|
Mustaqil ish topshirdi: Jo’rayev A. Qabul qildi: Xasanov D. Toshkent 023 lcm = (3,12,12,24,6) t = 1) 1-0-1-0, 2) 1-0-2 = -1, 3)3-0-2-1 t = 2)1-1-2-2 ) 3-1-2=0
|
| Sana | 05.01.2024 | | Hajmi | 33.93 Kb. | | #130677 |
Bog'liq xudo xoxlasa tushadi99%, 3-labarotoriya ishi Saralash usul va algoritmlarini tadqiq qilis, cmd buyruqlari, Incremental model nima, 1matematik, word sAM 1 savol, Документ Microsoft Word (4), Ma\'ruzalar (2), ЛАБОРАТОРНАЯ РАБОТА N1, Dasturlash 2, Ariza, Qalandarova Gulshoda, 1648631455, 1650692784, 1651669892 (2)
O‘ZBEKISTON RESPUBLIKASI AXBOROT TEXNOLOGIYALARI VA
KOMMUNIKATSIYALARINI RIVOJLANTIRISH VAZIRLIGI
MUHAMMAD AL-XORAZMIY NOMIDAGI
TOSHKENT AXBOROT TEXNOLOGIYALARI UNIVERSITETI
Dasturiy injinering fakulteti
Real vaqt tizimlari fani
Mustaqil ish 1
Topshirdi: Jo’rayev A.
Qabul qildi: Xasanov D.
Toshkent 2023
LCM = (3,12,12,24,6)
t = 0 1) 1-0-1-0, 2) 1-0-2 = -1, 3)3-0-2-1
t = 1 2)1-1-2-2 3) 3-1-2=0
t = 2 x
t = 3 3) 1-0-1-0 4) 5-3-3=-1
t = 4 4)5 - 4 - 3 = - 2 5 ) 1-9-12-4
t = 5 5 - 5 - 3 = - 3
t = 6 1-0-1=0 1-0-1=0
t = 9 1) 1 - o - 1 = 0
t=12 1) 1 - 13 - 2 = - 14 2 ) 1-0-2=-1 3)3-0-2=1 5)1-0-1=0
t = 18 1 - o - 1 = 0 5)1-0-1=0
LCM (3, 12, 12, 24, 6) = 24
t=0. Vaqt=0 bo'lganda, P1 jarayoni keladi. P1 jarayon vaqt=1 bo’lguncha bajariladi.
t=1. P1 jarayon bajarilib bo’ldi va P2, P3, P4 jarayonlar yetib keldi. P2,P3 va P4 jarayonlar uchun Slack Time hisoblanadi.
Slack Time = Deadline - Time` - Execution Time`
P2(st) = 1 - 1 - 2 = -2
P3(st) = 3 - 1 - 2 = 0
P4(st) = 2 - 1 - 3 = -2
P2 va P4 jarayonlarning slack time i teng va eng kichik chiqdi, lekin P2 jarayon birinchi kelgani uchun P2 jarayon bajariladi.
t=2.
P2(st) = 1 - 2 - 1 = -2
P3(st) = 3 - 2 - 2 = -1
P4(st) = 2 - 2 - 3 = -3
t=3.
P1(st) = 2 - 0 - 1 =1
P2(st) = 1 - 3 - 1 = -3
P3(st) = 3 - 3 - 2 = -2
P4(st) = 2 - 3 - 2 = -3
t=4.
P1(st) = 2 - 1 - 1 = 0
P3(st) = 3 - 4 - 2 = -3
P4(st) = 2 - 4 - 2 = -4
P5(st) = 5 - 4 - 1 = 0
t=5.
P1(st) = 2 - 2 - 1 = -1
P3(st) = 3 - 5 - 2 = -4
P4(st) = 2 - 5 - 1 = -4
P5(st) = 5 - 5 - 1 = -1
t=6.
P1(st) = 2 - 0 - 1 = 1
P3(st) = 3 - 6 - 1 = -4
P4(st) = 2 - 6 - 1 = -5
t=7.
P1(st) = 2 - 1 - 1 = 0
P3(st) = 3 - 7 - 1 = -5
t=8.
P1(st) = 2 - 1 - 1 = -1
t=9.
P1(st) = 2 - 0 - 1 = 1
t=10.
P5(st) = 5 - 4 - 1 = 0
t=11. t=12.
P1(st) = 2 - 0 - 1 = 1
t=13.
P2(st) = 1 - 1 - 2 = -2
P3(st) = 3 - 1 - 2 = 0
t=14.
P2(st) = 1 - 2 - 1 = -2
P3(st) = 3 - 2 - 2 = -1
t=15.
P1(st) = 2 - 0 - 1 = 1
P3(st) = 3 - 3 - 2 = -2
t=16.
P1(st) = 2 - 1 - 1 = 0
P3(st) = 3 - 4 - 1 = -2
P5(st) = 5 - 4 - 1 = 0
t=17.
P1(st) = 2 - 2 - 1 = -1
P5(st) = 5 - 5 - 1 = -1
t=18.
P1(st) = 2 - 0 - 1 = 1
P5(st) = 5 - 6 - 1 = -2
|
|
Bosh sahifa
Aloqalar
Bosh sahifa
Mustaqil ish topshirdi: Jo’rayev A. Qabul qildi: Xasanov D. Toshkent 023 lcm = (3,12,12,24,6) t = 1) 1-0-1-0, 2) 1-0-2 = -1, 3)3-0-2-1 t = 2)1-1-2-2 ) 3-1-2=0
|
