RFID & Item-Level Information Visibility

Transcription

1 Department of Information System & Operations Management University of Florida Gainesville, Florida December 13, 2008

2 1. Manufacturing an Engine Engine Builder Replenish Policy {A1,A2} A {B1,B2} B B 1 = 10cm & B 2 = 10.2cm T 1 = 10.2cm & T 2 = 10.4cm L(B 1,T 1 ) = L(B 2,T 2 ) = 10 years L(B 1,T 2 ) = L(B 2,T 1 ) = 9 years L(A,B) Link: Shandong Province Adopts RFID For Quality Control

3 Engine Builder Replenish Policy Building one engine: Without RFID visibility E[L] = E[E(B T)] = 9.5 years. With RFID visibility L = max{l(b 1, T 1 ), L(B 2, T 2 ), L(B 1, T 2 ), L(B 2, T 1 )} = 10 years Building two engine: Without RFID visibility 19 years. With RFID visibility 20 years

4 2. Replenish Retail Shelf Engine Builder Replenish Policy There s only one client in the store The client has 50 % possibility to buy a bottle of shampoo Original stock: 20 bottles Remaining number of shampoo can be any number in {0,1,2,,20} If the shelf is empty, the client leaves without buying the shampoo.

5 Engine Builder Replenish Policy Without RFID visibility Expected sale = 50% = With RFID visibility store clerk replenish the shelf once it s empty Expected sale = 0.5 bottle

6 Notations Motivating examples One Component Multiple Components Theorems {X 1,X 2,,X m }: m components {X i x i1,x i2,,x ini }: tagged information of cases in a component X f x (X): statistical property of the tagged information y i = g(x i ): production function of x i O: the outcome without RFID visibility Õ: the outcome with RFID visibility

7 Model Setup Motivating examples One Component Multiple Components Theorems X f x (X) x 1 g(x 1 ) {X x 1,x 2,x 3,...,x n } x 2 g(x 2 ) y = g(x i ) X x 3 g(x 3 ) O : O = E[g(X)] Õ : Õ = max{g(x)} f(x i ) x n g(x n )

8 One Component Multiple Components Theorems The p.d.f. of the sample without information visibility is: f y (y) = f x (g 1 (y))g (y) The p.d.f. of the sample with information visibility is: ( y ) n 1 fn y (y) = n f x(g 1 (y))g (y) f x (g 1 (y))g (y)dy

9 One Component Multiple Components Theorems The difference (benefit) of having information visibility is: δ = = Õ O yu (nu n 1 1)dy y where u = y f x (g 1 (y)g (y)dy

10 Best k outcomes Motivating examples One Component Multiple Components Theorems The p.d.f. of the kth best production is: f y k:n (y) = n! (k 1)!(n k)! uk 1 (1 u) n k f y (y) Hence the benefit of having information visibility is: δ = = k (E[y i:n ] E[y]) i=1 y (nf binomial(n 1,u) (k) k) yf y (y)dy

11 Setup Motivating examples One Component Multiple Components Theorems Two components: n components: x 1 y 1 z = f x,y ) 1 ( 1 1 X 2 x 2 y 2 z = f x,y ) 2 ( 2 2 X 1 X 3 x 1 y 1 z = f x,y ) 3 ( 2 1 G(X 1,X 1,...X n ) x 2 y 2 z = f x,y ) 4 ( 1 2 X 4 X n

12 One Component Multiple Components Theorems Assume that we have m components: {X X 1,X 2 X m } {X 1 x 11,x 12,x 13 x 1n1 } {X 2 x 21,x 22,x 23 x 2n2 }. {X m x m1,x m2,x m3 x mnm } X 1,X 2 X m follow joint distribution f X1,X 2 X m (X 1,X 2 X m )

13 One Component Multiple Components Theorems Y = g(x 1,X 2 X m ) The c.d.f. of the sample with information visibility is: F N y (y) = Pr[g(X 1,X 2 X m ) y] = g(x 1,X 2 X m) y f X 1,X 2 X m (X 1,X 2 X m )dx 1 dx 2 dx m The difference (benefit) with information visibility is: δ = n 1 n 2 n m u (u n 1n 2 n m 1 1)ydy where u = y f X1,X 2 X m (X 1,X 2 X m )dx 1 dx 2 dx m g(x 1,X 2 X m) y

14 The k best outcomes Motivating examples One Component Multiple Components Theorems f y k:n (y) = N! (k 1)!(N k)! uk 1 (1 u) N k f y (y) k E[y i:n ] = i=1 = k i=1 y y y N! (i 1)!(N i)! ui 1 (1 u) N i f y (y)dy nf binomial(n 1,u) (k) yf y (y)dy Hence the benefit of having information visibility, δ is: δ = (NF binomial(n 1,u) (k) k) yf y (y)dy y

15 One Component Multiple Components Theorems Theorem (1) The lower bound for δ is:. δ 0

16 One Component Multiple Components Theorems Theorem (2) When k n, δ is upper bounded as: (n k)2 2 δ (ne n k) yf y (y)dy (n k) yf y (y)dy y Proof. Hoeffding s inequality & Chernoff s inequality: (n k)2 2 F binomial (k;n,u) e n 1 y

17 One Component Multiple Components Theorems Theorem (3) If k = n, δ = 0 Theorem (4) Larger the sample space variance, larger the benefit δ. Theorem (5) if Y is positive, δ is a monotonic increasing function of n. Theorem (6) If all the resources are used, the benefit of having information visibility is zero if there s only one component; the benefit is greater than zero if there are multiple components.

18 Control Function Motivating examples Control Function Finite Horizon Infinite Horizon { 19 if X = 0 ψ(x) = X otherwise X ψ(x) Y = g(ψ(x)) δ = Ω (k,x,ψ(x),g(x))

19 Finite Horizon Motivating examples Control Function Finite Horizon Infinite Horizon Using Bellman s Method, the recursive function is: V (X T k,k) = max vt k {δ T k (X T k,v T k ) + V (X T k+1,k 1)} δ T k (X T k,ψ T k (X T k ))+V (ξ T k (X T k,ψ T k (X T k ),k 1) subject to: 1. X T k+1 = ξ T k (X T k,v T k,g(v T k )) 2. X 0 = X 0 3. v T k = ψ T k (X T k ) 4. v t Θ for all t = 0,1,,T 1

20 Infinite Horizon Motivating examples Control Function Finite Horizon Infinite Horizon subject to: V t (X t ) = max v t { βδ(xt,v t ) + V t+1 (X t+1) } 1. X t+1 = ξ 0 (X t,v t ) 2. X 0 = X 0 3. v t = ψ(x t ) (1) where β is the discount factor and 0 β 1. Sargent (1987) and Stokey (1989): starting from any bounded continuous W 0 will cause W to converge as t.

21 Simulation Simulation of Fabrication Quality Control X 1 Uniform ( 0.5,0.5) X 2 Uniform ( 0.5,0.5) G(x 1,x 2 ) = e x 1 x 2 ψ(x) = X

22 Simulation unifor(0,1) normal(0,0.1) normal(0,0.2) normal(0,0.3) 3 benefits due to Informatioin visibility sample size Figure: Benefit of having information visibility for fabrication quality control.

23 Analysis Motivating examples Simulation The results show: The benefit of information visibility increases if the scale of the manufacturing increases The benefit function is concave and bounded. If the samples are more volatile, more is the benefit of having RFID information visibility.

24 Simulation Merci!