Supply Chain Logistics Worksheet-Rutgers University .

MIE 597SL Supply Chain Logistics Prof. Ana Muriel Homework # 4 Inventory Decisions with Fixed Costs and Constant Demand Q1: Economic Order Quantity Decisions Artaic Innovative Mosaic purchases most of its tile from a supplier in China. Each order is ocean-shipped in a container at a fixed cost of K=$1,000. The tile costs c=$5/lb and Artaic estimates its annual inventory cost rate to be i=20%. The production process uses d=800lbs of tile a week. a) What is the annual ordering and inventory cost if they placed an order for Q=1,600lbs every T= 2weeks? b) Is this optimal? Explain intuitively why. c) What are the optimal order amount and frequency? d) Managers would like to order every quarter (i.e., every 13 weeks) because it helps keep a smooth budget from quarter to quarter. How much extra cost would that entail? Q2: Economic Order Quantity Decisions – Adapted from Factory Physics Quarter-inch stainless-steel bolts, 1 ½ inches long, are consumed in a factory at a fairly steady rate of 60 per week. The bolts cost the plant 2 cents each. It costs the plant $12 to initiate an order, and holding costs are based on annual interest rate of 25%. a) Determine the optimal number of bolts for the plant to purchase and the time between replacement orders. b) What is the yearly holding and setup cost for this item? c) Suppose instead of small bolts we were talking about a bulky item, such as packaging materials. What problem might there be with our analysis in practice? Q3 EOQ Robustness – Adapted from Factory Physics Consider the setting in the previous problem. Suppose that although we have estimated demand to be 60 per week, it turns out that it is actually 120 per week (i.e. we have a 100% forecasting error). a) If we use the lot size calculated in the previous problem (i.e. using the erroneous demand estimate), what will the setup plus holding cost be under the true demand rate? b) What would the cost be if we had used the optimum lot size? c) What percentage increase in cost was caused by the 100 percent demand forecasting error? What does this tell you about the sensitivity of the EOQ model to errors in the data? MIE 597SL Supply Chain Logistics Prof. Ana Muriel Q4 Economic Production Quantity Filter Systems produces air filters for domestic and foreign cars. One filter, part number JJ39877, is supplied on an exclusive contract basis to Oil Changers at a constant 200 units monthly. Filter Systems can produce this filter at a rate of 50 per hour. Setup time to change the settings on the equipment is 1.5 hours. Worker time (including overhead) is charged at the rate of $55 per hour and plant idle time during setups is estimated to cost the firm $100 per hour in lost profit. Filter Systems has established a 22 percent annual interest charge for determining holding cost. Each filter costs the company $2.50 to produce; they are sold for $5.50 each to Oil Changers. Assume 6-hour days, 20 working days per month, and 12 months per year for your calculations. a) How many JJ39877 filters should Filter Systems produce in each production run of this particular part to minimize annual holding and setup costs? b) Assuming that it produces the optimal number of filters in each run, what is the maximum level of on-hand inventory of these filters that the firm has at any point in time? c) What percentage of the working time does the company produce these particular filters, assuming that the policy in part (a) is used? Q5: Dynamic Programming Consider a manufacturing firm that supplies high-pressure valves for ship builders across the globe. Customer contracts drive the demand that needs to be satisfied over the next 12 weeks, which is given in file Homework 4_Q5.xlsx. There is a fixed cost of A=$100 every week that production is started. The valve costs c=$20/unit to produce. The annual inventory cost rate is estimated to be i=20%. How many units should the firm schedule for production each week to balance fixed production setup and inventory costs? Hint:You can calculate it either 1) by hand as we did in the slides, 2)using the AMPL computer code I provided in week 7, or 3) writing your own code in your language of choice . MIE 597SL Supply Chain Logistics Prof. Ana Muriel Appendix: Pseudo-Code if you want to write your own code for Q5 1. Define and read input parameters; define needed variables a. T= number of time periods =12 b. D: demand vector, as given in Assignment8.xlsx, D(i), i=1,2,…,T c. A: fixed cost = 100 d. h: inventory cost per unit per week e. C: matrix where C(i,j)= fixed and inventory costs of producing at i enough to satisfy demands for periods i through j. f. f: vector where f(i) is the cost to go from period i until T+1 g. next: vector where next(i) keeps track of which is the next production period after period i h. Production: vector where Production(i) represents the amount to be produced at period i. 2. Calculate the cost of producing at time i enough to satisfy demand up to j Observe that I’m defining C(i,j) as the cost from i all the way to j (rather than taking the definition in the lectures notes of up to j-1). This is because I found it easier in coding to work with the indexes this way in the 12 by 12 matrix C. ” ” ” 𝑐!” = 𝐴 + 𝑐! %& 𝐷! ( + ℎ! % & 𝐷# ( + ℎ!%& % & 𝐷# ( + ⋯ + ℎ”(& +𝐷) , #$! #$!%& #$!%’ We can simplify this calculation for the case where the costs are the same over time. The cost of purchasing is irrelevant when it is the same over time since we need to purchase all the units to satisfy the demand anyways. The inventory cost can be calculated recursively, adding the cost of holding the demand for the last period, j-1, for (j-1-i) periods. for i=1: T C(i,i)= A for j=i+1..T C(i,j)= C(i,j-1)+(j-i)*h*D(j) end end 3. Recursively calculate the optimal costs to satisfy demand from time period i until the end of the horizon: Initialize: f(T+1)=0 Recursion: for i=T:-1:1 /*let’s calculate f(i)= minj>=i {C(i,j)+f(j+1)}*/ f(i)=C(i,i)+f(i+1); /*Initialize*/ next(i)=i+1; /*Initialize to keep track of next production period*/ MIE 597SL Supply Chain Logistics Prof. Ana Muriel for j=i+1:T temp=C(i,j)+f(j+1); if tempD c unit production cost, not counting setup or inventory costs (dollars per unit) A fixed or setup cost to place an order (dollars) h holding cost (dollars per year); if the holding cost is consists entirely of interest on money tied up in inventory, then h = ic where i is an annual interest rate. Q the unknown size of the production lot size decision variable 11 Inventory vs Time in EPL Model Production run of Q takes Q/P time units Inventory (P-D)(Q/P) -D P-D (P-D)(Q/P)/2 Time 12 Solution to EPL Model Annual Cost Function: Y (Q ) = AD h(1 – D / P )Q + + Dc Q 2 setup holding production Solution (by taking derivative and setting equal to zero): Q* = 2 AD h (1 – D / P ) • tends to EOQ as Pè ∞ • otherwise larger than EOQ because replenishment takes longer 13 Sensitivity of EOQ Model to Quantity Optimal Unit Cost: hQ * A Y = Y (Q ) = + 2 D Q* * * = = We neglect unit cost, c, since it does not affect Q* h 2 AD h A + 2D 2 AD h 2A 2 AD h Optimal Annual Cost: Multiply Y* by D and simplify, Annual Cost = 2 ADh 14 Sensitivity of EOQ Model to Quantity (cont.) Annual Cost from Using Q’: Y (Q¢) = Ratio: hQ¢ AD + 2 Q¢ Cost (Q¢) Y (Q¢) hQ¢ 2 + AD Q¢ 1 é Q¢ Q * ù = = = ê *+ ú Cost (Q * ) Y (Q * ) 2 ëQ Q¢ û 2 ADh Example: If Q’ = 2Q*, then the ratio of the actual to optimal cost is (1/2)[2 + (1/2)] = 1.25 15 Sensitivity of EOQ Model to Quantity (cont.) Total Cost is not particularly sensitive to the optimal order quantity Order Quantity Cost Increase 50% 80% 90% 100% 110% 120% 150% 200% 125% 103% 101% 100% 101% 102% 108% 125% The EOQ can be used to get one in the ballpark even when the assumption of constant demand is not accurate 16 Sensitivity of EOQ Model to Order Interval Order Interval: Let T represent time (in years) between orders (production runs) Q T= D Optimal Order Interval: Q* T = = D * 2 AD h = 2A D hD 17 Sensitivity of EOQ Model to Order Interval (cont.) Ratio of Actual to Optimal Costs: If we use T’ instead
of T* annual cost under T ¢ 1 é T ¢ T * ù = ê + ú annual cost under T * 2 ë T * T ¢ û Powers-of-Two Order Intervals: The optimal order interval, T* must lie within a multiplicative factor of √2 of a “power-of-two.” Hence, the maximum error from using the best power-of-two is 1é 1 ù 2 + = 1.06 ú 2 êë 2û 18 The “Root-Two” Interval 2m T1* divide by less than √2 to get to 2m 2m 2 T2* 2 m +1 multiply by less than √2 to get to 2m+1 19 Medequip Example Optimum: Q*=169, so T*=Q*/D =169/1000 years = 62 days Y (Q*) = hQ * AD 35(169) 500(1000) + = + = $5,916 2 Q* 2 169 Round to Nearest Power-of-Two: 62 is between 32 and 64, but since 32√2=45.25, it is “closest” to 64. So, round to T’=64 days or Q’= T’D=(64/365)1000=175. Y (Q ‘ ) = hQ ‘ AD 35(175) 500(1000) + = + = $5,920 2 Q’ 2 175 Only 0.07% error because we were lucky and happened to be close to a powerof-two. But we can’t do worse than 6%. 20 Powers-of-Two Order Intervals Order Interval 1= 2 0 Week 0 1 2 3 4 5 6 7 8 2 = 21 4 = 22 8 = 23 21 Joint Replenishment Model EOQ Assumptions 1. Instantaneous production. 2. Immediate delivery. 3. Deterministic demand. 4. Constant demand. 5. Known fixed setup costs. 6. Single product or separable products. JR model relaxes this one 22 Joint Replenishment Problem There is a fixed setup cost A0 incurred every time an order is placed independent of which products are purchased. Assumptions for each item i, i=1,2,…n: • • • • • • • Demand for each item i is known and constant, di There is a fixed cost Ai when item i is ordered The holding cost rate for item i is hi Cost per unit is constant Lead times are known and constant Infinite planning horizon No shortages are allowed Objective: Select reorder intervals so as to minimize long-run average costs 23 Observe Optimal solution if items are ordered individually and there is no additional fixed cost EOQ i Q • where 2 Ai Di = hi gi = ⇒ Ti EOQ QiEOQ = = Di 2 Ai = hi Di hi Di 2 Ai gi Natural Reorder Interval Given the additional fixed cost… EOQ i Q 2( A0 + Ai )Di = hi ⇒ Ti EOQ = A0 + Ai gi 24 Solution Procedure: Powers of Two Intervals •Rank items so that A1 A2 A ≤ ≤ … ≤ n g1 g 2 gn •Let i* be the largest index i such that Ai A0 + A1 + …+ Ai ≤ g i g1 + g 2 + …+ g i •Then for i≥i*, the natural replenishment time is optimal. •For i=0; /* Number of units of product i produced*/ maximize profit: sum{i in I} P[i]*X[i]; subject to capacity {j in J}: sum{i in I} R[i,j]*X[i]