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## Why Wait?!?

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**Why Wait?!?**Bryan Gorney Josh Staidl Matt Boche Joe Walker Dave Mertz**Purpose: To minimize costs of businesses with waiting**lines, taking into consideration the cost of servers and the long-run loss of revenue by making customers wait too long. Outline: • Rudiments of Queuing Theory • a. 1 Customer, 1 Server • b. Many Customer, No Server • c. Many Customers, 1 Server • Concept of a Stationary Distribution • Traffic Intensity • a. Average Queue Length • b. Little’s Principle • c. Average Waiting Time • Store Profit Maximization**1 Customer, 1 Server**• N(t) = number of individuals at checkout counter at time t. • N(t) has only two possible values of 0 or 1. • “being served” corresponds to N(t) = 1. • “finished being served” corresponds to N(t) = 0.**(**) p(t) q(t) To obtain the time-dependent transition probabilities for this Markov chain, let Xt = be the time-dependent distribution vector for the states “being served”and“finished being served”. That is, p(t) = P[N(t) = 1] and q(t) = P[N(t) = 0]**(**) t 0 1 t 1 mDt 0 1 - mDt • q(t) = 1 – p(t) • lim p(t) = 0 lim Xt = State diagram • P[service is completed in Dt] = mDt • P[service is not completed in Dt] = 1 – mDt 1**(**) p(t) q(t) ( ( ) ) 1 – mDt 0 mDt 1 1 – mDt 0 mDt 1 ( ) p(t + Dt) q(t + Dt) = • The transition matrix for each Dt time step is given by A = • Thus, Xt+Dt = AXt which in matrix form gives:**Performing matrix multiplication yields:**p(t + Dt) = (1 – mDt)p(t) • Rearranging so difference quotient is on the left-hand side: • By letting Dt go to zero, the difference quotient becomes a derivative, and p(t + Dt) – p(t) Dt = -mp(t) p(t + Dt) – p(t) Dt p’(t) = lim = -mp(t) Dt 0**This is the exponential differential equation with solution:**p(t) = P[N(t) = 1] = p0e-mt • Since we assumed that there is an individual initially being served, N(0) = 1. This implies that p(0) = 1 which gives p0 = 1. Finally we have: p(t) = P[N(t) = 1] = e-mt • Since q(t) = 1 – p(t), the probability of completing the service is: q(t) = 1 – e-mt**Next let T denote the time at which service is completed.**It is considered to be the time until transition from one state to another. • T is a continuous random variable with range 0 < T < . • P[T > t] = P[N(t) = 1]. • P[T < t] = 1 – e-mt (complement). • Left-hand side = cumulative distribution function of the continuous random variable T. • Right-hand side = exponential distribution.**** 0 0 1 m 1 m = Mean Time Until Transition • P[T > s + t given T > s] = P[T > t] (Memoryless property) for all s,t > 0. • Density function is the derivative of the cumulative distribution which is: f(t) = P[T < t] = me-mt, 0 < t < . • E(T) = tf(t)dt = tme-mtdt = • Thus the average value of T is . We interpret the service rate m as: d dt 1 m**Many Customers, No Server**Let N(t) be a time dependent function that denotes the number of customer arrivals in a given interval [0,t], and assume the following: • 1. The probability that an arrival occurs in a short interval • [t, t+t] is proportional to the length of the interval t. • In symbols: • P[ N(t+t) - N(t) = 1] = t • for some constant >0.**2. The Memoryless Property:**• The probability that an arrival occurs in [t, t+t] does • not depend on the time of previous arrivals: • P[ N(t+t) – N(t) = n | N(s) = m ] = P[ N(t+t) – N(t) = n] • for all 0 s t. • Occurrences of arrivals in non-overlapping intervals are • independent. • The Probability of two or more arrivals in [t, t+t] is negligible.**(**) 1-t 0 0 0 . . . t 1-t 0 0 . . . 0 t 1- t 0 . . . Assume Pn(t) = P[N(t) = n] N(t) has possible values 0,1,2,… since it is possible to have any number of customers in line. Considering a time step of size t, N(t) forms a Markov Chain, with a transition matrix of: A =**t**t t 1-t 1-t 1-t ( ) P0(t) P1(t) : Pn(t) : Xt = State Diagram of Transition Matrix Xt+t = AXt is the state equation for each time step size t, where:**After Matrix Multiplication: we have the following**system of equations: P0(t+t) = (1-t)P0(t) P1(t+t) = tP0(t)+(1-t)P1(t) Pn(t+t) = tPn-1(t)+(1-t)Pn(t) These equations can then be rearranged to make difference quotients.**Letting t go to zero we obtain a system of differential**equations. For n 1, • Pn’(t) = limt0 = [Pn-1(t) – Pn(t)], and for n = 0, P0’(t) = -P0(t) • This equation is the exponential differential equation with initial condition that there are no arrivals (P0(0) = 1). This gives: P0(t) = e-t Pn(t+ t) – Pn(t) t**d**dt [etP1(t)] = so that Similarly, letting n=1, P1’(t) = [P0(t)- P1(t)] = [e-t - P1(t)] This is a first-order linear differential equation which can be solved by multiplying by the integrating factor et. • etP1(t) = t+c • The initial condition P0(0) =1 implies Pn(0) = 0 • for all n 1, thus: • P1(t) = te-t**It can be shown by induction that the system of**• differential equations has solution: • Pn(t) = e-t(t)nfor t0, n = 0,1,2,… n! • Known as a homogeneous Poisson process • with rate . ( does not depend on time variable • t) • Mean is t (t customers during time interval • of length t) • Ex. = 2 customers/minute, t =5 minutes • t = 5*2 = 10 customers over 5 minutes**Many Customers, 1 Server**• Let N(t) = number of individuals at checkout counter at time t. • N(t) = any of the integer values 0, 1, 2, … at time t. • Probability of arrival in Dt is lDt. • Probability of departure in Dt is mDt.**lDt**lDt lDt 0 1 2 mDt mDt mDt 1 – lDt 1 – lDt - mDt 1 – lDt - mDt • Three mutually exclusive events: I. Exactly one arrival in (t,t + Dt) II. Exactly one departure in (t,t + Dt) III. No arrivals or departures in (t,t + Dt).**.**. . . . ) ( • The time step for this Markov chain is Dt and the transition matrix is: 1-lDt mDt 0 0 0 … lDt 1 – mDt – lDt mDt 0 0 … 0 lDt 1 – mDt – lDt mDt 0 … A = . . . . . . . . . .**(**) P0(t) P1(t) Pn(t) . . . . Xt = . . • Let Pn(t) = P[N(t) = n] be the probability that there are n customers in the queue at time t. Thus letting**.**. . . . . • We obtain the equation Xt + Dt = AXt for the single-server queue. • The state equation gives the following (infinite) system of equations for the single-server queue: P0(t + Dt) = (1 – lDt)P0(t) + mDtP2(t) P1(t + Dt) = (1 – lDt - mDt)P1(t) + lDtP0(t) + mDtP2(t) Pn(t + Dt) = (1 – lDt - mDt)Pn(t) + lDtP0 - 1(t) + mDtPn + 1(t)**lim**Dt 0 Pn(t + Dt) – Pn(t) P’n(t) = • System of differential equations by letting Dt go to zero; more specifically, for n > 1, Dt = -(l + m)Pn(t) + lPn – 1(t) + mPn – 1(t) And, P0’(t) = -lP0(t) + mP1(t).**Many classical methods are available for the solving the**system: Pn’(t) =-(l + m)Pn(t) + lPn – 1(t) + mPn – 1(t) P0’(t) = -lP0(t) + mP1(t) However, they involve ideas beyond the scope of our analysis - instead, we will be using the system of equations to obtain steady-state (time-independent) behavior.**Stationary Distribution**Looking for whether the system of time-dependent probabilities settles down and displays no more “transient” behavior - analogous to finding the fixed points for deterministic systems. In this case, the fixed points will be “stationary” distributions. The system of differential equations: Pn’(t) =-(l + m)Pn(t) + lPn – 1(t) + mPn – 1(t) P0’(t) = -lP0(t) + mP1(t) has a fixed point or stationary distribution provided its rates of change are zero, that is: Pn’(t) = 0 for n 0.**lim Pn’(t) = 0 for n 0, is equivalent to assuming**that lim Pn(t) = Pn exists, where Pn does not depend upon t. Applying Pn’(t) = 0 to the system of equations, we obtain: 1.) 0 = -lP0 + mP1 and 2.) 0 = -(l + m)Pn + lPn-1 + mPn+l, for n > 1. Solving for P1, using equation 1: P1 = (l / m) P0 Solving for P2, using P1 and equation 2: P2 = (l / m)2 P0 Continuing on, using induction, shows: Pn = (l / m)n P0 t t **Queue size must be a non-negative integer, hence,**P0 + P1 + P2 + … = 1 So, P0 + (l/m)P0 + (l/m)2P0+ … = 1 Which is, P0Σ (l/m)n = 1. This sums to 1/(1-(l/m)) provided that (l/m) < 1, which is the condition for the stationary distribution. Using this sum, P0 = 1 – (l/m), so the stationary distribution for the single-server queue is: Pn = (l/m)n(1-(l/m)) n = 0**Traffic Intensity**• Average Queue Length • Little’s Principle • Average Waiting Time**Average Queue Length**Let ρ = (λ / μ). (called the traffic intensity of the queue) By assumption, ρ < 1, as λ < μ. Using ρ in: Pn = (λ / μ)n (1 – (λ / μ)), Pn = ρn (1 – ρ).**** n = 0 n = 0 n = 0 n = 0 n = 0 n = 0 d dρ d dρ d dρ ρ ρ - 1 Obtaining the average number Na of customers in the queue: Na = E(N) = Σ nPn = Σ n(ρn(1 – ρ)) = (1 – ρ) Σ nρn = (1 – ρ) Σ ρnρn-1 = (1 – ρ)ρ Σ = (1 – ρ)ρ Σ = (1 – ρ)ρ = (1 – ρ)ρ(1 – ρ)-2 = ρn ρn (1 / (1 – ρ))**ρ**ρ - 1 Na =**Average Waiting Time**Little’s Principle – the average number of customers in a queueing system is equal to the average arrival rate of customers to that system times the average time spent in that system. Na = λTa Ta = Na / λ = (ρ / ((1 – ρ)λ) = ((λ / μ) / ((1 – (λ / μ))λ)) = (1 / μ) / (1 – (λ / μ)) = (1 / μ) / (1 – ρ) = 1/(μ - λ)**{**0 for ŧ Ta0 C(ŧ) = ATa0 for ŧ Ta0 Model Formulation of Store Profit Maximization C(ŧ), which is the Customer Attrition Function, models the loss of profit per day and is given below: • Ta0 represents the threshold where customers stop returning to the store due to long waiting times.**Graph of Customer Attrition Function**C(ŧ) ŧ Ŧ0 • C(ŧ) = C(1/(μ-λ)) for one server • C(ŧ) = C(1/(μ-λ)) for servers • This models the loss of profit due to customers having to wait in lines**Customer waiting costs plus employee costs**• J() = C(1/(μ-λ)) +K 1 < • = # number of employees (checkers) K = cost associated with retaining those employees**Conclusion**The object of the manager is to minimize J() by finding the number of employees that eliminates the attrition factor by decreasing costs.**References**A Course in Mathematical Modeling, by Douglas Mooney and Randall Swift Dr. Steve Deckelman