2123 - 5 This is a transportation problem 3 2-8 1 5 4 4-7 2 3 7 5 6 5 6 5 5-5 1 7 4 4 2 3 8 5-7 5 3 8 5 9 4 0 6 Chapter 5 Network Modeling S-5 12

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5. This is a transportation problem. 3 2 -8 1 5 4 4 -7 2 3 7 5 6. +5 6 +5 +5 -5 1 7 4 +4 2 3 8 5 -7 5 3 +8 5 9 4 +0 6 Chapter 5 - Network Modeling : S-5 ------------------------------------------------------------------------------------------ 12. The LP model is: MIN ST 12 X12 + 8 X13 + 15 X14 + 9 X23 + 16 X25 + 6 X34 + 7 X35 + 12 X54 -X12 - X13 - X14 = -15 -X23 - X25 = -15 +X13 + X23 -X34 -X35 = 0 +X14 + X34 + X54 = 20 +X25 +X35 -X54 = 10 0 d Xij d 10 The solution is: X13 = 5, X14 = 10, X23 = 5, X25 = 10, X34 = 10 Minimum total cost = $455 See file: Prb5_20.xls 13. The optimal solution is: X14=X49=X9,12= 1, Minimum cost = $8 million See file: Prb5_32.xls 14. a. This is a transportation problem. Note that demand exceeds supply by 20 units. Warehouses Stores 1 5 +20 -30 1 4 6 5 2 3 6 +25 -30 2 4 4 4 3 2 3 +30 -30 3 4 b. MIN 5 X11 + 4 X12 + 6 X13 + 5 X14 +3 X21 + 6 X22 + 4 X23 + 4 X24 +4 X31 + 3 X32 + 3 X33 + 2 X34 -X11 - X12 - X13 - X14 = -30 -X21 - X22 - X23 - X24 = -30 -X31 - X32 - X33 - X34 = -30 +X11 + X21 + X31 + XD1 d +20 +X12 + X22 + X32 + XD2 d +25 +X13 + X23+ X33 + XD3 d +30 +X14 + X24+ X34 + XD4 d +35 Xij t 0 +35 ST Chapter 5 - Network Modeling : S-11 ------------------------------------------------------------------------------------------ 24. a. Chic. 6 17 NY 7 +1 18 18 25 24 95 LA 2 32 Denver 3 50 27 19 30 14 14 24 35 St Lou 4 Memp 5 5 13 35 105 45 San Diego 1 -1 b. MIN 5 X12 + 13 X13 + 45 X15 + 105 X17 + 27 X23 + 19 X24 + 50 X25 + 95 X27 + 14 X34 + 30 X35 + 32 X36 + 14 X43 + 35 X45 + 24 X46 + 35 X54 + 18 X56 + 25 X57 + 24 X64 + 18 X65 + 17 X67 -X12 - X13 - X15 - X17 = -1 +X12 - X23 - X24 - X25 - X27 = 0 +X13 + X23 + X43- X34 - X35 - X36 = 0 +X24 + X34 + X54 + X64 - X43 - X45 - X46 = 0 + X15 + X25 + X35 + X45 + X65 - X54 - X56 - X57 = 0 + X36 + X46 + X56 - X64 - X65 - X67 = 0 + X17 + X27 + X57 + X67 = +1 Xij t 0 ST c. See file: Prb5_14.xls The solution is: X13=X36=X67=1 with a minimum total cost of $62. Chapter 5 - Network Modeling : S-18 ------------------------------------------------------------------------------------------ 33. a. -30 Doha 0 L.B. = 7.5 ($0, 999) Tanke r ($1.20, 999) ($1.40, 999) Rotter dam +6 5 1 ($1.35, 999) ($0.35, 22.5) Toulon ($0.25, 999) 6 Damietta +15 +0 4 ($0.20, 999) ($0.15, 999) ($0.16, 15) Suez +0 2 ($0.20, 999) +0 b. c. See file: Prb5_22.xls Total Cost = $25.43 million U.B . 22.5 999 999 999 999 999 15 999 999 999 999 999 999 Unit Cost ($1,000,000s ) $0.35 $0.00 $1.20 $1.40 $1.35 $0.20 $0.16 $0.27 $0.28 $0.19 $0.25 $0.20 $0.15 ($0.27, 999) ($0.28, 999) P. Said 3 ($0.19, 999) Palermo +9 7 Legend: (cost, upper bound) Flow 22.5 7.5 6 1.5 0 7.5 15 0 0 7.5 0 13.5 1.5 L.B. 0 7.5 0 0 0 0 0 0 0 0 0 0 0 From 0 Doha 0 Doha 1 Tanker 1 Tanker 1 Tanker 2 Suez 2 Suez 3 Port Said 3 Port Said 3 Port Said 4 Damietta 4 Damietta 4 Damietta To 2 1 5 6 7 3 4 5 6 7 5 6 7 Suez Tanker Rotterdam Toulon Palermo Port Said Damietta Rotterdam Toulon Palermo Rotterdam Toulon Palermo Chapter 5 - Network Modeling : S-19 ------------------------------------------------------------------------------------------ 34. a. -30 Reg 1 Reg 2 Reg 3 Reg 4 Reg 5 $6.50 $7.50 $7.00 $8.00 $8.25 $7.25 $6.75 $7.75 $7.00 $7.50 Eustis 7 Pine Hills 6 -40 60 -25 -35 70 -33 40 $6.75 Sanfor d 8 b. c. See file Prb5_9.xls 20,000 from Region 1 to Pine Hills, 10,000 from Region 1 to Eustis, 40,000 from Region 2 to Pine Hills, 25,000 from Region 3 to Eustis, 35,000 from Region 4 to Sanford, 25,000 from Region 5 to Eustis, 5,000 from Region 5 to Sanford. Total cost $1,132,500. ...
View Full Document

This note was uploaded on 05/29/2010 for the course BAS 52545 taught by Professor Asdasd during the Summer '04 term at 카이스트, 한국과학기술원.

Ask a homework question - tutors are online