Modeling & Decision Analysis

Transkripsi

1 11/5/ Modeling & Decision Analysis DR. MOHAMMAD ABDUL MUKHYI, SE., MM. 1 Metode Kuantitatif Spektrum situasi keputusan: - Terstruktur/terprogram - Tak terstruktur/tak terprogram - Sebagian terstruktur Pembagian masalah : - Certainty : semua alternatif tindakan diketahui dan hanya terdapat satu konsekuansi untuk masing-masing tindakan. - Risk : apabila terdapat lebih dari satu konsekuensi atau alternatif dan pengambil keputusan mengetahui probabilitas konsekuensinya. - Ucertainty :apabila jumlah kemungkinan konsekuensi tidak diketahui oleh pengambil keputusan. 2 1

2 11/5/ We face numerous decisions in life & business. We can use computers to analyze the potential outcomes of decision alternatives. Spreadsheets are the tool of choice for today s managers. 3 What is Management Science? A field of study that uses computers, statistics, and mathematics to solve business problems. Also known as: Operations research Decision science 4 2

3 11/5/ Introduction We all face decision about how to use limited resources such as: Oil in the earth Land for dumps Time Money Workers 5 Home Runs in Management Science Motorola Procurement of goods and services account for 50% of its costs Developed an Internet-based auction system for negotiations with suppliers The system optimized multi-product, multivendor contract awards Benefits: $600 million in savings 6 3

4 11/5/ Home Runs in Management Science Waste Management Leading waste collection company in North America 26,000 vehicles service 20 million residential & 2 million commercial customers Developed vehicle routing optimization system Benefits: Eliminated 1,000 routes Annual savings of $44 million 7 Mathematical Programming MP is a field of management science that finds the optimal, or most efficient, way of using limited resources to achieve the objectives of an individual of a business. Optimization 8 4

5 11/5/ Applications of Optimization Determining Product Mix Manufacturing Routing and Logistics Financial Planning 9 What is a Computer Model? A set of mathematical relationships and logical assumptions implemented in a computer as an abstract representation of a real-world object of phenomenon. Spreadsheets provide the most convenient way for business people to build computer models. 10 5

6 11/5/ The Modeling Approach to Decision Making Everyone uses models to make decisions. Types of models: Mental (arranging furniture) Visual (blueprints, road maps) Physical/Scale (aerodynamics, buildings) Mathematical (what we ll be studying) 11 Characteristics of Models Models are usually simplified versions of the things they represent A valid model accurately represents the relevant characteristics of the object or decision being studied 12 6

7 11/5/ Benefits of Modeling Economy - It is often less costly to analyze decision problems using models. Timeliness - Models often deliver needed information more quickly than their real-world counterparts. Feasibility - Models can be used to do things that would be impossible. Models give us insight & understanding that improves decision making. 13 Example of a Mathematical Model Profit = Revenue - Expenses or Profit = f(revenue, Expenses) or Y = f(x 1, X 2 ) 14 7

8 11/5/ A Generic Mathematical Model Y = f(x 1, X 2,, X n ) Where: Y = dependent variable (aka bottom-line performance measure) X i = independent variables (inputs having an impact on Y) f(. ) = function defining the relationship between the X i & Y 15 Mathematical Models & Spreadsheets Most spreadsheet models are very similar to our generic mathematical model: Y = f(x 1, X 2,, X n ) Most spreadsheets have input cells (representing X i ) to which mathematical functions ( f(. )) are applied to compute a bottom-line performance measure (or Y). 16 8

9 11/5/ Categories of Mathematical Models Model Independent OR/MS Category Form of f(. ) Variables Techniques Prescriptive known, known or under LP, Networks, IP, well-defined decision maker s CPM, EOQ, NLP, control GP, MOLP Predictive unknown, known or under Regression Analysis, ill-defined decision maker s Time Series Analysis, control Discriminant Analysis Descriptive known, unknown or Simulation, PERT, well-defined uncertain Queueing, Inventory Models 17 The Problem Solving Process Identify Problem Formulate & Implement Model Analyze Model Test Results Implement Solution unsatisfactory results 18 9

10 11/5/ The Psychology of Decision Making Models can be used for structurable aspects of decision problems. Other aspects cannot be structured easily, requiring intuition and judgment. Caution: Human judgment and intuition is not always rational! 19 Anchoring Effects Arise when trivial factors influence initial thinking about a problem. Decision-makers usually under-adjust from their initial anchor. Example: What is 1x2x3x4x5x6x7x8? What is 8x7x6x5x4x3x2x1? 20 10

11 11/5/ Framing Effects Refers to how decision-makers view a problem from a win-loss perspective. The way a problem is framed often influences choices in irrational ways Suppose you ve been given $1000 and must choose between: A. Receive $500 more immediately B. Flip a coin and receive $1000 more if heads occurs or $0 more if tails occurs 21 Framing Effects (Example) Now suppose you ve been given $2000 and must choose between: A. Give back $500 immediately B. Flip a coin and give back $0 if heads occurs or give back $1000 if tails occurs 22 11

12 11/5/ A Decision Tree for Both Examples Alternative A Payoffs $1,500 Initial state Alternative B (Flip coin) Heads (50%) Tails (50%) $2,000 $1, Good Decisions vs. Good Outcomes Good decisions do not always lead to good outcomes... A structured, modeling approach to decision making helps us make good decisions, but can t guarantee good outcomes

13 11/5/ 2. Perumusan PL Ada tiga unsur dasar dari PL, ialah: 1. Fungsi Tujuan 2. Fungsi Pembatas (set ketidak samaan/pembatas strukturis) 3. Pembatasan selalu positip. Bentuk umum persoalan PL Cari : x 1, x 2, x 3,, x n. Fungsi Tujuan : Z = c 1 x 1 + c 2 x 2 + c 3 x c n x n optimum (max/min) (srs) Fungsi Kendala : a 11 x 1 + a 12 x 2 + a 13 x a 1n x n >< h 1. (F. Pembatas) a 21 x 1 + a 22 x 2 + a 23 x a 2n x n >< h 2. (dp) a 31 x 1 + a 32 x 2 + a 33 x a 3n x n >< h a m1 x 1 + a m2 x 2 + a m3 x a mn x n >< h m. x j > 0 j = 1, 2, 3 n nonnegativity consraint 25 srs : sedemikian rupa sehingga dp : dengan pembatas ada n macam barang yg akan diproduksi masing2 sbesar x 1,x 2,, x n x j : banyaknya barang yang diproduksi ke j, j = 1, 2, 3,., n c j : harga per satuan barang ke j, j = 1, 2, 3,., n ada m macam bahan mentah, masing2 tersedia h 1, h 2, h 3,., h m h i : banyaknya bahan mentah ke i, i = 1, 2, 3,., m a ij : banyaknya bahan mentah ke i yg digunakan utk memproduksi 1 satuan barang ke j x j > 0, j = 1, 2,,n ; cj tdk boleh neg, paling kecil 0 (nonnegativity consraint) Maksimum dp < h 1 artinya, pemakaian input tidak boleh melebihi h 1 Minimum dp > h 1 artinya, pemakaian input paling tidak dipenuhi h

14 11/5/ 3. Langkah-langkah dan teknik pemecahan Dasar dari pemecahan PL adalah suatu tindakan yang berulang (Inter-active search) dengan sekelompok cara untuk mencapai suatu hasil optimal. Tidakan dilakukan dengan cara sistimatis. Selanjutnya langkah2 dari tindakan berulang adl sebagai: 1. Tentukan kemungkinan2 kombinasi yang baik dari sumber daya al-am yang terbatas atau fasilitas yang tersedia, yang disebut se-bagai initial feasible solution. 2. Selesaikan persamaan pembatasan strukturil untuk mendapatkan titik2 ekstreem (dis sebagai basic feasible solution ). 3. Tentukanlah nilai dari titik2 ekstreem yang akan merupakan nilai2 pilihan, yang telah disesuaikan dengan nilai tujuan dari permasalahan. 4. Ulanglah langkah 3 hingga tercapai tujuan optimal (hanya satu yang bernilai tertinggi atau terendah). 27 Ada 3 (tiga) cara pemecahan PL 1. Cara dengan menggunakan grafik 2. Cara dengan substitusi (cara matematik/aljabar) 3. Cara simplex 28 14

15 11/5/ General Form of an Optimization Problem MAX (or MIN): f 0 (X 1, X 2,, X n ) Subject to: f 1 (X 1, X 2,, X n )<=b 1 : f k (X 1, X 2,, X n )>=b k : f m (X 1, X 2,, X n )=b m Note: If all the functions in an optimization are linear, the problem is a Linear Programming (LP) problem 29 Linear Programming (LP) Problems MAX (or MIN): c 1 X 1 + c 2 X c n X n Subject to: a 11 X 1 + a 12 X a 1n X n <= b 1 : a k1 X 1 + a k2 X a kn X n >=b k : a m1 X 1 + a m2 X a mn X n = b m 30 15

16 11/5/ An Example LP Problem Blue Ridge Hot Tubs produces two types of hot tubs: Aqua-Spas & Hydro-Luxes. Aqua-Spa Hydro-Lux Pumps 1 1 Labor 9 hours 6 hours Tubing 12 feet 16 feet Unit Profit $350 $300 There are 200 pumps, 1566 hours of labor, and 2880 feet of tubing available Steps In Formulating LP Models: 1. Understand the problem. 2. Identify the decision variables. X 1 =number of Aqua-Spas to produce X 2 =number of Hydro-Luxes to produce 3. State the objective function as a linear combination of the decision variables. MAX: 350X X

17 11/5/ 5 Steps In Formulating LP Models (continued) 4. State the constraints as linear combinations of the decision variables. 1X 1 + 1X 2 <= 200 9X 1 + 6X 2 <= X X 2 <= 2880 } pumps } labor } tubing 5. Identify any upper or lower bounds on the decision variables. X 1 >= 0 X 2 >= 0 33 LP Model for Blue Ridge Hot Tubs MAX: 350X X 2 S.T.: 1X 1 + 1X 2 <= 200 9X 1 + 6X 2 <= X X 2 <= 2880 X 1 >= 0 X 2 >=

18 11/5/ Solving LP Problems: An Intuitive Approach Idea: Each Aqua-Spa (X 1 ) generates the highest unit profit ($350), so let s make as many of them as possible! How many would that be? Let X 2 = 0 1st constraint: 1X 1 <= 200 2nd constraint: 9X 1 <=1566 or X 1 <=174 3rd constraint: 12X 1 <= 2880 or X 1 <= 240 If X 2 =0, the maximum value of X 1 is 174 and the total profit is $350*174 + $300*0 = $60,900 This solution is feasible, but is it optimal? No! 35 Solving LP Problems: A Graphical Approach The constraints of an LP problem defines its feasible region. The best point in the feasible region is the optimal solution to the problem. For LP problems with 2 variables, it is easy to plot the feasible region and find the optimal solution

19 11/5/ X 2 Plotting the First Constraint (0, 200) boundary line of pump constraint X 1 + X 2 = (200, 0) X 1 37 X Plotting the Second Constraint (0, 261) boundary line of labor constraint 9X 1 + 6X 2 = (174, 0) X

20 11/5/ X Plotting the Third Constraint (0, 180) boundary line of tubing constraint 12X X 2 = Feasible Region (240, 0) X 1 39 X 2 Plotting A Level Curve of the Objective Function (0, ) objective function 350X X 2 = (100, 0) X

21 11/5/ X 2 A Second Level Curve of the Objective Function (0, 175) objective function 350X X 2 = objective function 350X X 2 = (150, 0) X 1 41 X 2 Using A Level Curve to Locate the Optimal Solution objective function 350X X 2 = optimal solution objective function 350X X 2 = X

22 11/5/ Calculating the Optimal Solution The optimal solution occurs where the pumps and labor constraints intersect. This occurs where: X 1 + X 2 = 200 (1) and 9X 1 + 6X 2 = 1566 (2) From (1) we have, X 2 = 200 -X 1 (3) Substituting (3) for X 2 in (2) we have, 9X (200 -X 1 ) = 1566 which reduces to X 1 = 122 So the optimal solution is, X 1 =122, X 2 =200-X 1 =78 Total Profit = $350*122 + $300*78 = $66, X 2 Enumerating The Corner Points obj. value = $54,000 (0, 180) Note: This technique will not work if the solution is unbounded obj. value = $64,000 (80, 120) obj. value = $66,100 (122, 78) 50 0 obj. value = $0 (0, 0) obj. value = $60,900 (174, 0) X

23 11/5/ Summary of Graphical Solution to LP Problems 1. Plot the boundary line of each constraint 2. Identify the feasible region 3. Locate the optimal solution by either: a. Plotting level curves b. Enumerating the extreme points 45 Special Conditions in LP Models A number of anomalies can occur in LP problems: Alternate Optimal Solutions Redundant Constraints Unbounded Solutions Infeasibility 46 23

24 11/5/ Example of Alternate Optimal Solutions X objective function level curve 450X X 2 = alternate optimal solutions X 1 47 Example of a Redundant Constraint X boundary line of tubing constraint boundary line of pump constraint boundary line of labor constraint 50 Feasible Region X

25 11/5/ Example of an Unbounded Solution X objective function X 1 + X 2 = 600 objective function X 1 + X 2 = 800 -X 1 + 2X 2 = X 1 + X 2 = X 1 49 X 2 Example of Infeasibility X 1 + X 2 = 200 feasible region for second constraint feasible region for first constraint X 1 + X 2 = X