09 cumcm
DESCRIPTION
09 CUMCM 医院眼科的最优病床安排方案的研究与求解TRANSCRIPT
- 1. 2009 . , A/B/C/D B () 1. 2. 3. () 2009 9 14
- 2. 2009
- 3. 1 ...................................................32 ....................................................33 ......................................................44 ......................................................55 ......................6 5.1 .........................................................................................................................................6 5.2 .....................................................................................................................................7 5.3 .........................................................................................8 6 ..........................8 6.1 .........................................................................................................................................8 6.1.1 .....................................................................................................8 6.1.2 .........................................................................................................9 6.1.3 ...........................................................................................................10 6.2 ...........................................................................................................................................10 6.2.1 ...............................................................................................................................................10 6.2.2 ............................................................................................11 6.3 PFCFS ..............................................................................12 6.3.1 ...............................................................................................................................................12 6.3.2 ...................................................................................................................................12 6.3.3 PFCFS ..................................................................................................................15 6.4 .......................................................................................................................................167 ...............................17 7.1 ...................................................................................17 7.1.1 ...........................................................................................................................17 7.1.2 ...............................................................................................................................18 7.2 ...................................................................................19 8 ..............20 8.1 .......................................................................................................................................................20 8.2 ...........................................................................................20 8.2.1 1 ............................................................................................................................21 8.2.2 2 ............................................................................................................................21 8.2.3 3 ............................................................................................................................21 8.3 ...........................................................................................................................................22 9 ...................22..........................................................26 1..................................................................................................................................................................26 2..................................................................................................................................................................3210 ...............................................23 10.1 .........................................................................................................................23 1
- 4. 10.2 .................................................................................23 10.3 .................................................................................................24 10.4 .................................................................................................................24 .....................................................25 Y1 Y2 Y3 100%8.44 10.50 79 Poisson M / M /1 PFCFS(Priority-First Come First Serve) 6.5 FCFS JAVA 10.19% 9.27 9.4510.30 M / M /1 LINGO 0.106 0.194 0.113 0.587 8 15 9 47 2
- 5. PFCFS BP 1 79 2008 7 13 2008 9 11 1 2 60% 2-3 FCFSFirst come, Firstserve ? 2 Y1 Y2 Y3 3
- 6. M / M /1 JAVA FCFS 7 M / M /1 LINGO 3 1 2 3 4 5 6 4
- 7. 4 Y1 Y2 Y y1 y2 bi i n T N in tj t j pi i ni i Ls Lq Ws Wq c i T Wsi i xi i tsq ts thg tz t ws t sqjd t sqdd t sb tdy 5
- 8. 5 5.1 [1] 1 Y1 Y2 100% 5.1 6
- 9. 5.1 n b i Y1 = i =1 Tn N in Y2 = T bi i n T N in Y Y Y Y= 1 2 y1 y2 y1 y2 , 89%, 21 [2]5.2 Y3 7
- 10. Y3 ni 5 (t j + t j w) Y3 = ( j =1 ) pi i =1 ni t j t j w( w > 1) pi i ni i 5.3 Excel FCFS 5.1 5.1 FCFS Y1 Y2 Y Y3 100% 8.44 2.71 10.5 6 6.1 6.1.1 6.1 8
- 11. 6.1 1. 2. 3. 6.1.2 1. 6 2. 6.2 6.2 9
- 12. 6.1.3 1. 79 2. 3. 12 60% 4. 5. 2-3 6. 7. FCFSFirst come, First serve 6.2 6.2.1 [3] 6.3 10
- 13. 6.3 6.2.2 1) 2) [t , t + t ] 1 t P (t , t + t ) = t + (t )3) [t , t + t ] 2 t n ( t ) n t Pn (t ) = e , t > 0, n = 0,1, 2,L n! (6.2) 1 2008-07-13 2008-09-11 11
- 14. 6.3 PFCFS PFCFS(Priority-First Come FirstServe) FCFS 6.3.1 1. 2. 3. 6.3.2 tsq t s thg () t z t z = tsq + ts + thg 6.1 6.1 t sq t s thg t z 1 1 3 5 1 3 3 7 1 1 6 8 2 1 9 12 6.2 6.2 1 3 12
- 15. 1 2 4 5 6 7 1. 2. 1 : 2 3. 1 : 2 4. 2 : 3 6.3 0 6.3 6 7 1 2 3 4 5 () 2 3 0 () 3 0 0 4 4 4 1 2 3 13
- 16. t ws t ws = 0 3 1. t sqjd t sqdd t sqjd + tsqjd tsqjd =0 2. t sqjd + tsqjd tsqjd =0 . t sqjd + tsqjd tsqjd 0) {if(waiWaitList.isEmpty()&&bai1WaitList.isEmpty()&&otherWaitList.isEmpty()&&bai2WaitList.isEmpty()) { System.out.println(""); break; } if (waiWaitList.isEmpty() == false) // { proccess(waiWaitList, currentTime, 8); } // if(leftOfEmptyBeds >0 && bai1WaitList.isEmpty() == false) { proccess(bai1WaitList, currentTime, 5); } // if(leftOfEmptyBeds >0 && otherWaitList.isEmpty() == false) { proccess(otherWaitList, currentTime, 12); } // if(leftOfEmptyBeds >0 && bai2WaitList.isEmpty() == false) { proccess(bai2WaitList, currentTime, 7); }if(waiWaitList.isEmpty()&&bai1WaitList.isEmpty()&&otherWaitList.isEmpty()&&bai2WaitList.isEmpty()) { break; } } } break; case 5: case 6: case 0:
- 32. { while (leftOfEmptyBeds > 0) {if(waiWaitList.isEmpty()&&bai1WaitList.isEmpty()&&otherWaitList.isEmpty()&&bai2WaitList.isEmpty()) { System.out.println(""); break; } if (waiWaitList.isEmpty() == false) // { proccess(waiWaitList, currentTime, 8); } // if(leftOfEmptyBeds >0 && otherWaitList.isEmpty() == false) { proccess(otherWaitList, currentTime, 12); } // if(leftOfEmptyBeds >0 && bai1WaitList.isEmpty() == false) { proccess(bai1WaitList, currentTime, 5); } // if(leftOfEmptyBeds >0 && bai2WaitList.isEmpty() == false) { proccess(bai2WaitList, currentTime, 7); }if(waiWaitList.isEmpty()&&bai1WaitList.isEmpty()&&otherWaitList.isEmpty()&&bai2WaitList.isEmpty()) { break; } } } break; case 2: case 1: { while (leftOfEmptyBeds > 0) {if(waiWaitList.isEmpty()&&bai1WaitList.isEmpty()&&otherWaitList.isEmpty()&&bai2WaitList.isEmpty()) { System.out.println(""); break; } if (waiWaitList.isEmpty() == false) // { proccess(waiWaitList, currentTime, 8); }
- 33. // if(leftOfEmptyBeds >0 && bai2WaitList.isEmpty() == false) { proccess(bai2WaitList, currentTime, 7); } // if(leftOfEmptyBeds >0 && bai1WaitList.isEmpty() == false) { proccess(bai1WaitList, currentTime, 5); } // if(leftOfEmptyBeds >0 && otherWaitList.isEmpty() == false) { proccess(otherWaitList, currentTime, 12); }if(waiWaitList.isEmpty()&&bai1WaitList.isEmpty()&&otherWaitList.isEmpty()&&bai2WaitList.isEmpty()) { break; } } } break; default: break; } } // public static int minSeeTime(ArrayList p){ int min = 10000; for (Patient patient : p) { if(patient.getSeeTime() < min) min = patient.getSeeTime(); } return min; } // public static int numOfPatient(ArrayList p,int seeTime){ int num = 0; for (Patient patient : p) { if (patient.getSeeTime() == seeTime) { num++; } } return num; } // public static void proccess(ArrayList p ,int currentTime,inttotalTime) { int j; for ( j = 0;j
- 34. if (bed[i].isEmpty() == true) { p.get(j).setInTime(currentTime);// p.get(j).setWaitTime(currentTime-p.get(j).getSeeTime());// p.get(j).setOutTime(currentTime+totalTime);//8 bed[i].setPatients(p.get(j)); bed[i].setEmpty(false); p.remove(j); //psize j--; if (leftOfEmptyBeds>0) { leftOfEmptyBeds--; // } else { break; } System.out.println(bed[i].getPatients().getDisease()+""); System.out.println(""+leftOfEmptyBeds); break; } } } } // public static void readPatient(String line,int p[])throws DataFormatException { StringTokenizer token = new StringTokenizer(line,"_"); token.nextToken(); int i = 0; while (token.hasMoreElements()) { p[i] = Integer.valueOf(token.nextElement().toString()); i++; } }} 2 lingo max=x/((79/8)*x-1.05)+y/((79/7)*y-2.2)+z/((79/5)*z-1.65)+k/((79/12)*k-3.867); x>0; y>0; z>0; k>0; (((79/8)*x-1.05))>0; (((79/7)*y-2.2))>0; (((79/5)*z-1.65))>0; (((79/12)*k-3.867))>0; x+y+z+k=1;