Period Order Quantity

Transcription

1 IE Chapter 5 Dynamic Lot Sizing Techniques 1 Dynamic Lot Sizing Methods Simple Rules Period order quantity Fixed period demand Lot for lot (L4L) Heuristic Methods Silver-Meal method (SM) Least Unit Cost (LUC) Part Period Balancing (PPB) Dynamic Programming (Optimum) Wagner-Whitin 2 1

2 A Prototype Example Suppose for a certain product type you need to produce weekly demand below: A = $50 per order H = $0.5 per unit per week Assumption: Lead time is known with certainty (fixed lead time) 3 Period Order Quantity The average lot size desired is divided by the average period demand For weekly demand given above evaluate POQ for Q = 140 units, and 275 units. 4 2

3 POQ-Example Solution Total demand over 8 periods = 975 units Average weekly demand = 975 / 8 = 122 units per week 5 POQ-Example Solution Continued Fixed period for order Q is determined as follows: Where, Q = desired order (lot) size = average demand over the planning period T = number of periods (time interval between orders) 6 3

4 POQ-Example Solution Continued For Q = 140 T = fixed period between orders = 140 /122 = 1.14 = 1 week Week Beginning Inventory Demand Order End Inventory Total Cost = 8 ($50) ($0.5) = $ POQ-Example Solution Continued For Q = 275 T = fixed period between orders = 275 /122 = 2.25 = 2 weeks Week Beginning Inventory Demand Order End Inventory Total Cost = 4 ($50) ($0.5) = $

5 Fixed Period Demand Ordering m periods of demand, m = selected fixed period For weekly demand given above evaluate FPD for T = 2 weeks, and 4 weeks. 9 FPD-Example Solution For T = 2 weeks Q1 = 175 units, Q3 = 375 units, Q5 = 250 units, Q7 = 175 units For T = 4 weeks Q1 = 550 units, Q5 = 425 unit 10 5

6 FPD-Example Solution for T=2 t Beginning Inventory Demand Q t End Inventory Total cost = 4 ($50) ($0.5) = $ FPD-Example Solution for T=4 t Beginning Inventory Demand Q t End Inventory Total cost = 2 ($50) ($0.5) = $

7 Lot For Lot Rule L4L The order quantity is always the demand for one period For weekly demand given above evaluate L4L rule 13 L4L-Example Solution Lot size per order: Q 1 = 100 units, Q 2 = 75 units, Q 3 = 175 units, Q 4 = 200 units Q 5 = 150 units, Q 6 = 100 units, Q 7 = 75 units, Q 8 = 100 units 14 7

8 L4L-Example Solution Continued t Beginning Inventory Demand Q t End Inventory Total cost = 8 ($50) + 0 ($0.5) = $ Silver-Meal Method Heuristic approach to aim at a low-cost solution that is not necessarily optimal Aim to achieve the minimum average cost per period for the m-period span. The average cost per period includes ordering and inventory holding costs 16 8

9 Silver-Meal Method The average cost per period is as follows: Where; m = number of demand periods to be ordered in the present time. A = fixed ordering cost per order H = inventory holding cost per unit per period K(m) = average cost per period during m periods 17 Silver-Meal Method Compute K(m) for m = 1,2,,m Stop when, K(m+1) > K(m), i.e. the period in which the average cost per period start to increase. Order the quantity equals to m periods demand. Q i = D 1 + D D m Q i is the quantity ordered in period i, and it covers m periods into the future. The process repeats at period (m+i) and continues through the planning horizon. 18 9

10 SM-Example Determine the order quantities for the following lumpy demands using Silver Meal algorithm A = $50 per order H = $0.5 per unit per week 19 For Q 1 : m=1, K(1) = 50 m=2, K(2) = 1/2 ( (75)) = < K(1) m=3, K(3) = 1/3 ( (75) + (2)(0.5)(175)) = 87.6 > K(2) STOP m=2 is selected for Q 1 Q 1 = D 1 + D 2 Q 1 = = 175 units Next order should arrive in week 3, So continue for Q

11 For Q 3 : m=1, K(1) = 50 m=2, K(2) = 1/2 ( (200)) = 75 > K(1) STOP m=1 is selected for Q 3 Q 3 = D 3 Q 3 = 175 unit Next order should arrive in week 4, So continue for Q 4 21 For Q 4 : Q 4 = D 4 Q 4 = 200 units 22 11

12 SM-Example Solution Continued t Beginning Inventory Demand Q t End Inventory Total cost = 3 ($50) + 75 ($0.5) = $ Least Unit Cost Similar to SM algorithm except for the total cost calculation K( m) A hd 2hD... ( m 1) hd 2 3 m D D D... D m The stopping rule: K(m+1) > K(m) Order for m periods into the future : Q i = D 1 + D D m Continue from period (m+i) : Q (m+i) 24 12

13 LUC-Example Determine the order quantities for the following lumpy demands using Least Unit Cost method A = $50 per order H = $0.5 per unit per week 25 Part Period Balance Part Period (PP) is defined as the number of the inventory carrying periods. PP balancing is the quantity ordered which balance the A and H. PP m) D 2D... ( m 1) ( 2 3 PPF A H D m Economic part period factor The stopping rule: PP(m+1) > PPF Order for m periods into the future : Q i = D 1 + D D m Continue from period (m+1) : Q (m+1) 26 13

14 PPB-Example Determine the order quantities for the following lumpy demands using Part Period Balancing method A = $50 per order H = $0.5 per unit per week 27 Wagner-Whitin Algorithm WW is an optimization procedure based on dynamic programming to find optimum order quantity policy Q i with a minimum cost solution. WW evaluates all possible ways of ordering to cover demand in each period of the planning horizon. Wagner-Whitin replaces EOQ for the case of lumpy demand

15 Wagner-Whitin Algorithm Cost of placing order: Where; K(t,m) = total cost of quantity ordered at period t for m periods A = ordering cost, H = inventory holding cost per unit per period D j = demand at period j t = 1,2,..,N and m = t,t+1,t+2,,n 29 Wagner-Whitin Algorithm For each period minimum cost is defined as: K*(m) = min t = 1,2,,m {K*(t-1) + K(t,m)} K*(0) = 0 K*(N) is defined as the least cost solution

16 WW-Example Solution A = $50 per order H = $0.5 per unit per week 31 WW-Example Solution For m=2 K*(2) = K*(0) = 0 For m=1 K*(1) = K*(0) + K(1,1) = 0 + A = 50 K*(0) + K(1,2) = 0 + (A+HD 2 ) = (75) = 87.5 min K*(1) + K(2,2) = 50 + A = = 100 K*(2) =

17 WW-Example Solution For m=3 K*(3) = min K*(3) = K*(0) + K(1,3) = 0+[50+0.5(75)+2(0.5)(175)]=262.5 K*(1) + K(2,3) = 50 +[50+0.5(175)]= K*(2) + K(3,3) = = K*(0) + K(1,m) K*(1) + K(2,m) should not be considered for m>3 33 WW-Example Solution For m=4 K*(2) + K(3,4) = [50+0.5(200)]= K*(4) = min K*(3) + K(4,4) = = K*(4) = K*(2) + K(3,m) should not be considered for m>

18 WW-Example Solution For m=4 K*(2) + K(3,4) = [50+0.5(200)]= K*(4) = min K*(3) + K(4,4) = = K*(4) = Q 4 = D 4 = 200 Continue with K*(3) 35 WW-Example Solution For m=3 K*(3) = min K*(3) = K*(0) + K(1,3) = 0+[50+0.5(75)+2(0.5)(175)]=262.5 K*(1) + K(2,3) = 50 +[50+0.5(175)]= K*(2) + K(3,3) = = Q 3 = D 3 = 175 Continue with K*(2) 36 18

19 WW-Example Solution For m=2 K*(2) = min K*(0) + K(1,2) = 0 + (A+HD 2 ) = (75) = 87.5 K*(1) + K(2,2) = 50 + A = = 100 K*(2) = 87.5 Q 1 = D 1 + D 2 = WW-Example Solution t Beginning Inventory Demand Q t End Inventory Total cost = 3 ($50) + 75 ($0.5) = $