【Abstract】Since the “918 Incident”， the TSA has stipulated that all package boxes should be opened for inspection. Due to the Chicago airport security system setup， there are often passenger jams. Therefore， a pre-check model is designed in the paper and better solved the problem of congestion. In major point of the model is regard the pre-check system as resistance factors to people flow. This model obtains the relationship between the bottleneck of the system and the proportion of the three channels. At the same time， the system bottleneck identification is completed and solved the problem of congestion better.
　　【Key words】resistance factor； airport security system； people flow
　　【作者簡介】Cao Yumeng（1998-）， female， School of Mathematics， Jilin University， Han nationality， Fushun， Liaoning， Bachelor， Research direction： Mathematical Statistics.
　　2. The Model
　　2.1 Explore the flow of passengers through a security check point and Identify Bottlenecks
　　First of all， we do not consider the case of pre-check， that is， assuming that the data given in the table is on average. Obviously， the time required for each of these checkpoints is applicable to a variety of situations， that is to say， it is not possible to optimize these times for the purpose of checking adequately. Moreover， we believe that once the passengers into the C zone， then no longer considered that he successfully passed the security device. According to the data， we can easily calculate：
　　The average time to check the file for the first time t1
　　The average time to examine the file for the second timet2
　　Therefore， the corresponding value of f1 should be a coefficient k times t2， f2.
　　is the value of the coefficient multiplied by t2， and so on……
　　We can get， fi=k×ti （1）
　　We have assumed that there is only one device to check the documents， examine the human body， and inspect the baggage. Analysis checks the system for the first person to produce the obstruction capacity. t1=f1 f2 f6 （2）.
　　Suppose an obstruction factor fx such that f6=f3 f4 f5 fx（3） is established. We can understand the meaning of fx as the time cost by passengers wearing shoes， taking off their shoes， and opening the luggage to check the consumption correspond to the resistance we mentioned above. Ideally， fx would be 9.92k for the average passengers， and this value would be small for a precheck passenger， and we could even idealize it as 0. 　　Now we add more than one check file， check the body， check the luggage channel， of course， in real life is the case. We assume that there are x channels for checking files， channels for baggage， and z channels for checking the passenger’s body. So， we can change our model to Ti=f1 f6 f2/X×（i-1）×（t2-Δt） max（f6/Y-f2/X（f’3 fx）/Z-f2/X）×（i-1）×（f’3 fx-Δt）-（i-1）×k×Δt（2）However， the fact is not so simple， because 45% of the passengers will be pre-check qualified. Here we believe that checking the luggage time longer than checking the passenger’s body， we verify this point through the actual situation and the data. The reason why f6>f3 f4 f5is passengers taking off and wearing shoes wasting the time.
　　Therefore， for those pre-check passengers， the wasting time will be greatly reduced， so that in the model we believe that the pre-check passengers after checking the body can be taken directly from the conveyor belt to enter their own Zone C， that is for the pre-check passengers f6=f3 f4 f5. Under this assumption， for pre-check passengers： Ti=f1 f3 f6 f2/X×（i-1）×（t2-Δt） （f’3/Z-f2/X）×（i-1）×（f’3 fx-Δt）-（i-1）×k×Δt（3）.
　　References：
　　[1]LIU laifu，YANG chun. Mark M.Meerschaert（American）（Write）.HUANG haiyang（Translate）Mathematical Modeling（Third Edition） China Machine Press Beijing May，2009.