Presentation on theme: "Alternative lot sizing schemes"— Presentation transcript:
1Alternative lot sizing schemes Anadolu UniversityIndustrial Engineering DepartmentSource: Nahmias, S., Production And Operations Analysis, McGraw-Hill /Irwin, Fifth Edition, 2005, ISBN:
2The three major control phases of the productive system Information requirements for each end item over the planning horizonMaster Production SchedulePhase 2Lot sizing rules and capacity planningMaterials Requirements PlanningPhase 3Requirements for raw materialDetailed shop floor schedule
3The explosion calculus Explosion calculus is as term that refers to the set of rules by which gross requirements at one level of product structure are translated into a production schedule at that level and requirements at lower level.Child level (level2)Parent level (level 1)End item levelEnd itemA(2) 1 weekC(1) 2 weeksD(2) 1 weekB(1) 2 weeksC(2) 2 weeksE(3) 2 weeks
4Example 7.1The Harmon Music Company produces a variety of wind instruments at its plant in Joliet, Illinois. Because the company relatively small, it would like to minimize the amount of money tied to inventory. For that reason production levels are set to match predicted demand as closely as possible. In order to achieve this goal, the company has adopted an MRP system to determine production quantities.One of the instruments produced is the model 85C trumpet. The trumpet retails for $800 and has been a reasonably profitable item for the company.Valve: ses dugmesiTrumpet (end item)Bell Assembly (1) Lead Time: 2 weeksValve casing (1) Lead times 4 weeksSlide assemblies (3) Lead Time : 2 weeksValves (3) Lead times : 3 weeksFigure gives the product structure diagram for the construction of the trumpet.
5Explosion of trumpet BOM Week234567891011121314151617Demand77423821261124576TrumpetWeek891011Scheduled receipts12623 trumpet at the end of the week 7week891011121314151617net predicted demand423226112457638Bell AssemblyWeek67891011121314151617Gross requirements423226112457638Net requirementsTime-phased net requirementsplanned order relase (lot-for-lot)Time-phased net requirements4232122611245147638planned order relase (lot-for-lot)Valve CasingAssemblyWeek234567891011121314151617Gross requirements12696367833613542228114Scheduled receiptsOn-han inventory1866030Net requirements66Time-phased net requirementsplanned order relase (lot-for-lot)valves
6Alternative Lot-Sizing Schemes EOQ lot sizingThe silver-meal heuristicLeast Unit CostPart Period Balancing
7EOQ Lot Sizing To apply the EOQ formula, we need three inputs: The average demand rate, The holding cost rate, hSetup cost, K
8EOQ Lot Sizing- Valve casing assembly in Ex.7.1 Suppose that the setup operation for the machinery used in this assembly operation takes two workers about three hours.The worker average cost: $22 per hourK= (22)(2)(3) = $132The company uses a holding cost on a 22 percent annual interest rate.Each valve casing assembly costs $ in materials and value added for labor,Holding cost, h = (141.82)(0.22) / 52 = 0.60Lot-for-lot policy requiresTotal holding cost forTotal setup cost= 0= (132)(10) = $1320
9EOQ Lot Sizing- Valve casing assembly in Ex.7.1 Week4567891011121314151617Gross requirements423226112457638Net requirementsTime-phased net requirementsPlanned order relase (EOQ)139Planned deliveriesEnding inventory97552312410692117Lot-for-lot policy requiresCumulative ending inventoryTotal holding cost for = (0.6)(653) = $391.8Total setup cost = (132)(4) )= $528The total cost of EOQ policy )= $528 + $391.8 = $ 919.897+55+23+11+124+12+106+92+16+117= 653
10The silver meal heuristic named for Harlan Meal and Edward Silveris a forward method that requires determining the average cost per period as a function of the number of periods the current order is to span,and stopping the computation when this function first increases.C(T) as the average holding and setup cost per period if the current orders spans the next T periods.
11The silver meal heuristic Let (r1,…,rn) be the requirements over n-period horizon.Consider period 1.If we produce just enough in period 1 to meet the demand in period 1, then just incur the order cost K.C(1) = KIf we order enough in period 1 to satisfy the demand both periods 1 and 2 then we must hold r2 for one period.C(2)= (K + hr2)/2SimilarlyC(3)= (K + hr2 + 2hr3)/3and in general,C(j)= (K + hr2 + 2hr3 + … + (j-1)hr3) /jOnce C(j)>C(j-1), we stop and set y1 = r1 + r2 + … + rn , and begin the process again starting at period j.
12The silver meal heuristic – Example 7.2 A machine shop uses the Silver-Meal heuristic to schedule production lot sizes for computer casing. Over the next five weeks the demands for the casing are r = (18, 30, 42, 5, 20). The holding cost is $2 per case per week, and the production setup cost is $80. Find the recommended lot sizing.Starting in period 1C(1) = 80C(2) =[80 + (2)(30)] /2 = 70C(3) = [80 + (2)(30)+ (2)(2)(42)] /3 = stop because C(3)>C(2).Set y1 = r1 + r2 = = 48Starting in period 3C(2) =[80 + (2)(5)] /2 = 45C(3) = [80 + (2)(5)+ (2)(2)(20)] /3 = stopSet y3 = r3 + r4 = = 47Because period 5 is final period in the horizon, we do not need to start the process again.Set y5 = r5 = 20y = (48, 0, 47, 0, 20)
13Least unit costThe least unit cost (LUC) heuristSimilar to Silver – Meal heuristic except instead of dividing the cost over j periods by the number of periods, j, we divide it by the total number of units demanded through period j, r1 + r2 + … + rj .Define C(T) as the average holding and setup cost per unit for a T-period order horizon. Then,C(1) = K/ r1C(2)= (K + hr2)/(r1 + r2)…C(j)= [K + hr2 + 2hr3 + … + (j-1)hr3] /(r1 + r2 + … + rj )As with the Silver-Meal heuristic, this computation is stopped when C(j)>C(j-1), and production level is set to r1 + r2 + … + rj-1 .The process is then repeated, starting at period j and continuing until the end of the planning horizon is reached.
14Least Unit Cost – Example 7.4 Assume the same requirements schedule and costs as given in Ex.7.2Starting in period 1C(1) = 80/18 = 4.44C(2) =[80 + (2)(30)] /(18+30) = 2.92C(3) = [80 + (2)(30)+ (2)(2)(42)] /( ) = 3.42 stop because C(3)>C(2).Set y1 = r1 + r2 = = 48Starting in period 3C(1) = 80 /42 = 1.90C(2) =[80 + (2)(5)] /(42+5) = 1.92 stopSet y3 = r3 = 42Starting in period 4C(1) = 80/5 = 16C(2) =[80 + (2)(20)] /(5+20) = 4.8As we reached the en of the horizon, we set y4 = r4 + r5 = =25. The solution obtain by LUC heuristic y = (48, 0, 42, 25, 0)
15Part Period BalancingAlthough Silver-Meal heuristic seems to give better results in a greater number of cases, part period balancing seems to be more popular in practice.The method is set the order horizon equal to the number of periods that most closely matches the total holding cost with the setup cost over that period.The order horizon that exactly equates holding and setup costs will rarely be an integer number of periods.
16Part Period Balancing- Example 7.5 Consider Example 7.2.Starting in period 1Because 228 exceeds the setup cost of 80, we stop. As 80 is closer to 60 than to 228, the first order horizon is two periods. y1 = r1 + r2 = = 48We start again in period 3We have exceeded the setup cost of 80, so we stop.Because 90 is closer to 80 than 10, the order horizon is three periods.y3 = r3 + r4 + r5 = = 67y = (48,0,67,0,0)Order horizonTotal holding cost1260(30)(2)3228(30)(2)+(2)(2)(42)Order horizonTotal holding cost1210(5)(2)390(5)(2)+(2)(2)(20)
20S-M policy requires Lot for lot : TCL4L=$1320 EOQ : TCEOQ=$919.8 Total holding cost for = (0.6)(424) = $254.4Total setup cost = (132)(3)= $396The total cost of EOQ policy )= $396 + $254.4 = $Lot for lot : TCL4L=$1320EOQ : TCEOQ=$919.8
21Lot sizing with capacity constraints Capacity constraint clearly makes the problem far more realistic.However it also makes the problem more complex.Even finding a feasible solution may not be obvious. Consider the Example: r=(52, 87, 23, 56) and capacity for each period c=(60, 60, 60, 60).First we must determine if the problem is feasibleOn the surface the problem looks solvable, 4*60 =240>218But problem is infeasible. ! 60+60=120<139Feasibility conditionEven feasibility condition is satisfied, it is not obvious how to find feasible solution.
23Improvement step Modified requirements schedule r’ is feasible. Is there another feasible policy that has lower cost.Variety of reasonable improvement rules can be usedFor each lot that is scheduledStart from the last and work backward to the beginningDetermine whether it is cheaper to produce the units composing that lot by shifting production to prior periods of excess capacityBy eliminating a lot, one reduces setup const in that period to zero, but shifting production to prior periods increases the holding cost.
25Example 7.8y=(100, 109, 200, 263, 0, 0 , 120, , 0)r=(100, 79 , 230, 105, 3, 10, 99, 126, 40)Setup cost = 5 x 450 = 2250Holding cost = 2 x ( ) = 2 x 694 = $ 1388Total cost of this policy $3638, ($4482 for initial feasible policy).123456789r'100109200105285012030c400300y26315810858Excess capacity20911372951421922422721234(4 period) ($2 per unit per period) (30 unit) =$240 <K=$450
26Optimal Lot-Sizing for the time Varying Demand Heuristic techniques (EOQ lot sizing, The silver-meal heuristic, Least Unit Cost, Part Period Balancing) are easy to use and give lot sizes with costs that are near the true optimum.Lot size problem can be expressed as a shortest path problem.The dynamic programming can be used to find shortest path.Assume that|Forecasted demands over the next n periods are known and given by vector r=(r1,r2,…,rn).Costs are charged against holding at $h per unit per period and $K per setup.
27Example 7A.1The forecast demand for an electronic assembly produced at Hi-tech, a local semiconductor fabrication shop, over the next four weeks is 52, 87, 23, 56. There is only one setup each week for production of these assemblies, and there is no back-ordering of excess demand. Assume that the shop has the capacity to produce any number of assemblies in a week.Let y1, y2, y3, y4 be the order quantities in each week.Clearly y152 , if we assume that ending inventory in week 4 is zero, then y1 218 ( )y1 can take any one of 167 possible valuesIt is thus clear that for even moderately sized problems the number of feasible solution is enormous.
28Wagner-Whitin Algorithm The Wagner – Whitin algorithm is based on the following observation:Result. An optimality policy has the property that each value of y is exactly the sum of a set of future demands(exact requirement policy).y1 =r1, or y1=r1+r2+…, or y1=r1+r2+…+rny2 =0, or y2=r2 or y2=r2+r3+…, or y2=r2+r3+…+rn…yn =0, or yn =rn.The number of exact requirement policy is much smaller than the total number of feasible policies.s
29Wagner-Whitin Algorithm Because y1 must satisfy exact requirements, it can take only values : 52, 139, 162, 218Ignoring value of y1, y2 can take 0, 87, 110, 166.Every exact policy is the form (i1,i2,…,in) , where ij are either 0 or 1.For example the policy (1,0,1,0) means that production occurs in period 1 and 3 only, that is y= (139, 0, 79, 0).For this example, there are exactly 23 = 8 distinct exact requirement policies.
30Wagner-Whitin Algorithm A convenient way to look at the problem is as a one-way network with the number of nodes equal to exactly one more than the number of periods.Every path through to network corresponds to a specific exact requirements policy.For any pair (i,j) with i<j, if the arc (i,j) is one path, it means that ordering takes place in period i and the order size is equal to the sum of requirements in periods i, i+1, i+2 ,…, j-1.Period j is the next period of ordering.All paths end at period n+1,The value of each arc, cij, is defined as setup and holding cost of ordering in period i to meet requirements through j-1.12345c12c23c34c45c13c14c15c24c25c35
31Example 7A.2r = (52, 87, 23, 56), h=$1, K=$75The first step compute cij for 1i4 and 2j5c12 =75 (setup cost)c13 = = 162c14 =75 + (23 x 2) + 87 = 208c14 =75 + (56 x 3) + (23 x 2) + 87 = 376c23 =75c24 = = 98c25 =75 + (56 x 2) + 23 = 210c34 =75c35 = = 131c23 =75 (setup cost)c45 =75PathCost3002481-2-52852813121-3-52931-4-52831-5376i \ j123457516220837698210131
32Solution by Dynamic Programming The total number of exact policies for a problem of n periods is 2n-1.As n gets large, total enumeration is not efficient.Dynamic programming is a recursive solution technique that can significantly reduce the number of computations required.Dynamic programming is based on the principle of optimality.Define fk as the minimum cost starting at node k.Initial condition is fn+1 =0