Chapter7 7.1.3-7.1.4
第七章 图论
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7.1 图及相关概念
7.1.5 子图
Graphs
图论
定义7-1.8 给定图G1=<V1,E1>和G2=<V2,E2> ， (1)若V1V2 ，E1E2 ，则称G1为G2的子图。 (2)若V1=V2 ，E1E2 ，则称G1为G2的生成子图。
上图中G1和G2都是G的子图，
但只有G2是G的生成子图。
chapter7
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7.1 图及相关概念
7.1.6 图的同构
Graphs
图论
【例4】 设G1,G2,G3,G4均是4阶3条边的无向简单图，则
它们之间至少有几个是同构的？ 解：由下图可知，4阶3条边非同构的无向简单图共有3个， 因此G1,G2,G3,G4中至少有2个是同构的。
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7.1 图及相关概念
7.1.3 完全图
Graphs
图论
【例2】证明在 n（n≥2 ）个人的团体中，总有两个人在 此团体中恰好有相同个数的朋友。 分析 :以结点代表人，二人若是朋友，则在结点间连上一 证明：用反证法。 条边，这样可得无向简单图G，每个人的朋友数即该结点 设 G 中各顶点的度数均不相同，则度数列为 0 ， 1 ， 2 ， …， 的度数，于是问题转化为： n 阶无向简单图 G中必有两个 n-1 ，说明图中有孤立顶点，与有 n-1 度顶点相矛盾（因 顶点的度数相同。 为是简单图），所以必有两个顶点的度数相同。
vV1
deg(v) deg(v) deg(v) 2 | E |
vV2 vV
由于 deg( v) 是偶数之和，必为偶数，
vV1
数据模型与决策(运筹学)课后习题和案例答案(6)
CHAPTER 7NETWORK OPTIMIZATION PROBLEMS Review Questions7.1-1 A supply node is a node where the net amount of flow generated is a fixed positive number.A demand node is a node where the net amount of flow generated is a fixed negativenumber. A transshipment node is a node where the net amount of flow generated is fixed at zero.7.1-2 The maximum amount of flow allowed through an arc is referred to as the capacity of thatarc.7.1-3 The objective is to minimize the total cost of sending the available supply through thenetwork to satisfy the given demand.7.1-4 The feasible solutions property is necessary. It states that a minimum cost flow problemwill have a feasible solution if and only if the sum of the supplies from its supply nodesequals the sum of the demands at its demand nodes.7.1-5 As long as all its supplies and demands have integer values, any minimum cost flowproblem with feasible solutions is guaranteed to have an optimal solution with integervalues for all its flow quantities.7.1-6 Network simplex method.7.1-7 Applications of minimum cost flow problems include operation of a distribution network,solid waste management, operation of a supply network, coordinating product mixes atplants, and cash flow management.7.1-8 Transportation problems, assignment problems, transshipment problems, maximum flowproblems, and shortest path problems are special types of minimum cost flow problems. 7.2-1 One of the company’s most important distribution centers (Los Angeles) urgently needs anincreased flow of shipments from the company.7.2-2 Auto replacement parts are flowing through the network from the company’s main factoryin Europe to its distribution center in LA.7.2-3 The objective is to maximize the flow of replacement parts from the factory to the LAdistribution center.7.3-1 Rather than minimizing the cost of the flow, the objective is to find a flow plan thatmaximizes the amount flowing through the network from the source to the sink.7.3-2 The source is the node at which all flow through the network originates. The sink is thenode at which all flow through the network terminates. At the source, all arcs point awayfrom the node. At the sink, all arcs point into the node.7.3-3 The amount is measured by either the amount leaving the source or the amount entering thesink.7.3-4 1. Whereas supply nodes have fixed supplies and demand nodes have fixed demands, thesource and sink do not.2. Whereas the number of supply nodes and the number of demand nodes in a minimumcost flow problem may be more than one, there can be only one source and only onesink in a standard maximum flow problem.7.3-5 Applications of maximum flow problems include maximizing the flow through adistribution network, maximizing the flow through a supply network, maximizing the flow of oil through a system of pipelines, maximizing the flow of water through a system ofaqueducts, and maximizing the flow of vehicles through a transportation network.7.4-1 The origin is the fire station and the destination is the farm community.7.4-2 Flow can go in either direction between the nodes connected by links as opposed to onlyone direction with an arc.7.4-3 The origin now is the one supply node, with a supply of one. The destination now is theone demand node, with a demand of one.7.4-4 The length of a link can measure distance, cost, or time.7.4-5 Sarah wants to minimize her total cost of purchasing, operating, and maintaining the carsover her four years of college.7.4-6 When “real travel” through a network can end at more that one node, a dummy destinationneeds to be added so that the network will have just a single destination.7.4-7 Quick’s management must consider trade-offs between time and cost in making its finaldecision.7.5-1 The nodes are given, but the links need to be designed.7.5-2 A state-of-the-art fiber-optic network is being designed.7.5-3 A tree is a network that does not have any paths that begin and end at the same nodewithout backtracking. A spanning tree is a tree that provides a path between every pair of nodes. A minimum spanning tree is the spanning tree that minimizes total cost.7.5-4 The number of links in a spanning tree always is one less than the number of nodes.Furthermore, each node is directly connected by a single link to at least one other node. 7.5-5 To design a network so that there is a path between every pair of nodes at the minimumpossible cost.7.5-6 No, it is not a special type of a minimum cost flow problem.7.5-7 A greedy algorithm will solve a minimum spanning tree problem.17.5-8 Applications of minimum spanning tree problems include design of telecommunicationnetworks, design of a lightly used transportation network, design of a network of high- voltage power lines, design of a network of wiring on electrical equipment, and design of a network of pipelines.Problems7.1a)b)c)1[40] 6 S17 4[-30] D1 [-40] D2 [60] 5 8S2 6[-30] D37.2a)supply nodestransshipment nodesdemand nodesb)[200] P1560 [150]425 [125][0] W1505[150]490 [100]470 [100][-150]RO1[-200]RO2P2 [300]c)510 [175]600 [200][0] W2390 [125]410[150] 440[75]RO3[-150]7.3a)supply nodestransshipment nodesdemand nodesV1W1F1V2V3W2 F21P1W1RO1RO2P2W2RO3[-50] SE3000[20][0]BN5700[40][0]HA[50]BE 4000 6300[40][30] [0][0]NY2000[60]2400[20]3400[10] 4200[80][0]5900[60]5400[40]6800[50]RO[0]BO[0]2500[70]2900[50]b)c)7.4a)LA 3100 NO 6100 LI 3200 ST[-130] [70] [30] [40] [130]1[70]11b)c) The total shipping cost is $2,187,000.7.5a)[0][0] 5900RONY[60] 5400[0] 2900 [50]4200 [80][0] [40] 6800 [50]BO[0] 2500LA 3100 NO 6100 LI 3200 ST [-130][70][30] [40][130]b)c)SEBNHABERONYNY(80) [80] (50) [60](30)[40] ROBO (40)(50) [50] (70)[70]11d)e)f) $1,618,000 + $583,000 = $2,201,000 which is higher than the total in Problem 7.5 ($2,187,000). 7.6LA(70) NO[50](30)LI (30) ST[70][30] [40]There are only two arcs into LA, with a combined capacity of 150 (80 + 70). Because ofthis bottleneck, it is not possible to ship any more than 150 from ST to LA. Since 150 actually are being shipped in this solution, it must be optimal. 7.7[-50] SE3000 [20] [0] BN 5700 [40][0] HA[50] BE4000 6300[40][0] NY2000 [60] 2400 [20][30] [0]5900RO [60]17.8 a) SourcesTransshipment Nodes Sinkb)7.9 a)AKR1[75]A [60]R2[65] [40][50][60] [45]D [120] [70]B[55]E[190]T [45][80] [70][70]R3CF[130][90]SE PT KC SL ATCHTXNOMES S F F CAb)Oil Fields Refineries Distribution CentersTXNOPTCACHATAKSEKCME c)SLSFTX[11][7] NO[5][9] PT[8] [2][5] CA [4] [7] [8] [7] [4] [6][8] CH [7][5][9] [4] ATAK [3][6][6][12] SE KC[8][9][4][8] [7] [12] [11]MESL [9]SF[15][7]d)3Shortest path: Fire Station – C – E – F – Farming Community 7.11 a)A70D40 60O60 5010 B 20 C5540 10 T50E801c)Shortest route: Origin – A – B – D – Destinationd)Yese)Yes7.12a)31,00018,000 21,00001238,000 10,000 12,000b)17.13a) Times play the role of distances.B 2 2 G5ACE 1 31 1b)7.14D F1. C---D: Cost = 14.E---G: Cost = 5E---F: Cost = 1 *choose arbitrarilyD---A: Cost = 4 2.E---G: Cost = 5 E---B: Cost = 7 E---B: Cost = 7 F---G: Cost = 7 E---C: Cost = 4 C---A: Cost = 5F---G: Cost = 7C---B: Cost = 2 *lowestF---C: Cost = 3 *lowest5.E---G: Cost = 5 F---D: Cost = 4 D---A: Cost = 43. E---G: Cost = 5 B---A: Cost = 2 *lowestE---B: Cost = 7 F---G: Cost = 7 F---G: Cost = 7 C---A: Cost = 5F---D: Cost = 46.E---G: Cost = 5 *lowestC---D: Cost = 1 *lowestF---G: Cost = 7C---A: Cost = 5C---B: Cost = 2Total = $14 million7.151. B---C: Cost = 1 *lowest 4. B---E: Cost = 72. B---A: Cost = 4 C---F: Cost = 4 *lowestB---E: Cost = 7 C---E: Cost = 5C---A: Cost = 6 D---F: Cost = 5C---D: Cost = 2 *lowest 5. B---E: Cost = 7C---F: Cost = 4 C---E: Cost = 5C---E: Cost = 5 F---E: Cost = 1 *lowest3. B---A: Cost = 4 *lowest F---G: Cost = 8B---E: Cost = 7 6. E---G: Cost = 6 *lowestC---A: Cost = 6 F---G: Cost = 8C---F: Cost = 4C---E: Cost = 5D---A: Cost = 5 Total = $18,000D---F: Cost = 57.16B 34 2E HA D 2 G I K3C F 12J34B41E6A C41G2 FD1. F---G: Cost = 1 *lowest 6. D---A: Cost = 62. F---C: Cost = 6 D---B: Cost = 5F---D: Cost = 5 D---C: Cost = 4F---I: Cost = 2 *lowest E---B: Cost = 3 *lowestF---J: Cost = 5 F---C: Cost = 6G---D: Cost = 2 F---J: Cost = 5G---E: Cost = 2 H---K: Cost = 7G---H: Cost = 2 I---K: Cost = 8G---I: Cost = 5 I---J: Cost = 33. F---C: Cost = 6 7. B---A: Cost = 4F---D: Cost = 5 D---A: Cost = 6F---J: Cost = 5 D---C: Cost = 4G---D: Cost = 2 *lowest F---C: Cost = 6G---E: Cost = 2 F---J: Cost = 5G---H: Cost = 2 H---K: Cost = 7I---H: Cost = 2 I---K: Cost = 8I---K: Cost = 8 I---J: Cost = 3 *lowestI---J: Cost = 3 8. B---A: Cost = 4 *lowest4. D---A: Cost = 6 D---A: Cost = 6D---B: Cost = 5 D---C: Cost = 4D---E: Cost = 2 *lowest F---C: Cost = 6D---C: Cost = 4 H---K: Cost = 7F---C: Cost = 6 I---K: Cost = 8F---J: Cost = 5 J---K: Cost = 4G---E: Cost = 2 9. A---C: Cost = 3 *lowestG---H: Cost = 2 D---C: Cost = 4I---H: Cost = 2 F---C: Cost = 6I---K: Cost = 8 H---K: Cost = 7I---J: Cost = 3 I---K: Cost = 85. D---A: Cost = 6 J---K: Cost = 4D---B: Cost = 5 10. H---K: Cost = 7D---C: Cost = 4 I---K: Cost = 8E---B: Cost = 3 J---K: Cost = 4 *lowestE---H: Cost = 4F---C: Cost = 6F---J: Cost = 5G---H: Cost = 2 *lowest Total = $26 millionI---H: Cost = 2I---K: Cost = 8I---J: Cost = 37.17a) The company wants a path between each pair of nodes (groves) that minimizes cost(length of road).b)7---8 : Distance = 0.57---6 : Distance = 0.66---5 : Distance = 0.95---1 : Distance = 0.75---4 : Distance = 0.78---3 : Distance = 1.03---2 : Distance = 0.9Total = 5.3 miles7.18a) The bank wants a path between each pair of nodes (offices) that minimizes cost(distance).b) B1---B5 : Distance = 50B5---B3 : Distance = 80B1---B2 : Distance = 100B2---M : Distance = 70B2---B4 : Distance = 120Total = 420 milesHamburgBostonRotterdamSt. PetersburgNapoliMoscowA IRFIELD SLondonJacksonvilleBerlin RostovIstanbulCases7.1a) The network showing the different routes troops and supplies may follow to reach the Russian Federation appears below.PORTSb)The President is only concerned about how to most quickly move troops and suppliesfrom the United States to the three strategic Russian cities. Obviously, the best way to achieve this goal is to find the fastest connection between the US and the three cities.We therefore need to find the shortest path between the US cities and each of the three Russian cities.The President only cares about the time it takes to get the troops and supplies to Russia.It does not matter how great a distance the troops and supplies cover. Therefore we define the arc length between two nodes in the network to be the time it takes to travel between the respective cities. For example, the distance between Boston and London equals 6,200 km. The mode of transportation between the cities is a Starlifter traveling at a speed of 400 miles per hour * 1.609 km per mile = 643.6 km per hour. The time is takes to bring troops and supplies from Boston to London equals 6,200 km / 643.6 km per hour = 9.6333 hours. Using this approach we can compute the time of travel along all arcs in the network.By simple inspection and common sense it is apparent that the fastest transportation involves using only airplanes. We therefore can restrict ourselves to only those arcs in the network where the mode of transportation is air travel. We can omit the three port cities and all arcs entering and leaving these nodes.The following six spreadsheets find the shortest path between each US city (Boston and Jacksonville) and each Russian city (St. Petersburg, Moscow, and Rostov).The spreadsheets contain the following formulas:Comparing all six solutions we see that the shortest path from the US to Saint Petersburg is Boston → London → Saint Petersburg with a total travel time of 12.71 hours. The shortest path from the US to Moscow is Boston → London → Moscow with a total travel time of 13.21 hours. The shortest path from the US to Rostov is Boston →Berlin → Rostov with a total travel time of 13.95 hours. The following network diagram highlights these shortest paths.-1c)The President must satisfy each Russian city’s military requirements at minimum cost.Therefore, this problem can be solved as a minimum-cost network flow problem. The two nodes representing US cities are supply nodes with a supply of 500 each (wemeasure all weights in 1000 tons). The three nodes representing Saint Petersburg, Moscow, and Rostov are demand nodes with demands of –320, -440, and –240,respectively. All nodes representing European airfields and ports are transshipment nodes. We measure the flow along the arcs in 1000 tons. For some arcs, capacityconstraints are given. All arcs from the European ports into Saint Petersburg have zero capacity. All truck routes from the European ports into Rostov have a transportation limit of 2,500*16 = 40,000 tons. Since we measure the arc flows in 1000 tons, the corresponding arc capacities equal 40. An analogous computation yields arc capacities of 30 for both the arcs connecting the nodes London and Berlin to Rostov. For all other nodes we determine natural arc capacities based on the supplies and demands at the nodes. We define the unit costs along the arcs in the network in $1000 per 1000 tons (or, equivalently, $/ton). For example, the cost of transporting 1 ton of material from Boston to Hamburg equals $30,000 / 240 = $125, so the costs of transporting 1000 tons from Boston to Hamburg equals $125,000.The objective is to satisfy all demands in the network at minimum cost. The following spreadsheet shows the entire linear programming model.HamburgBoston Rotterdam St.Petersburg+500-320Napoli Moscow A IRF IELDSLondon -440Jacksonville Berlin Rostov+500-240Istanbul The total cost of the operation equals $412.867 million. The entire supply for SaintPetersburg is supplied from Jacksonville via London. The entire supply for Moscow is supplied from Boston via Hamburg. Of the 240 (= 240,000 tons) demanded by Rostov, 60 are shipped from Boston via Istanbul, 150 are shipped from Jacksonville viaIstanbul, and 30 are shipped from Jacksonville via London. The paths used to shipsupplies to Saint Petersburg, Moscow, and Rostov are highlighted on the followingnetwork diagram.PORTSd)Now the President wants to maximize the amount of cargo transported from the US tothe Russian cities. In other words, the President wants to maximize the flow from the two US cities to the three Russian cities. All the nodes representing the European ports and airfields are once again transshipment nodes. The flow along an arc is againmeasured in thousands of tons. The new restrictions can be transformed into arccapacities using the same approach that was used in part (c). The objective is now to maximize the combined flow into the three Russian cities.The linear programming spreadsheet model describing the maximum flow problem appears as follows.The spreadsheet shows all the amounts that are shipped between the various cities. The total supply for Saint Petersburg, Moscow, and Rostov equals 225,000 tons, 104,800 tons, and 192,400 tons, respectively. The following network diagram highlights the paths used to ship supplies between the US and the Russian Federation.PORTSHamburgBoston Rotterdam St.Petersburg+282.2 -225NapoliMoscowAIRFIELDS-104.8LondonJacksonvilleBerlin Rostov +240 -192.4Istanbule)The creation of the new communications network is a minimum spanning tree problem.As usual, a greedy algorithm solves this type of problem.Arcs are added to the network in the following order (one of several optimal solutions):Rostov - Orenburg 120Ufa - Orenburg 75Saratov - Orenburg 95Saratov - Samara 100Samara - Kazan 95Ufa – Yekaterinburg 125Perm – Yekaterinburg 857.2a) There are three supply nodes – the Yen node, the Rupiah node, and the Ringgit node.There is one demand node – the US$ node. Below, we draw the network originatingfrom only the Yen supply node to illustrate the overall design of the network. In thisnetwork, we exclude both the Rupiah and Ringgit nodes for simplicity.b)Since all transaction limits are given in the equivalent of $1000 we define the flowvariables as the amount in thousands of dollars that Jake converts from one currencyinto another one. His total holdings in Yen, Rupiah, and Ringgit are equivalent to $9.6million, $1.68 million, and $5.6 million, respectively (as calculated in cells I16:K18 inthe spreadsheet). So, the supplies at the supply nodes Yen, Rupiah, and Ringgit are -$9.6 million, -$1.68 million, and -$5.6 million, respectively. The demand at the onlydemand node US$ equals $16.88 million (the sum of the outflows from the sourcenodes). The transaction limits are capacity constraints for all arcs leaving from thenodes Yen, Rupiah, and Ringgit. The unit cost for every arc is given by the transactioncost for the currency conversion.Jake should convert the equivalent of $2 million from Yen to each US$, Can$, Euro, and Pound. He should convert $1.6 million from Yen to Peso. Moreover, he should convert the equivalent of $200,000 from Rupiah to each US$, Can$, and Peso, $1 million from Rupiah to Euro, and $80,000 from Rupiah to Pound. Furthermore, Jake should convert the equivalent of $1.1 million from Ringgit to US$, $2.5 million from Ringgit to Euro, and $1 million from Ringgit to each Pound and Peso. Finally, he should convert all the money he converted into Can$, Euro, Pound, and Peso directly into US$. Specifically, he needs to convert into US$ the equivalent of $2.2 million, $5.5 million, $3.08 million, and $2.8 million Can$, Euro, Pound, and Peso, respectively. Assuming Jake pays for the total transaction costs of $83,380 directly from his American bank accounts he will have $16,880,000 dollars to invest in the US.c)We eliminate all capacity restrictions on the arcs.Jake should convert the entire holdings in Japan from Yen into Pounds and then into US$, the entire holdings in Indonesia from Rupiah into Can$ and then into US$, and the entire holdings in Malaysia from Ringgit into Euro and then into US$. Without the capacity limits the transaction costs are reduced to $67,480.d)We multiply all unit cost for Rupiah by 6.The optimal routing for the money doesn't change, but the total transaction costs are now increased to $92,680.e)In the described crisis situation the currency exchange rates might change every minute.Jake should carefully check the exchange rates again when he performs thetransactions.The European economies might be more insulated from the Asian financial collapse than the US economy. To impress his boss Jake might want to explore other investment opportunities in safer European economies that provide higher rates of return than US bonds.。
形式语言与自动机Chapter7练习参考解答
Chapter 7 练习参考解答Exercise 7.1.3 从以下文法出发：S → 0A0 | 1B1 | BBA → CB → S | AC → S | εa) 有没有无用符号？如果有的话去除它们。
b) 去除ε-产生式。
c) 去除单位产生式。
d) 把该文发转化为乔姆斯基范式。
参考解答：a)没有无用符.b) 所有符号S,A,B,C都是可致空的，消去ε-产生式后得到新的一组产生式:S → 0A0 | 1B1 | BB | B | 00 | 11A → CB → S | AC → Sc) 单元偶对包括：（A,A）,（B,B）,（C,C）,（S,S）,（A,C）,（A,S）,（A,B）,（B,A）,（B,C）,（B,S）,（C,A）,（C,B）,（C,S）,（S,A）,（S,B）,（S,C）,消去单元产生式后得到新的一组产生式S → 0A0 | 1B1 | BB | B | 00 | 11A → CB → S | AC → SS → 0A0 | 1B1 | BB | 00 | 11A → 0A0 | 1B1 | BB | 00 | 11B → 0A0 | 1B1 | BB | 00 | 11C → 0A0 | 1B1 | BB | 00 | 11d)先消去无用符号C，得到新的一组产生式:S → 0A0 | 1B1 | BB | 00 | 11A → 0A0 | 1B1 | BB | 00 | 11B → 0A0 | 1B1 | BB | 00 | 11引入非终结符C，D，增加产生式C → 0和D → 1，得到新的一组产生式:S → CAC | DBD | BB | CC | DDA → CAC | DBD | BB | CC | DDB → CAC | DBD | BB | CC | DDC → 0D → 1引入非终结符E，F，增加产生式E → CA和F → DB，得到满足Chomsky范式的一组产生式:S → EC | FD | BB | CC | DDA → EC | FD | BB | CC | DDB → EC | FD | BB | CC | DDE → CAF → DBC → 0D → 1Exercise 7.2.1(b)用CFL泵引理来证明下面的语言都不是上下文无关的：b) {a n b n c i | i ≤n}。
真核基因表达调控模式
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7.1.3 真核生物DNA水平上的基因表达调控
分子生物学的最新研究表明，在个体发育过程中，用来合成 RNA的DNA模板也会发生规律性变化，从而控制基因表达和 生物体的发育。
高度重复基因的形成通常与个体分化阶段DNA的某些变化
有关。例如，一个成熟的红细胞能产生大量的可翻译出成熟 珠蛋白的mRNA，而其前体细胞却不产生珠蛋白。许多情况
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⑥ 真核生物的RNA在细胞核中合成，只有经转运 穿过核膜，到达细胞质后，才能被翻译成蛋白质， 原核生物中不存在这样严格的空间间隔。
⑦ 许多真核生物的基因只有经过复杂的成熟和剪接 过程（maturation and splicing），才能顺利地翻 译成蛋白质。
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在真核生物中，前rRNA 转录产物的分子量为45S，包 括18S，28S和5.8S三个主要 rRNA分子。前rRNA分子中 至少有100处被甲基化（主要 是核糖的2-OH甲基化），原 始转录产物也被特异性RNA 酶切割降解，产生成熟rRNA 分子。5S rRNA作为一个独立 的转录单位，由RNA聚合酶 III（而ห้องสมุดไป่ตู้是聚合酶I）完成1转7 录。
胞人工染色体(MAC)就是以此为基础, 再加自主复制序列
(ARS20)2、1/8/1选7 择标记和插入位点组建的。
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(2) DNA顺序重复
轻度、中度、高度重复序列三种：
轻度重复序列：单拷贝基因；一个基因组中有一个或几个拷贝 的序列；例如结构基因基本上属于不重复序列，如蛋清蛋白、 蚕的丝心蛋白等。
基因的编码序列在DNA分子上是不连续的，为不编码的序列所隔开。
不连续基因是通过mRNA和DNA杂交试验发现的。
现代物流的配送及业务管理
定量配送：按规定的批量在一个指定的时间范围中进行配送。
定时定量配送：按照规定配送时间和配送数量进行配送。
定时、定路线配送：在规定的运行路线上制定到达时间表， 按运行时间表进行配送，用户可按规定路线及规定时间接货和提 出配送要求。
即时配送：完全按用户要求的时间和数量进行配送的方式。
这种方式是以某天的任务为目标，在充分掌握这一天需要地点、
需要量及种类的前提下，及时安排最优的配送路线并安排相应的
配202送3/6车/18 辆，实行配送。
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7.1.3.4其它配送方式
❖共同配送〔joint distribution〕
几个配送中心联合起来，共同制定 方案，共同对某一地区用户进行配送， 具体执行时共同使用配送车辆，这种方
式称共同配送。
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重点掌握：配送的定义、特点、方式、配送中 心的功能，两种配货作业方法，节约里程法，提高 配送经济效益的途径。
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配送及配送业务管理：
第一节 配送概念 第二节 配送业务 第三节 配送管理
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配送是现代物流系统的一个子 系统。
配送〔distribution〕是按用 户的订货要求，在物流据点进行分 货、配货等工作，并将配好的货物 按时送达指定的地点和收货人的物 流活动。
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7.1 配送概念
7.1 Conception of Distribution
7.1.1 配送的特点
1、配送是从物流据点至用户的一种特殊送货形式。
2、从事送货的是专职流通企业〔配送〕。
3、配送不是单纯的运输或输送，而是运输与其它活动 共同构成的组合体。
4、配送是以供给者送货到户式的服务性供给。
英语词汇学课件chapter-7-Changes-in-Word-Meaning
▪ 7.1.1 Extension of meaning
Extension of meaning, also known as generalization, is the name given to the widening of meaning which some words undergo. It is a process by which a word which originally had a specialized meaning has now become generalized or has extended to cover a broader and often less definite concept. e.g. manuscript, fabulous, barn, thing, feedback, Sandwich.
▪ 7.1.5 Transference of Meaning
Some words which were used to designate one thing but later changed to mean something else have experienced the process of semantic transfer, known as transference or transfer of meaning. E.g. paper.
▪ According to Bernard and Delbridge (1980), there is a mechanism that facilitates the change in word meaning.
Each word is likely to have certain associative connotations, the total of which forms its associative field. This field embraces all the concepts that may occur to a mind when stimulated by that word. Peculiar or usual, there will normally be a connection of some sort between the given word and each of the other words in its associative field. The associations are dependent on one of the four types.
语言学教程Chapter Seven
language and culture, Language and society,
and language and cross-cultural communication.
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7.1 language and culture What is culture? Broadly speaking, it means the total way of life of a people, including the patterns of belief, customs, objects, institutions, techniques, and language. In a narrow sense, it refers to local or specific practice, beliefs or customs. 7.1.1 The relationship between L & C There exists a close relationship between language and culture, generally, a relation of part to whole, for L is part of C. Language is an indispensible carrier of culture. The knowledge and beliefs that constitute a people’s culture are habitually encoded and transmitted in L. Culture finds a better representation through language use. Language and culture correlate with each other at different levels of linguistic structure.
高中生物chapter7线粒体和过氧化物酶体
高中生物chapter7线粒体和过氧化物酶体第七章线粒体和过氧化物酶体7.1 线粒体的形态结构1850年，德国生物学家Rudolph K?lliker第一系统的研究了线粒体。
(肌细胞) 1900年，Leonor Michaelis 氧化还原反应1943年，Arbert Claude采用盐法分离技术分离到线粒体1948年George Hogeboom等采用蔗糖介质分离有活性的线粒体，确定了能量转换的部位。
逐步证明了线粒体具有Krebs循环、电子传递、氧化磷酸化的作用，从而证明了线粒体是真核生物进行能量转换的主要部位。
7.2 线粒体结构与化学组成线粒体膜通透性实验将线粒体放在100 mM蔗糖溶液中，蔗糖穿过外膜进入线粒体的膜间间隙；然后将线粒体取出测定线粒体内部蔗糖的平均浓度，结果只有50 mM，比环境中蔗糖的浓度低。
据此推测:线粒体外膜对蔗糖是通透的，而内膜对蔗糖是不通透的。
线粒体组分的分离首先将线粒体置于低渗溶液中使外膜破裂，此时线粒体内膜和基质(线粒体质)仍结合在一起，通过离心可将线粒体质分离。
用去垢剂毛地黄皂苷处理线粒体质，破坏线粒体内膜，释放线粒体基质，破裂的内膜重新闭合形成小泡，其表面有F1颗粒。
1. 外膜(outer menbrane)标志酶:单胺氧化酶;外膜含有较大的通道蛋白:孔蛋白;2. 内膜(inner membrane)线粒体进行电子传递和氧化磷酸化的部位，通透性差;含有大量的心磷脂(cardiolipin)，心磷脂与离子的不可渗透性有关;内膜的标志酶：细胞色素氧化酶;按作用分3类酶：运输酶类、合成酶类、电子传递合成和ATP合成的酶类。
3. 膜间隙(intermenbrane space)标志酶：腺苷酸激酶功能：建立电化学梯度4. 线粒体基质(matrix)标志酶：苹果酸脱氢酶功能：进行氧化反应，主要是三羧酸循环7.3 前导肽与线粒体蛋白定位线粒体各部分的蛋白质来自何方?定位机理如何?7.3.1 前导肽(leading peptide)蛋白质的两种运输方式：细胞质中的核糖体在合成蛋白质时有两种可能的存在状态：游离核糖体(free ribosome)：在蛋白质合成的全过程一直保持游离状态膜结合核糖体(membranous ribosomes)：核糖体在合成蛋白质的初始阶段处于自由状态，但是随着肽链的合成核糖体被引导到内质网上与内质网结合在一起1. 蛋白质的两种运输方式A翻译后转运：游离核糖体上合成的蛋白质释放到胞质溶胶后被运送到不同的部位，即先合成，后运输。
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7.1.4 To Which Extent Do We Need Culture in Our2 “Watergate”
The “Watergate” example:
This suffix –gate enjoys a rich productivity in American English. Words derived from this source inevitably take on a culturally pejorative implication to refer to “the disclosures of misconduct in high places”(被揭发出来的高层丑闻).
3)The semantic implicature it has will stay with the word for a long time.
THANKS
Conclusions:
1)As a word taking on a pejorative implicature, “Watergate” will stay in English for a long time. 2)Its structural status in the language becomes rather stable through the rich derivational processes it has undergone in word-formation.
Chapter 7
7.1.3 Case Studies 7.1.4 To Which Extent Do We Need Culture in Our Linguistic Study?
钟文苑 张煜文 赵宝瑾 赵春招 1405021184 1405021178 1405021179 1405021180
杨永林(1993) made a several case studies through his personal exposure to the American culture. (1)When you get your hand dirty, it does not necessarily mean in American culture that you’ve done some manual work and need to wash your hand. (practice something / be engaged in doing something) (2) When you have enough dumbbells, it does not necessarily mean that you keep pairs of this instrument for regular physical practices. (have lots of stupid guys)
“Billygate” (比利门) “Debategate” (辩论门) “Cattlegate” (牲畜门) “Ricegate” (大米门) British Watergate (英国水门） French Watergate（法国水门） Muldergate (South Africa)（南非水门） Koreagate (South Korea)（韩国水门）
7.1.3 Case Studies (个案研究）
A constellation of examples can be from different levels of language analysis to illustrate the interplay of language and culture and this selection may range from textual structure to phonological variation. Kaplan(1996): The structural organization of a text tends to be culturally specific. RP speakers and GA speakers produce rather different social connotations. Nida(1998): Words are sometimes “idiomaticallygoverned” and “culturallyspecific”.