packet is allowed to do so. Individual communication links can not support more than one packet at the same time. Time is divided in slots in such a way that at most one packet can traverse any link in each time slot.The remaining packets are sent out over other free edges,without wait-ing for the preferred outbound link.Each packet routed,is constantly transferred until it reaches its?nal predetermined destination.So for a packet,if its favorable direction(s) is(are)already taken by other packet(s)of the current node,it is forced to take a"not requested"direction,then it will be sent"towards"a node that is not on the shortest path over its destination node,we will say whereas it is"de?ected".A packet that arrives at a node other than its destination during one slot is transmitted to another node during the next slot.It goes the same way for all the packets.They are always forced to move, and to keep moving through the network,sometimes further away from their?nal des-tination,until they can reach it.Of course,de?ected packets do not advance directly towards their destination,and it is possible that they remain moving without ending up never reaching it,what becomes the well-known livelock problem.It refers to packets circulating in the network,"wandering the network",without making any progress to-wards their destination.Another problem of very similar sort,but totally opposed sense, is the also well-known,deadlock.Deadlock is the blockage of the system since multiple packets expect mutually,they require mutually dependent channel resources then pack-ets are each waiting for the other one to liberate the busy resource to proceed.The result
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is that packets do not move.In general,the process cannot conclude because it waits for the resources that will never be assigned to him.Nowadays,recent investigations have demonstrated that these last mentioned problems,are excluded of this type of strategies and the?rst ones do not represent a serious inconvenience[11].
In this paper we present the performance evaluation of a de?ection routing strategy for the emerging generation of networks,all-optical networks.Our investigation is di-rected to evaluate performance of a de?ection routing method,called"Scale Routing", on a synchronous network in2D grid.We have used this algorithm to try to?nd an op-timal routing.Here,when we refer to optimal route,it does not mean the shortest path, (although in many cases it comes near to it),nor the path with the minimum number of de?ections.But it means the route that provides to each step,in each time unit or slot,in every node,the biggest number of selection possibilities.The packet"will always pre-fer"the route that allows it to have greater amount of options to choose.Nevertheless, as we will see later,the most important characteristic of this type of routing is that in every node the best decision is always taken,which means to send the packet towards the route that makes it closer to its destination.
The?rst part of this article shows the model designed,then the scale routing https://www.doczj.com/doc/be18819401.html,ter some heuristics were introduced to choose the packet to be de?ected,when any customer has to be de?ected,based on its remaining distance until destination node and the number of de?ections that it has already had.We have also tested the incidence on the routing by changing the traf?c of the network,modifying the destination given to the packets.It was considered as an uniform distribution,then a non uniform distri-bution directed towards a corner and lastly a non uniform distribution directed towards the center of the grid.Finally the simulations results and some statistics and curves are presented to be analyzed and to compare the performances of the different heuristics applied to the routing strategy.
In section2.we can see the simulators design,in section2.1.some speci?cations on the model are presented,in section3.1.we describe the routing strategy,Scale Routing. Section3.1.1.describes the de?ection algorithm used.Section3.2.shows a related ex-ample and how the algorithm https://www.doczj.com/doc/be18819401.html,ter in section3.we describe the heuristics added to choose the customers to be de?ected.Section4.shows the effect on the routing by changing the traf?c in the network and also in section5.4.the diverses performances with all the changes in the selection of the packet to be de?ected,(implemented heuris-tics),and are also presented the results obtained by simulation,curves and statistics. Finally section6.presents some conclusions and discussions about future works.
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2.Simulators Design
We have implemented it with QNAP2II modelling tool.We have chosen QNAP2for the implementation of this algorithm,because it is a simulator especially developed for the description,the manipulation and the resolution of queuing systems.However the use of this tool makes easier the modelling and the simulation of a synchronous network (as the one that we treated here)than the one of a general language.Objects represented in QNAP are queues and customers[3].In our model we use different queues:one containing the customers coming from the network which have not been routed yet, another one containing the customers coming from outside and also before the routing takes place and another one that places in stock the customers after the routing.Then we use a global queue to interact with the mentioned before,which is a clock that allows the whole work to have a true synchronous system.
In the implemented algorithm,a packet is sent to the most convenient node,accord-ing to the Scale routing.We must emphasize on the fact that every customer contains information de?ning its origin and destination coordinates and the number of phases that it has already done in the network as well as the number of times it has been de?ected. The generation of customers is done according to a Poisson process.In the next section we will show some speci?cations of the designed simulator.
2.1.Model
This model has been created for a2D grid network,with a number N of nodes,equal to100,being the maximal distance among its nodes,a diameter D of the network,equal to10.And for the model,we have used a saturation strategy,(as soon as a customer leaves the system,a new customer enters it instantaneously),it means that the external line has contained waiting customers permanently.It is often interesting to study the system saturation,it means,when there are permanently waiting processes on the exter-nal storage unit.It gives,an upper bound on the throughput of the system:maximum number of process capable to come out of the system by time unit at maximum?ow of the system,[4].The network is time slotted.It means that,in this model,the time is discrete,it is divided in slots in such a way that at most one packet can cross any link in each time slot.Each packet may pass through several nodes on the path to its?nal destination,but it can not be stored at the intermediate nodes between time steps,be-cause of all-optical networks properties.Every vertex of the grid is called node and the packets moving are called customers.Every customer has an origin and a destination node,that are always distinct.The model always veri?es that the destination node is different from the source.We will call current node of a packet at a given time,the node where the packet is at the beginning of the corresponding slot at that date.We call fa-vorable direction for a customer,a routing direction that brings closer the customer to its
217 destination.We call available direction,a routing direction not yet assigned.A customer can have1or2favorable directions,(2being the maximum).The customer has only one favorable direction if,its current node is on the same line or on the same column as its destination node.It will have2favorable directions in all other cases.for this?rst approach we have used,a2D grid network considering its simplicity,and because of its good approximation to a real topology.
3.Heuristics and Implementation
The principal question was,which customer must be de?ected,then we have intro-duced some heuristics to choose the customers to be de?ected.The?rst heuristics we have used is based on the distance separating the current node to its?nal destination.In the second heuristic,the choice of the customer is based on the number of de?ections it already had.We observe that in order to do the tests by simulation we always work with a saturated network,with the objective of always having customers entering every node, at each time slot.When a customer leaves the network,another one is immediately gen-erated to take its place.The fact remains that a customer is always place in an available position.
We speak about heuristics because,it belongs to the process to know that it has to learn by trying rather than following some pre-established formula.This term has two interpretations,[4]:
1.Describing an approach to learn by trying without necessarily having an organized
hypothesis,or way of demonstrating that the results prove or disprove the hypoth-esis.Referring to a method of solving problems by intelligent trial and error. 2.It belongs to the general knowledge base resulting from experience or arguments
derived from experience.
In general,"heuristic",is a word that refers to a technique or proposal which is used as a tool,for?nding out more about the?eld of study in question.In fact, we have used it as an instrument to look for a better route for the packets and to investigate on the subject of de?ection routing.We will describe the heuristics used,in the same order that we have applied them.
The?rst one that we have applied,has been the remaining distance and after we have immediately added the number of de?ections.
3.1.Scale Routing
This kind of method is an example of dynamic(or not oblivious)routing strategy. Which means that,as in all de?ection routing strategy,other packets in the network can
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affect the path ultimately traveled by a packet in the network.This routing strategy sends the customers to their destination in such a way that it advantages the selection of the favorable direction which adjust the most square?gure possible,formed by the axes,the current and destination nodes.It means that it sends the customer towards the direction which promotes the formation of a square.In other words,to the nodes that contribute to create the?gure with the biggest surface area.
destination node possible destination next slot
current node
(i)(ii)(iii)
Fig.1.Scale Routing
We will try to explain better this fact in?gure1.The part(i)of this?gure shows the case of a customer at the i?th slot,with all its possibles directions,favorable and not favorable,and the respective possible positions at the slot i+1.It means,if the customer is not de?ected it will take the favorables directions East or North,otherwise it will be forced to take the not favorables directions W est or South.The part(ii)shows the case without de?ection.The customer should have reached the a or b positions.The Scale Routing favors the East direction because the?gure created by the points adeg comes closer to a square,and with a bigger surface area than the form constituted by the points bcef.The part(iii)shows the case with de?ection,and for the same reasons the Scale Routing favors the South direction.The?gure adeg forms a square and the?gure formed by the points bcef does not do it.Furthermore,the?gure adeg has a bigger area than?gure bcef,and once again we can see that this property is veri?ed.Then using the Scale Routing,if there is not de?ection the customer will take the East direction and the South direction if it is de?ected.
3.1.1.Scale Routing Algorithm
This is a distributed algorithm,it must be applied in every node and at each slot. Packets are sent to the most convenient node,according to the Scale Routing.The implemented algorithm tries to?nd the best route for the packets.In this sense,we can say that this routing algorithm is greedy,because it takes the best immediate,or local,
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decision.Every node forwards each packet closer to its destination whenever possible (that is,whenever the desired edges are not already assigned to other advancing packets).In general this algorithm operates for D directions and C customers,C ≤D .Every customer has at most 2favorable directions.We do not have shown the demonstration of this algorithm because we have considered that it is not in the scope of the present work.
Assume a packet a is in the router with coordonates (x 0,y 0).Let a a packet inside
this router and let (x (d )a ,y (d )
a )the ?nal destination for packet a .De?ne:
x a =x (d )
a ?x 0
y a =y (d )
a ?y 0
Assume that for a packet a ,the shortest remaining distance is given by the coordi-nates (x a ,y a ).
position possible next slot
current node destination node
Fig.2.Scale Routing Algorithm
Now,we describe how the algorithm works.In each node,it begins with the ?rst packet in the queue,and it keeps on assigning route to the packets until there is no more in the node.As we can observe in the ?gure 2,if a packet is located in the "current "node,(noted by a square in the ?gure),and it wants to go to the "destination "node,(noted by a black circle),in a slot,therefore it has two favorable directions that are North and East .In the following slot,if it is not de?ected,(and both directions are available),it will be in the positions "a "or "b ",East or North ,respectively.The algorithm calculates the following quotients:for the East direction,x/(x +y ),and for
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the North one,y/(x+y),like a way to assigns priorities to the packets.Since these values are compared and the direction with the bigger one is favored.It will be the one assigned to the packet.Here,the East direction is selected.Otherwise,if the packet is de?ected,directions South and W est are then evaluated,because it would be in the"c" or"d"position,(in both cases at a distance one unit more far from the desired position). The quotients are calculated,(x+1)/[(x+1)+(y+1)],for"c",and(y+1)/[(x+1)+ (y+1)],for"d".They are changed of sign,so that the comparison is favorable,in this case,to the address with the smaller value.So,the South direction is favored that is the chosen one.This is carried out for each packet.At?rst for the favorable directions.If the selected direction have already been assigned to another packet,it follows the remaining direction.Else,it is de?ected and the respective evaluation is made,and of equal way,if the chosen one is not available,the other one is assigned.Once routed all the packets of the present node,one goes to the next node of the network.
3.2.Examples
The example we present here can have at most four customers at each node and for every one of them,four possible directions,(N,S,E,W).
This example shows customers having1or2favorable directions and a case with de?ection,to try to explain better this algorithm.
The relative information to the favorable directions are noted in an array named T. This array will contains on the?rst line the number of requests for every direction.The remaining lines will contain the evolution of these requests according to the algorithm as well as the directions to which are sent the customers.
Example1Let us assume that we are in one node,and we are going to apply there the algorithm.This is in the particular queue"current node",(noted by a square in the?g-ure).We also assume that there are three customers,that have as destinations the nodes 1,2and3respectively.In ahead we will identify the packets with the corresponding numbers of its favorite destinations.So,the packet that wants to go to node1,will be named1,and so on.The customer1has the North(N)and the East(E)directions as favorables.The customer2wants only the East direction,and the customer3wants to take the South,(S)or West(W)direction.Therefore North is favorable for1customer, East for2,South for1and West for1.The customer2has only one favorable direction, because its destination is on the same line as the one of the current node.There is a de?ection in this example.Figure3shows the destination of the three customers from the queue current node at one date.It agrees to clarify that this it is not necessarily the origin node or the source of these packets.
We will show now the case of the application of the routing strategy for a packet. We will begin by the?rst packet in the queue,in the same way that the algorithm does.
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current node
destination node
N
S
E
W
Fig.3.A node of the network
Nevertheless we must remember that for the remaining packets,the method works in similar sort.Then we show the application of the algorithm to the?rst packet,marked with the number1.The algorithm calculates the values of the quotient for the favorable directions East and North,represented in the?gures(4and5)by the positions a and b respectively(called possible position favorable).In the array of?gure4we can see all these quotient values.Some of them,like a and b,are calculated in case that the packet is favorably addressed.Only in case of de?ection,quotients c and d are calculated.But in this work,they are shown in the same array,to illustrate the way the algorithm operates.In the mentioned?gures,positions c and d represent W est and South directions respectively.In these?gures c and d are the possible positions if the packet is de?ected.In?gure5,North direction,or b position,is added as a possible position if the packet is de?ected.Finally,in?gure6the situation is reversed.Now, c and d are the possible positions if the packet is well or favorably directed and the possible position if packet is de?ected are a and b.
As we can see in?gure4,the direction with the highest value is the East one,and it is the chosen direction when the packet is not de?ected.Once the direction is chosen, it is assigned to the packet1and it is marked as not available.In case the packet should be de?ected to position d,the chosen direction should be South,because it has a bigger quotient value than W est,or less negative.
Then,to address the next packet in the queue,the number2,the algorithm goes on the following way.Although it is advisable to stand out,that in this type of routing,it is in certain way favorable for a packet,to advance towards its?nal destination doing a kind of"zigzag"instead of going towards it in straight line.This trajectory has the form of the stairs and has originated the name of this strategy.Since in this way,the packet in each node where it arrives,has less options to be directed,because it will always have a single favorable direction,which could make its route more fragile,the packet is in
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Poss. posit. (fav.)
Poss. posit. (deflected)
Current Node Destination Node
Node
Quot.
(3,4)??(12,9)??(4,4)East Y es (3,5)North Y es (2,4)W est Y es (3,3)
South
Y es
Fig.4.Packet 1
certain way,more susceptible to be de?ected.
As shown in the ?gure 5,the quotient for the favorable direction East is calculated,but it is not available,then the algorithm calculates the quotient value for the remaining directions,it selects South direction,and it is immediately marked as non available.
As we can see in ?gure 6,the algorithm proceeds to assign the route to the third packet in the node.After that it calculates the values for the favorable directions,in this case,South and W est .But the South direction is not available,then the assigned direction is W est ,and consequently marked as non available.
The array 6shows to us the ?nal routing in the node,with all the assigned directions to the packets.Then we can see that East direction has been assigned to packet 1.Packet 2has been de?ected,and its assigned direction has been South ,and ?nally,packet 3has been directed to W est .
3.3.Remaining Distance
We consider the remaining distance of the customer from its current node to its ?nal destination in every slot,to choose the customer to de?ect.Then we have advantaged the customer which is farther to its destination.It means,if there are two or more clients at the same node which plan to take the same direction,the priority is given to the one
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Poss.posit. (fav.)
Current Node Destination Node
Poss. posit. (deflected)
Node
Quot.
(3,4)??(9,4)??(4,4)East No (3,5)North Y es (2,4)W est Y es (3,3)
South
Y es
Fig.5.Packet 2
whose distance to its destination is the longest.The algorithm revises all the packets in the node.It assigns bigger weight to the packet with the biggest distance from its desti-nation node.The packet to route is chosen and the best direction is assigned according to the established strategy.
In ?gure 9we can see some results of the application of this heuristic and the com-parison concerning the other ones.
3.4.Number of De?ections
We can say that the number of de?ections that the packet has already had,has helped us to establish a hierarchy among the packets,to assign a priority to the customers or packets that have had a bigger quantity of de?ections,until the moment to assign them the route.In the case of having two or more clients with the same distance until their destination node,then we observe the number of de?ections that they have had,and we favor the one that has undergone more de?ections.
3.5.Implementation
We will expose in this aside the techniques used with the purpose of implementing the model and the heuristics explained before.We have dealed with the routing problem
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Current Node Destination Node Poss. posit. (deflected)
Poss. posit. (fav.)
Node Coord.
Quot.
????East No North Y es W est Y es South
No
Fig.6.Packet 3
Packet De?ected 1
No
d
South
3
No
Fig.7.Final Routing Exemple 1
like a problem of graph theory in which we have to ?nd the maximal coupling between the nodes corresponding to the packets and its favorable directions.In this sense,we have assigned weights to the differents directions with the objective to evaluate later all the pairs packet-direction and to ?nd that which has the maximal evaluation.In the case of Scale Routing,we assign the weight in the following way:to x coordinate of the favorable direction,we give the value (y/(x +y )and to y coordinate the value (x/(x +y )with the intention to advantage to the best direction and then we give values with negative sign to the remaining directions.In the case of the other heuristics,packets with a bigger de?ection number are multiplied by a factor in order to increase its weight.In the same way,we treat the packets with the longest distance to its destination https://www.doczj.com/doc/be18819401.html,ter,all the packets are evaluated and the route with the highest combination of weights,is chosen.
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4.Traf?c in the Network
Finally we have tested the incidence on the routing of changes in the traf?c of the networks,and we have got by simulation some results that we will show later.The traf-?c in the network have been modi?ed by changing the customer destinations between: uniform,non uniform to the corner and non uniform to the centre.
4.1.Uniform Destination
We tested?rst with uniform traf?c,that is to say,the destinations given to the packets were uniformly distributed for the whole net.In this type of traf?c,a packet that has its origin in a node anyone of the network,can have its destination node in another point anyone of the network,generated randomly and without another restriction that to be different from its origin.
4.2.Towards1/4th Corner Destination
The following simulations were carried out by sending the packets towards a corner of the network.In the way we will explain below:the net was divided in four equal parts, and all the packages were sent towards only one of these four quadrants,established for all previously.In this type of traf?c,a packet that has its origin in a node anyone of the network,will have always its destination node in a point that will be within the established area as destination for all the packets;this is,the corner that represents the fourth part of the network,in other words,the part that represents a quadrant of the plane. This destination is generated randomly to each packet and it will be,of course,always different from the origin.
4.3.Towards1/4th Center Destination
In this section we present the simulations that we have done for last kind of traf?c in the network referred before.This time we have sent the packets to a part of the network which we have denominated the fourth quarter.The network has been divided in four parts,but this time the division was made in such a way that all the packets go to the fourth central quarter.It is to say,to the delimited space that corresponds approximately to the fourth part located in the centre of the network.In this type of traf?c,we have restricted the packet destination in a similar way that in the precedent type of traf?c presented.Packets have the origin node always different to its destination nodes,and in addition,destinations nodes are in the established area,1/4centre of the network.
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5.Analysis of Results
5.1.Uniform Destination Results
Figures 8and 9show,the results of the simulation for this type of traf?c,jointly with the application of the remaining heuristics.The ?rst curve we will comment is the one identi?ed as dist _d fx _E of the ?gure number 9,which represents the application of the Scale Routing only.This is a decreasing curve that shows the frequency of de?ections for the packets.When the number of de?ections increases,the amount of packets in every point decreases progressively.We can appreciate that the maximum point is zero,which obviously indicates,that the number of de?ections appearing as the most frequent between all the packets,is zero.Nevertheless,we can appreciate also that there are a lot of packets who have undergone a number of de?ections suf?ciently great.For instance,there are packets that have had 25de?ections or even more.The described situation can be appreciated better in ?gure 9.
0.050.10.150.20.250.3
0.350.40.450.5051015
202530354045
F r e q u e n c y
Number of deflection
Uniform Destination (TMAX = 10^3)
’dist_dfx_E_3.res’’dist_dfx_DfE_3.res’’dist_dfx_DfDiE_3.res’
Fig.8.Uniform Destination (i)
As we can see in the curve identi?ed as dist _d fx _DfE )of the ?gure number 9,when we add the application of the heuristic "number of de?ections that have undergone the packet",to choose the packet to be de?ected,the form of the curve has a very impor-tant change,respect to the curve analysed before.It passes from a constantly decreasing curve to a curve resembling to a parabola.Now the maximum value is moved,even
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1e-05
0.00010.0010.010.1
1
051015
202530354045
F r e q u e n c y
Number of deflection
Uniform Destination (TMAX = 10^3)
’dist_dfx_E_3.res’’dist_dfx_DfE_3.res’’dist_dfx_DfDiE_3.res’
Fig.9.Uniform Destination (ii)
though zero continues to be a very frequent value.The value two is now the maximum of the curve.That is to say,the number of de?ections most frequently repeated is two.With respect to the previous curve,the frequency of occurrence of the zero value de-creases slightly,but the values one,two and three have a bigger frequency in this curve than in the previous one.An important annotation is that in this case,packets with a number of de?ections bigger than eight,approximately,do not exist.We can then appre-ciate that the curve dist _d fx _E presents bigger values than the curve dist _d fx _DfE for the de?exion of the packets.
It means,we have favored the packet that has had more de?exions.If two packets want to take the same direction then the packet with bigger number of de?ections is routed ?rst.
The curve identi?ed as dist _d fx _DfDiE represents the application of the two ?rst heuristics with the inclusion of the last one mentioned.When "distance to cross until the destination"is taken into account to choose the packet to de?ect,we can appreciate that the curve becomes a little more closed,more pointed and less extended than the previous ones.The frequency of zero still decreases,but the number of packets with two de?ections is even so increased.
On the other hand,when this last heuristic is added,the curve does not take values bigger than six,then we can say that it does not exist any packet which undergoes more
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than six de?ections.The maximum number of de?ections that a packet anyone can undergone,when all the heuristics are applied,is six.
We have considered,to de?ect a packet,at ?rst the distance to cross until the desti-nation,then the number of de?exion it have had.It means,if two packets want to take the same direction then the packet with the longest distance to its destination is chosen.If the packets have the same distance to its destinations,then the one with the greatest number of de?ections is routed ?rst.
Statistics relatives to these curves will be shown and discussed in section 5.4..
5.2.Towards 1/4th Corner Destination Results
The ?gures 10and 11show results of the simulation for this traf?c function,jointly with the application of the different heuristics mentioned before.
0.1
0.2
0.3
0.4
0.5
0.6
050100150
200250300350400450
F r e q u e n c y
Number of deflection
1/4th corner Destination (TMAX = 10^3)
’NU_dist_dfx_E_3.res’’NU_dist_dfx_DfE_3.res’’NU_dist_dfx_DfDiE_3.res’
Fig.10.1/4Corner Destination (i)
In a similar way that in the precedent case of Uniform destination,the curve NU _dist _d fx _E shows the application of Scale Routing only,the curve identi?ed by NU _dist _d fx _DfE ,presents the results of applying additionally the advantage to packets with a bigger number of de?ections,and the curve (NU _dist _d fx _DfDiE shows the case in which,besides the previous heuristics,it is added the distance to cross by the packet until its destination.But,of course,all these cases are treated with this
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1e-05
0.00010.0010.010.1
1
050100150
200250300350400450
F r e q u e n c y
Number of deflection
1/4th corner Destination (TMAX = 10^3)
’NU_dist_dfx_E_3.res’’NU_dist_dfx_DfE_3.res’’NU_dist_dfx_DfDiE_3.res’
Fig.11.1/4Corner Destination (ii)
particular kind of traf?c,towards 1/4th Corner Destination.
The curve of Scale Routing,has a similar behaviour to the one of Uniform destina-tion,although its values are bigger than those in the previous one.It is also a decreasing curve,with the maximum point in zero.In the same way,the value that presents the frequency of the highest occurrence of de?ection,is the zero.As in the precedent case,there are packets that have had a great amount of de?ections through their route,but this time there are even packets with more than 400de?ections.
Curves (NU _dist _d fx _DfE )and (NU _dist _d fx _DfDiE ),have both a quite similar irregular behaviour.They present,like the ?rst one,a high frequency in point zero,but in addition presents other peaks,maximum and minimum.Nevertheless,these curves show in general,smaller amplitude than the curve of scale routing,in the sense that the biggest number of de?ections that a packet can undergo is smaller than 50,whereas in the ?rst curve,as we have said,it even can be superior to 400.
In section 5.4.,will be displayed and discussed statistics and results relative to these curves.
230
5.3.Towards 1/4th Center Destination Results
As we can observe in the ?gures number 12and 13,the behaviour of the Scale Routing curve,is similar to the previous cases shown,an always decreasing curve with the greatest value of frequency in zero,although with values of de?ection a little smaller than the case of 1/4corner.Nevertheless,the curves of the application of the other heuristics,
0.050.10.150.20.250.3
0.350.40.45050100
150200250300350
F r e q u e n c y
Number of deflection
1/4th center Destination (TMAX = 10^3)
’NUc_dist_dfx_E_3.res’’NUc_dist_dfx_DfE_3.res’’NUc_dist_dfx_DfDiE_3.res’
Fig.12.1/4Center Destination (i)
NUC _dist _d fx _DfE ,(number of de?ections that packet has undergone),and NUC _dist _d fx _DfDiE ,(distance to cross until the destination),are not as irregu-lar as in the previous case,because even though the curve presents peaks,these are less numerous than in the previous case.
Moreover the biggest value for the number of de?ection in these last curves,is smaller than the registered one in the curve NUC _dist _d fx _E ,which is further of 300,while that in the other curves,it is at most 25.It means,curves NUC _dist _d fx _DfE ,and NUC _dist _d fx _DfDiE ,are less extended than NUC _dist _d fx _E .
We will observe and discuss about the statistics relates to these curves,in next sec-tion.
231
1e-05
0.00010.0010.010.1
1
050100
150200250300350
F r e q u e n c y
Number of deflection
1/4th center Destination (TMAX = 10^3)
’NUc_dist_dfx_E_3.res’’NUc_dist_dfx_DfE_3.res’’NUc_dist_dfx_DfDiE_3.res’
Fig.13.1/4Center Destination (ii)
5.4.Other Curves ans Statistical Results
In this aside,we show some simulations results.We must refer here to the measure of peakedness or sharpness of the distributions,known as Curtosis.As it is known,it is an algebraically treatable and geometrically interpretable measurement.When we calculate the coef?cient of curtosis of one distribution,it gives a result that may corresponds to a platicurtic,mesocurtic or leptocurtic curve.A leptocurtic distribution has the greater part of its measurements concentrated in the center,a mesocurtic one resembles to a Gauss curve,and a platicurtic curve is more plain.Dispersion measures are smaller in a leptocurtic curve that in a platicurtic one.
In array 16,we can observe the results of the statistics calculated with the corre-sponding data to each type of traf?c.At ?rst,we show the statistics related to the uni-form traf?c.We can see that in the case of application of Scale Routing,the mean of de?ections is slightly superior to the cases in which the application of the heuristic ones are introduced.The average of de?ections increases when is privileged the packet that has undergone more de?ections and increases a little more when is favored the packet that is farther from its destination.
Nevertheless it happens the opposite with the Dispersion measurements,Variance and Standard Deviation.In the case of Scale Routing,these values are greater than in
232
the case in which we take into account the?rst heuristic,(number of de?ections that packet has undergone),and it increases a little when we introduce the next heuristic, (distance until the destination node).That is to say,when the?rst heuristic is introduced, the curve becomes less ample,and when introduced the next one,the curve stays more or less in the same form.There is not a big difference between these last curves.In terms of Curstosis,for the?rst case,the curve is?atter,(platicurtic)and when a new hierarchisation is applied it becomes more pointed,more acute,(leptocurtic).
In a similar way,in array16we can observe the statistics for the traf?c we have called Towards1/4th Corner Destination.When only scale routing is applied,the mean of de-?ections is small,but it increases considerably,almost the double,when the Heuristic Deflec is applied and it continues to increase,but just a little,when the next heuristic, Dist,is applied.Nevertheless the values of Variance and Standard Deviation are exag-gerated high respect to the ones obtained with the application of the heuristics.Then,it means that the distribution that represents the case of scale routing is very?at,(platicur-tic),whereas,the form of the distributions that represent the application of the heuristics is acute respect to the precedent,(leptocurtic).
As far as these last ones,when we just favor the packets with greater amount of de?ections,a greater average to the previous one is obtained,and when in addition we assign priorities to the packets farther from their destinations,the value of the average is increased slightly,but it is with the application of the three heuristics jointly that we obtain smaller values for the dispersion measures.
In the case of the curve1/4center is observed a behavior very similar to the previous cases,the smaller value of the average of the de?ections have been the obtained for the application of routing scale that as well has the greatest values for the dispersion mea-sures.Of equal way,as a heuristic one is added new,the value of the average increases and the values of Variance and Standard Deviation diminish,being the values obtained with the three heuristics meetings,the smallest for these measures.
Finally,?gures14and15,(normal and logarithmic scales)present the curves for scale routing and the two heuristics applied simultaneously,with each one of the three used types of traf?c in this work.The curve dist_d fx_DiDfE is the one that corre-sponds to the Uniform destination,the curve with abbreviations NU_dist_d fx_DiDfE corresponds to traf?c not uniform directed to the corner and the curve with the Nuc_dist_d fx_DiDfE abbreviations corresponds to traf?c not uniform directed to the center.As we can appreciate in the?gures,the curve that presents less irregularities is the one of uniform destination.The curve dist_d fx_DiDfE presents only one pointe, une maximum,whereas the other ones show at least two of them.By the other hand,the maximum number of deviations that can have the curve dist_d fx_DiDfE is inferior to 10,in curve Nuc_dist_d fx_DiDfE,this number is almost15,while that in the curve NU_dist_d fx_DiDfE is very near to30.Therefore we can appreciate that the curve
10. 大还是小 教学过程 第一课时 【课时目标】 1.会认“时、候”等11个字，会写“自、己、衣”3个字，认识双立人、竖心旁2个偏旁。 2.正确、流利地朗读课文。 【教具准备】 课件、生字卡片 【教学过程】 一、激趣导入，引入课题
1.（课件出示2）出示鸡蛋（一大一小）图片。 同学们，这是什么？（鸡蛋），你发现了什么 呢？（这两个鸡蛋一大一小） 2.师板书“大”和“小”。你认为自己是大还 是小呢？说说原因。 3.有一位小朋友，他自己很矛盾，有时候觉得 自己很大，有时候又觉得自己很小，到底怎么回事 呢？ 今天，我们一起学习《大还是小》一课，一起 去了解、感受这位小朋友的想法。 （板书课题：大还是小）齐读课题。 二、初读课文，检查预习 下面就让我们一起先来看看小作者认为自己是大 的还是小的。 1.自由读课文，注意读准字音，读通句子，做到“三 不”：不错字，不添字，不漏字。 2.要想读好课文，就必须先认识这些生字朋友。 （课件出示3） shí hou jué de zì jǐ hěn kuài chuān yī fu 时候觉得自己很快穿衣服 你认识它们吗？自己试着读一下。 指正：“自”是平舌音，“时、穿”是翘舌音。 “时候”中的“候”，“衣服”中的“服”在这里都 读轻声。 （1）谁能来当小老师带领大家读一读？其他同学 （2）这些字去掉拼音你还认识吗？ （课件出示4） 时候觉得自己很快穿衣 服 我们来开火车读一读。
（3）识记生字： 本课生字以合体字为主，可以运用多种方法帮助学生识记字形、理解字义。 （课件出示5）出示会意字图片：学习会意字“穿”。教师出示老鼠挖掘洞穴的图片，告诉学生：上面是一个“穴”，表示的是野兽居住的洞穴；下面是“牙”，表示野兽用自己的牙齿来挖掘洞穴，是凿通、凿穿的意思。 用熟字组成新词：时间、感觉、得到、很多、大自然、穿过。 小结：识字的时候，我们不仅可以用加一加、换一换的方法，还可以用猜字谜的方法，但要注意编的字谜要合理。 指名认读，齐读。 3.认识了生字朋友，读课文就更容易了。下面我请几位同学接读课文，其他同学边听边想，本文介绍了“我” 什么时候感觉自己很大，什么时候感觉自己很小？ 预设：“我”自己穿衣服和系鞋带的时候感觉自己很大。 预设：4.教师评价学生的朗读。 三、观察生字，指导书写 1.出示生字:自、己、衣（课件出示6） 观察字形，记住它们在田字格中位置。 2.指导写字规律。 自：横平竖直，中间几横之间的间距要均匀。 己：整个字上窄下宽，竖弯钩要圆转。 衣：整个字的重心落在田字格的正中，撇捺舒展，呈三角形；注意笔顺，最后一笔是长捺。
精品教学资料，欢迎老师您参考使用！ 《大还是小》教学反思 孩子们都希望自己快快长大，成为一个独立的人。与此同时，他们也离不开父母的呵护。《大还是小》这篇课文通过3个“有时候”和“更多的时候”把文章紧密地串联起来，形成一个有机整体。课文多处运用对比的方式来展现儿童的世界，儿童的内心是矛盾的，又是充满趣味的。第二自然段的“大”，第四自然段的“小”，就是这种矛盾的具体体现。教学重点为认识“时”“候”等11个生字和双人旁、竖心旁两个偏旁；会写“自”“己”等3个生字；正确、流利地朗读课文，结合插图，体会“我”自相矛盾的内心世界。结合生活体验，说说什么时候觉得自己很大，什么时候觉得自己很小。上完课后，教学效果感觉良好，也有许多的感受、体会。回顾整堂课的教学，总结如下： 一、教学效果 本节课围绕着教学目标，我取得了以下效果： 1.为了实现教学目标，我的教学思路主要还是提示学生读准字音。为了激发学生的识字兴趣用图片的方式来学习会意字“穿”。出示老鼠挖掘洞穴的图片，告诉学生，上面是一个穴表示的是野兽居住的洞穴，下面是“牙”表示野兽用自己的牙齿来挖掘洞穴，是凿通、凿穿的意思。因为学生认知事物的方式不同，鼓励学生用自己喜欢的方式进行识字。组内交流汇报识字方法，效率高。在写字教学中，采取对比学习方式“自”和“白”；“己”“衣”引导观察笔画互相衔接的位置。 2.朗读指导。采取男女生对读、同桌之间对读的形式，引导孩子读出内心成长的感受，体会“大”和“小”的情感变化，当自己觉得很大时，读出自豪之感；当自己觉得自己很小时，读出一种儿童依赖大人的感觉。在熟读的基础上，通过指导读好几个“有时候”和“更多的时候”，读出文章的结构的特点。第一个“有时候”要读出内心的自豪感；第二个“有时候”朗读时语调要有变化，相较于第一个“有时候”在语调上稍微短一点，读出“我觉得自己很小”中的“很小”。学生在朗读中体会到“我”内心世界的自相矛盾。 3.理解运用。从题目入手，学生说说对大和小的理解，能否用到一个人身上，激发学生的学习兴趣。 4.说一说。结合生活实例，将学生带入文本，加深对课文内容的理解。借助句式“有时候，我觉得自己（）。（）的时候，（）的时候，我觉得自己（）”引导学生说感受，把语言学习和内容理解有机结合。 二、成功之处 《语文课程标准》中提出语文课程是一门学习语言文字运用的综合性、实践性课程。读写不分家，学生初步了解课文的基础上，结合生活实例，将学生带入文本，借助句式“有时候，我觉得自己（）。（）的时候，（）的时候，我觉得自己（）”练
13、an en in un ün 第一课时 教学目的： 1．学会前鼻韵母an 、en 2个复韵母，读准字音，认清形，能在四线三格中正确书写。 2．学习声母与an 、en 组成的音节，准确拼读音节，读准三拼音节，复习ü上两点省写规则。 3．学习整体认读音节yuan 。 教学重点： 1．学会韵母an 、en 2个复韵母，读准音，认清形，能正确书写。 2．学会声母与an 、en 组成的音节和整体认读音节。 教学难点： 学会介母是ü的三拼音节，读准音节juan 、quan 、xuan 。 教学过程： 一、谈话导入 我们到目前为止学习了哪些复韵母？能按顺序说说吗？ （ai 、ei 、ui 、ao 、ou 、iu 、ie 、üe、er ）。今天我们一起来学习第13课，再认识几个韵母朋友，请同学们打开书看看。这课书的内容比较多，有信心学好吗？下面我们先来学习前2个韵母及音节。 板书：13 an en 二、看图学习韵母an 、en 1．学习韵母 an （1）出示an 图，问：图上画的是什么？ （2）自己试着发an （安） （3）教师指导发音：把嘴张大，摆好a 的口形，让气 流从前鼻腔里出来，也就是n 的尾音。 （4）学生练习读，体会前鼻韵母的发音方法。 （5）同桌同学互读，纠正发音。 （6）指名读，开火车读。 2．学习韵母 en （1）出示en 图，问：你们看这个人在干什么？ （2）借助“摁”的第四声交成第一声学生练习发en 的音。 （3）en 是由哪两个字母组成的？（e 和n ）发音时，先发e ， 嘴半闭，舌尖抬起抵住上牙床快速读，鼻子出气，一口气读出en 的音。 三、书写韵母an 和en
10 大还是小 教材解读： 《大还是小》是一篇富有儿童情趣的文章，内容浅显易懂，同时富有教育意义。孩子们都希望自己快快长大，成为一个独立的人。与此同时，他们也离不开父母的呵护。课文通过 3 个“有时候”和“更多的时候”把文章紧密地串联起来，形成一个有机整体。课文多处运用对比的方式来展现儿童的内心世界，儿童的内心世界是矛盾的，又是充满趣味的。第二自然段的“大”，第四自然段的“小”，就是这种矛盾的具体体现。课文配有一幅插图，“大”和“小”的行为都在其上，可以借助课文插图来展开教学。 教学目标： 1.认识“时、候”等 11 个生字和双立人、点横头、竖心旁 3 个偏旁；会写“自、己” 等3个字。 2.正确、流利地朗读课文。结合插图，体会“我”自相矛盾的内心世界。 3.结合生活体验，说说什么时候觉得自己很大，什么时候觉得自己很小。 教学重点、难点： 教学重点：正确、流利地朗读课文。 教学难点：体会“我”自相矛盾的内心世界；会写“己、衣”等 字。 第一课时 一、课时目标： 1.认识“时、候”等 11 个生字，能用不同的识字方法进行识记。学习双立人、点横头、竖心旁 3 个偏旁；会写“自、己”两个生字。 2.正确、流利地朗读课文，初读课文学习质疑并能通过自读自悟解读疑问。 二、教学过程 （一）激趣导入，引出课题，启发质疑 1．出示字卡“大”。
大声读这个字。说说和它意思相反的字是什么吗？ 2．出示字卡“小”。 小声读这个字。(生读：小) 提醒：上课时，回答问题声音不能太小，否则别人就听不到了。老师要看看这节课谁的表现最棒。 3.同时出示字卡“大小”，一起来读一读。 4．质疑：你认为自己是大还是小呢？能说说为什么吗？(指名回答) 5.过渡：有一个小朋友也遇到了这个问题，他是怎么回答的呢？我们这节课就来学习《大还是小》。(板书课文题目) 6．读了课文题目，你有什么疑问？(师生梳理出主要问题) （二）自主探究学习 1.教师出示自读要求 (1)自由朗读课文，遇到不认识的字，借助拼音多读几遍。把词语读正确，句子读通顺。 (2)拼读课前圈画的生字，要读准字音，想办法记住这些生字。 2.根据自探提示先自主学习，然后在小组长的组织下在小组内交流。 （三）初读课文，学习生字 1.检查自主学习情况。 (1)我会读 课件出示词语： 时候觉得穿衣服自己很小快点儿 ①指名开火车朗读，师生正音。 ②齐读。 ③自主选择一个词语说一句话。 ④去掉拼音指名读，齐读。 (2)我会认 ①这些词中有些生字需要我们记住，瞧，它们已经从词中跳出来了，你还能认出它们吗？ 课件出示生字，指名读。
10.大还是小 同学们，你们愿意快快长大，还是永远做一个孩子呢？《大还是小》这篇课文的小作者有时候觉得自己很小，有时候觉得自己很大，怎么回事呢，我们一起走进课文看看吧！ 学习目标—要知道 1.能正确流利、有感情地朗读课文。 2.会正确认读“候、穿”等12个生字，学会写“自”等4个字，认识“ㄔ”等个部首。 3.感受小作者要长大心情，能自己的事情自己做。 字词详解—要掌握 2.会认的字
3.多音字 ào （睡觉）de （写得） ? （感觉）d ěi (得学会) 运用：我宁愿睡觉也不看这部电影。我觉得你应该走快点。 他的作文写得很华丽。他们得学会尊重并欣赏这一点。 4.近义词 觉得——感觉 陪——伴 照顾——照看 盼着——希望 5.反义词 大——小 多——少 快——慢 运用：①妈妈买了两个西瓜，一个大一个小。 ②我们班的学生多，二班的学生少。 ③散步时奶奶的脚步总是很慢，而我总是走得飞快。 6.词语听写 自己 妈妈 妹妹 7.一词多义 8.词语拓展 反义词：前——后 左——右 上——下 外——内
表示动作的词语：散步漫步飞跑张望倾听大哭 课文内容详解 导读：每个小朋友都有自己的梦想。有的小朋友希望自己快快长大，成为很厉害的人。但是他们有时候又离不开父母的呵护。课文以简洁、生动、形象的语言写出了一个小朋友心里的真实想法。本文共八个自然段，通过生活中的几件小事写出了一个渴望长大的孩子的心情。 课文详解—要领悟 1.概述内容 《大还是小》这篇课文，以简洁、生动、形象的语言写出了“我”在做不同的事的时候，会有不同的想法和感受，表达了自己想要长大的心情。 2理清层次
教学目标： 1．学会前鼻韵母an、en 2个复韵母，读准字音，认清形，能在四线三格中正确书写。 2．学习声母与an、en组成的音节，准确拼读音节，读准三拼音节，复习ü上两点省写规则。 3．学习整体认读音节yuan。 教学重点： 1．学会韵母an、en 2个复韵母，读准音，认清形，能正确书写。 2．学会声母与an、en组成的音节和整体认读音节。 教学难点：学会介母是ü的三拼音节，读准音节juan、quan、xuan。 教学过程： （一）复习检查。 1、你们好，我是喜羊羊。新的一天又开始了，你们还记得昨天学的字母宝宝吗？让我来考考你们！ 看哪个小朋友能读得又准确又响亮。 （卡片认读复韵母：ai ei ui ao ou iu ie üe er .） 教师小结：咱们班的小朋友可真会学习！。 （二）教前鼻韵母an和整体认读音节yuan。 老师呢就奖励你们一幅好看的图画，好不好！？ 【出示天安门图片】 1.小朋友看，这是什么？ 说说你知道的有关天安门的知识。 （天安门在北京，北京是首都，等等） 2、教学an的发音。 真棒，知道了那么多它的知识，小朋友一定能读准它。跟老师读读看，an （1）讲解发音要领：把a和-n合在一起，先发a，口不宜张得太大，马上用舌尖顶住上腭的前部，使气流从鼻孔出来，要念成一个音。 （2）教师范读、领读、指名读、开火车读、齐读。 3、an的四声练习：ān（天安门）ǎn（俺家）àn（黑暗） 4、教学整体认读音节yuan，【出示画了“圆”的图片】，图上画着什么？ 5、教学发音，yuan是整体认读音节，板书yuan。教师范读、领读。 6、yuan的声调标在a上，进行四声练习： yuān（冤家）yuán（原因）yuǎn（遥远）yuàn（庭院）（三）教学前鼻韵母en。 1、读得可真棒，来！把掌声送给自己！好，你们小嘴巴很厉害，那老师考考你们的小眼睛，翻开书，看看第三幅图上画着什么？！在什么情况下摁门铃？ 摁门铃可是文明礼貌的行为，但是不能乱摁，应该怎么做呢！？ 学生发言，表演示范。 教师总结 2、教学en的发音：启发学生用学习an的方法练习en 的发音，提示，先发e，马上用舌尖顶住上腭的前部，使气流从鼻孔中出来。
活动名称：复韵母an en in 教学目标：1.情感目标：培养幼儿学习拼音的兴趣。 2.态度目标：能够认真的学习拼音。 3.知识目标：学习an en in的认读、四声调及书写格式。 4.能力目标：能掌握声母与an en in拼读。 教学重点：学习an en in的认读、四声调及书写格式。 教学难点：能掌握声母与an en in拼读。 教学准备：课件、拼音卡片。 教学过程： 一、开始部分 1.复习：（课件出示蓝猫形象）小朋友们，你们知道这是谁吗？师：对了，这是小朋友 们最喜欢的蓝猫，蓝猫对小朋友们说：“你们还记得以前学过的拼音吗？它们是谁呢？”出示卡片。 2.小朋友们认得都很棒。今天蓝猫又给你们带来了几位新朋友。请看画面 二、基本部分 （一）认读复韵母an。 （1）出示第一幅图（天安门），“a和n合在一起我们读an，a在前，n在后，我们读，anan an。an在发音时：先发a的音，然后舌尖逐步抬起，顶住上牙床发n的音。 模仿读，个别读，开火车读，齐读。 （2）我们编成了一句识记儿歌便于小朋友记忆“天安门天安门anan an”分组读或请个别幼儿重复读。 （3）an的四声练习。 （4）an的书写格式。（占中格） （二）认读复韵母en 课件出示蓝猫图像：“小朋友们真厉害！认识了一个新朋友，蓝猫送给大家热烈的掌声。现在我们来看看第二个新朋友，出示第二幅图（有关点头的画面）小朋友们看，它们在做什么，仿佛在说什么？ （2）它们仿佛在说：“恩，小朋友们真行！”）那么我们今天所学的新朋友就是复韵母“en” “e和n做朋友时，我们读en，e在前，n在后，en在发音时：先发e的音，然后舌面抬高，气流从鼻腔泄出，发n的音。 我们也编成了一句话识记儿歌“点点头enenen” （3）en的四声练习。 （4）en的书写格式。（占中格） （三）认读复韵母in （1）课件出示蓝猫图像：“小朋友们真厉害！又认识了一个新朋友，蓝猫送给大家热烈的掌声。现在我们来认识最后一个新朋友，出示印章的画面。这个印章里in的发音就是我们今天要学习的。 （2）我们读印章in时，i在前，n在后，我们读in”in在发音时：先发i的音，气流从鼻腔泄出，然后发n的音。 in的前面加一个y就是印章的印的读音完整的样子了，变成了“yin”，它是一个整体认读音节，什么叫整体认读音节呢？添加一个声母后读音仍然和原来的韵母一样的音节（也就是指不用拼读可以直接认读的音节，叫整体认读音节。） 这里我们也有一句识记儿歌是“印章印章in in in” （3）in的四声练习。 （4）in的书写格式。（占中上格） （四）an en in与声母的拼读。
新人教版部编小学一年级语文上册教案及反思 10、大还是小 教学目标： 知识与技能 1.会认“时”“候”“觉”等11个生字，会写“自”“己”“衣”3个生字。掌握3种偏旁“彳”“亠”“忄”。 2.正确、流利地朗读课文，读准字音。 情感态度与价值观 引导学生要学会正视自己，知道什么时候自己很大，什么时候自己又很小。 教学重难点： 1.掌握本课所学生字，能够按笔顺准确、规范地书写生字。 2.正确、流利地朗读课文，读准字音。 教学课时：2课时 第一课时 教学过程： 一、谈话导入 １.师：同学们，每天早上爸爸妈妈送你们上学的时候，当你们看到高年级的哥哥姐姐们自己来上学，有没有很羡慕呢？（生答：有） 2.师：那个时候，你心里是怎么想的呢？（要是我也像哥哥姐姐一样大多好啊！） 3.师：为什么呢？（因为我再大一些，爸爸妈妈就不用每天辛苦送我上学了。）
4.师：有一个小朋友啊，他也和你们一样，有时候能自己系鞋带、穿衣服时，他觉得自己很大；但是有时候呢，够不到按钮、害怕打雷时，他又觉得自己很小。这节课我们一起去认识这位小朋友吧！ 二、看图读文，整体感知 1.让学生自由朗读课文，画出课文生字词，多读几遍。 2.读一读，标出自然段序号。 3.看图，说说图上画了什么。你能根据课文内容说一说吗？（引导学生结合插图说一说。） 4.不明白的地方用横线画下来，并向老师请教。 三、动动脑筋，学习生字 1.看拼音读词语。（课件出示重点词语） 2.课件出示课文生字（去拼音），指名读，开火车读。 3.认读这些生字，并给这些生字找朋友。（口头扩词练习） 4.巧识字形。 （1）师：你们有什么好办法能很快记住这些字的字形吗？ （2）四人小组讨论识记方法。（鼓励学生结合字形和字义巧识巧记。）（3）同桌之间互相说一说你是如何记住的，汇报交流识记方法。 （比一比：己—已自—目） 第二课时 教学过程： 一、创设情境，激发兴趣 1.出示本课生字词卡片，检查学生认读情况。 2.出示课件“图图上小学了”。
部编版一年级上册语文10大还是小 1.认识“时、候”等11个生字和双人旁、竖心旁2个偏旁；会写“自、己”等3个生字。 2.正确、流利地朗读课文。结合插图，体会“我”自相矛盾的内心世界。 3.结合生活体验，说说什么时候觉得自己很大，什么时候觉得自己很小。 4.在仿说中迁移运用文中句式。 重点 1.正确、流利地朗读课文。 2.体会“我”自相矛盾的内心世界。 难点 在仿说中迁移运用文中句式，让学生与主人公的心理产生共鸣。 1.字词教学。 生活识字：在教学“穿”时联系学生生活，说说自己还会穿什么，如“穿鞋子、穿裤子”等，在拓展中认读生字，识记生字。 对比识字：“双人旁”是本课新学的偏旁。教学时可引导学生与“单人旁”进行辨析，将“得、很”组合教学，并给“很”组词，通过对“很大、很小、很多、很少”等词语的对读，体会“很”是表示程度的加深，也为后面的朗读指导打好基础。 字义识字：学习“竖心旁”时结合“心”字说说演变过程，从而理解带有“竖心旁”的字大多跟心情有关。 书写生字：“自、己、衣”在写字板块需重点指导。指导“自”时可采用加一加的办法记住字形 “目+ ”；“己”的书写要点是笔顺及书写最后一笔的位置；“衣”要让学生观察笔画的细节，以及几个笔画相互衔接的位置。 2.朗读教学。 本课的语言兼有散文和诗歌的特点，语句有长有短，长句由结构重复的短句组成，如“______的时候，_____的时候，我觉得自己_____”。在指导朗读时，可以采用范读法，让学生能明显判断停顿的位置，进行标注后再反复练习。如：“我自己/穿衣服的时候，我自己/系鞋带的时候，我觉得/自
己很大。”还可采用对比朗读法，一年级的孩子读书时最容易出现拖音拖气的现象，教师通过示范形成对比，让学生正确朗读。评价法也是指导学生朗读的有效途径。如“我自己穿衣服的时候，我自己系鞋带的时候……”教师评价：“你把‘我自己’读得那么响亮，教师听出了你的自豪。” 3.迁移运用。 课文第1、2自然段与第3、4自然段用了以下句式：“有时候，我觉得自己很_____。我自己的时候，我自己______的时候，我觉得自己很_____。”教学时可引导学生联系自身经历进行仿说。不仅做到句式上的迁移运用，还让学生与课文中的主人公产生共鸣。 教学准备 1.借助拼音读课文，认读本课生字。 2.多媒体课件。 教学课时 2课时 第1课时 1.认识“候、得”等生字，注意读准轻声。 2.认识“双人旁”，会写“衣”字。 3.正确、流利地朗读课文。 一、创设情境，引出课题，启发质疑。 1.教师讲故事：在森林王国里住了小白兔一家，一天小白兔欢欢跑到妈妈跟前对妈妈说：“妈妈，妈 妈，你看我长高啦！我是个大孩子了！”同学们，你们同意欢欢的说法吗？说说你的看法。（生自由讨论） 2.质疑：你认为自己是大还是小呢？能说说为什么吗？（生自由回答） 3.有一个小朋友也遇到了这个问题，他是怎么回答的呢？我们这节课就来学习《大还是小》。 4.给课题加上问号，指导学生读出疑问的语气。
部编新人教版一年级语文上册第10课《大还是小》课堂教学实录 本帖最后由 ljalang 于 XX-10-29 11:31 编辑 10 大还是小 名师教学设计片段 ◆激发识字兴趣，运用多种方法识字(教学难点) 师：看到小朋友们能正确地读出这篇课文，老师心里可高兴了。不仅我高兴，连藏在这篇课文里的十一个生字娃娃也为你们高兴呢。瞧它们出来了！(课件出示：时、候、觉、得、穿、衣、服、自、己、很、快) 师：没有拼音，你们还能叫出它们的名字吗？别急，试着读一读吧。 (学生认读，教师巡视) 师：刚才老师看到有几位小朋友皱起了眉头，看来是碰到了不会认的字，不过没有关系，这很正常。只要我们动动脑筋，想想办法，就一定能认识它们。请你想一想：如果碰到了不会认的字，你会怎么办呢？ 生1：我会看书上的拼音。
生2：我会去查字典。 师：对，字典的确是一位好老师。 生3：我会举手问老师，还可以问旁边的同学。师：你们真聪明，一下子想出了这么多好办法，有了这么多好办法，老师相信你们一定都能跟这些生字娃娃交朋友。好，赶快行动，想办法和这些生字娃娃交朋友吧。 (学生自由认读生字，有的大声认读，有的在看书上的拼音，还有的问起了旁边的同学，老师一会儿看看这个小朋友，一会儿问问那个小朋友。) 师：你们学得这么认真，一定都和生字娃娃交了朋友。老师想来检查一下，请把书合上。看，生字娃娃都跑到老师这里来了。(出示生字卡片)能叫出它们名字的请举手。 师：这些生字娃娃的名字会认了，那这些生字娃娃的模样我们又该怎么记住呢？ (课件出示“很、得”) 师：你有什么发现吗？ 生：它们很像。 师：是呀，这两个生字娃娃长得这么像，怎么把它们区分开呢？
《大还是小》说课稿 今天我说课的内容是小学语文一年级上册第七单元第10课《大还是小》。下面我将从教材、学情、教学目标、教法学法、教学过程、教学板书六个方面作简单的说明。 一、说教材 《大还是小》是一篇富有儿童情趣的文章，内容浅显易懂，同时富有教育意义。课文用简洁平实的语言写了“我”有时候觉得自己很大，有时候又觉得自己很小，并举了一些具体的事例，让孩子们意识到自己的事情应该自己做。 二、说学情 一年级的小朋友对“长大”非常向往，但是有时候又有一些力不从心的事情，让他们意识到自己还没有长大。这篇课文能让他们对“长大”有更确切的认识，能够让他们意识到自己的事情应该自己做。 三、说教学目标 依据教材和学情，我设定以下教学目标： 1.认识11个生字，会写3个生字。 2.能正确、流利、有感情地朗读课文。 3.理解课文内容，体验长大的快乐。 本课的教学重点：学会本课生字词，能正确、流利、有感情地朗读课文。 教学难点：理解课文内容，体验长大的快乐。 四、说教法学法 语文课程标准提出：“努力建设开放而有活力的语文课程。”所以本节课主要采用媒体演示、自主读书，自主识字、合作学习、合作解疑的方法。学生在教师的引导下动脑、动手、动口。通过自己的劳动获取知识，变被动学习为主动学习。体现“以教师为主导，学生为主体，训练为主线”的原则。 五、说教学过程 （一）谈话导入，激发兴趣 “兴趣是最好的老师。”“兴趣是求知获艺的先导。”因此，上课伊始，我先出示“大”和“小”两个字，让孩子们说说觉得自己是大还是小，为什么？顺势引出课题，激发学生探究课文的兴趣。 （二）初读感知，学习生字
1、老师先范读课文。 2、孩子们借助拼音自由读课文，圈出不认识的字，向小组内其他的同学请教后，多读几遍。 3、学习生字词。首先，课件出示词语，再出示生字进行认读，接着让学生先小组合作交流识记方法，再全班交流，认识双人旁，竖心旁。并利用游戏的方式激发学生的识字兴趣。然后，把生字词放到句中读，再放到文中分段朗读课文。这样，一层层的推进，集中识字与随文识字相结合，字不离词，词不离句，只有将汉字及时纳入词中、句中，并在语言环境中会认、会读，才算真正“会认”，这样的识字也才是有意义。 4、指导写字。出示生字，让学生先自己观察，说说书写时要注意什么，强调笔画顺序。然后老师范写，学生在练习。 新课标提出：要让学生初步感受汉字的形体美。一年级学生是训练写字的关键时期，上课时有选择地渗透一些书法知识可为学生以后的书写奠定良好的基础。 【设计意图：在这一环节中，我从学生实际出发，以“扎实、朴实”为目标，利用课件，认读字词，努力在字词上抓落实，为深入学习课文打下坚实的基础。】（三）合作探究，细读体悟 《语文课程标准》明确指出：阅读是学生的个性化行为，不应以教师的分析代替学生的阅读实践，要十分重视培养学生的自学能力。因此在教学中，要特别注重学法的指导和渗透。 1、自由朗读课文，勾画语句，思考：什么时候觉得自己“很大”，用“____”画出来；“我”什么时候觉得自己“很小”，用“﹏”画出来。在小组内交流，想一想为什么。在汇报交流中，课件相机出示句子引导理解，并进行朗读指导。 2、仿照课文的句式说一说：你什么时候觉得自己很大？什么时候觉得自己很小？ 【设计意图：授人以鱼，不如授人以渔。在这一环节，让学生自己探究并找到答案。，以“画一画”“说一说”“读一读”的方式培养学生的动口、动手、动脑的学习习惯，充分激发了学生的主动意识和进取精神，加深了学生对文本的理解。同时，合作学习的方式，又培养了学生的合作意识和能力。】 （四）联系生活，拓展升华 《语文课程标准》指出：教学要结合课内外资源，多途径的提高学生的语文素养。 1、说一说，你是盼望长大，还是希望一直这样小小的？为什么。在全班交流中对学生进行情感教育，体验长大的快乐。
10.大还是小
课文10 大还是小 【教学目标】 1. 认识“时、候”等 11 个生字和双立人、点横头、竖心旁 3 个偏旁；会写“自、己”等 3 个字。 2. 正确、流利地朗读课文。结合插图，体会“我”自相矛盾的内心世界。 3. 结合生活体验，说说什么时候觉得自己很大，什么时候觉得自己很小。 【教学重点】 正确、流利地朗读课文。 【教学难点】 体会“我”自相矛盾的内心世界；会写“己、衣”等字。 【课前准备】 1．制作多媒体课件，准备生字词卡片。(教师) 2．借助拼音自主朗读课文，预习课文，标出自然段，圈画生字，拼读生字，记忆生字。(学生) 【课时安排】 2课时 【教学过程】 第一课时 一、激趣导入，引出课题，启发质疑 1．出示字卡“大”。 师：同学们，请大声地读这个字。(生读：大) 师：上课时，回答问题的声音要大。你知道和它意思相反的字是什么吗？(生答：小) 2．出示字卡“小”。 师：请小声地读这个字。(生读：小)上课时，回答问题声音不能太小，否则别人就听不到了。老师要
看看这节课谁的表现最棒。(同时出示字卡“大小”)现在，我们一起来读一读。(生读：大小) 3．质疑：你认为自己是大还是小呢？能说说为什么吗？(指名回答) 师：有一个小朋友也遇到了这个问题，他是怎么回答的呢？我们这节课就来学习《大还是小》。(板书课文题目) 4．读了课文题目，你有什么疑问？ (师生梳理出主要问题) 5．自主探究学习。 教师出示自探提示一。 (1)自由朗读课文，遇到不认识的字，借助拼音多读几遍。把词语读正确，句子读通顺。 (2)拼读课前圈画的生字，要读准字音，想办法记住这些生字。 6．根据自探提示先自主学习，然后在小组长的组织下在小组内交流。 二、初读课文，学习生字 检查自主学习情况。 (1)我会读 时候觉得穿衣服 自己很小快点儿 (2)我会认 ①这些词中有些生字需要我们记住，瞧，它们已经从词中跳出来了，你还能认出它们吗？ 时候觉得自己很穿衣服快 ②识记生字： ③我来考考大家： “我在洞穴里发现了一颗牙。”(穿) 这是我们的识字办法之一——编谜语，猜谜语。接下来要看你们的本领了，说说你们的识字办法吧！(学生自由选择生字说说自己的识字方法。) ④小结：识字的时候，我们不仅可以用加一加、换一换的方法，还可以用猜字谜的方法，但要注意编的字谜要合理。 ⑤指名认读，齐读。 三、写字指导(自、己) 1．交流谈话。 师：你觉得在这十一个生字中哪个字最简单？(己)组一个词好吗？(自己)现在我们就来写好下面这两
12. an en in un ün 教学目标： 1、学会前鼻韵母an en in un ün，读准音，认清形，能正确书写，学会整体认读音节yuan、yin、yun。 2、掌握关于本节课韵母的简单两拼、三拼音节，能自主拼读 教学重点： 1、对课本导引图片的探索与发现（开发同学们的思维探索能力） 2、掌握本节课韵母 新授课过程： 一、复习回忆 回忆学过的声母、单韵母、复韵母（整体背诵） 二、观察拼音，找到共同点 1.引入：今天，老师请来了五个拼音娃娃，它们是——an、en、in、un、ün，请你们仔细观察它们，你们有什么发现？ 2.学生发现都有n。 3.师：同学们观察的很仔细，这后面都有一个相同的字母n，但是这个小尾巴，在这儿不读n，那么这个音该怎么读呢，请小朋友看老师的手势（这是小朋友上面一排牙齿，这是上齿背、舌头这是舌尖）读的时候要把舌尖轻轻地顶上去，顶到上齿背（做两遍）请小朋友听老师的发音，舌尖顶住，请大家跟老师读一读。发这个音的时候气流要从鼻子里出来，对，轻轻地。谁来试试看。（提问并表扬，再一起来）再次强调（舌尖抵住上齿背，气流从鼻子里出来）设计意图：通过学生自己对拼音的观察，从而发现前鼻韵母的规律，这样能激发学生的学习兴趣，并把韵母记得更牢。 像这样有一个小尾巴n的韵母，我们叫它们前鼻韵母，来跟我说。 这节课我们一起来学这五个前鼻韵母，请大家看图，图上画着什么呢。 三、观察图片，学习韵母和整体认读音节 (一)教学an 1.出示图片（天安门）引导发音 师：图上画着什么呢(生：天安门城楼)对了，这个天安门的安，跟我们这节课要学习的第一个前鼻韵母的发音是一样的。大家仔细看，这个前鼻韵母是由a和n组成的。在读的时候，我们先要做好a的口型，先把嘴巴张大，再把舌尖顶住上齿背，请小朋友们听老师是怎么读的。An an an，跟我来学一学。（生：an an an）谁来试试？ 2.多种方式练读（多次练读）
12 an en in un ?n 教案设计 设计说明 《语文课程标准》指出：学生是学习和发展的主体，语文课程必须根据学生身心发展和语文学习的特点，关注学生的个体差异和不同的学习需求，爱护学生的好奇心，充分激发学生的主动意识和进取精神，倡导自主、合作、探究的学习方式。因此，围绕重点，教学设计遵循了趣味性、活动性和开放性三个原则。意在以活动和游戏为主，使儿童在愉快的教学环境中学习拼音，在多种多样的儿童喜闻乐见的活动中提高拼读能力，从而充分发挥汉语拼音帮助识字、学习普通话的作用。 课前准备 1.制作关于an、en、in、un、?n的多媒体课件。(教师) 2.制作关于前鼻韵母an、en、in、un、?n，整体认读音节yin、yun、yuan的音节卡片。(教师) 课时安排 2课时。 教学过程 第一课时 一、观察拼音，尝试发音 1.引入：今天，老师请来了五个拼音娃娃，它们是——an、en、in、un、?n，请你们仔细观察它们，你们有什么发现？ 2.学生发现都有n，尝试发音。 设计意图：通过学生自己对拼音的观察，从而发现前鼻韵母的规律，这样能激发学生的学习兴趣，并把韵母记得更牢。 二、观察图片，学习韵母和整体认读音节 (一)教学an和整体认读音节yuan。 1.出示图片引导发音。 师：请同学们跟随拼音娃娃一起去它们家里看看吧！看，这一家人都在干什么呢？(生：看电视呢)电视上演的是哪儿啊？(生：天安门)天安门的“安”就是旁边这个韵母的发音。谁来试试？发这个音的时候，先做好ɑ的口形，舌头再慢慢往上抬起，感觉音是从鼻子里面发出来的，请看我的口形……谁看清楚了？ 2.多种方式练读。 学生自由练读、指读、齐读、开火车读。 3.探究发音方法。 学生编个顺口溜来记这个韵母的发音，拍着小手一起说说。
复韵母“an en in”的拼读 ān án ǎn àn bān bán bǎn bàn pān pán pǎn pàn mān mán mǎn màn fān fán fǎn fàn dān dán dǎn dàn tān tán tǎn tàn nān nán nǎn nàn lān lán lǎn làn gān gán gǎn gàn kān kán kǎn kàn hān hán hǎn hàn zhān zhán zhǎn zhàn
chān chán chǎn chàn shān shán shǎn shàn zān zán zǎn zàn cān cán cǎn càn sān sán sǎn sàn yān yán yǎn yàn wān wán wǎn wàn rān rán rǎn ràn ēn ?n ěn an bēn b?n běn ban pēn p?n pěn pan mēn m?n měn man fēn f?n fěn fan
dēn d?n děn dan nēn n?n něn nan gēn g?n gěn gan kēn k?n kěn kan zhēn zh?n zhěn zhan chēn ch?n chěn chan shēn sh?n shěn shan zēn z?n zěn zan cēn c?n cěn can sēn s?n sěn san wēn w?n wěn wan rēn r?n rěn ran
īn ín ǐn ìn bīn bín bǐn bìn pīn pín pǐn pìn mīn mín mǐn mìn nīn nín nǐn nìn līn lín lǐn lìn jīn jín jǐn jìn qīn qín qǐn qìn xīn xín xǐn xìn 整体认读音节： yīn yín yǐn yìn
an en in an en in 教学目标 1．学会前鼻韵母an、en、in和整体认读音节yin及其四声，读准音，认清形，正确书写。 2．正确认读由声母和an、en、in组成的音节，会读拼音词和拼音句子。教学准备 情境图，歌曲磁带《我爱北京天安门》，。 课时安排：2课时 第一课时 课前游戏 1、做你指我猜的游戏（ai ui iu ye yue ei ao） 2、背儿歌读韵母 一、看图导入 1、小朋友们可真厉害，为了奖励你们，我们一起来看电视。出示情景图 2、谁会开电视机？（请同学开电视） 每个小朋友都跃跃欲试，想来开电视。 3、说说电视里在放什么？ 4、电视里的电视机里在放什么？
基本上能够说完整。 5、听小朋友们在唱什么歌？播放《我爱北京天安门》。 6、学习语境歌：有一首儿歌写的就是这幅图的内容，我们一起来学习。 7、听读儿歌：遥控器，摁一摁，荧屏出现天安门，色彩鲜艳音乐美，小朋友越看越开心。（跟老师读） 8、导出韵母，板贴：an en in 二、指导发音 1．学习韵母an。 (1)教师板书an，过渡叙述：这是天安门中“安”的拼音，去掉声调符号就是我们今天要学习的第一个韵母an。 (2)认清形。(教师指着a。)这个单韵母认识吗?在单韵母a的后面加了个鼻音做尾巴，这样组成的韵母就叫前鼻韵母（跟老师读两遍）。(3)指导读准音。教师告诉学生读这个韵尾要用舌尖抵住上齿龈，不留缝隙，鼻子出气。教师做口形，让学生体会舌尖顶住上齿龈的感觉。教师领读，指名读、学生自由练读，齐读，“开火车”读。 (4)指导读好an的四声。出示ān、án、ǎn、àn，请同学自由读，同桌互读，指名读。 2．学习韵母en。 (1)请学生做一做摁遥控器的动作。 (2)板书en，指导读好音。把“摁”改成第一声会读吗?
《an en in》教学设计 教学要求 1．学会前鼻韵母an、en、in和整体认读音节yin、yuan及其四声，读准音，认清形，正确书写。 2．正确认读由声母和an、en、in组成的音节，会读拼音词。 教学准备 情境图，复韵母卡片、课件 教学过程 第一课时 课前复习： 学过哪些单韵母？( a o e i u ü ) 6个 学过哪些复韵母？( ai ei ui ao ou iu ie ue er ) 8+1个 今天再来学习三个复韵母跟一个整体认读音节。 一、看图导入 1．教师出示情境图，引导学生观察：图画上的小兔正在干什么? 屏幕上出现了什么? 2．学习语境歌。 (1)、同时诵读情境歌。 (遥控器，摁一摁，荧屏映出天安门。色彩鲜艳音乐美，小朋友越看越开心。) (2)、请学生跟着老师念两遍情境歌，引出：天安门、摁、音。 3．教师引导过渡：今天我们就学习an、en、in这几个韵母。 同学们观察这三个复韵母。这些复韵母都是在一个单韵母后面加一个鼻音做尾巴，这样组成的韵母就叫鼻韵母。再说一遍读法。指导读前鼻音。 二、指导发音 1．学习韵母an。 (1)教师领读天安门，过渡叙述：“安”去掉声调要学习的an。领读。 （指导读法：先发a音，舌头迅速抵住上牙龈，鼻子发音） (2)指名读、学生自由练读，齐读，“开火车”读。 (3)老师教an写法。写三个，读三遍。 (4)指导读好an的四声。出示ān、án、ǎn、àn，请同学自由读，同桌互读，指名读。 课件出示简单的拼读。领读，小老师领读，自由读。 2．学习韵母en。 (1)我这里有个遥控器，我请一个同学来摁一下，引出en。 (2)领读：en，指导读好音。（先发e音，舌头迅速抵住上牙龈，鼻子发音） (3) 学生自由练读，齐读，“开火车”读。 (4) 老师教en写法。加声调的方法，加上四个声调。 (5)指导读好en的四声。请同学自由读，同桌互读，指名读。 3．学习韵母in和整体认读音节yin。 (1)同学们看这个韵母（板书in）指导学生自学：你们能运用前面学到的发音方法读这个复韵母吗? (2)检查自读情况，纠正问题。
汉语拼音12 an en in un ?n 教师：龙昌小学张顺英概述： 本课有四部分内容。第一部分是五个前鼻韵母ɑn、en、in、un、?n和三个整体认读音节yuɑn、yin、yun，配有图画。第一幅图是天安门，用“安”提示ɑn的音。第二幅图是圆圈，用“圆”提示yuɑn 的音。第三幅图是一个人在摁门铃，用“摁”提示en的音。第四幅图是小兔在树阴下乘凉，用“阴”提示in和yin的音。第五幅图是蚊子，用“蚊”提示un的音。第六幅图是白云，用“云”提示?n和yun的音。 第二部分是拼音练习。包括两项内容：（1）声母与ɑn、en、in、un、?n的拼音，巩固新学的韵母，复习j、q、x跟?组成音节省写?上两点的规则。（2）看图读音节词语，培养学生认识事物、准确拼音的能力。 第三部分是看图借助汉语拼音认字读韵文。配有一幅山区田园图，图画表现了韵文的意思。韵文中有许多含前鼻韵母的音节，还有要认的五个生字。 第四部分是儿歌，配有图画。儿歌中有多个含前鼻韵母的音节。随儿歌学习“半、云、她”三个字。 学情分析： 针对孩子们的年龄特点，在教学中要力求做到趣味性、富有童趣，让学生在游戏活动中能正确认读五个前鼻韵母和它们组成的音节。另外在学习本课之前，学生已学过9个复韵母，会读声母和韵母组成的音节，能按两相拼的方法拼读音节。对三相拼的拼读方法掌握起来比较困难，老师要加强指导。学生单独拼读一个音节容易，如果连成一句话就困难些，老师可以示范教读，然后叫小朋友们模仿读。 教学目标： 知识与技能： 1.学会前鼻韵母an en in un ?n和整体认读音节yuan yin yun，读准音，记清形，正确书写。 2.学习声母与前鼻韵母组成的音节，准确拼读音节，读准三拼音节，复习?上两点省写规则。 3.能够看图说话，根据音节拼读词语和句子。