11CHAPTER 13(Utilization of a Constrained ResourcePages 594-597)2Constrained 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 costs3• If there is only one constraint, this is easy– Max Find profit/CM per unit of constraint for each alternative• Pick highest– Min Find cost per unit of constraint for each alternative• Pick lowest24• E.g., CM is $4 for Product A & $9 for Product B– You can make 3 Product As in 1 hour– You can make 1 Product B in 1 hour– Only 100 hours are available• 1stCalculate the CM for 1 hour with each Product:• CM of $12 per hour for Product A• CM of $9 per hour for Product B– Only make Product A.5• How do you:– Maximize Sales Revenue/CM, or– Minimize costs – When dealing with multiple constraints • We will discuss 2 ways to solve these problems:– Graphical approach– Excel6Graphical Approach• 1ststate 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 produce37• 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 produced8• 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 produce9• Assume each bolt must pass through following machines, & time required on each machine differs, as shown below: Machine I Machine II Machine IIIA Bolt .1 min .1 min .1 minB 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?410• Add constraints to OB: Description ProblemOB: 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 11• 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: 12• 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 inequality513•To graph 1stconstraint (.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 = 240A = 240/.1A = 2400• So, line crosses A-axis at (2400, 0)14• 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 = 240B = 240/.1B = 2400• So, line crosses B-axis at (0, 2400)15• With these 2 points, you can now graph the line:616• 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 Origin0 ≤ 240 True Statement17• So, Feasible Region for 1stconstraint includes side of line that contains origin• Feasible Region that satisfies 1st& last constraint consists of:18• To graph 2ndconstraint (.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 = 720A = 720/.1A = 7200719• Line crosses B-axis at (0,1800): .1A + .4B = 720.1(0) + .4B = 720.4B = 720B = 720/.4B = 180020• By testing origin, we see that side that contains origin satisfies inequality:.1A + .4B ≤ 720 Inequality.1(0) + .4(0) ≤ 720 Test the Origin0 ≤ 720 True Statement21• Feasible Region for 2ndconstraint includes side of line that contains origin• Feasible Region that satisfies 1st, 2nd& last constraints consists of following:822• To graph 3rdconstraint (.1A+.5B ≤ 160)– Graph line, .1A+.5B = 160– Line crosses A-axis at (1600, 0):.1A + .5B = 160.1A + .5(0) = 160.1A = 160A = 160/.1A = 160023• Line crosses B-axis at (0, 320):.1A + .5B = 160.1(0) + .5B = 160.5B = 160B = 160/.5B = 32024• By testing origin, we see side that contains origin satisfies inequality:1A + .5B ≤ 160 Inequality1(0) + .5(0) ≤ 160 Test the Origin0 ≤ 160 True Statement925• Feasible Region for 3rdconstraint includes side of line that contains origin• Feasible Region that satisfies all of constraints is:26• The fact that Feasible Region no longer touches lines for 1st& 2nd constraints tells you that 1st& 2ndconstraints are not “binding”• Time on Machines I and II could be unlimited and it would not affect our production possibilities27• 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
View Full Document
Unlocking...