1. Why Wait?!? Bryan Gorney Joe Walker Dave Mertz Josh Staidl Matt Boche

2. 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: I.Rudiments of Queuing Theory a. 1 Customer, 1 Server b. Many Customer, No Server c. Many Customers, 1 Server II.Concept of a Stationary Distribution III.Traffic Intensity a. Average Queue Length b. Little’s Principle c. Average Waiting Time IV.Store Profit Maximization

3. 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.

4. To obtain the time-dependent transition probabilities for this Markov chain, let X t = 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] p(t) q(t) )(

5. q(t) = 1 – p(t) lim p(t) = 0 lim X t = State diagram  P[service is completed in  t] =  t  P[service is not completed in  t] = 1 –  t t  t  10  t  t ( 0101 ) 1

6.  The transition matrix for each  t time step is given by A =  Thus, X t+  t = AX t which in matrix form gives: ( 1 –  t 0  t 1 ) p(t +  t) q(t +  t) () = ( 1 –  t 0  t 1 ) p(t) q(t) )(

7.  Performing matrix multiplication yields: p(t +  t) = (1 –  t)p(t)  Rearranging so difference quotient is on the left-hand side:  By letting  t go to zero, the difference quotient becomes a derivative, and p(t +  t) – p(t)  t = -  p(t) p(t +  t) – p(t)  t tt 0 limp’(t) = = -  p(t)

8.  This is the exponential differential equation with solution: p(t) = P[N(t) = 1] = p  e -  t  Since we assumed that there is an individual initially being served, N(0) = 1. This implies that p(0) = 1 which gives p 0 = 1. Finally we have: p(t) = P[N(t) = 1] = e -  t  Since q(t) = 1 – p(t), the probability of completing the service is: q(t) = 1 – e -  t

9. 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 -  t (complement). Left-hand side = cumulative distribution function of the continuous random variable T. Right-hand side = exponential distribution.

10.  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] =  e -  t, 0 < t < .  E(T) = tf(t)dt = t  e -  t dt =  Thus the average value of T is. We interpret the service rate  as: d dt 0 0 1  1 1 Mean Time Until Transition  =

11. Many Customers, No Server 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. 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:

12. 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. 3.Occurrences of arrivals in non-overlapping intervals are independent. 4.The Probability of two or more arrivals in [t, t+  t] is negligible.

13. Assume P n (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 = ( 1-  t 0 0 0...  t 1-  t 0 0... 0  t 1-  t 0... )

14. tt tt tt 1-  t State Diagram of Transition Matrix X t+  t = AX t is the state equation for each time step size  t, where: ( P 0 (t) P 1 (t) : P n (t) : ) X t =

15. After Matrix Multiplication: we have the following system of equations: P 0 (t+  t) = (1-  t)P 0 (t) P 1 (t+  t) =  tP 0 (t)+(1-  t)P 1 (t) P n (t+  t) =  tP n-1 (t)+(1-  t)P n (t) These equations can then be rearranged to make difference quotients.

16. Letting  t go to zero we obtain a system of differential equations. For n  1,  P n ’(t) = lim  t  0 = [P n-1 (t) – P n (t)], and for n = 0, P 0 ’(t) = - P 0 (t)  This equation is the exponential differential equation with initial condition that there are no arrivals (P 0 (0) = 1). This gives: P 0 (t) = e - t P n (t+  t) – P n (t)  t

17. Similarly, letting n=1, P 1 ’(t) = [P 0 (t)- P 1 (t)] = [e - t - P 1 (t)] This is a first-order linear differential equation which can be solved by multiplying by the integrating factor e t. d dt [ e t P 1 (t) ] = so that e t P 1 (t) = t+c The initial condition P 0 (0) =1 implies P n (0) = 0 for all n  1, thus: P 1 (t) = te - t

18. It can be shown by induction that the system of differential equations has solution:  P n (t) = e - t ( t) n for t  0, 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

19. 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  t is  t. Probability of departure in  t is  t.

20. Three mutually exclusive events: I. Exactly one arrival in (t,t +  t) II. Exactly one departure in (t,t +  t) III. No arrivals or departures in (t,t +  t). 012 tttt tttt tttt  t 1 –  t -  t 1 –  t 1 –  t -  t

21. The time step for this Markov chain is  t and the transition matrix is: 1-  t  t 0 0 0 …  t 1 –  t –  t  t 0 0 …  t 1 –  t –  t  t 0 0 … 0  t 1 –  t –  t  t 0 … 0  t 1 –  t –  t  t 0 …............... ( ) A =

22. Let P n (t) = P[N(t) = n] be the probability that there are n customers in the queue at time t. Thus letting P 0 (t) P 1 (t) P n (t)...... () Xt =Xt =Xt =Xt =

23. We obtain the equation X t +  t = AX t for the single-server queue. The state equation gives the following (infinite) system of equations for the single- server queue: P 0 (t +  t) = (1 –  t)P 0 (t) +  tP 2 (t) P 1 (t +  t) = (1 –  t -  t)P 1 (t) +  tP 0 (t) +  tP 2 (t) P n (t +  t) = (1 –  t -  t)P n (t) +  tP 0 - 1 (t) +  tP n + 1 (t)......

24.  System of differential equations by letting  t go to zero; more specifically, for n > 1,lim  t 0 P n (t +  t) – P n (t) tttt P’ n (t) = = -(  P n (t)  P n – 1 (t) +  P n – 1 (t) And, P 0 ’(t) = - P 0 (t) +  P 1 (t).

25. Many classical methods are available for the solving the system: P n ’(t) =-(  P n (t)  P n – 1 (t) +  P n – 1 (t) P n ’(t) = -(  P n (t)  P n – 1 (t) +  P n – 1 (t) P 0 ’(t) = - P 0 (t) +  P 1 (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.

26. 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: P n ’(t) =-(  P n (t)  P n – 1 (t) +  P n – 1 (t) P n ’(t) = -(  P n (t)  P n – 1 (t) +  P n – 1 (t) P 0 ’(t) = - P 0 (t) +  P 1 (t) has a fixed point or stationary distribution provided its rates of change are zero, that is: n ’ P n ’(t) = 0 for n  0.

27. lim P n ’(t) = 0 for n  0, is equivalent to assuming that lim P n (t) = P n exists, where P n does not depend upon t. Applying P n ’(t) = 0 to the system of equations, we obtain: 1.) 0 = - P 0 +  P 1 and 2.) 0 = -(  +  )P n + P n-1 +  P n+l, for n > 1. Solving for P 1, using equation 1: P 1 = (  P 0 Solving for P 2, using P 1 and equation 2: P 2 = (  2 P 0 Continuing on, using induction, shows: P n = (  n P 0 t  t 

28. Queue size must be a non-negative integer, hence, P 0 + P 1 + P 2 + … = 1 So, P 0 + (  )P 0 + (  2 P 0 + … = 1 Which is, P 0 Σ (  n = 1. This sums to 1/(1-(  )) provided that (  ) < 1, which is the condition for the stationary distribution. Using this sum, P 0 = 1 – (  so the stationary distribution for the single-server queue is: P n = (  ) n (1-(  )) n = 0 

29. Traffic Intensity Average Queue Length Little’s Principle Average Waiting Time

30. Average Queue Length Let ρ = (λ / μ). (called the traffic intensity of the queue) By assumption, ρ < 1, as λ < μ. Using ρ in: P n = (λ / μ) n (1 – (λ / μ)), P n = ρ n (1 – ρ).

31. Obtaining the average number N a of customers in the queue: N a = E(N) = Σ nP n = Σ n(ρ n (1 – ρ)) = (1 – ρ) Σ nρ n = (1 – ρ) Σ ρnρ n-1 = (1 – ρ)ρ Σ = (1 – ρ)ρ = (1 – ρ)ρ(1 – ρ) -2 = n = 0   ddρddρ ρnρn    ddρddρ  ddρddρ ρnρn (1 / (1 – ρ)) ρ ρ - 1

32. ρ ρ - 1 N a =

33. 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. a = N a = λT a a / T a = N a / λ = (ρ / ((1 – ρ)λ) = ((λ / μ) / ((1 – (λ / μ))λ)) = (1 / μ) / (1 – (λ / μ)) = (1 / μ) / (1 – ρ) = 1/(μ - λ)

34. T a = 1/ (μ – λ )

35. 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: C(ŧ) = { 0 for ŧ  T a0 AT a0 for ŧ  T a0  T a0 represents the threshold where customers stop returning to the store due to long waiting times.

36. Ŧ0Ŧ0 ŧ C(ŧ) Graph of Customer Attrition Function 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

37. Customer waiting costs plus employee costs J(  ) = C(1/(  μ-λ)) +K  1   <   = # number of employees (checkers) K = cost associated with retaining those employees

38. Conclusion The object of the manager is to minimize J(  ) by finding the number of employees that eliminates the attrition factor by decreasing costs.

39. References A Course in Mathematical Modeling, by Douglas Mooney and Randall Swift Dr. Steve Deckelman