求前辈给解答下这时序分析道题

Hold Violation，1Hz都跑不了

您好，希望给出计算公式。还有题目中parametersgiven ocv +/-5% for clock是什么意思

hold violation= capture time - lauch time
=[0.4* ( 1.05) -0.1 -0.05] -[ (0.5+ 0.01) * 1.1 ] = - 0.2811 ns
Tmax= [ 0.1*1.05 + 0.01 + 0.5] - [ ( 0.1 + 0.4 ) * 0.95 - 0.2 - 0.25] = 0.59 ns

hold按照悲观分析（0.5+0.01）不是应该乘以0.9吗，为什么乘1.1这样hold更好满足啊

hold_vio = data_path - (clock_path+clock_uncerainty + hold_req_time) = (0.01+0.5) * 0.9 - (0.4*1.05+0.05+0.1) = 0.459 - 0.57 = -0.111
setup_slack = Tcycle + clock_path - clock_uncerainty - setup_req_time - data_path > 0
Tcycle + 0.4*0.95 - 0.25 - 0.2 - (0.01+0.5) > 0,Tcycle > 0.58
clock path上的公共路径0.1ns在OCV计算时要剔除掉

多谢

是我搞错啦, sorry