1 / 43

Linear Programming

Linear Programming. Linear Programming. A mathematical technique to help plan and make decisions relative to the trade-offs necessary to allocate resources Will find the minimum or maximum value of the objective Guarantees the optimal solution to the model formulated.

brittney-foley
Download Presentation

Linear Programming

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Linear Programming

  2. Linear Programming • A mathematical technique to help plan and make decisions relative to the trade-offs necessary to allocate resources • Will find the minimum or maximum value of the objective • Guarantees the optimal solution to the model formulated

  3. Requirements of an LP Problem LP problems seek to maximize or minimize some quantity (usually profit or cost) expressed as an objective function The presence of restrictions, or constraints, limits the degree to which we can pursue our objective

  4. Requirements of an LP Problem There must be alternative courses of action to choose from The objective and constraints in linear programming problems must be expressed in terms of linear equations or inequalities

  5. Formulating LP Problems The product-mix problem at Shader Electronics • Two products • Shader Walkman, a portable CD/DVD player • Shader Watch-TV, a wristwatch-size Internet-connected color TV • Determine the mix of products that will produce the maximum profit

  6. Hours Required to Produce 1 Unit Walkman Watch-TVs Available Hours Department (X1)(X2) This Week Electronic 4 3 240 Assembly 2 1 100 Profit per unit $7 $5 Formulating LP Problems Decision Variables: X1 = number of Walkmans to be produced X2 = number of Watch-TVs to be produced

  7. Formulating LP Problems Objective Function: Maximize Profit = $7X1 + $5X2 • There are three types of constraints • Upper limits where the amount used is ≤ the amount of a resource • Lower limits where the amount used is ≥ the amount of the resource • Equalities where the amount used is = the amount of the resource

  8. Electronic time used Assembly time used Assembly time available Electronic time available is ≤ is ≤ Formulating LP Problems First Constraint: 4X1 + 3X2 ≤ 240 (hours of electronic time) Second Constraint: 2X1 + 1X2 ≤ 100 (hours of assembly time)

  9. Graphical Solution • Can be used when there are two decision variables • Plot the constraint equations at their limits by converting each equation to an equality • Identify the feasible solution space • Create an iso-profit line based on the objective function • Move this line outwards until the optimal point is identified

  10. X2 – – 80 – – 60 – – 40 – – 20 – – – Assembly (constraint B) Number of Watch-TVs Electronics (constraint A) | | | | | | | | | | | 0 20 40 60 80 100 X1 Number of Walkmans Graphical Solution Feasible region

  11. X2 – – 80 – – 60 – – 40 – – 20 – – – $210 = $7X1 + $5X2 Number of Watch-TVs (30, 0) | | | | | | | | | | | 0 20 40 60 80 100 X1 Number of Walkmans Graphical Solution (0, 42)

  12. X2 – – 80 – – 60 – – 40 – – 20 – – – $210 = $7X1 + $5X2 $350 = $7X1 + $5X2 $280 = $7X1 + $5X2 Number of Watch-TVs $420 = $7X1 + $5X2 | | | | | | | | | | | 0 20 40 60 80 100 X1 Number of Walkmans Graphical Solution

  13. X2 – – 80 – – 60 – – 40 – – 20 – – – Maximum profit line Optimal solution point (X1 = 30, X2 = 40) Number of Watch-TVs $410 = $7X1 + $5X2 | | | | | | | | | | | 0 20 40 60 80 100 X1 Number of Walkmans Graphical Solution

  14. X2 – – 80 – – 60 – – 40 – – 20 – – – Number of Watch-TVs 2 3 1 | | | | | | | | | | | 0 20 40 60 80 100 X1 4 Number of Walkmans Corner-Point Method

  15. Solving Minimization Problems • Formulated and solved in much the same way as maximization problems • In the graphical approach an iso-cost line is used • The objective is to move the iso-cost line inwards until it reaches the lowest cost corner point

  16. Minimization Example X1 = number of tons of black-and-white chemical produced X2 = number of tons of color picture chemical produced Minimize total cost = 2,500X1 + 3,000X2 Subject to: X1 ≥ 30 tons of black-and-white chemical X2 ≥ 20 tons of color chemical X1 + X2≥ 60 tons total X1, X2≥ $0 nonnegativity requirements

  17. X2 60 – 50 – 40 – 30 – 20 – 10 – – X1 + X2= 60 X1= 30 X2= 20 | | | | | | | 0 10 20 30 40 50 60 X1 Minimization Example Table B.9 Feasible region b a

  18. Minimization Example Total cost at a = 2,500X1 + 3,000X2 = 2,500 (40) + 3,000(20) = $160,000 Total cost at b = 2,500X1 + 3,000X2 = 2,500 (30) + 3,000(30) = $165,000 Lowest total cost is at point a

  19. Department Product Wiring Drilling Assembly Inspection Unit Profit XJ201 .5 3 2 .5 $ 9 XM897 1.5 1 4 1.0 $12 TR29 1.5 2 1 .5 $15 BR788 1.0 3 2 .5 $11 Capacity Minimum Department (in hours) Product Production Level Wiring 1,500 XJ201 150 Drilling 2,350 XM897 100 Assembly 2,600 TR29 300 Inspection 1,200 BR788 400 LP Applications Production-Mix Example

  20. LP Applications X1 = number of units of XJ201 produced X2 = number of units of XM897 produced X3 = number of units of TR29 produced X4 = number of units of BR788 produced Maximize profit = 9X1 + 12X2 + 15X3 + 11X4 subject to .5X1 + 1.5X2 + 1.5X3 + 1X4 ≤ 1,500 hours of wiring 3X1 + 1X2 + 2X3 + 3X4 ≤ 2,350 hours of drilling 2X1 + 4X2 + 1X3 + 2X4 ≤ 2,600 hours of assembly .5X1 + 1X2 + .5X3 + .5X4 ≤ 1,200 hours of inspection X1 ≥ 150 units of XJ201 X2 ≥ 100 units of XM897 X3 ≥ 300 units of TR29 X4 ≥ 400 units of BR788

  21. The Simplex Method • Real world problems are too complex to be solved using the graphical method • The simplex method is an algorithm for solving more complex problems • Developed by George Dantzig in the late 1940s • Most computer-based LP packages use the simplex method

  22. NLP in Facility Location • Consider an existing network with m facilities • It is desired to add n new facilities to the network • Let’s • (ai, bi) denote the coordinates of existing facility ith • (Xi, Yi) denote the coordinates of the to-be-found new facility iththatminimize the total distribution cost

  23. NLP in Facility Location (cont.) • Let’s • gij denote the load or flow of activity from a new facility ith to anexistingfacility jth • fij denote the load or flow of activity between new facilities ith and jth • cijdenote the cost per unit travel between new facilities • dij denote the cost per unit travel between new facilities ith to anexistingfacility jth

  24. NLP in Facility Location (cont.) • NLP Model

  25. Exampleof NLP Application • บริษัทหนึ่งมีศูนย์บริการ 4 แห่ง และ Warehouse 1 แห่ง ตั้งอยู่ที่จุด Coordinate (X, Y) คือ (8,20),(8,10),(10,20),(16,30) และ (35,20) ตามลำดับ บริษัทต้องการสร้าง Warehouse อีก 2 แห่ง ซึ่งต้องตั้งห่างกันตามแนวแกน Y และ X ไม่น้อยกว่า 5 หน่วย ปริมาณงานระหว่างนับเป็น Trip ระหว่าง Facilities มีดังตาราง และต้นทุนการขนส่งระหว่าง Warehouse ที่สร้างใหม่เท่ากับ 5 ต่อระยะทาง 1 หน่วย และ ระหว่าง Warehouse ที่สร้างใหม่กับ Facilities เดิมเท่ากับ10 ต่อระยะทาง 1 หน่วย จงกำหนดตำแหน่งที่ตั้งของ Warehouse ใหม่ทั้ง 2 แห่ง

  26. Exampleof NLP Application

  27. Transportation Models

  28. Transportation Modeling • An interactive procedure that finds the least costly means of moving products from a series of sources to a series of destinations • Can be used to help resolve distribution and location decisions

  29. Transportation Modeling • A special class of linear programming • Need to know • The origin points and the capacity or supply per period at each • The destination points and the demand per period at each • The cost of shipping one unit from each origin to each destination

  30. Transportation Problem

  31. Boston (200 units required) Cleveland (200 units required) Des Moines (100 units capacity) Albuquerque (300 units required) Evansville (300 units capacity) Fort Lauderdale (300 units capacity) Transportation Problem

  32. Des Moines capacity constraint Factory capacity To Albuquerque Boston Cleveland From $5 $4 $3 100 Des Moines Cell representing a possible source-to-destination shipping assignment (Evansville to Cleveland) $8 $4 $3 300 Evansville $9 $7 $5 300 Fort Lauderdale Warehouse requirement 300 200 200 700 Cost of shipping 1 unit from Fort Lauderdale factory to Boston warehouse Clevelandwarehouse demand Total demand and total supply Transportation Matrix Figure C.2

  33. Northwest-Corner Rule • Start in the upper left-hand cell (or northwest corner) of the table and allocate units to shipping routes as follows: • Exhaust the supply (factory capacity) of each row before moving down to the next row • Exhaust the (warehouse) requirements of each column before moving to the next column • Check to ensure that all supplies and demands are met

  34. Northwest-Corner Rule Assign 100 tubs from Des Moines to Albuquerque (exhausting Des Moines’s supply) Assign 200 tubs from Evansville to Albuquerque (exhausting Albuquerque’s demand) Assign 100 tubs from Evansville to Boston (exhausting Evansville’s supply) Assign 100 tubs from Fort Lauderdale to Boston (exhausting Boston’s demand) Assign 200 tubs from Fort Lauderdale to Cleveland (exhausting Cleveland’s demand and Fort Lauderdale’s supply)

  35. To (A) Albuquerque (B) Boston (C) Cleveland Factory capacity From $5 $4 $3 100 100 (D) Des Moines $8 $4 $3 200 100 300 (E) Evansville $9 $7 $5 100 200 300 (F) Fort Lauderdale Warehouse requirement 300 200 200 700 Means that the firm is shipping 100 bathtubs from Fort Lauderdale to Boston Northwest-Corner Rule

  36. Route From To Tubs Shipped Cost per Unit Total Cost D A 100 $5 $ 500 E A 200 8 1,600 E B 100 4 400 F B 100 7 700 F C 200 5 $1,000 Total: $4,200 Northwest-Corner Rule Computed Shipping Cost This is a feasible solution but not necessarily the lowest cost alternative

  37. Intuitive Lowest-Cost Method Identify the cell with the lowest cost Allocate as many units as possible to that cell without exceeding supply or demand; then cross out the row or column (or both) that is exhausted by this assignment Find the cell with the lowest cost from the remaining cells Repeat steps 2 and 3 until all units have been allocated

  38. To (A) Albuquerque (B) Boston (C) Cleveland Factory capacity From $5 $4 $3 100 100 (D) Des Moines $8 $4 $3 300 (E) Evansville $9 $7 $5 300 (F) Fort Lauderdale Warehouse requirement 300 200 200 700 Intuitive Lowest-Cost Method First, $3 is the lowest cost cell so ship 100 units from Des Moines to Cleveland and cross off the first row as Des Moines is satisfied

  39. To (A) Albuquerque (B) Boston (C) Cleveland Factory capacity From $5 $4 $3 100 100 (D) Des Moines $8 $4 $3 100 300 (E) Evansville $9 $7 $5 300 (F) Fort Lauderdale Warehouse requirement 300 200 200 700 Intuitive Lowest-Cost Method Second, $3 is again the lowest cost cell so ship 100 units from Evansville to Cleveland and cross off column C as Cleveland is satisfied

  40. To (A) Albuquerque (B) Boston (C) Cleveland Factory capacity From $5 $4 $3 100 100 (D) Des Moines $8 $4 $3 200 100 300 (E) Evansville $9 $7 $5 300 (F) Fort Lauderdale Warehouse requirement 300 200 200 700 Intuitive Lowest-Cost Method Third, $4 is the lowest cost cell so ship 200 units from Evansville to Boston and cross off column B and row E as Evansville and Boston are satisfied

  41. To (A) Albuquerque (B) Boston (C) Cleveland Factory capacity From $5 $4 $3 100 100 (D) Des Moines $8 $4 $3 200 100 300 (E) Evansville $9 $7 $5 300 300 (F) Fort Lauderdale Warehouse requirement 300 200 200 700 Intuitive Lowest-Cost Method Finally, ship 300 units from Albuquerque to Fort Lauderdale as this is the only remaining cell to complete the allocations

  42. To (A) Albuquerque (B) Boston (C) Cleveland Factory capacity From $5 $4 $3 100 100 (D) Des Moines $8 $4 $3 200 100 300 (E) Evansville $9 $7 $5 300 300 (F) Fort Lauderdale Warehouse requirement 300 200 200 700 Intuitive Lowest-Cost Method Total Cost = $3(100) + $3(100) + $4(200) + $9(300) = $4,100

  43. To (A) Albuquerque (B) Boston (C) Cleveland Factory capacity From $5 $4 $3 100 100 (D) Des Moines $8 $4 $3 200 100 300 (E) Evansville $9 $7 $5 300 300 (F) Fort Lauderdale Warehouse requirement 300 200 200 700 Intuitive Lowest-Cost Method This is a feasible solution, and an improvement over the previous solution, but not necessarily the lowest cost alternative Total Cost = $3(100) + $3(100) + $4(200) + $9(300) = $4,100

More Related