a) Give an example of a case when an (s,s) policy is not the same as an (R,Q) policy. (2p)

Transcription

1 roblem a) Give an example of a case wen an (s,s) policy is not te same as an (R,) policy. (p) b) Consider exponential smooting wit te smooting constant α and moving average over N periods. Ten, tese two tecniques are comparable if α = /(N+). In wat way are tey comparable? (3p) c) Wat problem does Moore s algoritm solve? (p) d) In connection wit a common cycle time in cyclic planning te following constraint (sv. bivillkor) sould be satisfied, T N i= s i N i= ρ. i Explain te background of tis constraint. (3p) roblem ssumetattedemandintefollowingeigtweeksis: d = (30,75,80,55,00,60,85,95). Te ordering cost is 500 SEK, and te olding cost per unit and week is SEK. a) Calculate: ) te minimal cost for te first four weeks. ) te minimal cost for te first six weeks given tat te last delivery was in week 3. 3) te minimal cost for te first five weeks given tat te last delivery is in week 5. (Use to cost you obtained in.a. for tis task.) (6p) b) Use te Silver-Meal euristic to solve te problem. Wat is te optimal ordering sequence according to tis euristic? How large is te relative cost increase by using tis euristic instead of te optimal metod? (4p)

2 roblem 3 ssume tat te customer demand in period t is x t = c+d t, were c and d are known constants. Start in period and use exponential smooting. Te smooting constant is 0 < α < and â 0 = 0. Sow tat te forecast error: â t x t+ d/α, wen t. (0p) (Tis problem illustrates wat will appen if te simple exponential smooting metod is applied wen demand as a trend pattern.) roblem 4 Consider a product wic is produced from raw material in two macines, see te figure below. In te beginning of a production cycle, a batc of units of raw material is M 3 M delivered to inventory. Te output from macine is immediately transfered to macine (continuous flow). In te same manner, a product is immediately available for a customer demand wen te product is finised in macine. We ave te following data: i = setup cost for macine i {,}, j = olding cost per unit and unit time for inventory j {,,3}, p i = production rate for macine i {,}, d = constant customer demand per unit of time. ssume tat p > p > d. No sortages are allowed. a) raw graps for te inventory levels as a function of time (for all tree inventories). lso, determine te maximum inventory level and te fraction of time wit positive stock on and, for eac inventory. (5p) b) ssume tat = = 000 SEK, = SEK/day, = SEK/day, 3 = 3 SEK/day, p = 50 per day, p = 00 per day, and d = 80 per day. Calculate te optimal order quantity. (5p)

3 roblem 5 Consider an inventory system for spare parts. Since spare parts often are low demanded we assume tat te ordering policy is a (S,S) policy (or sometimes called a base stock policy). Te leadtime is constant and equal to L. Furtermore, we assume tat te probability for k customer demands during a period ofttimeunitsisoissondistributed,wicmeanstat((t) = k) = e λt (λt) k /k!,were (t) is te stocastic demand during te time interval t. Moreover, E((t)) = λt and V ar((t)) = λt. ll unsatisfied customer demands are backordered (sv. restnoterade). ssume S = 5, λ = and L = 4. a) Calculate te average inventory level. (p) b) Calculate te average stock on and (fysisk medellagernivå). (4p) c) Calculate te average number of backorders. (p) d) Calculate te steady state probability of aving exactly two units outstanding (i.e., exactly units wic ave not yet arrived in stock). (p) roblem 6 ssume tat a single inventory at a company is controlled by a continuous review (R,) policy. ssume tat te demand during te leadtime follows a normal distribution. Te company as decided tat te fraction of demand tat can be satisfied immediately from stock on and sould be 99%, and tat te probability of no stockout per order cycle sould be 85%. Te company is, at te moment, using R = 50 and = 00 in order to acieve tese target levels. a) Information about te average demand during te leadtime, and te standard deviation of te demand during te leadtime ave for some unfortunate reason been deleted in te IT-system. You ave been asked by te logistics manager to calculate tese missing values, given te information above. (8p) b) ssume tat you know te average demand during te leadtime, and te standard deviation of te demand during te leadtime. ssume also tat you know te average waiting time to satisfy a demand. Ten, ow would you calculate te average number of backorders in te system? (p) 3

4 roblem a) See xsäter (006), pp b) See xsäter (006), pp c) See Namias (009), pp d) See xsäter (006), p. 58. roblem Wagner Witin yields: eriod emand a) From te table we obtain Te minimal cost for te first four periods = 60 Te minimal cost for te first six periods given tat te last delivery is in period 3 = 00 Te minimal cost for te first five periods given tat te last delivery is in period 5 = = 760. b) Silver Meal: C = cost per period. elivery in period : periods: C ( ) / , 3 periods: C ( 500 (75 80)) / , 4 periods: C ( 500 ( )) / elivery in period 4: periods: C ( ) / , 3 periods: C ( 500 (00 60)) / , 4 periods: C ( 500 ( )) / elivery in period 7 periods: C ( ) / Total cost = 600, wic means tat te relative cost increase is.% (te minimal cost is 570 according to Wagner Witin).

5 roblem 3 Set t = â t x t+. Tis gives directly, t = â t x t+ = ( α)â t +α(c+d t) (c+d(t+)) = ( α)(â t (c+d t))+α( d α ) = ( α) t +α( d α ) Hence, tis is exponential smooting for te constant demand d/α. Te desired result is ten clear. Remark: if tis is not clear, note tat wen t, we must ave t = t =. Tis gives wen t. = ( α)+α( d α ) = d/α,

6 roblem 4 a) Below te inventory levels for inventories, and 3 are sown. Here, we directly see tat te fraction of time wit positive stock on and is / for inventory, / for inventory, and for inventory 3. b) Te total cost becomes: ) ( 3 C. ifferentiate C wrt and set te derivative equal to zero. Tis gives / 3 * ) (. Since C is convex in tis minimum is te global minimum. Wit numbers, 438 *. / / (-/) / / (-/)

7 roblem 5 S = 5, λ =, L = 4. a) E(IL) = S λl = 3. b) E(IL + ) = S k= k(il = k) = S k= ke λl (λl) S k /(S k)! = c) E(IL ) = E(IL + ) E(IL) = d) Let V denote te number of outstanding orders in stationarity. Ten, (V = k) = e λl (λl) k /k! = roblem 6 S = 0.85, S = 0.99, R = 50, = 00. a) Tis gives, R = µ +.04σ. ( ) R µ S = Φ = 0.85 R µ =.04. σ σ S = σ ( G ( ) ( )) R µ R+ µ G = σ σ Since is large we get te approximate relation (ceck tat tis is ok!), σ G(.04) = 0.99 σ =.95 µ = b) ssume tat te mean waiting time, W, is known. Te leadtime and average demand during te leadtime is also assumed to be known. Ten, due to Little s formula. Here, µ = µ /L. E(IL ) = µw,