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ReduceAlongDimension - Maple Help

ArrayTools

 ReduceAlongDimension
 reduce the elements of an Array by a function

 Calling Sequence ReduceAlongDimension(fcn,A,dim,opts)

Parameters

 fcn - function to reduce along a dimension to a single value A - Array dim - (optional) dimension to add along opts - (optional) one or more options

Description

 • ReduceAlongDimension(fcn,A,dim) reduces with function fcn the elements in dimension dim of an N-dimensional Array to produce an Array with (N-1) dimensions.
 • When dim is not specified, the first non-singleton dimension is used for reduction. If the dimension specified is a higher dimension than the number of dimensions in the input, the input itself is returned.
 • If ReduceAlongDimension is applied to a Vector, a scalar is returned. If applied to a Matrix, a Vector is returned.
 • By default, the datatype of the result Array is the same as that of input Array A. Depending on fcn, this may not be suitable. For example, if A has datatype integer[1], and fcn adds up values, the result may be too large to store in that datatype.
 The option ':-datatype'=dt, where dt is a valid Array datatype, can be used to override the default datatype for the result.
 • If option evalhf or evalhf=true is passed, the arguments passed to each call to fcn are first evaluated to hardware floating-point using evalhf, as is the result returned by fcn. Furthermore, unless the datatype option was specified, the result Array will have datatype float[8].
 • This function is part of the ArrayTools package, so it can be used in the short form ReduceAlongDimension(..) only after executing the command with(ArrayTools). However, it can always be accessed through the long form of the command by using ArrayTools[ReduceAlongDimension](..).

Examples

 > $\mathrm{with}\left(\mathrm{ArrayTools}\right):$

Collapse a Matrix into a Vector of row or column totals.

 > $M≔\mathrm{Matrix}\left(\left[\left[a,b,c\right],\left[d,e,f\right]\right]\right)$
 ${M}{≔}\left[\begin{array}{ccc}{a}& {b}& {c}\\ {d}& {e}& {f}\end{array}\right]$ (1)
 > $\mathrm{ReduceAlongDimension}\left(\mathrm{+},M,1\right)$
 $\left[\begin{array}{ccc}{a}{+}{d}& {b}{+}{e}& {c}{+}{f}\end{array}\right]$ (2)
 > $\mathrm{ReduceAlongDimension}\left(\mathrm{+},M,2\right)$
 $\left[\begin{array}{c}{a}{+}{b}{+}{c}\\ {d}{+}{e}{+}{f}\end{array}\right]$ (3)

Find the maximum per column or row of a Matrix.

 > $N≔\mathrm{Matrix}\left(\left[\left[3,1,4\right],\left[1,5,9\right],\left[2,6,5\right]\right]\right)$
 ${N}{≔}\left[\begin{array}{ccc}{3}& {1}& {4}\\ {1}& {5}& {9}\\ {2}& {6}& {5}\end{array}\right]$ (4)
 > $\mathrm{ReduceAlongDimension}\left(\mathrm{max},N,1\right)$
 $\left[\begin{array}{ccc}{3}& {6}& {9}\end{array}\right]$ (5)
 > $\mathrm{ReduceAlongDimension}\left(\mathrm{max},N,2\right)$
 $\left[\begin{array}{c}{4}\\ {9}\\ {6}\end{array}\right]$ (6)

Compute sums along each dimension of a three-dimensional Array.

 > $A≔\mathrm{Array}\left(1..2,1..2,1..2,\left(x,y,z\right)↦a‖x‖y‖z\right)$
 ${A}{≔}\begin{array}{c}\left[\begin{array}{cc}{\mathrm{a111}}& {\mathrm{a121}}\\ {\mathrm{a211}}& {\mathrm{a221}}\end{array}\right]\\ \hfill {\text{slice of 2 × 2 × 2 Array}}\end{array}$ (7)
 > $A\left[1\right],A\left[2\right]$
 $\left[\begin{array}{cc}{\mathrm{a111}}& {\mathrm{a112}}\\ {\mathrm{a121}}& {\mathrm{a122}}\end{array}\right]{,}\left[\begin{array}{cc}{\mathrm{a211}}& {\mathrm{a212}}\\ {\mathrm{a221}}& {\mathrm{a222}}\end{array}\right]$ (8)
 > $\mathrm{ReduceAlongDimension}\left(\mathrm{+},A,1\right)$
 $\left[\begin{array}{cc}{\mathrm{a111}}{+}{\mathrm{a211}}& {\mathrm{a112}}{+}{\mathrm{a212}}\\ {\mathrm{a121}}{+}{\mathrm{a221}}& {\mathrm{a122}}{+}{\mathrm{a222}}\end{array}\right]$ (9)
 > $\mathrm{ReduceAlongDimension}\left(\mathrm{+},A,2\right)$
 $\left[\begin{array}{cc}{\mathrm{a111}}{+}{\mathrm{a121}}& {\mathrm{a112}}{+}{\mathrm{a122}}\\ {\mathrm{a211}}{+}{\mathrm{a221}}& {\mathrm{a212}}{+}{\mathrm{a222}}\end{array}\right]$ (10)
 > $\mathrm{ReduceAlongDimension}\left(\mathrm{+},A,3\right)$
 $\left[\begin{array}{cc}{\mathrm{a111}}{+}{\mathrm{a112}}& {\mathrm{a121}}{+}{\mathrm{a122}}\\ {\mathrm{a211}}{+}{\mathrm{a212}}& {\mathrm{a221}}{+}{\mathrm{a222}}\end{array}\right]$ (11)

The datatype option can be used if the input datatype would not be suitable for the result.

 > $A≔\mathrm{Array}\left(\left[\left[1,40,80\right],\left[30,40,50\right]\right],\mathrm{datatype}=\mathrm{integer}\left[1\right]\right)$
 ${A}{≔}\left[\begin{array}{ccc}{1}& {40}& {80}\\ {30}& {40}& {50}\end{array}\right]$ (12)
 > $\mathrm{ReduceAlongDimension}\left(\mathrm{+},A,1\right)$
 > $\mathrm{ReduceAlongDimension}\left(\mathrm{+},A,1,\mathrm{datatype}=\mathrm{integer}\left[2\right]\right)$
 $\left[\begin{array}{ccc}{31}& {80}& {130}\end{array}\right]$ (13)

Compatibility

 • The ArrayTools[ReduceAlongDimension] command was introduced in Maple 2021.


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October  2019, 15(4): 1631-1651. doi: 10.3934/jimo.2018115

Coordinating a multi-echelon supply chain under production disruption and price-sensitive stochastic demand

 1 Department of Mathematics, Jadavpur University, Kolkata, India 2 Department of Mechanical and Industrial Engineering, Louisiana State University, Baton Rouge LA 70803, USA

Received  February 2017 Revised  March 2018 Published  August 2018

This paper considers a three-echelon supply chain system with one raw-material supplier, one manufacturer and one retailer in which both the manufacturer and the raw-material supplier are exposed to the risk of production disruptions. The market demand is assumed to be uncertain but sensitive to the retail price. The objective is to determine the optimal lot sizes of the supplier and the manufacturer, and the selling price of the retailer when the wholesale prices of the upstream entities are prescribed and the retailer's order quantity is chosen before the actual demand is realized. As the benchmark case, the expected total profit of the centralized channel is maximized. The decentralized supply chain is coordinated under pairwise and spanning revenue sharing mechanisms. Numerical study shows that disruptions have remarkable impact on supply chain decisions.

Citation: Bibhas C. Giri, Bhaba R. Sarker. Coordinating a multi-echelon supply chain under production disruption and price-sensitive stochastic demand. Journal of Industrial & Management Optimization, 2019, 15 (4) : 1631-1651. doi: 10.3934/jimo.2018115
References:

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References:
Impact of $\alpha$ on the manufacturer's decisions
Impact of $\beta$ on the supplier's profit
Impact of $\beta$ on the supply chain's total profit
Effects of $\alpha$ and $\beta$ on the manufacturer's and the supplier's decentralized decisions
 when $p^d$ (= 24.60) is known when $Q^d$ (= 19.42) is known $\alpha$ $Q^d$ $\Pi_m$ $\beta$ $R^d$ $\Pi_s$ 0.0 16.79 50.36 0.0 19.42 38.84 0.2 19.42 38.33 0.2 22.62 29.94 0.4 21.28 29.82 0.4 25.20 25.45 0.6 22.03 21.98 0.6 26.27 21.91 0.8 22.43 14.33 0.8 26.87 18.65 1.0 22.69 6.76 1.0 27.24 15.51
 when $p^d$ (= 24.60) is known when $Q^d$ (= 19.42) is known $\alpha$ $Q^d$ $\Pi_m$ $\beta$ $R^d$ $\Pi_s$ 0.0 16.79 50.36 0.0 19.42 38.84 0.2 19.42 38.33 0.2 22.62 29.94 0.4 21.28 29.82 0.4 25.20 25.45 0.6 22.03 21.98 0.6 26.27 21.91 0.8 22.43 14.33 0.8 26.87 18.65 1.0 22.69 6.76 1.0 27.24 15.51
Optimal results for different values of $\xi$ and $\eta$ in the decentralized system
 $\xi$ $\eta$ $\tilde{p}^d\tilde{\Pi}_r$ $\tilde{Q}_d\tilde{\Pi}_m$ $\tilde{R}^d\tilde{\Pi}_s$ Total profit 0.95 0.90 23.58 89.98 24.12 42.50 30.11 31.18 163.66 0.92 23.58 89.98 24.60 46.76 30.71 27.24 163.98 0.94 23.58 89.98 25.12 51.07 30.71 23.06 164.11 0.97 0.90 23.03 96.38 25.89 39.40 32.32 32.78 168.56 0.92 23.03 96.38 26.41 43.83 32.97 28.80 169.01 0.94 23.03 96.38 26.97 48.32 33.67 24.74 169.44 0.99 0.90 22.49 103.11 27.80 35.89 34.71 34.48 173.48 0.92 22.49 103.11 28.35 40.51 35.39 30.35 173.97 0.94 22.49 103.11 28.96 45.19 36.16 26.14 174.44
 $\xi$ $\eta$ $\tilde{p}^d\tilde{\Pi}_r$ $\tilde{Q}_d\tilde{\Pi}_m$ $\tilde{R}^d\tilde{\Pi}_s$ Total profit 0.95 0.90 23.58 89.98 24.12 42.50 30.11 31.18 163.66 0.92 23.58 89.98 24.60 46.76 30.71 27.24 163.98 0.94 23.58 89.98 25.12 51.07 30.71 23.06 164.11 0.97 0.90 23.03 96.38 25.89 39.40 32.32 32.78 168.56 0.92 23.03 96.38 26.41 43.83 32.97 28.80 169.01 0.94 23.03 96.38 26.97 48.32 33.67 24.74 169.44 0.99 0.90 22.49 103.11 27.80 35.89 34.71 34.48 173.48 0.92 22.49 103.11 28.35 40.51 35.39 30.35 173.97 0.94 22.49 103.11 28.96 45.19 36.16 26.14 174.44
A comparison of results in different scenarios of pairwise RS contract
 Decentralized model Retailer's profit Manufacturer's profit Supplier's profit Total profit without RS contract 87.03 38.33 27.48 152.84 Pairwise RS contract $\xi =0.95, \eta = 0.90$ 89.98 42.50 31.18 163.66 $\xi =0.95, \eta = 1.0$ 89.98 64.38 11.21 165.57 $\xi =1.0, \eta = 0.90$ 106.60 33.96 35.35 175.91
 Decentralized model Retailer's profit Manufacturer's profit Supplier's profit Total profit without RS contract 87.03 38.33 27.48 152.84 Pairwise RS contract $\xi =0.95, \eta = 0.90$ 89.98 42.50 31.18 163.66 $\xi =0.95, \eta = 1.0$ 89.98 64.38 11.21 165.57 $\xi =1.0, \eta = 0.90$ 106.60 33.96 35.35 175.91
Optimal results in Example 2
 Model scenario Retailer's profit Manufacturer's profit Supplier's profit Total profit Centralized - - - 463.60 Decentralized without RS contract 167.97 73.05 52.35 293.37 Decentralized with pairwise RS $\xi = 0.95, \eta = 0.90$ 172.28 81.26 58.84 312.38 Decentralized with spanning RS $\xi_1 = 0.05, \xi_2 = 0.02$ 178.30 81.92 62.27 322.49
 Model scenario Retailer's profit Manufacturer's profit Supplier's profit Total profit Centralized - - - 463.60 Decentralized without RS contract 167.97 73.05 52.35 293.37 Decentralized with pairwise RS $\xi = 0.95, \eta = 0.90$ 172.28 81.26 58.84 312.38 Decentralized with spanning RS $\xi_1 = 0.05, \xi_2 = 0.02$ 178.30 81.92 62.27 322.49
Optimal results for different values of $e$ in the decentralized system
 $e$ Retailer($p^d, \Pi_r^d$) Manufacturer($Q^d, \Pi_m^d$) Supplier($R^d, \Pi_s^d$) Total profit 3.0 (31.5,167.97) (37.01, 73.05) (46.21, 52.35) 293.37 3.1 (31.0,119.06) (27.54, 54.37) (34.38, 38.96) 212.39 3.2 (30.55, 84.52) (20.47, 40.41) (25.31, 28.68) 153.61 3.3 (30.13, 60.08) (15.22, 30.05) (19.0, 21.53) 111.66 3.4 (29.75, 42.77) (11.31, 22.32) (14.12, 16.0) 81.09 3.5 (29.40, 30.48) (8.39, 16.58) (10.47, 11.87) 58.92
 $e$ Retailer($p^d, \Pi_r^d$) Manufacturer($Q^d, \Pi_m^d$) Supplier($R^d, \Pi_s^d$) Total profit 3.0 (31.5,167.97) (37.01, 73.05) (46.21, 52.35) 293.37 3.1 (31.0,119.06) (27.54, 54.37) (34.38, 38.96) 212.39 3.2 (30.55, 84.52) (20.47, 40.41) (25.31, 28.68) 153.61 3.3 (30.13, 60.08) (15.22, 30.05) (19.0, 21.53) 111.66 3.4 (29.75, 42.77) (11.31, 22.32) (14.12, 16.0) 81.09 3.5 (29.40, 30.48) (8.39, 16.58) (10.47, 11.87) 58.92
Optimal results for different values of $a$ in the decentralized system
 $a$ Retailer($p^d, \Pi_r^d$) Manufacturer($Q^d, \Pi_m^d$) Supplier($R^d, \Pi_s^d$) Total profit 5000 (31.5,167.97) (37.01, 73.05) (46.21, 52.35) 293.37 6000 (31.5,201.56) (44.41, 87.66) (55.45, 62.83) 352.05 7000 (31.5,235.16) (51.81,102.28) (64.68, 73.30) 410.74 8000 (31.5,268.75) (59.21,116.89) (73.92, 83.77) 469.41 9000 (31.5,302.34) (66.61,131.50) (83.16, 94.24) 528.08 10000 (31.5,335.94) (74.01,146.11) (92.40,104.71) 586.76
 $a$ Retailer($p^d, \Pi_r^d$) Manufacturer($Q^d, \Pi_m^d$) Supplier($R^d, \Pi_s^d$) Total profit 5000 (31.5,167.97) (37.01, 73.05) (46.21, 52.35) 293.37 6000 (31.5,201.56) (44.41, 87.66) (55.45, 62.83) 352.05 7000 (31.5,235.16) (51.81,102.28) (64.68, 73.30) 410.74 8000 (31.5,268.75) (59.21,116.89) (73.92, 83.77) 469.41 9000 (31.5,302.34) (66.61,131.50) (83.16, 94.24) 528.08 10000 (31.5,335.94) (74.01,146.11) (92.40,104.71) 586.76
Optimal results for different values of $\sigma$ in the decentralized system
 $\sigma$ Retailer's profit Manufacturer's profit Supplier's profit Total profit 51 167.97 73.05 52.35 293.37 53 160.95 69.96 50.14 281.05 55 153.97 66.81 47.88 268.66 57 147.04 63.75 45.69 256.48 59 140.19 60.74 43.53 244.46
 $\sigma$ Retailer's profit Manufacturer's profit Supplier's profit Total profit 51 167.97 73.05 52.35 293.37 53 160.95 69.96 50.14 281.05 55 153.97 66.81 47.88 268.66 57 147.04 63.75 45.69 256.48 59 140.19 60.74 43.53 244.46
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Abstract Game Theory pioneers J. von Neumann and O. Morgenstern gave most of their attention to the cooperative side of the subject. "Bezalel Peleg and Peter Sudhölter's book is a welcome addition to any game theory library. 28 0 obj Every coalition Shas a set of available actions A S. A set of players N. A coalition S is a subset of N. ... Two Player unanimity game: An Almost Trivial Example . But cooperative game theory has had relatively little effect on economics. 6th Summer School AACIMP - Kyiv Polytechnic Institute (KPI) - National Technical University of Ukraine, 8-20 August 2011 IntroductionIntroduction Cooperative game theory deals with coalitions who coordinate their actions and pool their winnings. Non-Cooperative Game Theory Having Fun with Strategic Games Wonbin Kang Ph.D. If we consider any permutation $$\pi$$ of $$[N]$$ then the marginal contribution of player $$i$$ with respect to $$\pi$$ is given by: We can now define the Shapley value of any game $$G=(N,v)$$. We defined matching games; 2. Nash Equilibrium is a game theory Game Theory Game theory is a mathematical framework developed to address problems with conflicting or cooperating parties who are able to make rational decisions.The concept that determines the optimal solution in a non-cooperative game in which each player lacks any incentive to change his/her initial strategy. In this essay, I suggest why that might be and what is needed for cooperative theory to become more relevant to economics. If we consider any permutation $$\pi$$ of $$[N]$$ then we denote by $$S_\pi(i)$$ the set of predecessors of $$i$$ in $$\pi$$: For example for $$\pi=(1,3,4,2)$$ we have $$S_\pi(4)=\{1,3\}$$. endobj zh{0d�liI�����'�oT�aqwvf��{)N�8�:�8�Sg�%����[���i�R�D��rH���T Our taxi example is monotone, however the $$G=(3,v_1)$$ with $$v_1$$ defined as: A characteristic function game $$G=(N,v)$$ is called superadditive if it satisfies $$v(C_1\cup C_2)\geq v(C_1)+v(C_2).$$ for all $$C_1\cap C_2=\emptyset$$. Here are the costs for each individual journey: To construct the characteristic function we first obtain the power set (ie all endobj Game Theory uses a special terminology to capture the distinction between strategic situations in which agreements are enforceable and those in which they are not. Example: Ahmed and Ali are buyer and seller respectively and they are bargaining over the price of a smartphone. Types of Game Theory Cooperative Game : It is an economic game played by firms in which players or firms can negotiate on binding contracts which allows them to make mixed or joint strategies. To define it we need two last definitions. 29 0 obj endobj Cooperative Game Theory The two branches of game theory Non-cooperative game theory No binding contracts can be written Players are individuals Nash equilibrium Cooperative game theory Binding contract can be written Players are individuals and coalitions of individuals Main solution concepts: Core Shapley value The focus of today! Game theory is divided into two parts: non-cooperative and cooperative. Transferable utility is a term used in cooperative game theory and in economics. endobj << /S /GoTo /D [34 0 R /Fit ] >> << /S /GoTo /D (Outline0.3) >> endobj endobj >> Game Theory Example: Steak Pricing Game A pricing game for steaks if shown in Figure $$\PageIndex{1}$$. cooperative game theory tries to characterize the set of outcomes that are, seen from a viewpoint of rationality, interesting. stream Note that the advertising game is not a prisoner’s dilemma, since there is no incentive to cheat once the cooperative solution has been achieved. x��XKo7��W�TH���^���I�&E� Cooperative game theory applies both to zero-sum and non-zero-sum games. Cooperative Game Theory. Game theory, branch of applied mathematics that provides tools for analyzing situations in which parties, called players, make decisions that are interdependent. (Introduction) Candidate, Political Economy and Government TEDy Wonbin Kang Game Theory. Another potential solution would be $$\lambda=(6,6,30)$$, however at this point sharing the taxi is of no benefit to player 1. The characteristic function is given below: A characteristic function game $$G=(N,v)$$ is called monotone if it satisfies $$v(C_2)\geq v(C_1)$$ for all $$C_1\subseteq C_2$$. endobj This interdependence causes each player to consider the other player’s possible decisions, or strategies, in formulating strategy. “3 players share a taxi. Biology. Our taxi example is not superadditive, however the $$G=(3,v_2)$$ with $$v_2$$ defined as: When talking about a solution to a characteristic function game we imply a payoff vector $$\lambda\in\mathbb{R}_{\geq 0}^{N}$$ that divides the value of the grand coalition between the various players. A game is cooperative if the players are able to form binding commitments.For instance the legal system requires them to adhere to their promises. Under cooperative games, players can coordinate their strategies and share the payoff. individuals:Non-transferable utility games. For some games, outcomes are speciﬁed in terms of the total amount of dollars or utility that a coalition can di-vide. endobj It is well known that the mathematical modeling of various real-world decision-making situations gives rise to combinatorial optimization problems. /Length 1319 The formation of coalitions or alliances is omnipresent in many applications. For example, scientists studying antibiotic resistance are using a game theory scenario called the snowdrift game, in which a player always benefits from cooperating. 13 0 obj Example 2.3 Three player zero-sum game We have three players in this game. Cooperative Game Theory Cooperative games are often deﬁned in terms of a charac-teristic function, which speciﬁes the outcomes that each coalition can achieve for itself. I believe cooperative game theory is a misnomer. %PDF-1.4 24 0 obj Thus $$\lambda$$ must satisfy: Thus one potential solution to our taxi example would be $$\lambda=(14,14,14)$$. Non-cooperative games (or broader, game theory) has also been applied to biology. Players negotiate and enter into a joint strategy in Cooperative Game theory, whereas players compete and reach an equilibrium in the Non-Cooperative Game theory. In the previous chapter: 1. (sometimes known as non-cooperative game theory) Coalitional Game with transferable payoffs. 16 0 obj Players 1 and 2. 2/38 COOPERATIVE GAMES MIHAI MANEA 1. Although most cooperative game theory deals with profit games, all concepts can easily be translated to the cost setting. We proved certain results regarding the Gale-Shapley algorithm. Player 3: 42 Normal Form and Extensive Form Games: Normal form games refer to the description of game in … Obviously this is not ideal for player 1 and/or 2: they actually pay more than they would have paid without sharing the taxi! How much should each individual contribute?”. through contract law). For example, in political games, parties, or individuals can form coalitions for improving their voting power. << /S /GoTo /D (Outline0.5) >> These are games with transferable utility. Before we talk about Game Theory, we need to first understand the idea behind oligopolies. Examples: linear production games, ﬂow games, market games. In particular, sets of players, called coalitions, can make binding agreements about joint strategies, pool their individual agreements and, redistribute the total in a speciﬁed way. Specialists in cooperative game theory study methods of complex decision-making in situations with multiple criteria. endobj 39 0 obj << According to Green , the non-cooperative game (unaided judgment) can be applied to forecasting market decisions. This is an example of a game without side payments, also called a game with non-transferrable utility (NTU). Given $$G=(N,v)$$ the Shapley value of player $$i$$ is denoted by $$\phi_i(G)$$ and given by: As an example here is the Shapley value calculation for our taxi sharing game: Thus the fair way of sharing the taxi fare is for player 1 to pay 2, player 2 to pay 5 and player 3 to pay 35. A coalition is a group of players SˆN. The key distinguishing feature is the absence of external authority to establish rules enforcing cooperative behavior. For the rest of this chapter we’ll deal with games with side payments, or transferrable << /S /GoTo /D (Outline0.2) >> We described the Gale-Shapley algorithm; 3. In this Chapter we’ll take a look at another type of game. We refer to Nas the grand coalition. endobj 25 0 obj Interesting for totally balanced games are population monotonic allocation schemes (pmas) introduced by Sprumont (GEB 1990). Similarly $$(0,12,30)$$ would have no incentive for player 2. endobj �ǬKQk����Ci�s�d\$E��S��%�ۋZV*N7�kN-Ӻ��g�>��'�㓟����,V�\p~떳 u���K�/ks3"�E�W��ٰ=U~h�wC܀�R&U���C�U*��ݏ�K��E:�d9U. $$2^{{1,2,3}}=\{\emptyset,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\Omega\}$$ (Imputations) (Constant-Sum Games) attempts to take into consideration the interactions between the participants and their behavior to study the strategic decision-making between rational individuals As another example, there is a branch of game theory known as ' cooperative game theory ', that has received only a fraction of the attention of its non-cooperative counterpart when it comes to climate change (e.g., Diamantoudi and Sartzetakis, 2006; Diamantoudi and Sartzetakis, 2014). << /S /GoTo /D (Outline0.6) >> 32 0 obj endobj All source files can be found at this github repository. Such analysis is very limited or non-existent in other books on game theory. In cooperative games, actions are taken by groups of agents, coalitions, and payo s are given to the group, that has to divided it among its members: Transferable utility games. ... For example, if the game you have deﬁned is a penalty kick, you could ask who will win the penalty kick by virtue of getting Download Cooperative Game Theory And Applications books, In this book applications of cooperative game theory that arise from combinatorial optimization problems are described. (If you’re stuck in your apartment building after a blizzard, you benefit by shoveling the driveway, but so does everyone else who lives there and doesn’t shovel.) A payoff vector possesses the additivity property if: We will not prove in this course but in fact there is a single payoff vector that satisfies these four properties. Game theory - Game theory - N-person games: Theoretically, n-person games in which the players are not allowed to communicate and make binding agreements are not fundamentally different from two-person noncooperative games. Definitions A coalitional (or cooperative) game is a model of interacting decision-makers that focuses on the behavior of groups of players. A characteristic function game G is given by a pair $$(N,v)$$ where $$N$$ is the number of players and $$v:2^{[N]}\to\mathbb{R}$$ is a characteristic function which maps every coalition of players to a payoff. Class website for my third year Game Theory course. 33 0 obj possible coalitions) In such a situation, groups … It is something that does not, to date, exist. Why You Should Understand & Study ‘Game Theory’ If You Are In Business? << /S /GoTo /D (Outline0.4) >> The book provides comprehensive coverage of the theory of cooperative games and, as such, would make useful reference book for researchers interested cooperative games. To find a “fair” distribution of the grand coalition we must define what is meant by “fair”. We proved certain results regarding the Gale-Shapley algorithm. 9 0 obj English examples for "cooperative game theory" - The main assumption in cooperative game theory is that the grand coalition N will form. In this Chapter we’ll take a look at another type of game. endobj << /S /GoTo /D (Outline0.1) >> They can form coalitions. This is however, less accurate than using the role playing method and should be used with care (Green, 2005). /Filter /FlateDecode Contrast to Nash equilibrium which focuses on what individuals can do acting alone. Figure 2: Simultaneous-Move Game (Payoff Matrix) Cooperative or Non-Cooperative. 12 0 obj (Coalitions) where $$\Omega$$ denotes the set of all players ($$\{1,2,3\}$$). Cooperative game theory model scenarios, where agents can bene t by cooperating, and binding agreements are possible. Page 1 of 3 Game Theory Professor Giacomo Bonanno COOPERATIVE GAMES: the SHAPLEY VALUE The description of a cooperative game is still in terms of a characteristic function which specifies for every group of players the total payoff that the members of S can obtain by 17 0 obj through credible threats). Ndenotes the set of players. John Harsanyi: An economist who won the Nobel Memorial Prize in 1994 along with John Nash and Reinhard Selten for his research on game theory, a … Equivalently, a game is totally balanced, if (the game and) all subgames have a nonempty core. In noncooperative games this is not possible. Contemporary social science is based on game theory, economics, and psychology in which mathematical logic is applied. In cooperative game theory the interest lies with understanding how coalitions form in competitive situations. In game theory, a cooperative game (or coalitional game) is a game with competition between groups of players ("coalitions") due to the possibility of external enforcement of cooperative behavior (e.g. In game theory, a non-cooperative game is a game with competition between individual players, as opposed to cooperative games, and in which alliances can only operate if self-enforcing (e.g. 4. We require four desirable properties: For $$G=(N,v)$$ a payoff vector $$\lambda$$ is efficient if: For $$G(N,v)$$ a payoff vector possesses the null player property if $$v(C\cup i)=v(C)$$ for all $$C\in 2^{\Omega}$$ then: For $$G(N,v)$$ a payoff vector possesses the symmetry property if $$v(C\cup i)=v(C\cup j)$$ for all $$C\in 2^{\Omega}\setminus\{i,j\}$$ then: For $$G_1=(N,v_1)$$ and $$G_2=(N,v_2)$$ and $$G^+=(N,v^+)$$ where $$v^+(C)=v_1(C)+v_2(C)$$ for any $$C\in 2^{\Omega}$$. Non-cooperative Games and Solving Technologies Dominance Criteria of NCGT. through credible threats).. (Outline) 21 0 obj (A Voting Game) Those are opposed to non-cooperative games in which there is either no possibility to forge alliances or all agreements need to be self-enforcing (e.g. 20 0 obj ( unaided judgment ) can be found at this github repository theory ’ if You in! Ntu ) incentive for player 2 non-zero-sum games Kang game theory and applications,! 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text:
# Shrink the entire beamer pdf by leaving a 1/2 inch margin on all four edges of each slide, would like to print each slide without notes

How do you shrink everything in a beamer such that it leaves 1/2 inch border around each slide? I would like to maximize its size and be able to print everything on each page, including the border.

\documentclass{beamer}

\begin{document}
\title{Beamer Class well nice}
\author{Author's name here}
\date{\today}

\begin{frame}
\titlepage
\end{frame}

\section{Section no. 2}
\subsection{Lists I}
\begin{frame}\frametitle{unnumbered lists}
\begin{itemize}
\item Introduction to  \LaTeX
\item Course 2
\item Termpapers and presentations with \LaTeX
\item Beamer class
\end{itemize}
\end{frame}

\subsection{Lists II}
\begin{frame}\frametitle{numbered lists}
\begin{enumerate}
\item Introduction to  \LaTeX
\item Course 2
\item Termpapers and presentations with \LaTeX
\item Beamer class
\end{enumerate}
\end{frame}

\end{document}

EDIT: I kept googling and have found what I've been looking for. Modify the lines below to shrink your pdf file. If someone posts this as an answer, I will accept.

\usepackage{pgfpages}

\pgfpagesuselayout{resize to}[letterpaper,border shrink=5mm,landscape]

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Post the answer yourself... –  karlkoeller Oct 21 '13 at 17:28
@karlkoeller since I'm a new user, it won't let me post answers. –  user38665 Oct 21 '13 at 17:42
I think you can after a certain time or when you get a little reputation. –  karlkoeller Oct 21 '13 at 17:45

\usepackage{pgfpages}


text:
# Problem: For the reaction, A → B, the rate constant is 0.0208 M-1 sec-1. How long would it take for [A] to decrease from 0.100 to 0.0450 M?a. 2.66 secb. 38.4 secc. 72.2 secd. 588 sece. 1072 sec

###### FREE Expert Solution

Determine the order of the reaction and calculate for the time it will take to decrease A from 0.10 to 0.450 M

In this case, follow the steps:

Step 1. Determine the order of the reaction using the units of rate constant as hint

Step 2. Calculate for the time using the integrated rate law equation from Step 1.

85% (429 ratings)
###### Problem Details

For the reaction, A → B, the rate constant is 0.0208 M-1 sec-1. How long would it take for [A] to decrease from 0.100 to 0.0450 M?

a. 2.66 sec

b. 38.4 sec

c. 72.2 sec

d. 588 sec

e. 1072 sec


text:
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The dyad (binary group) is a fundamental kinematic chain with two links $(n=2)$ and three one degree of freedom joints $\left(c_{5}=3\right.$ ). Figure $1.7$ depicts different types of dyads: rotation rotation rotation (dyad RRR) or dyad of type one, Fig. 1.7a; rotation rotation translation (dyad RRT) or dyad of type two, Fig.1.7b; rotation translation rotation (dyad RTR) or dyad of type three, Fig. $1.7 \mathrm{c}$; translation rotation translation (dyad TRT) or dyad of type four, Fig. $1.7 \mathrm{~d}$; rotation translation translation (dyad RTT) or dyad of type five, Fig. 1.7e. The advantage of the group classification of a mechanical system is in its simplicity. The solution of the whole mechanical system can be obtained by adding partial solutions of different fundamental kinematic chains [55-57].

The number of DOF for the mechanism in Fig. $1.8$ is $M=1$. If $M=1$, there is one driver link (one actuator). The rotational link 1 can be selected as the driver link. If the driver link is separated from the mechanism the remaining moving kinematic chain (links $2,3,4,5$ ) has the number of DOF equal to zero. The dyad is the simplest system group with two links and three joints. On the contour diagram, the links 2 and 3 form a dyad and the links 4 and 5 represent another dyad. The mechanism has been decomposed into a driver link (link 1) and two dyads (links 2 and 3, and links 4 and

5). The dyad with the links 2 and 3 is a RTR dyad and the dyad with the links 4 and 5 is also a RTR dyad. The whole mechanism can be symbolized as a R-RTR-RTR mechanism.

For planar mechanisms the two degrees of freedom joints can be substituted and mechanisms with one degree of freedom joints are obtained. The transformed mechanism has to be equivalent with the initial mechanism from a kinematical point of view. The number of degrees of freedom of the transformed mechanism has to be equal to the number of degrees of freedom of the initial mechanism. The relative motion of the links of the transformed mechanism has to be the same as the relative motion of the links of the initial mechanism.

A planar straight link with the end nodes at $A$ and $B$ is considered. The coordinates of the joint $A$ are $\left(x_{A}, y_{A}\right)$ and the coordinates of the joint $B$ are $\left(x_{B}, y_{B}\right)$. The length $A B=l_{A B}$ is constant and the following relation can be written
$$\left(x_{B}-x_{A}\right)^{2}+\left(y_{B}-y_{A}\right)^{2}=A B^{2},$$

The angle of the link $A B$ with the horizontal axis $O x$ is $\phi$ and the slope $m$ of $A B$ is
$$m=\tan \phi=\frac{y_{B}-y_{A}}{x_{B}-x_{A}} .$$
The equation of the straight link is $y=m x+n$, where $x$ and $y$ are the coordinates of any point on this link and $n$ is the intercept of $A B$ with the vertical axis $O y$.
The R-RRT (slider-crank) mechanism shown in Fig. 1.10a has the dimensions: $A B=0.5 \mathrm{~m}$ and $B C=1 \mathrm{~m}$. The driver link 1 makes an angle $\phi=\phi_{1}=60^{\circ}$ with the horizontal axis. The positions of the joints and the angles of the links with the horizontal axis will be calculated. A Cartesian reference frame $x y$ is selected. The joint $A$ is in the origin of the reference frame, $A \equiv O$,
$$x_{A}=0, y_{A}=0 .$$
The coordinates of the joint $B$ are
\begin{aligned} &x_{B}=A B \cos \phi=(0.5) \cos 60^{\circ}=0.250 \mathrm{~m} \ &y_{B}=A B \sin \phi=(0.5) \sin 60^{\circ}=0.433 \mathrm{~m} \end{aligned}
The coordinates of the joint $C$ are $x_{C}$ and $y_{C}$. The joint $C$ is located on the horizontal axis $y_{C}=0$. The length of the segment $B C$ is constant
$$\left(x_{B}-x_{C}\right)^{2}+\left(y_{B}-y_{C}\right)^{2}=B C^{2},$$
or
$$\left(0.250-x_{C}\right)^{2}+(0.433-0)^{2}=1^{2} .$$
The two solutions for $x_{C}$ are:
$$x_{C_{1}}=1.151 \mathrm{~m} \text { and } x_{C_{2}}=-0.651 \mathrm{~m}$$
To determine the position of the joint $C$, for the given angle of the link 1 , an additional condition is needed. For the first quadrant, $0 \leq \phi \leq 90^{\circ}$, one condition can be $x_{C}>$ $x_{B}$. The $x$-coordinate of the joint $C$ is $x_{C}=x_{C_{1}}=1.151 \mathrm{~m}$. The angle of the link 2 (link $B C$ ) with the horizontal is
$$\phi_{2}=\arctan \frac{y_{B}-y_{C}}{x_{B}-x_{C}} .$$
The numerical solutions for $\phi_{2}$ is $154.341$ degrees. The MATLAB code is:
clear; clc; close all;
뭉 Input data
clear; clc; close
ㅇㅎㅇ Input data
$\mathrm{AB}=0.5 ;$ 황 $(\mathrm{m}}$
$A B=0.5 ;$ 핳 $(\mathrm{m})$

## 数学代写|matlab仿真代写simulation代做|Velocity and Acceleration Analysis for Rigid Body

The motion of a rigid body $(R B)$, with respect to a fixed reference frame, is defined by the position, velocity and acceleration of all points of the rigid body. A fixed orthogonal Cartesian reference frame $x_{0} y_{0} z_{0}$ with the constant unit vectors $\mathbf{1}{0}, \mathbf{J}{0}$, and $\mathbf{k}_{0}$, is shown in Fig. 1.12. The body fixed (mobile or rotating) orthogonal Cartesian

reference frame $x y z$ with unit vectors $\mathbf{1}, \mathbf{J}$ and $\mathbf{k}$ is attached to the moving rigid body. The unit vectors $\mathbf{1}{0}, \mathbf{J}{0}$, and $\mathbf{k}{0}$ of the primary reference frame are constant with respect to time and the unit vectors $\mathbf{1}, \mathbf{J}$, and $\mathbf{k}$ are functions of time, $t$. The unit vectors $\mathbf{1}, \mathbf{J}$, and $\mathbf{k}$ of the body-fixed frame of reference rotate with the body-fixed reference frame. The origin $O$ is arbitrary. The position vector of any point $M, \forall M \in(R B)$, with respect to the fixed reference frame $x{0} y_{0} z_{0}$ is $\mathbf{r}{M}=\mathbf{r}{O_{0} M}$ and with respect to the rotating reference frame $O x y z$ is denoted by $\mathbf{r}=\mathbf{r}{O M}$. The position vector of the origin $O$ of the rotating reference frame with respect to the fixed point $O{0}$ is $\mathbf{r}{o}=\mathbf{r}{{ }{0}} o$. The position vector of $M$ is $$\mathbf{r}{M}=\mathbf{r}{O}+\mathbf{r}=\mathbf{r}{O}+x \mathbf{1}+y \mathbf{J}+z \mathbf{k}$$
where $x, y$, and $z$ represent the projections of the vector $\mathbf{r}=\mathbf{r}{O M}$ on the rotating reference frame $$\mathbf{r}=x \mathbf{1}+y \mathbf{j}+z \mathbf{k}$$ The distance between two points of the rigid body $O$ and $M$ is constant, $O \in(R B)$, and $M \in(R B)$. The components $x, y$ and $z$ of the vector $\mathbf{r}$ with respect to the rotating reference frame are constant. The unit vectors $\mathbf{1}, \mathbf{J}$, and $\mathbf{k}$ are time-dependent vector functions. For the unit vectors of an orthogonal Cartesian reference frame, 1, Jand $\mathbf{k}$, there are the following relations $$\mathbf{1} \cdot \mathbf{1}=1, \quad \mathbf{J} \cdot \mathbf{J}=1, \quad \mathbf{k} \cdot \mathbf{k}=1, \quad \mathbf{1} \cdot \mathbf{J}=0, \quad \mathbf{j} \cdot \mathbf{k}=0, \quad \mathbf{k} \cdot \mathbf{1}=0 .$$ The velocity of an arbitrary point $M$ of the rigid body with respect to the fixed reference frame $x{0} y_{0} z_{0}$, is the derivative with respect to time of the position vector $\mathbf{r}{M}$ \begin{aligned} \mathbf{v}{M} &=\frac{d \mathbf{r}{M}}{d t}=\frac{d \mathbf{r}{O_{o} M}}{d t}=\frac{d \mathbf{r}{O}}{d t}+\frac{d \mathbf{r}}{d t} \ &=\mathbf{v}{O}+x \frac{d \mathbf{1}}{d t}+y \frac{d \mathbf{j}}{d t}+z \frac{d \mathbf{k}}{d t}+\frac{d x}{d t} \mathbf{1}+\frac{d y}{d t} \mathbf{j}+\frac{d z}{d t} \mathbf{k} \end{aligned}
where $\mathbf{v}{O}=\dot{\mathbf{r}}{O}$ is the velocity of the origin of the rotating reference frame $O x y z$ with respect to the fixed reference frame. Because all the points in the rigid body maintain their relative position, their velocity relative to the rotating reference frame $x y z$ is zero, i.e., $\dot{x}=\dot{y}=\dot{z}=0$.
The velocity of point $M$ is
$$\mathbf{v}{M}=\mathbf{v}{O}+x \frac{d \mathbf{1}}{d t}+y \frac{d \mathbf{j}}{d t}+z \frac{d \mathbf{k}}{d t}=\mathbf{v}_{O}+x \mathbf{i}+y \mathbf{j}+z \dot{\mathbf{k}}$$
The derivative of the Eq. (1.11) with respect to time gives
$$\frac{d \mathbf{1}}{d t} \cdot \mathbf{1}=0, \quad \frac{d \mathbf{J}}{d t} \cdot \mathbf{J}=0, \quad \frac{d \mathbf{k}}{d t} \cdot \mathbf{k}=0$$

## matlab代写

5）。具有链接 2 和 3 的对子是 RTR 对子，具有链接 4 和 5 的对子也是 RTR 对子。整个机制可以用R-RTR-RTR机制来表示。

(X乙−X一种)2+(是乙−是一种)2=一种乙2,

X一种=0,是一种=0.

X乙=一种乙因φ=(0.5)因60∘=0.250 米 是乙=一种乙罪φ=(0.5)罪60∘=0.433 米

(X乙−XC)2+(是乙−是C)2=乙C2,

(0.250−XC)2+(0.433−0)2=12.

XC1=1.151 米 和 XC2=−0.651 米

φ2=反正切是乙−是CX乙−XC.

clear; cl；关闭所有;
뭉 输入数据

ㅇㅎㅇ 输入数据

## 数学代写|matlab仿真代写simulation代做|Velocity and Acceleration Analysis for Rigid Body

d1d吨⋅1=0,dĴd吨⋅Ĵ=0,dķd吨⋅ķ=0

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。


text:
# Subtraction of rational numbers (Math 0)

So far, we have met the idea of a rational number, treating them as chunks of apples, and how to add them together. Now we will discover how the idea of the anti-apple (by analogy with the integers’ anti-cow) must work.

# The anti-apple

Just as we had an anti-cow, so we can have an anti-apple. If we combine an apple with an anti-apple, they both annihilate, leaving nothing behind. We write this as $$1 + (-1) = 0$$.

A very useful thing for you to ponder for thirty seconds (though I will give you the answer soon): given that $$\frac{a}{n}$$ means “divide an apple into $$n$$ equal pieces, then take $$a$$ copies of the resulting little-piece”, what would $$\frac{-1}{n}$$ mean? And what would $$-\frac{1}{n}$$ mean?

$$\frac{-1}n$$ would mean “divide an apple into $$n$$ equal pieces, then take $$-1$$ copies of the resulting little-piece”. That is, turn it into an anti-little-piece. This anti-little-piece will annihilate one little-piece of the same size: $$\frac{-1}{n} + \frac{1}{n} = 0$$.

$$-\frac{1}{n}$$, on the other hand, would mean “divide an anti-apple into $$n$$ equal pieces, then take $$1$$ copy of the resulting little-anti-piece”. But this is the same as $$\frac{-1}{n}$$: it doesn’t matter whether we do “convert to anti, then divide up the apple” or “divide up the apple, then convert to anti”. That is, “little-anti-piece” is the same as “anti-little-piece”, which is very convenient.

What about chunks of apple? If we combine half an apple with half an anti-apple, they should also annihilate, leaving nothing behind. We write this as $$\frac{1}{2} + \left(-\frac{1}{2}\right) = 0$$.

How about a bit more abstract? If we combine an apple with half an anti-apple, what should happen? Well, the apple can be made out of two half-chunks (that is, $$1 = \frac{1}{2} + \frac{1}{2}$$); and we’ve just seen that half an apple will annihilate half an anti-apple; so we’ll be left with just one of the two halves of the apple. More formally, $$1 + \left(-\frac{1}{2}\right) = \frac{1}{2}$$; or, writing out the calculation in full, $$1 + \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{2} + \left(-\frac{1}{2}\right) = \frac{1}{2}$$$Let’s go the other way round: if we combine an anti-apple with half an apple, what happens? It’s pretty much the same as the opposite case except flipped around: the anti-apple is made of two anti-half-chunks, and the half apple will annihilate one of those chunks, leaving us with half an anti-apple: that is, $$(-1) + \frac{1}{2} = -\frac{1}{2}$$. We call all of these things subtraction: “subtracting” a quantity is defined to be the same as the addition of an anti-quantity. # General procedure Since we already know how to add, we might hope that subtraction will be easier (since subtraction is just a slightly different kind of adding). In general, we write $$-\frac{1}{n}$$ for “the $$\frac{1}{n}$$-sized building block, but made by dividing an anti-apple (instead of an apple) into $$n$$ equal pieces”. Remember from the pondering above that this is actually the same as $$\frac{-1}{n}$$, where we have divided an apple into $$n$$ equal pieces but then taken $$-1$$ of the pieces. Then in general, we can just use the instant addition rule that we’ve already seen. noteRecall: this was $$\frac{a}{m} + \frac{b}{n} = \frac{a\times n + b \times m}{m \times n}$$. Remember that the order of operations in the integers is such that in the numerator, we calculate both the products first; then we add them together. In fact, $$\frac{a}{m} - \frac{b}{n} = \frac{a}{m} + \left(\frac{-b}{n}\right) = \frac{a \times n + (-b) \times m}{m \times n} = \frac{a \times n - b \times m}{m \times n}$$$

Why are we justified in just plugging these numbers into the formula, without justification? You’re quite right if you are dubious: you should not be content merely to learn this formulanoteThis goes for all of maths! It’s not simply a collection of arbitrary rules, but a proper process that we use to model our thoughts. Behind every pithy, unmemorable formula is a great edifice of motivation and reason, if you can only find it.. In the rest of this page, we’ll go through why it works, and how you might construct it yourself if you forgot it. I took the choice here to present the formula first, because it’s a good advertisement for why we use the $$\frac{a}{n}$$ notation rather than talking about “$\frac{1}{n}$-chunks” explicitly: it’s a very compact and neat way of expressing all this talk of anti-apples, in the light of what we’ve already seen about addition.

Very well: what should $$\frac{a}{m} - \frac{b}{n}$$ be? We should first find a smaller chunk out of which we can build both the $$\frac{1}{m}$$ and $$\frac{1}{n}$$ chunks. We’ve seen already that $$\frac{1}{m \times n}$$ will work as a smaller chunk-size.

Now, what is $$\frac{a}{m}$$ expressed in $$\frac{1}{m \times n}$$-chunks? Each $$\frac{1}{m}$$-chunk is $$n$$ of the $$\frac{1}{m \times n}$$-chunks, so $$a$$ of the $$\frac{1}{m}$$-chunks is $$a$$ lots of “$n$ lots of $$\frac{1}{m \times n}$$-chunks”: that is, $$a \times n$$ of them.

Similarly, $$\frac{b}{n}$$ is just $$b \times m$$ lots of $$\frac{1}{m \times n}$$-chunks.

So, expressed in $$\frac{1}{m \times n}$$-chunks, we have $$a \times n$$ lots of positive chunks, and $$b \times m$$ lots of anti-chunks. Therefore, when we put them together, we’ll get $$a \times n - b \times m$$ chunks (which might be negative or positive or even zero—after all the annihilation has taken place we might end up with either normal or anti-chunks or maybe no chunks at all—but it’s still an integer, being a number of chunks).

So the total amount of apple we have is $$\frac{a \times n - b \times m}{m \times n}$$, just like we got out of the instant formula.

the rest of the page, including examples; make an exercises page, including subtracting negatives and so on

Parents:


text:
# Does my Linklist implementation have memory leaks?

Are there any memory leaks in this linklist implementation.Also is the implementation correct?Can the time complexity be optimized?

#include <iostream>
#include <cstdlib>
#include <cstdio>
using namespace std;

struct node
{
int n;
node* point;
};

node* first=NULL;

void search(int n);
void insert(int n);
void del(int n);
void print();

{
int s,n;
cout<<"Enter 1 to Search"<<endl;
cout<<"Enter 2 to Insert"<<endl;
cout<<"Enter 3 to Delete"<<endl;
cout<<"Enter 4 to Print"<<endl;
cout<<"Enter 5 to Exit"<<endl;

cin>>s;

switch(s)
{
case 1: cout<<"Enter number to Search: ";
cin>>n;
search(n);
break;

case 2: cout<<"Enter number to Insert: ";
cin>>n;
insert(n);
break;

case 3: cout<<"Enter number to Delete: ";
cin>>n;
del(n);
break;

case 4: print();
break;

case 5: exit(0);

default: cout<<"Wrong Input"<<endl<<endl<<endl;
}
}

int main()
{
while(1)
}

void search(int n)
{
node* ptr=first;

while(ptr!=NULL)
{
if(ptr->n==n)
{
cout<<endl<<n<<" Found!"<<endl<<endl<<endl;
return;
}
else
ptr=ptr->point;

}

return;
}

void insert(int n)
{
node* prev=NULL;
node* next=first;

if(first!=NULL and first->n>n)
{
first=new node;
first->n=n;
first->point=next;
cout<<endl<<endl;
return;
}

while(next!=NULL)
{
if(next->n>n)
{
prev->point=new node;
(prev->point)->n=n;
(prev->point)->point=next;
cout<<endl<<endl;
return;
}
else
{
prev=next;
next=next->point;
}
}
if(prev==NULL)
{
first=new node;
first->n=n;
first->point=NULL;
}
else
{
prev->point=new node;
(prev->point)->n=n;
(prev->point)->point=NULL;

}
cout<<endl<<endl;
return;

}

void print()
{
node* next=first;

while(next!=NULL)
{
cout<<next->n<<" ";
next=next->point;
}
cout<<endl<<endl<<endl;
return;
}

void del(int n)
{
node* prev=NULL;
node* next=first;

while(next!=NULL)
{
if(next->n==n)
{
if(prev!=NULL)
{
prev->point=next->point;
delete next;
}
else
{
first=next->point;
delete next;
}
cout<<n<<" Deleted!";
cout<<endl<<endl;
return;
}
else
{
prev=next;
next=next->point;
}
}

cout<<"Number not in list"<<endl<<endl<<endl;
return;
}

• Welcome to Code Review! It looks like you're unsure whether this code works. Please see the How To Ask section of the FAQ. Questions about whether code works or not are not in the scope of this site. Nov 25, 2015 at 15:20
• Try using valgrind if you're not on Windows. Nov 25, 2015 at 15:40
• First comment is that this is C not C++ Nov 25, 2015 at 20:56
• Code compiles and seems to work as expected. Voting to re-open. Nov 25, 2015 at 21:01
• @LokiAstari This is C++....I've just included the cstdlib and cstdio for the exit function to work. Nov 26, 2015 at 7:57

It does not appear to be leaking memory (except on exit). However, there are a number of things you can do to improve this code.

## Fix the formatting

Code that has poor formatting is hard to read, understand, and maintain. For that reason, you should strive to have nicely formatted code. In this case, that means fixing indentation and inserting whitespace. For instance, instead of this:

cout<<"Number not in list"<<endl<<endl<<endl;

write this:

cout << "Number not in list" << endl << endl << endl;

## Don't abuse using namespace std

Putting using namespace std at the top of every program is a bad habit that you'd do well to avoid.

## Separate I/O from core routines

It's often better to have input and output operation in a single place (or a few select places) rather than having them scattered in several places. For this reason, I'd recommend having the search, insert and del routines simply be responsible for their corresponding linked list function and not print anything. Printing can be done instead in the calling routine if needed.

## Use object orientation

Because you're writing in C++, it would make sense to have the methods that operate on a class (such as node) be member functions rather than separate functions. Objects and classes are one of the main strengths of C++ and something you should learn soon if you haven't already. Encapsulate related code in objects where it makes sense to do so.

## Eliminate global variables

Eliminating global variables will make your code more readable and maintainable, both of which are important characteristics of well-written code. Global variables introduce messy linkages that are difficult to spot and error-prone. Global variables can be altered by literally any code, making it difficult to understand how they're being used, and making it all too easy to inadvertently create an inconsistent state. See "how bad are global variables?" for more. The variable first should instead be passed explicitly to each routine that needs it, or you can use an object as mentioned in the previous suggestion.

## Return something useful from functions

All of your functions currently return void, but the code could be improved by making some of them return a bool indicating success or failure. For instance, del could return false if the number is not in the list or true if it was found and deleted.

## Use nullptr rather than NULL

Modern C++ uses nullptr rather than NULL. See this answer for why and how it's useful.

## Use for instead of while where appropriate

In almost every place in the code that uses a while loop, it would be much more idiomatic C++ if a for loop were used instead. For example, the print routine is currently this:

void print()
{
node *next = first;

while (next != NULL) {
cout << next->n << " ";
next = next->point;
}
cout << endl << endl << endl;
return;
}

It would be better written like this:

void print(node *head)
{
std::cout << head->n << " ";
}
std::cout << "\n\n\n";
}

Note that the head parameter is being passed in rather than using a global variable, and that the namespace is explicitly used. Also, note that I've used the string literal "\n\n\n" instead of three repetitions of std::endl. This is more efficient because it doesn't force the buffer to be flushed repeatedly. Finally, I've removed the return because it is not needed for a void function.

## Rethink the interface

It would be simpler and more logical for the search routine to return a pointer to the node or nullptr if it is not found. Here's one way to write it:

node* search(node *head, int n)
{
}
}

## Sanitize user input

Right now, if the user inputs anything that is not a number, the program stays in a loop forever. Instead of extracting an int, get a std::string and use std::stoi() to convert it to an integer.

## Use helper routines

Right now, there is a lot of code duplication. The search, insert and delete routines all search for the node first. If you extracted this code into a helper routine to find the node before a match, returning a node *, you could reuse searchPrev within insert and delete instead of duplicating it in every function:

node* searchPrev(node *head, int n)
{
}
}

• People have told me earlier to avoid global variable and pass them as arguments to each function but isn't using global variable much less tedious especially when I have multiple classes and functions in one class operate on objects of other classes too.Can you provide me some examples where using global variables makes the program error prone? Nov 26, 2015 at 8:08
• I've added a bit more about why one should avoid using global variables. Nov 26, 2015 at 13:02

While your code doesn't leak right now, it will leak as soon as you replace int as a value type with a more complex or generic object type. Then for example in lines like this:

first->n=n;

exceptions can occur and you will leak memory. This is subtle and difficult to get right. Therefore always* use smart pointers to express ownership, in this case std::unique_ptr.

*: unless absolutely necessary and you know what you are doing.


text:
## Discussion on the Calculation of Rate Constant and Activation Energy of Second-Order Reaction

Li Qibiao, Hao Yajuan,

Abstract

In this paper, we discussed several problems on calculation of reaction rate constant in physical chemistry reference books. When calculating the rate constant k of the second-order reaction, we should make the half-life formula consistent with the reaction stoichiometric equation. And when calculating the activation energy with Arrhenius formula for ideal gas-phase reaction, attention should be paid to identifying the difference between the reaction rate constant kc and kp, as well as the difference between corresponding activation energy Eac and Eap. It is beneficial for students to have a correct understanding in calculation.

Keywords： Reaction rate constant ; Half-life ; Gas-phase reaction ; Activation energy

Li Qibiao. Discussion on the Calculation of Rate Constant and Activation Energy of Second-Order Reaction. University Chemistry[J], 2020, 35(9): 205-208 doi:10.3866/PKU.DXHX201910055

## 2 反应速率常数计算问题示例

${k_{p, 970{\rm{K}}}} = \frac{1}{{{p_{{{\rm{A}}_{\rm{0}}}}}{t_{1/2}}}} = \frac{1}{{39.2 \times {{10}^3}{\rm{ Pa}} \times 1529{\rm{ s}}}} = 1.7 \times {10^{ - 8}}{\rm{ P}}{{\rm{a}}^{ - {\rm{1}}}} \cdot {{\rm{s}}^{ - {\rm{1}}}}$

${k_{p, 1030{\rm{K}}}} = \frac{1}{{{p_{{{\rm{A}}_{\rm{0}}}}}{t_{1/2}}}} = \frac{1}{{48.0 \times {{10}^3}{\rm{ Pa}} \times 212{\rm{ s}}}} = 9.8 \times {10^{ - 8}}{\rm{ P}}{{\rm{a}}^{ - {\rm{1}}}} \cdot {{\rm{s}}^{ - {\rm{1}}}}$

${k_{p, 1030{\rm{K}}}} = \frac{1}{{2{p_{{{\rm{A}}_{\rm{0}}}}}{t_{1/2}}}} = 4.9 \times {10^{ - 8}}{\rm{ P}}{{\rm{a}}^{ - {\rm{1}}}} \cdot {{\rm{s}}^{ - {\rm{1}}}}$

$\ln \frac{{{k_p}_{(1030{\rm{K}})}}}{{{k_p}_{(970{\rm{K}})}}} = \frac{{{E_{\rm{a}}}}}{R}(\frac{1}{{970{\rm{ K}}}} - \frac{1}{{1030{\rm{ K}}}})$

$\ln \frac{{9.8 \times {{10}^{ - 8}}{\rm{ P}}{{\rm{a}}^{ - {\rm{1}}}} \cdot {{\rm{s}}^{ - {\rm{1}}}}}}{{1.7 \times {{10}^{ - 8}}{\rm{ P}}{{\rm{a}}^{ - {\rm{1}}}} \cdot {{\rm{s}}^{ - {\rm{1}}}}}} = \frac{{{E_{\rm{a}}}}}{{8.314{\rm{ J}} \cdot {\rm{mo}}{{\rm{l}}^{ - {\rm{1}}}} \cdot {{\rm{K}}^{ - {\rm{1}}}}}}(\frac{1}{{970{\rm{ K}}}} - \frac{1}{{1030{\rm{ K}}}})$

$\ln \frac{{{k_c}_{(1030{\rm{K}})}}}{{{k_c}_{(970{\rm{K}})}}} = \ln \frac{{{k_p}_{(1030{\rm{K}})} \cdot R \times 1030{\rm{ K}}}}{{{k_p}_{(970{\rm{K}})} \cdot R \times 970{\rm{ K}}}} = \frac{{{E_{{\rm{a}}c}}}}{R}(\frac{1}{{970{\rm{ K}}}} - \frac{1}{{1030{\rm{ K}}}})$

$\ln \frac{{{k_{c, {T_2}}}}}{{{k_{c, {T_1}}}}} = \frac{{{E_{{\rm{a}}c}}}}{R}(\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}})$

$\ln \frac{{{k_{p, {T_2}}}}}{{{k_{p, {T_1}}}}} = \frac{{{E_{{\rm{a}}p}}}}{R}(\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}})$

${E_{{\rm{a}}c}} - {E_{{\rm{a}}p}} = (n - 1)R\frac{{{T_2}{T_1}}}{{{T_2} - {T_1}}} \cdot \ln \frac{{{T_2}}}{{{T_1}}}$

(1) ${K_p} = \frac{{{k_{p, 1}}}}{{{k_{p, - 2}}}} = \frac{{{p_{\rm{B}}}{p_{\rm{C}}}}}{{{p_{\rm{A}}}}} = \frac{{0.21{\rm{ }}{{\rm{s}}^{ - {\rm{1}}}}}}{{5 \times {{10}^{ - 9}}{\rm{ P}}{{\rm{a}}^{ - {\rm{1}}}} \cdot {{\rm{s}}^{ - {\rm{1}}}}}} = 4.2 \times {10^7}{\rm{ Pa}}$

(2)由于${E_{{\rm{a}}, 1}}{\rm{ = }}{E_{{\rm{a}}, - 2}}{\rm{ = }}{E_{\rm{a}}}$

$\ln \frac{{{k_{{T_2}}}}}{{{k_{{T_1}}}}}{\rm{ = }}\ln 2 = \frac{{{E_{\rm{a}}}}}{R}(\frac{1}{{298{\rm{ K}}}} - \frac{1}{{310{\rm{ K}}}})，{E_{\rm{a}}} = 44.36{\rm{ kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}。$

(3)根据平衡常数与温度的关系式，在相同的升温区间内正、逆反应速率常数的变化值相同，所以

$\frac{{{\rm{d}}\ln {K_p}}}{{{\rm{d}}T}} = \frac{{{\Delta _{\rm{r}}}{H_{\rm{m}}}}}{{R{T^2}}} = 0，{\Delta _{\rm{r}}}{H_{\rm{m}}} = {\rm{0}}$

${\Delta _{\rm{r}}}{U_{\rm{m}}} = {\Delta _{\rm{r}}}{H_{\rm{m}}} - \sum {{\nu _{\rm{B}}}RT}$

T = 298 K时，ΔrUm = −2.48 kJ·mol−1T = 308 K时，ΔrUm = −2.56 kJ·mol−1

$\ln \frac{{{k_{{c_1}, 310{\rm{K}}}}}}{{{k_{{c_1}, 298{\rm{K}}}}}} = \ln \frac{{{k_{{p_1}, 310{\rm{K}}}}}}{{{k_{{p_1}, 298{\rm{K}}}}}} = \ln 2 = \frac{{{E_{{\rm{a}}c, 1}}}}{R}(\frac{1}{{298{\rm{ K}}}} - \frac{1}{{310{\rm{ K}}}})$

$\ln \frac{{{k_{{c_{ - 2}}, 310{\rm{K}}}}}}{{{k_{{c_{ - 2}}, 298{\rm{K}}}}}} = \ln \frac{{{k_{{p_{ - 2}}, 310{\rm{K}}}}R \times 310{\rm{ K}}}}{{{k_{{p_{ - 2}}, 298{\rm{K}}}}R \times 298{\rm{ K}}}} = \ln 2 + \ln \frac{{310{\rm{ K}}}}{{298{\rm{ K}}}} = \frac{{{E_{{\rm{a}}c, - 2}}}}{R}(\frac{1}{{298{\rm{ K}}}} - \frac{1}{{310{\rm{ K}}}})$

/

 〈 〉


text:
HomeEnglishClass 12MathsChapterCurve Tracing

Draw the graph of the function...

# Draw the graph of the function f(x) = (1+1/x)^(2)

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Updated On: 29-4-2021

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Text Solution

Solution :
f(x) = (1+1/x)^(x),f(x) is defined if 1+1/x gt 0
 rArr (x+1)/x gt 0 rArr (-infty, -1) cup (0 , infty)
 Now f^(')(x) =(1+1/x)^(x) ["ln "(1+1/x)+x/(1+1/x)-1/x^(2)]
 =(1+1/x)(x) ["ln "(1+1/1x)-1/(x+1)]
 Now (1+1/x)^(x) is always positive, hence the sign of the f^(')(x) depends on the sign of "ln "(1+1/x)-1/(x+1)
 Let g(x) = "ln "(1+1/x)-1/(x+1)
 g^(')(x) =1/(1+1/x) -1/x^(2)+1/(x+1)^(2) =-1/(x(x+1)^(2)).
 For x in (0, infty), g^(')(x) lt 0.
 So g(x) is monotonically decreasing for x in (0,infty). Thus,
 g(x) gt underset(x to infty)"lim"g(x)
 rArr g(x) gt 0
 rArr f^(')(x) gt 0
 For x in (-infty, -1), g^(')(x) gt 0.
 So g(x) is monotonically increasing in its domain.
 Also underset( x to +-infty)"lim"(1+1/x)^(x) =e
 underset(x to +-infty(1+1/x)^(x)=e
 underset(x to 0^(+))"lim"(1+1/x)^(x)=1 and underset(x to 1) "lim"(x+1/x)^(x)=infty
 From the above discussion the graph of the function is as shown in the following figure.

 Range is y in (1, infty)- {e}

Transcript

TimeTranscript
00:00 - 00:59hello students today's question is for the graph of the function f of x is equal to 1 + 1 by X whole square so that is begin with the solution the question we are given that f of x is equal to 1 + 1 by X whole square now at x is equal to zero this function will not be defined as the denominator will become zero so the domain the domain of this function will be our -0 this will be the dominant now for increasing decreasing nature latest first stash of xeof4 of X will be equal to 2 into 1 + 1 by X in 20 - 1 by
01:00 - 01:59x square ok so from here we get the value as minus 2 into X + 1 divided by x cube now we have to put a dash of 0f dash of x is equal to zero so it will be equal to minus 2 after sex sequel 20 so it will be minus 2 into X + 1 divided by x cube is equal to zero from here we get two values for x first is X is equal to minus of 1 and second first year equation the numerator with 0 this is just for finding the points ok so was there in creating the numerator with zeros we get the value of excess -1
02:00 - 02:59and then very quiet and denominator with zeros in getting the value of X 400 now we will plot these on the number line show some hear this is -1 this is zero so what will be - as a Coefficient is -2 so it starts from - + - it is increasing and decreasing ok so for a 40 in the functions of define solar lot pot for a 40 will start with F of 14 - 1 sweat at 4 - 1 will get 1 + 1 upon 2 minus 1 whole square this is equal to zero ok so this equation is of P by Q form so have to take the limit so we are taking
03:00 - 03:59the limit where limit X tends to infinity into 1 + 1 by X whole square Sohail put the limits to the power 1 + 1 by infinity which will equal to 1 now for limit exchange 2 - of infinity 1 + 1 by X whole square it will be equal to 1 + 1 upon minus infinity is equal to 1 letter draw the graph for this function graph here we have y-axis and this is our x-axis so we will draw a
04:00 - 04:59line which is a Y is equal to one have we will draw the line for Y is equal to 1 this side then why is equal to what 280 the function is not defined the standing to is infinity so it will become like this and it will come down ok like this and this is about point -10 So has a graph is going like ok so this is a required answer thank you