Administracion De Operaciones Heizer Pdf

Administracion De Operaciones Heizer Manual En Lnea#

Rating: 5 ADMINISTRACION DE OPERACIONES PRODUCCION DIRECCION DE LA PRODUCCIN Y OPERACIONES: DECISIONES ESTRATEGICAS - JAY HEIZER y BARRY RENDER Reviewed by. Principios De Administracion De Operaciones Heizer PdfIn this textbook, Heizer (business administration, Texas Lutheran U.) and Render (operations management, Rollins College) provide a broad introduction to the field of operations management.

Administracion De Operaciones Heizer Manual En Lnea

Any further distribution of this work must maintain attribution to the author(s) and the title of the work, journal citation and DOI. Puede descargar versiones en PDF de la gua, los manuales de usuario y libros electrnicos sobre administracion de operaciones heizer pdf, tambin se puede encontrar y descargar de forma gratuita un manual en lnea gratis (avisos) con principiante e intermedio, Descargas de documentacin, Puede descargar archivos PDF (o DOC y PPT) acerca. Principios de administracion de operaciones jay heizer barry render. Libros Gratis en PDF de Tercera Parte: Administracin de operaciones Captulo 11.

The problem can then be solved using the minimization algorithm. Question 15.4 Because this is a maximization problem, create an opportunity loss table by subtracting each number in the table from 11 (the highest number in the table): Original problem:Job 1 to Machine D Job 2 to Machine A Job 3 to Machine C Job 4 to Machine B(b) Total production = 10 + 10 + 9 + 11 = 40 BUS P301:01Question 15.9 Because this is a maximization problem, subtract each number from 95 to create an opportunity loss table. Optimal assignment: Taxi at post 1 to customer C Taxi at post 2 to customer B Taxi at post 3 to customer A Taxi at post 4 to customer D Total distance traveled = 4 + 4 + 6 + 4 = 18 miles. CHAPTER 15: SHORT-TERM SCHEDULING вЂ“ Suggested Solutions to Selected QuestionsQuestion 15.3 Original distance minimization problem:A minimum of four lines was used вЂ“ implying existence of an optimal assignment. For this purpose an analysis of the production processes of the line was made and the aggregate plan was defined evaluating two strategies: Exact Production Plan (Zero Inventory) and Constant Workforce Plan (Vary Inventory) by studying the costs of both inventory maintenance and workforce. It was also determined how the installed capacity was used with the standards of the rubber line and measures for decreasing production costs were proposed.

(a) Dispatching Rule EDD SPT LPT FCFSJob Sequence CXвЂ“BRвЂ“SYвЂ“DEвЂ“RG BRвЂ“CXвЂ“SYвЂ“DEвЂ“RG RGвЂ“DEвЂ“SYвЂ“CXвЂ“BR CXвЂ“BRвЂ“DEвЂ“SYвЂ“RGStarting day number: 241 (i.e., work can be done on day 241) Method: SPTвЂ”Shortest processing time Processing TimeCX-01 25 270 2 40 BR-02 15 300 1 15 DE-06 35 320 4 105 SY-11 30 310 3 70 RG-05 40 360 5 145 Total 145 375 Average 75 Sequence: BR-02,CX-01,SY-11,DE-06,RG-05 Average # in system = 2.586 = 375/145Method: LPTвЂ”Longest processing time Processing TimeCX-01 25 270 4 130 BR-02 15 300 5 145 DE-06 35 320 2 75 SY-11 30 310 3 105 RG-05 40 360 1 40 Total 145 495 Average 99 Sequence: RG-05,DE-06,SY-11,CX-01,BR-02, Average # in system = 3.414 = 495/145Method: Earliest due date (EDD) earliest to latest date Processing TimeCX-01 25 270 0 1 BR-02 15 300 0 2 DE-06 35 320 0 4 SY-11 30 310 0 3 RG-05 40 360 0 5 Total 145 Average Sequence: CX-01,BR-02,SY-11,DE-06,RG-05 Average # in system = 2.655 = 385/145Method: First come, first served (FCFS) Processing TimeCX-01 25 270 0 1 BR-02 15 300 0 2 DE-06 35 320 0 3 SY-11 30 310 0 4 RG-05 40 360 0 5 Total 145 Average Sequence: CX-01,BR-02,DE-06,SY-11,RG-05, Average # in system = 2.69 = 390/145Either of these choices could be supported. Question 15.13 This is a job-sequencing problem comparing performance under different dispatching rules. Scheduling RuleSPT is best on all criteria. Original problem: Today is day 205.Days Late 0 2 Number of jobs in system= 41/20 = 2.05 4 14 Total: 20 days Average lateness = 20/4 = 5.0Days Late 2 6 Number of jobs in system= 59/20 = 2.95 13 15 Total: 36 days Average lateness = 36/4 = 9.0Days Late 0 1 Number of jobs in system= 43/20 = 2.15 8 12 Total: 21 days Average= 21/4 = 5.25(212 вЂ“ 205)/6 = 1.17 (209 вЂ“ 205)/3 = 1.33 (208 вЂ“ 205)/3 = 1.00 (210 вЂ“ 205)/8 = 0.63Need date today s date Days required to complete jobA minimum total lateness of 20 days seems to be about the least we may achieve. Comparing the scheduling efficiency of the several algorithms presented in terms of lateness. Total rating remains 335.Question 15.12 This is a job-sequencing problem whose objective here is to minimize total lateness.