更正：3.26题
Homework 3：
Questions from Data Networks:
3.12 Let τ1 and τ2 be two exponentially distributed, independent random variables with means 1/λ1 and 1/λ2. Show that the random variable min{τ1, τ2} is exponentially distributed with mean 1/(λ1+λ2) and that P{τ1<τ2}=λ1/(λ1+λ2). Use these facts to show that the M/M/1 queue can be described by a continuous-time Markov chain with transition rates q n(n+1)=λ, q (n+1)n =μ, n=0,1,…. Solution:
{}()()()()()x x x e e e x P x P x x P x P 2121212121,,min λλλλττττττ+---==>>=>>=≥
因此{}21,min ττ服从指数分布，均值为
2
11
λλ+
()
()()
()2
1121202210121212λλλλλλλττλλλλλ+=
∞???? ??++-=-=<+---∞
-?x x x x e e dx
e e P
3.13 Persons arrive at a taxi stand with room for W taxis according to a Poisson process with rate λ. A person boards a taxi upon arrival if one is available and otherwise waits in a line. Taxis arrive at the stand according to a Poisson process with rate μ. An arriving taxi that finds the stand full departs immediately; otherwise, it picks up a customer if at least one is waiting, or else joins the queue of waiting taxis.
(a) Use an M/M/1 queue formulation to obtain the steady-state distribution of the person ’s queue. What is the steady-state probability distribution of the taxi queue size when W = 5 and λ and μ are equal to 1 and 2 per minute, respectively? (Answer: Let p i = Probability of i taxis waiting. Then p 0 = 1/32, p 1 = 1/32, p 2 = 1/16, p 3 = 1/8, p 4 = 1/4, p 5 = 1/2.)
(b) In the leaky bucket flow control scheme to be discussed in Chapter 6, packets arrive at a network entry point and must wait in a queue to obtain a permit before entering the network. Assume that permits are generated by a Poisson process with given rate and can be stored up to a given maximum number; permits generated while the maximum number of permits is available are discarded. Assume also that packets arrive according to a Poisson process with given rate. Show how to obtain the occupancy distribution of the queue of packets waiting for permits. Hint: This is the same system as the one of part (a).
(c) Consider the flow control system of part (b) with the difference that permits are not generated according to a Poisson process but are instead generated periodically at a given rate. (This is a more realistic assumption.) Formulate the problem of finding the occupancy distribution of the packet queue as an M/D/1 problem. Solution ：
（a ）以(n,m)系统表示系统状态,n 为车的数目,m 为人
的数目.可能的状态为(根据题目所描述的规则，不会存在车和人同时排队的情况): (W,0),(W-1,0),…(0,0),(0,1),…,(0,n),….
状态转移率为:
(i,0)－>(i-1,0)为λ,
(0,n)－>(0,n ＋1)也是λ, (i-1,0)－>(i,0)为μ, (0,n)－>(0,n-1)也是μ.
车的队列的分布（可根据马氏链解出P(W,0)=1-ρ .其他任意状态的概率可由P(W,0)来表示。）
P[0辆车]＝∑n ≥0ρW+n (1- ρ)=ρW . P[i 辆车]= ρW-i (1- ρ), 1≤i ≤W
同样可以算出人的队列的分布: P[无人]=∑0≤i ≤W P(i,0), P[i 个人]=P(i,0). 3.13(b)是的3.13(a)应用
3.14（此题给出的题目有误，不算分。正确的题目及答案如下）：A communication node A receives Poisson packet traffic from two other nodes, 1 and 2, at rates λ1 and λ2, respectively, and transmits it, on a first-come first_serve basis, using a link with capacity C bits/sec. The two input streams are assumed independent and their packet lengths are identically and exponentially distributed with mean L bits. A packet from node 1 is always accepted by A. A packet from node 2 is accepted only if the number of packets in A (in queue or under transmission) is less than a given number K > 0; otherwise, it is assumed lost.
(a) What is the range of values of λ1 and λ2 for which the expected number of packets in A will stay bounded as time increases?
(b) For λ1 and λ2 in the range of part (a) find the steady-state probability of having n packets in A (∞<≤n 0). Find the average time needed by a packet from source 1 to clear A once it enters A, and the average number of packets in A from source 1. Repeat for packets from source 2. Solution:
(a)因为当A 内的节点数≥K 时，来自2 的数据包即被丢弃，所以主要考虑来自节点1的数据包。节点A 接受所有来自节点1的包,所以当λ1<μ=C/L 时系统是稳定的, 即节点A 内期望的packet 数<∞.
(b)马氏链的状态转移率:
q i,i+1=λ1+λ2, 0≦i
列出DBE,解出平稳分布,并计算系统内平稳客户数N.
源1的packet 从到达系统到离开的平均时间T 1=1/μ+N(1/μ)= (1+N) (1/μ); 源1的packet 在系统内的平均数为: N 1=λ1T 1 下面考虑源2
设N’为系统内客户数 源2的packet 从到达系统到离开的平均时间T 2= (1+N’) (1/μ); ∑ ∑-=-=='1010/K n n K n n p np N 实际进入节点A 的源2的到达率λ’2等于 λ2(1-P[节点A 内packet 数≥K]) 所以源2的packet 在系统内的平均数N2=λ’2T 2. 3.15 Consider a system that is identical to M/M/1 except that when the system empties out, service does not begin again until k customers are present in the system (k is given). Once service begins it proceeds normally until the system becomes empty again. Find the steady-state probabilities of the number in the system, the average number in the system, and the average delay per customer. [Answer: The average number in the system is N = ρ/(1-ρ)+(k-1)/2.] Solution: The Markov chain is 由GBE ，可得： ()()' -'-' ''===122110k k p p p p p p λλλλλλ ()'-''====? 1210k p p p p 且有 ()() ()() () 0111011120 1p p p p p p p p p p p p k k k k +=+=+=+==-'--'λλμλλμλμ ()k i p p i i ≥=+λμ1 即 () ()k i p p i i ≤≤+++=-110 1ρρρ () ()k i p p k k i i >+++=--+0 111ρρρ 由 10 1 =+∑∑∞ =='i i k i i p p 得 ()k p ρ-=10 系统平均顾客数： ()()ρρ-+-=+=∑∑∞ =='1210 1 k ip ip N i i k i i 平均时间： λN T = 3.16 M/M/1-Like System with State-Dependent Arrival and Service Rate. Consider a system which is the same as M/M/1 except that the rate λn and service rate μn when there are n customers in the system depend on n. Show that ()001p p n n ρρ =+ Where 1+=k k k μλρ and ()1 0001-∞ =?? ????+=∑k k p ρρ Solution: The markov chain is 根据此链写出细节平衡方程（DBE ），易证所给结论。 3.18 Empty taxis pass by a street corner at a Poisson rate of 2 per minute and pick up a passenger if one is waiting there. Passengers arrive at the street corner at a Poisson rate of 1 per minute and wait for a taxi only if there are fewer than four persons waiting; otherwise, they leave and never return. Find the average waiting time of a passenger who joins the queue. (Answer: 13/15 min.) Solution: 汽车站内至多有4个客户等待, 系统状态n ∈{0,1,2,3,4}; λ=1, μ=2. 马氏链同于M/M/1/4, 平均waiting 时间为M/M/1/4系统客户平均延迟T=N/λ’. λ’=λ(1-P[系统内有4个客户]) 实际上等车队列的队头客户处于抽象排队系统的服务器中. 3.22 An athletic facility has five tennis courts. Players arrive at the courts at a Poisson rate of one pair per 10 min and use a court for an exponentially distributed time with mean 40 min. (a) Suppose that a pair of players arrives and finds all courts busy and k other pairs waiting in queue. How long will they have to wait to get a court on the average? (b) What is the average waiting time in queue for players who find all courts busy on arrival? Solution: (a) 相当于M/M/5系统server 全忙的情形, 需要等(k+1)(1/m μ). (相当于服务速率是m μ的一个单服务器系统) (b)问题要求计算条件期望值:在所有server 都忙时客户在队列内的等待时间（可使用P175，式3.37）. 因为在所有server 都忙时队列内平均客户数N’等于ρ/(1-ρ), ρ=λ/m μ. 应用Lilltle’s 定理得到 W=N’/λ=ρ/λ(1-ρ). 或使用类似(P-K)公式的分析,得到: W=(N’+1)(1/m μ),where N’=N Q /P Q . 3.23 Consider an M/M /∞ queue with servers numbered 1,2,… There is an additional restriction that upon arrival a customer will choose the lowest-numbered server that is idle at the time. Find the fraction of time that each server is busy. Will the answer change if the number o f servers is finite? Hint: Argue that in steady-state the probability that all of the first m servers are busy is given by the Erlang B formula of the M/M/m/m system. Find the total arrival rate to servers (m+1) and higher, and from this, the arrival rate to each server. Solution ： 如果单独看前m 个server ，与一个M/M/m/m 系统是一致的。因为： 1、前m 个server 有空闲时，到达客户一定会进入前m 个server 。 2、前m 个server 都忙时，到达客户进入后面的server ，相当于丢弃。 因此前m 个server 忙的概率由Erlang-B 公式给出： ()()∑==m n n m m n m p 0 ! ! μλμλ m+1及以后所有server 的总到达率： λm m p r = server m 的到达率： ()λλm m m m m p p r r -=-=--11 server m 的利用率为 μ λm 其中r m 表示前m 个server 都忙时，从server m 到达server m+1及其他server 的到达率 λm 表示实际到达server m 的到达率。 3.24 M/M/1 shared Service System. Consider a system which is the same as M/M/1 except that whenever there are n customers in the system they are all served simultaneously at an equal rate 1/n per unit time. Argue that the steady-state occupancy distribution is the same as for the M/M/1 system. Note: It can be shown that the steady-state occupancy distribution is the same as for M/M/1 even if the service time distribution is not exponential (i.e., for an M/B/1 type of system). Solution ： 参照PPT-3中的无穷小分析方法： P(恰有一个离开) = P(有一个离开)P(无到达) 其中：P(无到达) = ()δλδλδo e --=-1 P(有一个离开) = ()δμδμδδμδμδμδμo e e i e C i i i i i i +=??? ? ??? ?=?????????? ? ????? ???? ??? ? ??-?? ? ??--?? ? ??-1 1 因此 P(恰有一个离开) = ()[]()[]()δμδδμδδλδo o o +=+--1 同理，P(恰有一个到达) = ()δλδo + 因此结果与M/M/1是一致的。 3.26 A facility of m identical machines is sharing a single repairperson. The time to repair a failed machine is exponentially distributed with mean 1/λ. A machine, once operational, fails after a time that is exponentially distributed with mean 1/μ. All failure and repair times are independent. What is the steady-state proportion of time where there is no operational machine? Solution ： 以系统内的能使用的机器数n 为状态, 做一马氏链. 该马氏链的转移率如下: n →n-1的转移率为n μ; n →n+1的转移率为λ. 列细节平衡方程，求0p 。 或者以系统内坏掉的机器数m 为状态，做一马氏链，求p m 。 3.28 In 3.11, verify the formula ()γμλσs f 21=. Hint: Write {} ?? ? ???????????????????? ??=????????????? ??=∑∑==n E E E f E n i i n i i |2 1212 γγ, And use the fact that n is Poisson distributed. Solution ：证明按照条件期望值展开进行。 塑性力学中本构关系的讨论 摘要：本构方程是塑性力学解决问题不同于弹性力学的一大不同点，本文从主要描述塑性变形问题的两个本构理论出发，借鉴现有理论和实验结果，对比增量理论和全量理论的优缺及各自在工程中的适用性。 关键词：塑性力学；增量理论；全量理论；有限元法 引言 塑性力学和弹性力学之间的根本差别在于弹性力学是以应力与应变成线性关系的广义胡克定律为基础的。而塑性力学研究范畴中，应力与应变一般成非线性关系，而这种非线性的特征又不能一概而论，对于不同的材料，在不同的条件下，都具有不同的规律。塑性变形的基本规律是建立在实验的基础上，根据实验结果简化抽象出塑性状态下应力与应变关系的特征。 与弹性力学比较，主要影响塑性力学本构方程的有以下几点： 应力与应变之间的关系是非线性的，其比例系数不仅与材料有关而且与塑性应变有关； 由于塑性变形的出现，弹塑性材料在卸载时，体元的应力-应变状态不能沿原来的加载路径返回，应力与应变之间不再存在一一对应的关系，而与加载历史有关； 变形体中可分为弹性区和塑性区，在弹性区，加载与卸载都服从广义胡克定律，在塑性区，加载过程服从塑性规律而卸载过程服从广义胡克定律。 因此在塑性力学发展初期，最初提出的是以增量方法来讨论应力增量与应变增量之间的关系，它不受加载条件的限制，但在实际计算过程中，需要按加载过程中的变形路径进行积分，计算比较复杂。Hencky于1924年提出的全量理论在实践中使用方便很多，但全量本构关系仅能应用于特定情况，及体元应力-应变过程为单调过程，不能描述弹塑性变形规律全貌。 1.增量理论 塑性应力应变关系的重要特点是非线性和非简单对应，非线性及应力与应变关系不是线性关系，非简单对应及应变不能由应力唯一确定。在材料变形的塑性阶段， 弹塑性本构关系的认识及其在钢筋 混凝土结构中的应用浅谈 摘要：本文首先对弹塑性本构关系和钢筋混凝土材料的本构模型作了简要概述，然后结合上课所学知识和自己阅读的几篇文章，从材料的屈服准则、流动准则、硬化准则和加载卸载准则等四个方面详细阐述了弹塑性本构关系。最后，结合上述准则简要论述了混凝土这一常用材料在地震作用下的弹塑性本构关系。 关键词：弹塑性本构关系，钢筋混凝土，地震 Understanding of Elastoplastic Constitutive Relation and a Brife Talk of Its Aapplication to Reinforced Concrete Structure Abstract：This paper firstly makes a brief overview about elastoplastic constitutive relation and reinforced concrete constitutive model. Then，elaborating the elastoplastic constitutive relation from the four aspects of material yield criterion，flow rule，hardening rule，loading and unloading criterion based on what I have learned in class and reading from a few articles. Lastly，a simply introduction on the elastoplastic constitutive of reinforced concrete under earthquake is demonstrated. Keywords：elastoplastic constitutive relation; reinforced concrete structure; earthquake 1 引言 钢筋混凝土结构材料的本构关系对钢筋混凝土结构有限元分析结果有重大的影响，如果选用的本构关系不能很好地反映材料的各项力学性能，那么其它计算再精确也无法反映结构的实际受力特征。所谓材料的本构关系，主要是指描述材料力学性质的数学表达式。用什么样的表达式来描述材料受力后的变化规律呢？不同的学者根据材料的性质、受力条件和大小、试验方法以及不同的理论模型等因素综合考虑，建立了许多种钢筋混凝土材料的本构关系表达式。 材料的本构关系所基于的理论模型主要有：弹性理论、非线性弹性理论、弹塑性理论、粘弹性理论、粘弹塑性理论、断裂力学理论、损伤力学理论、内时理论等。迄今为止，由于钢筋混凝土材料的复杂因素，还没有一种理论模型被公认为可以完全描述钢筋混凝土材料的 岩土类材料的弹塑性力学模型及本构方程 摘要：本文主要结合岩土类材料的特性，开展研究其在受力变形过程中的弹性及塑性变形的特点，描述简化的力学模型特征及对应的适用条件，同时在分析研究其弹塑性力学模型的基础上，探究了关于岩土类介质材料的各种本构模型，如M-C、D-P、Cam、D-C、L-D及节理材料模型等，分析对应使用条件，特点及公式，从而推广到不同的材料本构模型的研究，为弹塑性理论更好的延伸发展做一定的参考性。 关键词：岩土类材料，弹塑性力学模型，本构方程 不同的固体材料，力学性质各不相同。即便是同一种固体材料，在不同的物理环境和受力状态中，所测得的反映其力学性质的应力应变曲线也各不相同。尽管材料力学性质复杂多变，但仍是有规律可循的，也就是说可将各种反映材料力学性质的应力应变曲线，进行分析归类并加以总结，从而提出相应的变形体力学模型。 第一章岩土类材料 地质工程或采掘工程中的岩土、煤炭、土壤，结构工程中的混凝土、石料，以及工业陶瓷等，将这些材料统称为岩土材料。 岩土塑性力学与传统塑性力学的区别在于岩土类材料和金属材料具有不同的力学特性。岩土类材料是颗粒组成的多相体，而金属材料是人工形成的晶体材料。正是由于不同的材料特性决定了岩土类材料和金属材料的不同性质。归纳起来，岩土材料有3点基本特性：1.摩擦特性。2.多相特性。3.双强度特性。另外岩土还有其特殊的力学性质：1.岩土的压硬性，2.岩土材料的等压屈服特性与剪胀性，3.岩土材料的硬化与软化特性。4.土体的塑性变形依赖于应力路径。 对于岩土类等固体材料往往在受力变形的过程中，产生的弹性及塑性变形具备相应的特点，物体本身的结构以及所加外力的荷载、环境和温度等因素作用，常使得固体物体在变形过程中具备如下的特点。 固体材料弹性变形具有以下特点：(1)弹性变形是可逆的。物体在变形过程中，外力所做的功以能量（应变能)的形式贮存在物体内，当卸载时，弹性应变能将全部释放出来，物体的变形得以完全恢复； (2)无论材料是处于单向应力状态，还是复杂应力状态，在线弹性变形阶段，应力和应变成线性比例关系；(3)对材料加载或卸载，其应力应变曲线路径相同。因此，应力与应变是一一对应的关系。 固体材料的塑性变形具有以下特点： (l)塑性变形不可恢复，所以外力功不可逆。塑性变形的产生过程，必定要消耗能量(称耗散能或形变功)； (2)在塑性变形阶段，应力和应变关系是非线性的。因此，不能应用叠加原理。又因为加载与卸载的规律不同，应力与应变也不再存在一一对应的关系，也即应力与相应的应变不能唯一地确定，而应当考虑到加载的路径(即加载历史)； (3)当受力固体产生塑性变形时，将同时存在有产生弹性变形的弹性区域和产生塑性变形的塑性区域。并且随着载荷的变化，两区域的分界面也会产生变化。【6A文】塑性力学增量和全量本构关系讨论
弹塑性本构关系的认识及其在钢筋混凝土中的应用浅谈_塑
岩土类材料弹塑性力学模型及本构方程
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