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T1: 5B + 8I + 2B,T2: 5B + 8I +2B,T3: 5B + 8I + 2B,CPU执行顺序似乎是：(1) CPU on T1的5B, so t= 5, 故T1开始IO;(2) CPU on T2的5B, so t=5+5 = 10, 故T2开始IO;(3) CPU on T3的3B,so t=10+3 = 13, 故T1完成 IO;(4) CPU on T1的2B,so t=13+2 = 15, 故T1完成;(5) CPU on T3的2B,so t=15+2 = 17, 故T3开始IO;(6) CPU 空闲1, ,so t=17+1 = 18, 故T2完成 IO;(7) CPU on... |
