1. The production inventory problem What is the expected inventory level? What is expected backorder level? What is the expected total cost? What is the optimal base-stock level?

2. I : level of finished goods inventory B : number of backorders (backorder level) IO : inventory on order. Three basic processes

3. Under a base-stock policy, the arrival of each customer order triggers the placement of an order with the production system  s = I + IO – B  s = E[ I ] + E[ IO ] – E[ B ] Three basic processes

4. I and B cannot be positive at the same time  I = max(0, s - IO ) = ( s – IO ) +  E[ I ] = E[( s – IO ) + ]  B = max(0, IO - s ) = ( IO - s ) +  E[ B ] = E [( IO - s ) + ] Three basic processes

5. The production system behaves like an M/M/1 queue, with IO corresponding to the number of customers in the system.

6. Expected backorder level

7. Expected inventory level

8. Expected cost

9. Optimal base-stock level

11. Queueing Theory

12. The study of queues – why they form, how they can be evaluated, and how they can be optimized. Building blocks – arrival process and a service process.

13. Arrival process – individually/in groups, independent/correlated, single source/multiple sources, infinite/finite population, limited/unlimited capacity. Service process – single/multiple servers, single/multiple stages, individually/in groups, independent/correlated, service discipline (FCFS/priority). Some characteristics of arrival and service processes

14. GX/GY/k/NGX/GY/k/N A Common notation G : distribution of inter- arrival times X : distribution of arrival batch (group) size G : distribution of service times Y : distribution of service batch size k : number of servers N : maximum number of customers allowed

15. Common examples M / M /1 M / G /1 M / M / k M / M /1/ N M X / M / 1 GI / M /1 M / M / k/k

16. Fundamental quantities L : expected number of customers in the system, L = E ( n ). L Q : expected number of customers waiting in queue. W : expected time a customer spends in the system. W Q : expected time a customer spends waiting in queue E [ S ]: expected time customer spends in service. : customer arrival rate, = lim t  ∞ N ( t )/ t, where N ( t ) is the number of arrivals up to time t.

17. Fundamental relationships L = L Q + N s W = W Q + E(S) L = W L Q = W Q N s = E(S) The relationship L = W is often referred to as Little’s law.

18. t 1 t 2 t 3 t 4 t 5 t 6 t 7 T Number in system 3 2 1 A heuristic proof

19. L = [1( t 2 - t 1 )+2( t 3 - t 2 )+1( t 4 - t 3 )+2( t 5 - t 4 )+3( t 6 - t 5 )+2( t 7 - t 6 )+1(T- t 7 )]/ T = (area under curve)/ T = (T+ t 7 + t 6 - t 5 - t 4 + t 3 - t 2 - t 1 )/ T W = [( t 3 - t 1 )+( t 6 - t 2 )+( t 7 - t 4 )+(T- t 5 )]/4 = (T+ t 7 + t 6 - t 5 - t 4 + t 3 - t 2 - t 1 )/4 = (area under curve)/ N(T)

20. L = (area under curve)/ T, W = (area under curve)/ N(T)  LT = WN ( T )  L = WN ( T )/T Since as T  ∞, N ( T )/ T , L = W as T  ∞. A similar heuristic proof can be used to show L Q = W Q and N s = E(S).

21. L = (area under curve)/ T, W = (area under curve)/ N(T)  LT = WN ( T )  L = WN ( T )/T Since as T  ∞, N ( T )/ T , L = W as T  ∞. A similar heuristic proof can be used to show L Q = W Q and N s = E(S).

22. For a single server queue:

23. Case 1 Customers arrive at regular & constant intervals Service times are constant Arrival rate < service rate ( <  ) W Q = 0 W = W Q + E(S) = E(S) L = W = E(S) W Q = W Q = 0 TH (output rate) = Why do queues form?

24. Case 1 Customers arrive at regular & constant intervals Service times are constant Arrival rate < service rate ( <  ) W Q = 0 W = W Q + E(S) = E(S) L = W = E(S) W Q = W Q = 0 TH (output rate) = Why do queues form?