Focus on minimizing costs EOQ Linear Programming. Two types of inventory costs (IC): Order/Setup Costs (OCs), and Carrying Costs (CCs) IC = OC + CC

Transcription

1 Focus on minimizing costs EOQ Linear Programming Economic Order Quantity (EOQ) model determines: Optimal amount of inventory to produce/purchase at given time Discussion applicable to production runs and orders Two types of inventory costs (IC): Order/Setup Costs (OCs), and Carrying Costs (CCs) IC = OC + CC OCs: Purchasing inventory all costs involved in placing order Manufacturing inventory cost of setting up production line CCs: All costs involved in holding inventory E.g., interest on inventory cost, insurance, breakage and storage cost We will use following variables: TC - Total Inventory Costs P - Cost to place one purchase order Q - Quantity - Units ordered in a single order; or D - Demand - number of inventory needed in year C - Carrying Cost for 1 unit for entire year Remember that ICs are divided into 2 parts: OCs CCs 1 st Focus on OCs 1

2 OCs: cost to place 1 order (P), multiplied by # of orders placed # of orders placed: All OCs in a year: (Cost to Place an Order) x (# of orders placed): P( D ) Q # of Orders = Annual Demand For Units = D Size of Order Q Annual CCs: (Annual Carrying Cost Per Unit) x (# of Units in Inventory) Is # of units in Inventory = units in order (Q)? NO! We buy units, sell them, buy some more: Average = ½ of amount of order: _Q_ 2 Carrying costs: C times the number of you units that your have on hand at any given time. On average, your firm has Q/2 units on hand at any given time. C ( _Q_ ) 2 P( TC: OCs + CCs _D Q_ Q ) + C( 2 ) 2

3 To get optimal amount to order to minimize ICs: Take 1 st derivative of equation with respect to Q Set derivative to 0 Solve for Q Rewrite Inventory Cost equation to make it easier to use Power Rule for taking derivatives: TC = PDQ -1 + ½CQ 1 / Q TC = -PDQ -2 + ½C 0 = -PDQ -2 + ½C -½C = -PDQ -2 -½CQ 2 = -PD Q 2 = _2PD_ C Q = / _2PD_ ( EOQ Formula ) C EOQ Formula gives optimal amount of units to order Lead Time is time it takes to obtain new inventory once order placed E.g., Lead Time is 2 days if: It takes 2 days for inventory to arrive once ordered Reorder Point is point in time where existing inventory is just sufficient to cover demand until new inventory arrives Reorder Point can be calculated as follows: Reorder Point = Rate of Usage x Lead Time E.g., we sell 10 units a day & it takes 2 days to receive new inventory once ordered Reorder Point is 20 units Firm should place order once inventory level drops to 20 units Assuming inventory sold/used at uniform rate, Rate of Usage is: E.g., Assume Rate of Usage = Annual Demand For Units Number of Working Days in Year Safety Stock is extra inventory carried as buffer against fluctuation in demand Reorder Point formula can be modified to reflect Safety Stock: Reorder Pt. = (Rate of Usage x Lead Time) + Sfty. Stk. Annual demand for units is 5,000 units (D) Annual Carrying Cost of 1 unit is $10 per year (C) Setup Cost is $1,000 (P) Co. scheduled 4 equal production runs for upcoming year Co. has 250 business days/year Sales occur uniformly Production takes 1 day 3

4 Using EOQ formula optimal run is 1,000 units: EOQ = 2(P)(D)/(C) = 1,000 units EOQ = [2($1,000)(5,000)]/$10 = 1,000 units Since Co. has 250 business days Co. sells 20 units/day (5,000/250). If Co. needs 1 day to produce units Co. should begin production run once inventory drops to 20 units If Co wants Safety Stock Add Safety Stock to 20-unit Reorder Point Co. saves following by using production runs of 1,000 units rather than the 1250-unit run (5,000/4) currently planned: Inventory Costs = P( _D CQ_ ) + ( Q Unit Run = [$1000 (5000/1250) + ½ (1250 x 10)] $10, Unit Run = [$1000 (5000/1000) + ½ (1000 x 10)] -10,000 Inventory Cost Savings = $250 ) Constrained Optimization We want to either: Maximize profits/cm, or Minimize costs This is easy, you make either infinity units to maximize profits/cm or zero units to minimize costs Problem have a constraint. Limited Capacity Max profits/cm Need to achieve goals Min costs How do you: Maximize Sales Revenue/CM, or Minimize costs When dealing with constraints We will discuss Graphical approach Graphical Approach 1 st state your goal E.g., Maximize CM / Minimize cost Mathematical formula for item being maximized or minimized Called Objective Function (OB) Variables in OB are your alternative courses of action What you can change E.g., # of each type of product that you will produce 4

5 Assume: Co can make 2 types of bolts: Bolt A or Bolt B Bolt A has CMU =10 Bolt B has CMU =12 Co wants to know # of each type of bolt to produce to maximize CM: OB: Maximize.10A +.12B Where A # of A Bolts produced B # of B Bolts produced Without more information, Co. would produce # of every bolt With this production, Co. would have CM In reality, there is limit to # of bolts that Co. can produce Assume each bolt must pass through following machines, & time required on each machine differs, as shown below: Machine I Machine II Machine III A Bolt.1 min.1 min.1 min B Bolt.1 min.4 min.5 min In 1 day, there are 240, 720, &160 minutes available, on Machine I, Machine II and Machine III, respectively How many of each type of bolts should Co. produce in 1 day? Add constraints to OB: Description Problem OB: Max.10(A) +.12(B) Constraints: subject to: Only 240 mins available on Machine I..1A +.1B 240 Only 720 mins available on Machine II..1A +.4B 720 Only 160 mins available on Machine III..1A +.5B 160 Can't manufacture negative # of bolts. A, B 0 Next, graph production area that meets all of constraints Feasible Region Graph consists of Cartesian axis with the OB Variables as x-axis & y-axis E.g., Last constraint alone means you are dealing with top right quarter of Cartesian axis: Now, graph other constraints, which are inequalities either or Area covered by such an inequality consists of 1 line that divides Cartesian plane & 1 side of that line. Formula of line is inequality formula with = substituted for or To graph inequality graph line that divides the Cartesian plane & chose side of line that satisfies inequality 29 5

6 To graph 1 st constraint (.1A+.1B 240) Graph line (.1A+.1B = 240) Easiest way to graph line is identify points where line crosses each axis We know that B=0 on any point on A-axis Therefore, point where line crosses A-axis has 0 as B coordinate. So, we replace B with 0 in equation & solve for A:.1A +.1B = 240.1A +.1(0) = 240.1A = 240 A = 240/.1 A = 2400 So, line crosses A-axis at (2400, 0) Where does line cross B-axis? We know that A=0 at any point on B-axis Therefore, point where line crosses B-axis has 0 as A coordinate So, we replace A with 0 in equation and solve for B:.1A +.1B = 240.1(0) +.1B = 240.1B = 240 B = 240/.1 B = 2400 So, line crosses B-axis at (0, 2400) 32 With these 2 points, you can now graph the line: Now, we need to pick side of line that satisfies inequality Easiest way to do this is to test 1point on 1 side of line You test point by substituting coordinates into inequality formula, & check if coordinates produce a value that satisfies inequality If point tested satisfies inequality, then every point on same side of line will satisfy inequality Easiest point to test is origin (0,0) because you are dealing with zeros as variables So, plug origin into inequality & see if inequality is true:.1a +.1B 240 Inequality.1(0) +.1(0) 240 Test the Origin True Statement So, Feasible Region for 1 st constraint includes side of line that contains origin Feasible Region that satisfies 1 st & last constraint consists of: To graph 2 nd constraint (.1A+.4B 720) Graph line,.1a+.4b = 720 We know that line crosses A-axis at (7200, 0):.1A +.4B = 720.1A +.4(0) = 720.1A = 720 A = 720/.1 A =

7 Line crosses B-axis at (0,1800):.1A +.4B = 720.1(0) +.4B = 720.4B = 720 B = 720/.4 B = 1800 By testing origin, we see that side that contains origin satisfies inequality:.1a +.4B 720 Inequality.1(0) +.4(0) 720 Test the Origin True Statement Feasible Region for 2 nd constraint includes side of line that contains origin Feasible Region that satisfies 1 st, 2 nd & last constraints consists of following: To graph 3 rd constraint (.1A+.5B 160) Graph line,.1a+.5b = 160 Line crosses A-axis at (1600, 0):.1A +.5B = 160.1A +.5(0) = 160.1A = 160 A = 160/.1 A = 1600 Line crosses B-axis at (0, 320):.1A +.5B = 160.1(0) +.5B = 160.5B = 160 B = 160/.5 B = 320 By testing origin, we see side that contains origin satisfies inequality:.1a +.5B 160 Inequality.1(0) +.5(0) 160 Test the Origin True Statement 7

8 Feasible Region for 3 rd constraint includes side of line that contains origin Feasible Region that satisfies all of constraints is: The fact that Feasible Region no longer touches lines for 1 st & 2nd constraints tells you that 1 st & 2 nd constraints are not binding Time on Machines I and II could be unlimited and it would not affect our production possibilities Once, you have identified Feasible Region that satisfies all constraints You have to decide which points within Feasible Region maximize CM Corners points of Feasible Region are most extreme points Therefore, corner points represent production levels that produce most extreme CM E.g., highest & lowest CM To find highest CM Plug each corner points into original OB [.10(A)+.12(B)] Check to see which point produces highest CM: Corner.10(A)+.12(B) CM (1600, 0).1 (1600) +.12 (0) = = $ (0, 320).1 (0) +.12 (320) = = (0, 0).1 (0) +.12 (0) = = 0 Co. generates highest CM by producing 1600 A Bolts and 0 B Bolts THE END Dr. Michael Constas