Chapter 10 Multicriteria DecisionMarking Models 1 Application Context

Chapter 10 Multicriteria Decision-Marking Models 1

Application Context è multiple objectives that cannot be put under a single measure; e. g. , èdistribution: cost and time èas a single objective function problem if time can be converted into cost èsupply cost chain: customer service and inventory 多目標而目標沒有 共同的衡量方式。 2

Chapter Summary è 10. 0 Scoring Model 評點法 è 10. 1 Weighting Method 權重法 è 10. 2 Goal Programming 目標規劃 è 10. 3 AHP (Analytical Hierarchy Process)層 級分析法 我們只學每個方法 基本的概念。 3

Motivation Problem èdinner, two factors to consider: distance and cost èthree restaurants èA: (2, 3) èB: (7, 1) èC: (4, 2) èwhich (distance from home, cost) (2, $$$) home (7, $) B A C (4, $$) one to choose? 4

Scoring Model 評點法 5

Scoring Model 評點法 (7, $) èa subjective method (2, $$$) B A C weights to each criterion home (4, $$) è assign a rating for each decision alternative on each criterion (distance from home, cost) è assign è Restaurant è min w 1 1 1 Selection Example: Version 1 w 1 (distance) + w 2 (cost) w 2 1 3 A (2, $$$) B (7, $) 5 8 11 10 C (4, $$) choice 6 A 10 B, C Version 1只需要決定各目標的權重。 6

Scoring Model 評點法 è Restaurant Selection Example: Version 2 èweights of criteria, and ratings on criteria for alternatives èExample: Tom dislikes walking and likes good food (from expensive restaurants) 每個選擇在每一項目標中有點數（ ratings, scores），而目標有各 自的權重（weight）。 7

Scoring Model 評點法 è Restaurant èweights Selection Example, Version 2 of walking and price by Tom: 1 to 1 èratings (scores) of each restaurant for walking and price: dislikes walking and likes expensive, good food criterion restaurant distance price w 1 = 1 w 2 = 1 A (2, $$$) 10 8 B (7, $) 3 2 C (4, $$) 6 5 objective of Tom: max w 1 (rating of distance) + w 2 (rating of price) 8

Scoring Model 評點法 èa subjective method on assigning è weights è ratings 9

Example 10 -5 Product Selection è to expand the product line by adding one of the following: microwave ovens, refrigerators, and stoves è decision criteria è è manufacturing capability/cost è market demand è profit margin è long-term profitability/growth è transportation costs è useful life assigning weights to the criteria and ratings to the three alternatives for each criterion è maximizing the total score 10

Example 10 -5 Product Selection manuf. cap. /cost market demand profit margin (long-term) prof. /growth Transp. costs useful life weight 4 5 3 5 2 1 microwave refers stoves 4 8 6 3 9 1 3 4 9 6 2 5 8 2 5 7 4 6 è Scoremicro = 4(4)+5(8)+3(6)+5(3)+2(9)+1(1) = 108 è Scorerefer = 4(3)+5(4)+3(9)+5(6)+2(2)+1(5) = 98 è Scorestove = 4(8)+5(2)+3(5)+5(7)+2(4)+1(6) = 106 any comments on the relative values? 11

Weighting Method 權重法 12

Weighting Method 權重法 èa form of scoring method ètransforming a multi- to a single-criterion objective function by finding the weights of the criteria 以目標的權重（weight） 將多目標的問題轉化為 單目標的問題。 13

Weighting Method 權重法 èmax Z(x) = [z 1(x), z 2(x), …, z. P(x)] s. t. x S èturning into a single-criterion objective function by weighting (with weights) èmax Z(x) = w 1 z 1(x)+w 2 z 2(x)+… +wpz. P(x) s. t. x S 14

Weighting Method 權重法 è criteria (i. e. , objectives) è max z 1(x) = 2 x 1+3 x 2 x 3 è min z 2(x) = 6 x 1 x 2 è max z 3(x) = 2 x 1+x 3 è constraints è x 1+x 2+x 3 15 è x 1+2 x 2+x 3 è x 3 20 2 è x 1, x 2, x 3 0 15

Weighting Method 權重法 è somehow got: w 1 = 1, w 2 = 2, w 3 = 4 = (2 x 1+3 x 2 x 3) 2(6 x 1 x 2) + 4( 2 x 1+x 3) = 18 x 1+5 x 2+3 x 3, è max z 1(x) 2 z 2(x)+4 z 3(x) è s. t. negative sign x 1+x 2+x 3 15 ; x 1+2 x 2+x 3 20; x 3 2; x 1, x 2, x 3 0. max z 1(x) = 2 x 1+3 x 2 x 3, min z 2(x) = 6 x 1 x 2, , max z 3(x) = 2 x 1+x 3, s. t. x 1+x 2+x 3 15 ; x 1+2 x 2+x 3 20; x 3 2; x 1, x 2, x 3 0. 16

Goal Programming 目標規劃 17

1/4: Introducing the Ideas of Goal Programming 18

Goal Programming è GP: priority + goal è priority of the goals (i. e. , of the criteria) è (saving) money is most important: B è (shortest) distance is most important: A è (best) food is the most important: A (7, $) (2, $$$) home B A C (4, $$) 19

Goal Programming èa goal èan objective with a desirable quantity èno good to be over and under this quantity v( ) over u( ) under goal 20

General Idea of Goal Programming èsuppose the goals are: 3 units for distance, and 2 units (i. e. , $$) for price A(2, 3) B(7, 1) C(4, 2) distance u(d, ) v(d, ) 1 0 0 4 0 1 price u(p, ) v(p, ) 0 1 1 0 0 0 (7, $) B A (2, $$$) C home (4, $$) 21

General Idea of Goal Programming èpriority A B C P 1 up > P 2 v d > P 3 ud > P 4 v p P 1 > P 2 > P 3 > … distance u(d, ) v(d, ) 1 0 0 4 0 1 price u(p, ) v(p, ) 0 1 1 0 0 0 P 1 up , P 2 vd , P 3 ud , P 4 vp P 1 up , P 2 ud , P 3 vd , P 4 vp P 1 vp , P 2 ud , P 3 vd , P 4 up A C C B is dominated by C, i. e. , C is optimal for any priority that B is optimal. 22

2/4 : A More General Goal Programming Approach 23

General Idea of Goal Programming èa goal program è parts like a linear program èwith decisions variables èwith hard constraints è parts unlike a linear program èwith soft constraints è expressed as goals to be achieved of constraints such as x 1 10 and x 1 7 in a GP if they are soft constraints è co-existence èwith the objective function in LP replaced by the priorities of goals in GP 24

Deviation Variables for a Soft Constraints è example: a soft constraint on labor hour è x 1 units of product 1, each for 4 labor hours è x 2 units of product 2, each for 2 labor hours è goal: 100 labor hours 人世間有不少soft constraints （可以 斟酌的限制式） èa soft constraint: 4 x 1+2 x 2 100 è 2 deviation variables u and v: 4 x 1+2 x 2 + u v = 100 è u: under utilization of labor è v: over utilization of labor 25

Example 10 -1: Formulation of a GP è three products, quantities to produce x 1, x 2, and x 3 è objectives in order of priority Suppose that the material availability is a hard constraint, i. e. , there is no way to get more material. è min overtime in assembly è min undertime in assembly è min sum of undertime and overtime in packaging product x 1 material 2 (lb/unit) assembly (min. 9 unit) packaging 1 (min/unit) x 2 x 3 availability 4 3 600 pounds 8 7 900 minutes 2 3 300 minutes 26

Example 10 -1: Formulation of a GP èmin P 1 v 1, P 2 u 1, P 3(u 2+v 2), ès. t. 2 x 1 + 4 x 2 + 3 x 3 600 (lb. , hard const. ) è 9 x 1 + 8 x 2 + 7 x 3 + u 1 v 1 = 900 (min. , soft const. ) è 1 x 1 + 2 x 2 + 3 x 3 + u 2 v 2 = 300 (min. , soft const. ) èall variables 0 è 27

3/4 : Solution of a Goal Program 28

Example: Solution of a GP è min P 1 u 1, P 2 u 2, P 3 u 3, è s. t. è 5 x 1 + 3 x 2 150 (hard const. ) (A) è 2 x 1 + 5 x 2 + u 1 v 1 = 100 (soft const. ) (1) è 3 x 1 + 3 x 2 + u 2 v 2 = 180 (soft const. ) (2) è x 1 + u 3 v 3 = 40 (soft const. ) (3) è all variables 0 29

Example: Solution of a GP x 2 min P 1 u 1, P 2 u 2, P 3 u 3, s. t. 5 x 1 + 3 x 2 150 2 x 1 + 5 x 2 + u 1 v 1 = 100 3 x 1 + 3 x 2 + u 2 v 2 = 180 x 1 + u 3 v 3 = 40 all variables 0 50 5 x 1 + 3 x 2 = 150 feasible solution space x 2 30 (A) (1) (2) (3) u 1 = 0, v 1 > 0 x 1 20 direction of improvement in P 1 2 x 1 + 5 x 2 = 100 u 1 > 0, v 1 = 0 50 x 1 30

Example: Solution of a GP x 2 50 ) (A ft Soft (3) P 3 t (1 ) 30 (2 P 1 Sof ) 50 60 x 1 ft So 20 P 1 So ) (A P 2 ) rd Ha optimal with (A), (1), (2), and (3) P 2 (A P 1 Actually at this point we know 50 that the point is optimal even with the third constraint added and the third goal considered. Why? Sof 20 t (1 ) x 2 60 30 50 x 1 50 rd Ha 20 optimal with (A), (1), and (2) x 2 60 region with u 1 = 0 t (1 ) (2 Sof 30 ) 50 60 x 1 31

Example 10 -2 x 2 region with v 1 = 0 50 P 1 40 x 1 min P 1 v 1, P 2 u 2, P 3 v 3, s. t. 5 x 1 + 4 x 2 + u 1 v 1 = 200 2 x 1 + x 2 + u 2 v 2 = 40 2 x 1 + 2 x 2 + u 3 v 3 = 30 all variables 0 x 2 50 50 40 40 P 1 15 P 2 P 3 20 (1) (2) (3) 40 x 1 1520 40 optimal, with v 1 = u 2 = v 3 = 0 x 1 32

4/4 : Another Form of Goal Programming 33

Another Form of GP: Weighted Goals ègoals èu 1 èthe with weights min P 1 u 1, P 2 u 2, P 3 u 3, s. t. 5 x 1 + 3 x 2 150 2 x 1 + 5 x 2 + u 1 v 1 = 100 3 x 1 + 3 x 2 + u 2 v 2 = 180 x 1 + u 3 v 3 = 40 all variables 0 (A) (1) (2) (3) = 30, u 2 = 20, v 2 = 20, u 3 = 20, v 3 = 10 GP expressed as LP min 30 u 1+20 u 2+20 v 3 +20 u 3 + 10 v 3 s. t. 5 x 1 + 3 x 2 150 (A) 2 x 1 + 5 x 2 + u 1 v 1 = 100 (1) 3 x 1 + 3 x 2 + u 2 v 2 = 180 (2) x 1 + u 3 v 3 = 40 (3) all variables 0 34

Assignment #4 è#1. Chapter 8, Problem 16 è(a). Find the maximal flow for this network. Show all the steps. è(b). Formulate this problem as a linear program. è#2. Chapter 10, Problem 1 è#2. Chapter 10, Problem 4 35