Chapter8-4

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分枝定界法
例 maxZ= 6x1 +5 x2
2x1 + x2 ≤9 5x1 +7 x2 ≤35 x1, x2 ≥0 x1, x2取整数
• 第一步,不考虑变量的整数约束,求相应LP(问题1)的最优解：
x1=28/9，x2 =25/9，Z1=293/9
• 第二步,定界过程
这个解不满足整数约束，这时目标函值Z1是整数规划的目标上界； 因为x1=x2=0是整数规划问题的可行解，所以下界为0。
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分枝定界法
问题6：maxZ= 6x1 +5 x2 2x1 + x2 ≤9 5x1 +7 x2 ≤35 x1≤3 x2 ≥3 x1≤2 x1, x2 ≥0 x1, x2取整数 问题7： maxZ= 6x1 +5 x2 2x1 + x2 ≤9 5x1 +7 x2 ≤35 x1 ≤3 x2 ≥3 x1 ≥ 3 x1, x2 ≥0 x1, x2取整数
• 第六步，定界过程
LP4的解满足整数约束，不必再分枝，它的目标函数值是28， 小于原有下界29，则下界仍为29； 现有上界为未分枝子问题中目标函数最大值，即为159/5。 LP5的解仍不满足整数约束的要求，它的目标函数值159/5大于 现有下界29，则应继续分枝。
• 第七步，分枝过程
将不满足整数约束的变量x1 进行分枝，构造两个新的约束条件： x1≤ [14/5]=2，x1≥ [14/5] +1=3
3 2 1
• • •
1
• • பைடு நூலகம் •
2x1 + 3 x2 =14.66
• •
• •
2x1 +3 x2 =14
2 +3 x =6 2x1 3 2 4

1 A B
2 C
3 n D
T2 = −m2 − m3 = −(4.78 + 4.78) = −9.56kN⋅ m
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③绘制扭矩图 Draw the torque diagram
T max = 9.56 kN ⋅ m
BC段为危险截面。 段为危险截面。 段为危险截面 Section BC is dangerous section .
目 的
①确定扭矩变化规律 To determine the rule of change of the torque |T| max 截面 的确定
purposes
To determine the location of dangerous section
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[例1]已知：一传动轴， n =300r/min，主动轮输入 P1=500kW，从动轮输出 P2=150kW， 例 已知 一传动轴， 已知： ， ， ， P3=150kW，P4=200kW，试绘制扭矩图。 ， ，试绘制扭矩图。 Example 1 The transmission shaft is shown as Fig.the input power of driver wheel C is 500KW The export powers of driven wheel A,Band D are 150KW,150KW,200KW.To draw the torque diagram of the transmission shaft .
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8–1 扭转的概念及外力偶矩的计算
Concepts of torsion and how to calculate the external couple moment 轴：工程中以扭转为主要变形的构件。如：机器中的传动轴、钻杆等。 工程中以扭转为主要变形的构件。 机器中的传动轴、钻杆等。 Shaft :the members create torsion deformation due to subjected to the external moments.such as the transmission shafts and drill pipes of the machines . 扭转的外力特点：外力的合力为一力偶， 扭转的外力特点：外力的合力为一力偶，且力偶的作用面与直杆的轴线 垂直。 垂直。 external forces:Two couples that have the same magnitude moment, the opposite direction and the plane of couples perpendicular to the axial line 变形特点： 变形特点：各横截面绕轴线发生相对转动 The deformation is that the external loads tend to twist one segment of the body with respect to the other

Linear City: Firm’s choice of product differentiation 1−a−b p2 − p1 + 2 2t(1 − a − b) 1−a−b p1 − p2 + 2 2t(1 − a − b)
D1 (p1 , p2 ) = x = a +
D1 (p1 , p2 ) = 1 − x = b +
C HAPTER 8: I NDUSTRY S TRUCTURE AND P ERFORMANCE (4
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P RODUCT D IFFERENTIATION : P RICE C OMPETITION AND N ON - PRICE C OMPETITION OVERVIEW T HE C IRCULAR C ITY PM AXIMAL OR M INI S PATIAL C OMPETITION OF SCP T HEORY OF RICE M ARKUPS AN
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P RODUCT D IFFERENTIATION : P RICE C OMPETITION AND N ON - PRICE C OMPETITION OVERVIEW T HE C IRCULAR C ITY PM AXIMAL OR M INI S PATIAL C OMPETITION OF SCP T HEORY OF RICE M ARKUPS AN
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C HAPTER 8: I NDUSTRY S TRUCTURE AND P ERFORMANCE (2
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P RODUCT D IFFERENTIATION : P RICE C OMPETITION AND N ON - PRICE C OMPETITION OVERVIEW T HE C IRCULAR C ITY PM AXIMAL OR M INI S PATIAL C OMPETITION OF SCP T HEORY OF RICE M ARKUPS AN

Innatist, represented by Chomsky,claims that human beings are biologically programmed for language and that the language develops in the child just as other biological functions such as walking.
Chapter 8 Language and Society
1. How is language related to society? [1] Language is used to establish and maintain social
relationships. Such as “Good morning!”, “Hi!”, “How’s your family?”, “Nice day today, isn’t it?” [2] the kind of language speakers chooses to use is in part determined by their social background. [3] the structure of lexicon, reflects both the physical and the social environments of a society.
On the one hand, language as an integral part of human being, permeates his thinking and way of viewing the world, language both expresses and embodies cultural reality. On the other, language, as a product of culture, helps perpetuate the culture, and the changes in language uses reflect the cultural changes in return.