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# review2_306 - Northwest Corner Solution Exercise 3a...

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Unformatted text preview: Northwest Corner Solution Exercise 3a Exercise WH WH Plant Plant A 1 1200 6 8 11 2 1900 1800 500 8 9 9 3 7 9 8 B 95 C - Dummy D E Capacity 7 6 5 8 1500 3500 0 1200 0 4200 0 5000 4 Demand 3100 1800 2000 4400 1100 12400 0 2000 900 1100 Northwest Corner Solution Exercise 3a Exercise Total cost Total \$9 (1200) = \$10,800 \$9 \$6 (1900) = \$11,400 \$8 (1800) = \$14,400 \$11 (500) = \$5,500 \$9 (1500) = \$13,500 \$5 (3500) = \$17,500 \$8 (900) = \$7,200 \$0 (1100) = \$ 0 \$80,300 North-South Rule Solution Exercise 3b Exercise WH WH Plant Plant A 1 B 95 6 100 C 11 9 8 Dummy D E Capacity 7 6 1100 3300 2000 Demand 3100 1800 2000 4400 1100 12400 2 3100 8 9 3 1700 7 9 4 8 5 8 0 1100 1200 0 4200 0 5000 0 2000 North-South Rule Solution Exercise 3b Exercise Total cost Total \$5 (100) = \$ 500 \$5 \$0 (1100) = \$ 0 \$6 (3100) = \$18,600 \$6 (1100) = \$6,600 \$9 (1700) = \$15,300 \$5 (3300) = \$16,500 \$8 (2000) = \$16,000 \$73,500 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 2 3 B 95 6 8 7 8 9 9 C 11 9 8 Dummy D E Capacity Diff. 7 6 5 8 4 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 3 1 10 Diff. 0 1200 5 0 4200 6 0 5000 5 0 2000 7 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 2 3 B 95 6 8 7 8 9 9 C 11 9 8 Dummy D E Capacity Diff. 7 6 5 8 4 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 3 1 10 Diff. 0 1200 5 0 4200 6 0 5000 5 0 2000 7 1100 1100 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 2 3 B 95 6 8 7 8 9 9 C 11 9 8 Dummy D E Capacity Diff. 7 6 5 8 4 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 3 1 10 Diff. 0 1200 5 0 4200 6 0 5000 5 0 2000 7 1100 1100 Difference Method SolutionExercise 3c Exercise C 11 9 8 WH WH Plant Plant A 1 2 3 B 95 6 8 7 8 9 9 Dummy D E Capacity Diff. 7 6 5 8 4 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 3 1 10 Diff. 0 1200 5 2 0 4200 6 0 0 5000 53 0 2000 71 1100 1100 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 2 3 B 95 6 8 7 8 9 9 C 11 9 8 Dummy D E Capacity Diff. 7 6 5 4400 4 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 3 1 10 Diff. 8 0 1200 5 2 0 4200 6 0 0 5000 53 0 2000 71 1100 1100 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 2 3 B 95 6 8 7 8 9 9 C 11 9 8 Dummy D E Capacity Diff. 7 6 5 4400 4 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 3 1 10 Diff. 8 0 1200 5 2 0 4200 6 0 0 5000 53 0 2000 71 1100 1100 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 2 3 B 95 6 8 7 8 9 9 C 11 9 8 Dummy D E Capacity Diff. 7 6 5 4400 4 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 3 1 10 Diff. 8 0 1200 5 2 4 0 4200 6 0 2 0 5000 5 31 0 2000 71 1100 1100 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 2 3 B 95 6 8 7 1200 C 11 9 8 Dummy D E Capacity Diff. 7 8 6 5 4400 9 9 4 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 3 1 10 Diff. 8 0 1200 5 2 4 0 4200 6 0 2 0 5000 5 31 0 2000 71 1100 1100 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 2 3 B 95 6 8 7 1200 C 11 9 8 Dummy D E Capacity Diff. 7 8 6 5 4400 9 9 4 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 3 1 10 Diff. 8 0 1200 5 2 4 0 4200 6 0 2 0 5000 5 31 0 2000 71 1100 1100 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 2 3 B 95 6 8 7 1200 C 11 9 8 Dummy D E Capacity Diff. 7 8 6 5 4400 9 9 4 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 31 1 10 Diff. 8 0 1200 5 2 4 0 4200 6 0 2 0 5000 5 31 0 2000 71 1100 1100 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 B 95 6 1200 C 11 9 8 Dummy D E Capacity Diff. 7 2 3100 8 3 7 4 8 6 5 4400 9 9 8 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 31 1 10 Diff. 0 1200 5 2 4 0 4200 6 0 2 0 5000 5 31 0 2000 71 1100 1100 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 B 95 6 1200 C 11 9 8 Dummy D E Capacity Diff. 7 2 3100 8 3 7 4 8 6 5 4400 9 9 8 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 31 1 10 Diff. 0 1200 5 2 4 0 4200 6 02 0 5000 5 31 0 2000 71 1100 1100 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 B 95 6 1200 C 11 9 8 Dummy D E Capacity Diff. 7 2 3100 8 3 7 4 8 6 5 4400 9 9 8 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 31 1 10 Diff. 0 1200 5 2 4 0 4200 6 02 3 0 5000 5 3 10 0 2000 71 1100 1100 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 B 95 6 1200 C 11 9 8 Dummy D E Capacity Diff. 7 2 3100 600 8 9 3 7 9 4 8 6 5 4400 8 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 31 1 10 Diff. 0 1200 5 2 4 0 4200 6 02 3 0 5000 5 3 10 0 2000 71 1100 1100 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 B 95 6 1200 C 11 9 8 Dummy D E Capacity Diff. 7 2 3100 600 8 9 3 7 9 4 8 6 5 4400 8 Demand 3100 1800 2000 4400 1100 12400 Diff. 1 31 1 10 Diff. 0 1200 5 2 4 0 4200 6 02 3 0 5000 5 3 10 0 2000 71 1100 1100 Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 B 95 6 1200 C 11 Dummy D E Capacity Diff. 7 900 Demand 3100 1800 2000 4400 1100 12400 2 3100 600 500 8 9 9 5 3 600 4400 7 9 8 8 4 1 31 1 1 8 6 0 1200 5 2 4 0 4200 6 02 3 0 5000 5 3 10 0 2000 71 1100 1100 0 Diff. Diff. Difference Method Solution Exercise 3c Exercise WH WH Plant Plant A 1 B 95 6 1200 C 11 Dummy D E Capacity Diff. 7 900 Demand 3100 1800 2000 4400 1100 12400 2 3100 600 500 8 9 9 5 3 600 4400 7 9 8 8 4 1 31 1 1 8 6 0 1200 5 2 4 0 4200 6 02 3 0 5000 5 3 10 0 2000 71 1100 1100 0 Diff. Diff. Difference Method Solution Exercise 3c Exercise Total cost Total \$5 (1200) = \$ 6,000 \$5 \$6 (3100) = \$18,600 \$8 (600) = \$ 4,800 \$11(500) = \$ 5,500 \$9 (600) = \$ 5,400 \$5 (4400) = \$22,000 \$8 (900) = \$7,200 \$0 (1100) = \$ 0 \$69,500 LP Formulation - Exercise 3d Exercise Min. Z = 9A1 + 5B1 + 7D1 + 6A2 + 8B2 + 11C2 + 6D2 + 8A3 + 9B3 + 9C3 + 11C2 5D3 + 7A4 + 9B4 + 8C4 + 8D4 Subject to: Subject LP Formulation - Exercise 3d Exercise A1 + B1 + D1 + E1 <=1200 A2 + B2 + C2 + D2 + E2 <=4200 Capacity A3 + B3 + C3 + D3 + E3 <=5000 A4 + B4 + C4 + D4 + E4 <=2000 A1 + A2 + A3 + A4 =>3100 B1 + B2 + B3 + B4 =>1800 C1 + C2 + C4 =>2000 Demand D1 + D2 + D3 + D4 =>4400 E1 + E2 + E3 + E4 =>1100 Optimal Solution (from LP) Exercise 3d Exercise WH WH Plant Plant A 1 B 95 6 1200 C 11 9 8 Dummy D E Capacity 7 6 5 8 2000 Demand 3100 1800 2000 4400 1100 12400 2 3100 600 8 9 3 7 9 4 8 4400 600 0 1200 0 500 4200 0 5000 0 2000 Optimal Solution - Exercise 3d Exercise Total cost Total \$5 (1200) = \$ 6,000 \$5 \$6 (3100) = \$18,600 \$8 (600) = \$ 4,800 \$8(2000) = \$16,000 \$5 (4400) = \$22,000 \$0 (500) = \$ 0 \$0 (600) = \$ 0 \$67,400 Optimal Solution (from LP) Exercise 3 extra credit Exercise WH WH Plant Plant A 1 B 95 6 1200 C 11 Dummy D E Capacity 7 6 2000 Demand 3100 1800 2000 4400 1100 12400 8 2 3100 600 8 9 3 7 9 4 + 9-5 8 8 4400 600 0 1200 - 0 4200 500 + 0 5000 +6 -5 +0 -0 0 2000 +1 Optimal Solution (from LP) Exercise 3 extra credit Exercise WH WH Plant Plant A 1 B 9 -5 6+8 1200 C 11 Dummy D E Capacity + 7 6 2000 Demand 3100 1800 2000 4400 1100 12400 2 3100 600 8 9 3 7 9 4 9-5 8 8 4400 +7 0 1200 -5 - 0 4200 +0 -0 500 + 0 5000 +8 -5 600 0 2000 +5 Exercise 5 - PERT/CPM Exercise Job A B C D E F G Predecessor Predecessor --------A C A B,D,E te 3 6 2 5 2 7 4 to 2 4 1 3 2 5 2 tp Crash cost Crash 4 8 3 7 3 8 6 \$5,000 \$6,000 \$2,500 \$4,000 --\$3,000 \$7,000 Exercise 5 - PERT diagram ) 2 D(5 (3 ) A B(6) 1 2) E( F(7) 3 G(4) 5 2 2)) C( C( 4 Paths and Durations - Exercise 5-1 Exercise Paths Paths A-F A-F A-D-G A-D-G B-G B-G C-E-G C-E-G Duration 10 weeks 10 12 weeks *Critical Path* 12 10 weeks 10 8 weeks Total Project Cost = \$61,000 Exercise 5 - 1 PERT/CPM Exercise Revise schedule to complete project in 10 Revise weeks: weeks: 1) Crash D 1 week 2) Crash D 1 week Total crash cost Total Original project cost Total cost Total \$4,000 \$4,000 \$8,000 \$8,000 \$61,000 \$69,000 \$69,000 Revised Paths and Durations Exercise 5-2 Exercise Paths Paths A-F A-F A-D-G A-D-G B-G B-G C-E-G C-E-G Duration 10 weeks 10 10 weeks 10 10 weeks 10 8 11 *Critical Path* Total Project Cost = \$69,000 Exercise 5-2 Exercise To keep on a 10-week schedule, we must To crash Path C-E-G down to 10 weeks crash C cannot be crashed; E cannot be crashed cannot Only choice is to crash G by one week Only Total cost = \$69,000 (after crashing D) \$ 7,000 (crashing G 1 wk) 7,000 \$ 4,500 (additional cost of C) 4,500 \$80,500 \$80,500 Expediting to 6 weeks Exercise 5-3 Exercise Paths Paths A-F A-F A-D-G A-D-G B-G B-G C-E-G C-E-G Duration 10 weeks *Critical Path* 10 10 weeks *Critical Path* 10 10 weeks *Critical Path* 10 8 Total Project Cost = \$69,000 Expediting to 6 weeks Expediting Activity A B C D E F G 2 4 1 3 2 5 2 to Exercise 5-3 Exercise te Cost Cost 3 6 2 5 2 7 4 (te - to) Weeks to crash 1 2 1 210 0 2 2 \$5,000 6,000 2,500 4,000 -3,000 7,000 Expediting to 6 weeks Exercise 5-3 Exercise 1. Crash F and G 1 week each Cost \$10,000 Expediting to 6 weeks Exercise 5-3 Exercise Paths Paths A-F A-F A-D-G A-D-G B-G B-G C-E-G C-E-G Duration 10 wks. 10 10 10 10 10 8 9 9 9 7 Expediting to 6 weeks Expediting Activity A B C D E F G 2 4 1 3 2 5 2 to Exercise 5-3 Exercise te Cost Cost 3 6 2 5 2 7 4 (te - to) Weeks to crash 1 2 1 2 0 2 2 \$5,000 6,000 2,500 4,000 -3,000 7,000 10 1 1 Expediting to 6 weeks Exercise 5-3 Exercise 1. Crash F and G 1 week each 2. Crash F and G 1 week each Cost \$10,000 10,000 10,000 Expediting to 6 weeks Exercise 5-3 Exercise Paths Paths A-F A-F A-D-G A-D-G B-G B-G C-E-G C-E-G Duration 10 wks. 10 10 10 10 10 8 9 9 9 7 8 8 8 6 Expediting to 6 weeks Expediting Exercise 5-3 Exercise Activity Activity A B C D E F G 2 4 1 3 2 5 2 to te 3 6 2 5 2 7 4 \$5,000 6,000 2,500 4,000 -3,000 7,000 Cost Cost (te - to) Weeks to crash 1 2 1 2 0 2 2 10 10 10 Expediting to 6 weeks Exercise 5-3 Exercise At this point, only A and B can be crashed At together, adding \$11,000 in crash costs. Since \$11,000 > \$10,000, we will accept \$11,000 we the penalty cost instead for the remaining 2 weeks. Expediting to 6 weeks Exercise 5-3 Exercise Total cost : \$61,000 8,000 10,000 10,000 20,000 \$ 109,000 Original project cost Crashing D 2 weeks Crashing F and G 1 wk. Crashing F and G 1 wk. Penalty for 2 wks. late Exercise 5-4 Estimating Probability Estimating Exercise What is the likelihood of completing the project in 9 weeks? 1. Determine the critical path and its duration C.P. = A-D-G = 12 weeks C.P. What is the likelihood of completing the project in 9 weeks? the 2. For each activity on CP (A-D-G): σΑ 2 = ( σD 2 = ( σG 2 = ( 4-2 6 7-3 6 6-2 6-2 6 )2 = )2 = )2 = 1 3. 9 Total Variance for C.P. Total 4 = 1.0 1.0 9 4 9 9 =1 9 What is the likelihood of completing What the project in 9 weeks? the 4. Determine Z-score Desired completion time completion x z Z= - µ Expected Expected completion time completion x-µ =σ Variance σ 9 - 12 12 1.0 1.0 z= = - 3.0 3.0 What is the probability of completing What the project in 9 weeks? the Prob. (z <= -3.0)= .00135 Prob. Prob. of completing project in 9 weeks = .00135 -3.0 0 Exercise 4a Exercise Chase Exercise 4a Exercise Exercise 4a Exercise Exercise 4a Exercise Chase Exercise 4a Exercise Exercise 4a Exercise Exercise 4b Exercise Level Exercise 4b Exercise Exercise 4b Exercise Methods to Evaluate Location Decisions Decisions 1. Center of Gravity Method 2. Factor Rating (Weighted Score) Method 3. Locational Breakeven Analysis Center of Gravity Method Center Y 20 B(retail) A B C Volume Cost (N-S) 12 A(WH) 5 T. 4 T. 1 T. \$10/T/mi \$7/T/mi \$5/T/mi 3 10 (Arbitrary) 18 28 C(retail) X (E-W) Center of Gravity Method Center We want to locate a fourth site (distribution We center) that is centrally located in terms of distance and cost and volume. and and Use Weighted Mean Distance: Z= Σ (ΤiViZi) (Τ i Σ (TiVi) (T i Calculate twice, once for each direction Center of Gravity Method Center East-West direction: ZE-W 10(5)(10) + 7(4)(18) + 5(1)(28) = = 13.8 10(5) + 7(4) + 5(1) North-South direction: 10(5)(12) + 7(4)(20) + 5(1)(3) ZN-S = = 14.2 10(5) + 7(4) + 5(1) Locate point (13.8, 14.2) Center of Gravity Method Center Y 20 B(retail) (N-S) 14.2 12 New D.C. A(WH) C(retail) 10 13.8 18 (Arbitrary) 28 3 X (E-W) Factor Rating Method Airport Location Decision Factors A(1/2 mi.) B(3 mi.) C(10 mi.) Noise 1 3 4 Convenience Safety Cost 4 2 3 2 3 4 1 4 2 Ratings: 1(worst) - 4(best) Factor Rating Method Wgts. Factors Airport Location Decision A(1/2 mi.) B(3 mi.) C(10 mi.) 1(.15)=.15 4(.15)=.6 2(.40)=.8 3(.30)=.9 3(.15)=.45 4(.15)=.6 2(.15)=.30 1(.15)=.15 3(.40)=1.2 4(.40)=1.6 4(.30)=1.2 2(.30)=.6 .15 Noise .15 Conven. .40 Safety .30 Cost 1.0 2.45 3.15 2.95 Locational Breakeven Analysis Locational Compares location alternatives on the basis of Fixed Cost and Variable Cost, basis and selects best location for given and level of Production. Location A B C D Fixed cost \$350,000 100,000 100,000 250,000 250,000 150,000 150,000 Variable cost \$5 40 40 10 10 23 23 Locational Breakeven Analysis Put costs into equations: A B C D Y= 5X + 350,000 Y= 40X + 100,000 Y= 10X + 250,000 Y= 23X + 150,000 Plot all of the lines together Locational Breakeven Analysis Locational B 5004003002001003 6 9 12 D C A 15 18 21 Locational Breakeven Analysis Locational Determine cutoff points for changing Determine location based on production level. Example: Find intersection of B and D Example 40x + 100,000 = 23x + 150,000 40x X = 2941 units If production <= 2941, choose Location B If If production > 2941, choose Location D Locational Breakeven Analysis Complete set of decision rules: If production is 0 2941 units 2941 7692 units 7692 20,000 units 20,000 Over 20,000 units Choose Choose Choose Choose Loc. B Loc. Loc. D Loc. Loc. C Loc. Loc. A Loc. Sequencing Rules Sequencing Job A B C Due Date Operating time Slack Due Operating Day 4 Day 7 Day 2 1 day 5 days 2 days 3 2 0 Sequencing Rules Sequencing 1) FCFS 1) FCFS Seq. A B C OT OT 1 5 2 Flow time time 1 6 Due Date Date 4 7 Days Late Late 0 0 6 6/3=2 days 8 2 15/3=5 days Sequencing Rules Sequencing Job A B C Due Date Operating time Slack Due Operating Day 4 Day 7 Day 2 1 day 5 days 2 days 3 2 0 Sequencing Rules Sequencing 2) SOT 2) SOT Seq. A C B OT OT 1 2 5 Flow time time 1 3 Due Date Date 4 2 Days Late Late 0 1 1 2/3=2/3 days 8 7 12/3=4 days Sequencing Rules Sequencing 3) LOT 3) LOT Seq. B C A OT OT 5 2 1 Flow time time 5 7 Due Date Date 7 2 Days Late Late 0 5 4 9/3=3 days 8 4 20/3=6.67 days Sequencing Rules Sequencing 4) Static Slack Seq. C B A OT OT 2 5 1 Flow time time 2 7 Due Date Date 2 7 Days Late Late 0 0 4 4/3=1.3 days 8 4 17/3=5.67 days Johnson’s Rule Johnson’s Job 1 2 3 4 5 Station A 27 min. 18 70 26 15 Station B 45 min. 33 30 24 10 5 Johnson’s Rule Johnson’s Job 1 2 3 4 5 2 Station A 27 min. 18 70 26 15 Station B 45 min. 33 30 24 10 5 Johnson’s Rule Johnson’s Job 1 2 3 4 5 2 Station A 27 min. 18 70 26 15 Station B 45 min. 33 30 24 10 4 5 Johnson’s Rule Johnson’s Job 1 2 3 4 5 2 Station A 27 min. 18 70 26 15 1 Station B 45 min. 33 30 24 10 4 5 Johnson’s Rule Johnson’s Job 1 2 3 4 5 2 Station A 27 min. 18 70 26 15 1 3 Station B 45 min. 33 30 24 10 4 5 Johnson’s Rule Johnson’s Job 1 2 3 4 5 2 Station A 27 min. 18 70 26 15 1 3 Station B 45 min. 33 30 24 10 4 5 Johnson’s Rule Johnson’s 0 18 2 18 1 2 51 45 3 1 96 115 115 4 3 141 5 156 4 5 169 179 A B 145 Total throughput time = 179 min. Total idle time = 19 min. Total idle Assignment Problem Assignment Resource Job A B C D E 1 30 90 110 60 30 2 50 40 60 100 50 3 40 30 80 40 60 4 80 50 100 120 40 5 20 70 90 50 90 Assignment Problem - Step 1 Step Resource Job A B C D E 1 30 90 110 60 30 2 50 40 60 100 50 3 40 30 80 40 60 4 80 50 100 120 40 5 20 -20 70 -30 90 -60 50 -40 90 -30 1) Subtract lowest cost in each row 1) from all others in that row. from Assignment Problem - Step 2 Step Assignment Resource Job A B C D E 1 10 60 50 20 0 2 30 10 0 60 20 3 20 0 20 0 30 4 60 20 40 80 10 5 0 40 30 10 60 -10 2) Subtract lowest cost in each column 2) from all others in that column. from Assignment Problem - Step 3 Job Job 1 2 3 4 Resource A 10 30 20 50 B C D E 60 50 20 0 10 0 60 20 0 20 0 30 10 30 70 0 5 0 40 30 10 60 3) Cover all “0”s with fewest # of horizontal 3) + vertical lines. Does # lines = # rows? Assignment Problem - Step 3 Job 1 2 3 4 Resource A 10 30 20 50 B C D E 60 50 20 0 10 0 60 20 0 20 0 30 10 30 70 0 5 0 40 30 10 60 # of lines = # of columns (or # of rows) of Assignment Problem - Step 4 Job 1 2 3 4 Resource A 10 30 20 50 B C D E 60 50 20 0 10 0 60 20 0 20 0 30 10 30 70 0 5 0 40 30 10 60 -10 4) Subtract lowest uncovered cost from all other 4) uncovered costs, add it to intersection cells. uncovered Assignment Problem - repeat Step 3 Job 1 2 3 4 5 Resource A 0 30 20 40 0 B C D E 50 40 10 0 10 0 60 30 0 20 0 40 0 20 60 0 40 30 10 70 3) Cover all “0”s with fewest # of horizontal 3) + vertical lines. Does # lines = # rows? Assignment Problem Job 1 2 3 Resource A 0 30 20 B C D E 50 40 10 0 10 0 60 30 0 20 0 40 4 40 0 20 60 0 5 0 40 30 10 70 # of lines = # of rows (or columns) Make optimal assignment by 0’s. Assignment Problem - Optimal Assignment Resource Job A B C D E 1 0 50 40 10 0 2 30 10 0 60 30 3 20 0 20 0 40 4 40 0 20 60 0 5 0 40 30 10 70 A-5 B-4 C-2 D-3 E-1 Assignment Problem - Optimal cost Optimal Resource Job A B C D E 1 30 90 110 60 30 2 50 40 60 100 50 3 40 30 80 40 60 4 80 50 100 120 40 5 20 70 90 50 90 Total cost = 200 ...
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