麻省理工 网络优化 课程课件08basicalgorithmsformaxflow

合集下载

麻省理工 网络优化 课程课件20lagrangian_relaxation_1

麻省理工 网络优化 课程课件20lagrangian_relaxation_1
A
0 or 1 for all ( i , j )
The Lagrangian relaxation was obtained by penalizing the constraint and then eliminating (relaxing) the constraint. Theorem. L( ) z( ) z*
j
( i , j ) A ij
j
xij
x ji
1 if i = s 1 if i = t 0 otherwise A
xij
0 or 1 for all ( i , j )
7
What is the shortest path from node 1 to node 6?
2
(1,10)
(1,1) (2,3)
c xij
j
j
xij
t xij
x ji
T
1 if i = s 1 if i = t 0 otherwise
Complicating constraint
( i , j ) A ij
xij
0 or 1 for all (i, j)
A
4
Example
Find the shortest path from node 1 to node 6 with a transit time at most 10 2
1
(10,3)
(10,1)
6
To reduce the transit time of the shortest path, we put a penalty proportional to transit time. Suppose that we charge a toll of $1 per unit transit time.

网络优化及实例ppt课件

网络优化及实例ppt课件


运价 20 23 26 29
精品课件
32

20
精品课件
21
钢管运输问题(CUMCM-
2000B)
常用解法: 二次规划
先计算最小运费矩阵
➢ 两种运输方式(铁路/公路)混合最短路问题 ➢ 是普通最短路问题的变种,需要自己设计算法
精品课件
22
钢管运输问题(CUMCM-
2000B)
fi表示钢厂i是否使用;xij是从钢厂i运到节点j的钢管量 yj是从节点j向左铺设的钢管量;zj是向右铺设的钢管量
Min
i, j
(
pi
c ij
) xij
0.1 2
15
[(1
j 1
y j ) y j (1 z j ) z j ]
15
s.t. 500 f i xij S i f i , i 1,..., 7. j 1
7
xij y j z j ,
j 1,..., 15 .
cumcm2000b.lg4
性规划模型除了可以利用数学软件求解 外,讨论问题推广时应设计快速近似算法 3.一题多解讨论算法性能比较与分析

精品课件
24
大规模数据处理是近年竞赛题的 倾如:向
1. 04年A题:奥运会临时超市网点设计 2. 05年A题:长江水质的评价和预测 3. 05年B题:DVD的在线租赁
难 度逐年增大
•单向? •双向?
精品课件
2
欧拉把哥尼斯堡七桥问题转化为一个 图论上的问题:
精品课件
3
七桥问题 的
顶因
答案是 否定的 点 为






精品课件

麻省理工学院开放课件-通信与信息-电气工程与计算机科学

麻省理工学院开放课件-通信与信息-电气工程与计算机科学

1.1 Shannon's grand challenge
The �eld of information theory and coding has a unique history, in that many of its ultimate limits were determined at the very beginning, in Shannon's founding paper [7]. Shannon's most celebrated result is his channel capacity theorem, which we will review in Chapter 3. This theorem states that for many common classes of channels there exists a channel capacity C such that there exist codes at any rate R � C that can achieve arbitrarily reliable transmission, whereas no such codes exist for rates R � C . For a band-limited AWGN channel, the capacity C in bits per second (b/s) depends on only two parameters, the channel bandwidth W in Hz and the signal-to-noise ratio SNR, as follows: C � W log2(1 + SNR) b�s: Shannon's theorem has posed a magni�cent challenge to succeeding generations of researchers. Its proof is based on randomly chosen codes and optimal (maximum likelihood) decoding. In practice, it has proved to be remarkably di�cult to �nd classes of constructive codes that can be decoded by feasible decoding algorithms at rates which come at all close to the Shannon limit. Indeed, for a long time this problem was regarded as practically insoluble. Each signi�cant advance toward this goal has been awarded the highest accolades the coding community has to o�er, and most such advances have been immediately incorporated into practical systems. In the next two sections we give a brief history of these advances for two di�erent practical channels: the deep-space channel and the telephone channel. The deep-space channel is an unlimited-bandwidth, power-limited AWGN channel, whereas the telephone channel is very much bandwidth-limited. (We realize that many of the terms used here may be unfamiliar to the reader at this point, but we hope that these surveys will give at least an impressionistic picture. After reading later chapters, the reader may wish to return to reread these sections.) Within the past decade there have been remarkable breakthroughs, principally the invention of turbo codes [1] and the rediscovery of low-density parity check (LDPC) codes [4], which have allowed the capacity of AWGN and similar channels to be approached in a practical sense. For example, Figure 1 (from [2]) shows that an optimized rate-1/2 LDPC code on an AWGN channel can approach the relevant Shannon limit within 0.0045 decibels (dB) in theory, and within 0.04 dB with an arguably practical code of block length 107 bits. Practical systems using block lengths of the order of 104 {105 bits now approach the Shannon limit within tenths of a dB.

MIT(麻省理工)信号与系统讲义-lecture2

MIT(麻省理工)信号与系统讲义-lecture2

Observation: Even if the independent variable is time, there are interesting and important systems which have boundary conditions.
6
Ex. #5
• A rudimentary “edge” detector
“Proof” a) Suppose system is causal. Show that (*) holds.
b) Suppose (*) holds. Show that the system is causal.
19
LINEAR TIME-INVARIANT (LTI) SYSTEMS • Focus of most of this course - Practical importance (Eg. #1-3 earlier this lecture are all LTI systems.) - The powerful analysis tools associated with LTI systems
16
LINEARITY
A (CT) system is linear if it has the superposition property: If x1(t) →y1(t) and x2(t) →y2(t) then ax1(t) + bx2(t) → ay1(t) + by2(t)
y[n] = x2[n] Nonlinear, TI, Causal y(t) = x(2t) Linear, not TI, Noncausal Can you find systems with other combinations ? -e.g. Linear, TI, Noncausal Linear, not TI, Causal

美国麻省理工——结构力学unit8

美国麻省理工——结构力学unit8
∇4φ = 0
homogeneous form
What happened to E, ν??
Paul A. Lagace © 2001
Unit 8 - p. 9
MIT - 16.20
Fall, 2002
⇒ this function, and accompanying governing equation, could be defined in any curvilinear system (we’ll see one such example later) and in plane strain as well.
These must be solved for a generic body under some generic loading subject to the prescribed boundary conditions (B.C.’s) There are two types of boundary conditions: 1. Normal (stress prescribed) 2. Geometric (displacement prescribed) --> you must have one or the other To solve this system of equations subject to such constraints over the continuum of a generic body is, in general, quite a challenge. There are basically two solution procedures: 1. Exact -- satisfy all the equations and the B.C.’s 2. Numerical -- come as “close as possible” (energy methods, etc.)

麻省理工大学算法导论lecture02

麻省理工大学算法导论lecture02

2
(
5 + 5 2 1 + 16 16
+L geometric series
L2.17
( ) +( )
)
Introduction to Algorithms
The master method
The master method applies to recurrences of the form T(n) = a T(n/b) + f (n) , where a ≥ 1, b > 1, and f is asymptotically positive.
Day 3
Introduction to Algorithms
L2.11
Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 T(n/16) T(n/8) (n/2)2 T(n/8) T(n/4)
Day 3
Introduction to Algorithms
(n/8)2

Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 (n/16)2 Θ(1)
Day 3 Introduction to Algorithms L2.16
n2
5 n2 16 25 n 2 256

(n/2)2 (n/8)2 (n/4)2
Pick c1 big enough to handle the initial conditions.
Day 3 Introduction to Algorithms L2.7

MIT基础数学讲义(计算机系)lecture8

MIT基础数学讲义(计算机系)lecture8

1 Annuities
Would you prefer a million dollars today or $50,000 a year for the rest of your life? This is a question about the value of an annuity. An annuity is a nancial instrument that pays out a xed amount of money at the beginning of every year for some speci ed number of years. In particular, an n-year, m-payment annuity pays m dollars at the start of each year for n years. In some cases, n is nite, but not always. Examples include lottery payouts, student loans, and home mortgages. There are even Wall Street people who specialize in trading annuities. A key question is what an annuity is worth. For example, lotteries often pay out jackpots over many years. Intuitively, $50,000 a year for 20 years ought to be worth less than a million dollars right now. If you had all the cash right away, you could invest it and begin collecting interest. But what if the choice were between $50,000 a year for 20 years and a half million dollars today? Now it is not clear which option is better. In order to answer such questions, we need to know what a dollar paid out in the future is worth today. We will to assume that money can be invested at a xed annual interest rate p. These days a good estimate for p is around 8% we'll use this value for the rest of the lecture. Here is why the interest rate p matters. Ten dollars invested today at interest rate p will become (1 + p) = $10:80 in a year, (1 + p)2 $11:66 in two years, and so forth. Looked at another way, ten dollars paid out a year from now are only really worth 1=(1 + p) $9:26 today. The reason is that if we had the $9:26 today, we could invest it and would have $10.00 in a year anyway. Therefore, p determines the value of money paid out in the future.

class08

class08

4.4 局域网组网技术
4.4.1 4.4.2 4.4.3 4.4.4 4.4.5 4.4.6 以太网 快速以太网 千兆位以太网 令牌环网络 FDDI光纤环网 ATM局域网
4.4.1 以太网
1.以太网 以太网(Ethernet)是由美国Xerox公司和 Stanford大学联合开发并于1975年提出的,目的 是为了把办公室工作站与昂贵的计算机资源连接 起来,以便能从工作站上分享计算机资源和其他 硬件设备。 1983年IEEE802委员会公布的802.3局域网络协 议(CSMA/CD),基本上和Ethernet技术规范一 致,于是,Ethernet技术规范成为世界上第一个 局域网的工业标准。
4.2 局域网介质访问控制方式
4.2.1 载波侦听多路访问/冲突检测法 4.2.2 令牌环访问控制方式 4.2.3 令牌总线访问控制方式
4.2.1 载波侦听多路访问/冲突 检测法
载波侦听多路访问(CSMA,Carrier Sense Multiple Access)是一种适合于总线结构的具有 信道检测功能的分布式介质访问控制方法,其控 制手段称之为“载波侦听”。 实际上,当一个站开始发送信息时,检测到本 次发送有无冲突的时间很短,它不超过该站点与 距离该站点最远站点信息传输时延的2倍。假设A 站点与距离A站最远B站点的传输时延为T(图4-1 所示),那么2T就作为一个时间单位。
FDDI的网络拓扑结构属于星形、环形结 构,如图4-16所示。
A类站2 主环 副环 A类 站1 集线器 路由器 WAN A类站3

图 FDDI
络 4 拓 扑 16 结 构 的
B类站6 B类站5 B类站4
2.FDDI介质访问控制方式
FDDI 所 采 用 的 介 质 访 问 控 制 方 式 与 IEEE802.5标准中的对应部分相似。所不 同的是802.5中采用的是单数据帧访问方 式;而在FDDI中则采用多数据帧访问方 式,即允许在环路中同时存在着多个数 据帧,可提高信道利用率。

麻省理工 网络优化 课程课件03graphsearchalgorithms

麻省理工 网络优化 课程课件03graphsearchalgorithms
4
Initialize
Initialize begin
unmark all nodes in N; mark node s; pred(s) = 0; {that is, it has no predecessor} next := 1; (next is a counter} order(s) := next; LIST = {s}
9
Next Steps
Prove a “cutset theorem” showing how the algorithm will terminate. A cutset is a partition of the nodes into two parts S and T = N\S. Show that it ends with the set of nodes reachable from the origin. Show that the distances from the origin node in a breadth first tree aref such a cutset exists, then there is no path from s to t.
11
Proof of “only if” part of the Cutset Theorem
6 1 3 t
S
5 s 2 4
T
Let S be the set of nodes reachable from s. Let T be the remaining nodes
end
Unmarking takes O(n) All else takes O(1)
8
Finding an admissible arc

麻省理工大学算法导论lecture08

麻省理工大学算法导论lecture08

of hash functions. Suppose h is used to hash n arbitrary keys into the m slotsen key x, we have E[#collisions with x] < n/m.
2001 by Charles E. Leiserson Introduction to Algorithms Day 12 L8.2
Universal hashing
Definition. Let U be a universe of keys, and let H be a finite collection of hash functions, each mapping U to {0, 1, …, m–1}. We say H is universal if for all x, y ∈ U, where x ≠ y, we have |{h ∈ H : h(x) = h(y)}| = |H|/m. That is, the chance of a collision between x and y is 1/m if we choose h randomly from H.
i =0 i =1
r
(mod m)
or r a0 ( x0 y0 ) + ∑ ai ( xi yi ) ≡ 0 which implies that a0 ( x0 y0 ) ≡ ∑ ai ( xi yi )
i =1
2001 by Charles E. Leiserson Introduction to Algorithms
y∈T {x}
∑ cxy .
2001 by Charles E. Leiserson

几种常见的优化方法ppt课件

几种常见的优化方法ppt课件
fast, this is relatively unimportant because the time
required for integration is usually trivial in comparison to
the time required for the force calculations.
Example of integrator for MD simulation
• One of the most popular and widely used integrators is
the Verlet leapfrog method: positions and velocities of
7
Continuous time molecular dynamics
1. By calculating the derivative of a macromolecular force
field, we can find the forces on each atom
as a function of its position.
11
Choosing the correct time step…
1. The choice of time step is crucial: too short and
phase space is sampled inefficiently, too long and the
energy will fluctuate wildly and the simulation may
– Rigid body dynamics
– Multiple time step algorithms

麻省理工 网络优化 课程课件06radixheapalgorithm

麻省理工 网络优化 课程课件06radixheapalgorithm

1
2
3
4
K
K+1
Each path from 1 to K+1 gives a production and inventory schedule. The cost of the path is the cost of the schedule. 1 6 8 11
K+1
Interpretation: produce in periods 1, 6, 8 and 11. Conclusion: The minimum cost path from node 1 to node K+1 gives the minimum cost lot-sizing solution.
22 2 22 0 1 41 3 41 31 5 14 26 3 27 45 4 29 6
If costs are larger, one can make larger buckets. But then, how does one do a “find min?” 60 69 70 79
0-9
10 19
0
1
12
Some General Comments
The Update Step seems fairly routine, but reinserting a node can take O(K) steps, where K is the number of buckets. The Find Min Step is complicated, and it seems more time intensive than anything else we have looked at. It looks like it could be O(n) per find min and thus O(n2) in total. Remarkable Fact. A relatively elementary implementation of this algorithm leads to an O(m + n log C) algorithm.

MIT本科计算机教材

MIT本科计算机教材

30 教材名称: COMPUTER SIMULATION OF LIOUIDS
作者: ALLEN
31 教材名称: CONTROL OF UNCERTAIN SYSTEMS
作者: DAHLEH
32 教材名称: CONTROL SYSTEM DESIGN
作者: FRIEDALND
作者: PERLMAN
62 教材名称: INTRO TO ALGORITHMS
作者: CORMEN
63 教材名称: INTRO.TO FORTRAN 90 F/ENGRS.+SCI.
作者: NYHOFF
64 教材名称: INTRO.TO MATLAB F/ENGRS+SCI.
作者: HOWE
76 教材名称: MICROSOFT ACCESS 2000 BIBLE
作者: PRAGUE
77 教材名称: MICROSYSTEM DESIGN
作者: SENTURIA
78 教材名称: MICRSFT,VISUAL BASIC:PROGRAMMER'S GDE
作者: MENEZES
59 教材名称: HOW COMPUTERS WORK,MILLENIUM ED.-W/CD
作者: WHITE
60 教材名称: HOW TO SET UP+MAINTAIN A WEB SITE-W/CD
作者: STEIN
61 教材名称: INTERCONNECTIONS:BRIDGES,ROUTERS…
作者: HERTZ
68 教材名称: Introduction to Random Signals & Applied Kalman Filtering 3/E (With Matlab Exercise & Solutions )

麻省理工 网络优化 课程课件15successiveshortestpath

麻省理工 网络优化 课程课件15successiveshortestpath

2
5 0
7
Update the Potentials
-7 2 7 0 0 1 0 6 3 -6 0 2 5 -8 5 6 0 2 1 43 -8 4 To obtain new node potentials, subtract the shortest path distances from the old potentials. The shortest path tree now has 0 reduced cost arcs.
Capacity Scaling
15
Augmentations Are Per Scaling Phase
Assume: there is a large capacity path from node 1 to each node and from each node to node 1. At the end of the ∆-scaling phase: a. There is no node i with excess e(i) ≥ ∆, or b. There is no node j with excess e(i) ≤ -∆. And each arc in G(x, ∆) has a non-negative reduced cost. The infeasibility at the end of the ∆-scaling phase is at most n∆. The infeasibility at the beginning of the ∆-scaling phase is O(m∆). The number of augmentations per scaling phase is O(m).
5

麻省理工 网络优化 课程课件07shortestpathslabelcorrectingalgo

麻省理工 网络优化 课程课件07shortestpathslabelcorrectingalgo

INPUT G = (N, A) with costs c Node 1 is the source node There is no negative cost cycle We will relax that assumption later
2
Optimality Conditions
Lemma. Let d*(j) be the shortest path length from node 1 to node j, for each j. Let d( ) be node labels with the following properties: d(j) ≤ d(i) + cij for i ∈ N for j ≠ 1. (1) d(1) = 0. (2) Then d(j) ≤ d*(j) for each j. Proof. Let P be any path from node 1 to node j, with length c(P), and suppose P has k arcs. Claim: d(j) ≤ c(P). Note: if P is the shortest path from 1 to j, then d(j) ≤ c(P) = d*(j), which is what we want to prove.
6
A Generic Shortest Path Algorithm
Algorithm LABEL CORRECTING; begin d(1) : = 0 and Pred(1) := ø; d(j) : = ∞ for each j ∈ N – {1}; while some arc (i,j) satisfies d(j) > d(i) + cij do begin d(j) : = d(i) + cij; Pred(j) : = i; end; end; Label correcting animation.

麻省理工 网络优化 课程课件13globalmincutalgorithm

麻省理工 网络优化 课程课件13globalmincutalgorithm


xij = 2 for each node i j
(1) (2) (3) (4)
4
∑ i∈S , j∈N \ S xij ≥ 2 for each node set S ≠ N
xij = x ji 0 ≤ xij ≤ 1 for each i , j xij integer for each i , j
2
Appli Problem
What is a minimum length tour that visits each point?
3
An integer programming formulation
Let xij = 1 if there is an arc (i,j) in the tour T xij = 0 otherwise. There are two arcs incident to node i For every cutset (S, N-S), there are two arcs from S to N-S.
15.082 and 6.855J April 1, 2003
The Global Min Cut problem
1
Global Min cut
INPUT: A network G = (N, A) OUTPUT: A cut (S, N\S) such that cap(S, N\S) is minimum. Note: We do not assume that there is a source node s and destination node t. Typically, but not always, the network is undirected.
8

MIT 麻省课件-08

MIT 麻省课件-08
Project
Main building
1st floor
Structures Plumbings Electrical HVAC
2nd floor
Service building
Structures Plumbings Electrical HVAC
Procedure Approach
Project
1.040/1.401
Project Management
Spring 2007 Planning
Dr. SangHyun Lee
Department of Civil and Environmental Engineering Massachusetts Institute of Technology
Construction force
Worker Subcontractor
Electrical Plumbing Insulation Tankage Roads
OBS/WBS Linkages
Linking the WBS and OBS.
Project People
WBS
OBS
Major Facility
Types of WBS
Project WBS
Operative tool for monitoring and control contractor work
Contractual WBS
Defines the level of reporting that the seller will provide the buyer May include less detail than Project WBS Agreed between owner and contractor
  1. 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
6
Transformation to a maximum flow problem
persons 1 1 1 1 1 2 3 4 6 7 1 8 tasks 5 1 1 1 t
s
Does the maximum flow from s to t have 4 units?
7
The Residual Network
10
Proof of Correctness of the Algorithm
Assume that all data are integral. Lemma: At each iteration all residual capacities are integral. Proof. It is true at the beginning. Assume it is true after the first k-1 augmentations, and consider augmentation k along path P. The residual capacity ∆ of P is the smallest residual capacity on P, which is integral. After updating, we modify residual capacities by 0, or ∆, and thus residual capacities stay integral.
s 6 9 1 3 7 t
not reachable from s in G(x)
2
13
Method 2: Cut Duality Theory
10, 9 s 6, 6 1 1,1 10,7 8,8 t
2
An (s,t)-cut in a network G = (N,A) is a partition of N into two disjoint subsets S and T such that s ∈ S and t ∈ T, e.g., S = { s, 1 } and T = { 2, t }. The capacity of a cut (S,T) is CAP(S,T) = ∑i∈S∑j∈Tuij
12
How do we know when a flow is optimal?
10, 9 s 6, 6 1 1,1 10,7 8,8 t
2
METHOD 1. There is no augmenting path in the residual network. reachable 1 8 1 from s in G(x)
v
Maximize Subject to
Σj xij
-
Σk xki
=
= 0
for each i ≠ s,t
Σj xsj
v
0 ≤ xij ≤ uij for all (i,j) ∈ A.
2
Maximum Flows
We refer to a flow x as maximum if it is feasible and maximizes v. Our objective in the max flow problem is to find a maximum flow.
1 1,1 6, 5 10,6 8,7 t
10, 8 s
2
A max flow problem. Capacities and a nonoptimum flow.
3
The feasibility problem: find a feasible flow
warehouses 6 5 4 5 4 1 2 3 4 5 6 7 8 9 6 7 6 5 retailers
11
Theorem. The Ford-Fulkerson Algorithm is finite
Proof. The capacity of each augmenting path is at least 1. The augmentation reduces the residual capacity of some arc (s, j) and does not increase the residual capacity of (s, i) for any i. So, the sum of the residual capacities of arcs out of s keeps decreasing, and is bounded below by 0. Number of augmentations is O(nU), where U is the largest capacity in the network.
2
If S = {s,1}, then the flow across (S, T) is 8+1+6 = 15.
10, 9 s 6, 6
1 1,1
8,8 t 10,7
2
If S = {s,2}, then the flow across (S, T) is 9 + 7 - 1 = 15.
16
More on Flows Across Cuts
persons 1 2 3 4 tasks 5 6 7 8
Is there a way of assigning persons to tasks so that each person is assigned a task, and each task has a person assigned to it?
18
Max Flow Min Cut Theorem
Theorem. (Optimality conditions for max flows). The following are equivalent. 1. A flow x is maximum. 2. There is no augmenting path in G(x). 3. There is an s-t cutset (S, T) whose capacity is the flow value of x. Corollary. (Max-flow Min-Cut). The maximum flow value is the minimum value of a cut. Proof of Theorem. 1 ⇒ 2. (not 2 ⇒ not 1) Suppose that there is an augmenting path in G(x). Then x is not maximum.
s
4 4 5 5 4 4
There is a 1-1 correspondence with flows from s to t with 24 units (why 24?) and feasible flows for the transportation problem.
5
The feasibility problem: find a matching
10, 9 s 6, 6
=
1
∑i∈S∑j∈T xij
8,8 t 10,7
-
∑i∈S∑j∈T xji If S = {s}, then the flow across (S, T) is 9 + 6 = 15.
15
1,1
2
Flows Across Cuts
10, 9 s 6, 6 1 1,1 10,7 8,8 t
10, 8 s 6, 5 1 1,1 10,6 uij - xij 2 s 8 1 5 2 1 1 1 7 4 6 t i xij j 8,7 t xij uij
i
j
2
The Residlet rij denote the residual capacity of arc (i,j)
Is there a way of shipping from the warehouses to the retailers to satisfy demand?
4
Transformation to a max flow problem
warehouses 6 6 5 5 1 2 3 4 5 6 7 8 9 6 6 7 7 t 6 6 5 5 retailers
2 s 8 1 5 2 1 1 1 7 4 6 6 2 t s
for (i,j) ∈ P.
2 8 1 1 8 4 6 9 t
The Ford Fulkerson Maximum Flow Algorithm
Begin x := 0; create the residual network G(x); while there is some directed path from s to t in G(x) do begin
8
A Useful Idea: Augmenting Paths
An augmenting path is a path from s to t in the residual network. The residual capacity of the augmenting path P is δ(P) = min{rij : (i,j) ∈ P}. To augment along P is to send d(P) units of flow along each arc of the path. We modify x and the residual capacities appropriately. rij := rij - δ(P) and rji := rji + δ(P)
let P be a path from s to t in G(x); ∆ := δ(P); send ∆ units of flow along P; update the r's;
相关文档
最新文档