Tarea8Eq13 - INSTITUTO TECNOLOGICO Y DE ESTUDIOS SUPERIORES...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
INSTITUTO TECNOLOGICO Y DE ESTUDIOS SUPERIORES DE MONTERREY UNIVERSIDAD VIRTUAL METODOS DE OPTIMIZACION PARA LA TOMA DE DECISIONES TAREA 8 EQUIPO 13: AYDE ORIHUELA LEYVA A00927380 LUIS ROMULO ROMERO GUADARRAMA A00927779 FECHA DE ENTREGA: 09 DE ABRIL 2012
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Instrucciones . Resolver los siguientes problemas. Problema 1 (extraído del libro de texto, problema 4, página 371) Steelco manufactures three types of steel at different plants. The time required to manufacture 1 ton of steel (regardless of type) and the costs at each plant are shown in Table 8. Each week, 100 tons of each type of steel (1, 2, and 3) must be produced. Each plant is open 40 hours per week. a Formulate a balanced transportation problem to minimize the cost of meeting Steelco’s weekly requirements. b Suppose the time required to produce 1 ton of steel depends on the type of steel as well as on the plant at which it is produced (see Table 9). Could a transportation problem still be formulated? Problema 2 (extraído del libro de texto, problema 2, página 398) Doc Councillman is putting together a relay team for the 400-meter relay. Each swimmer must swim 100 meters of breaststroke, backstroke, butterfly, or freestyle. Doc believes that each swimmer will attain the times given in Table 51. To minimize the team’s time for the race, which swimmer should swim which stroke? Solución: Min z = 54 gf + 51 mf + 50 jf + 56 cf + 54 gb + 57 mb + 53 jb + 54 cb + 51 gfl + 52 mfl + 54 jfl + 55 cfl + 53 gba + 52 mba + 56 jba + 53 cba Sujeto a:
Background image of page 2
gf + gb + gfl + gba = 1 mf + mb + mfl + mba = 1 jf + jb + jfl + jba = 1 cf + cb + cfl + cba = 1 gf + mf + jf + cf = 1 gb + mb + jb + cb = 1 gfl + mfl + jfl + cfl = 1 gba + mba + jba + cba = 1 Al resolver con Lindo; obtenemos lo siguiente: LP OPTIMUM FOUND AT STEP 8 OBJECTIVE FUNCTION VALUE 1) 207.0000 VARIABLE VALUE REDUCED COST GF 0.000000 4.000000 MF 0.000000 0.000000 JF 1.000000 0.000000 CF 0.000000 4.000000 GB 0.000000 2.000000 MB 0.000000 4.000000 JB 0.000000 1.000000 CB 1.000000 0.000000 GFL 1.000000 0.000000 MFL 0.000000 0.000000 JFL 0.000000 3.000000 CFL 0.000000 2.000000 GBA 0.000000 2.000000 MBA 1.000000 0.000000 JBA 0.000000 5.000000 CBA 0.000000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 2.000000
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/09/2012 for the course INDUSTRIAL 102 taught by Professor Ramonurbina during the Spring '12 term at ITESM.

Page1 / 8

Tarea8Eq13 - INSTITUTO TECNOLOGICO Y DE ESTUDIOS SUPERIORES...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online