运筹学讲义 =p(Z<△)=1-e=1-[1-4M+o(-△)=uM-0(-△)=△+o(△),M→0; p(A|B2)=p{在时刻t+M时,排队系统内有0个顾客|在时刻t时,排队系统内有0个顾客} =p{在时间[t,t+△门]内,无顾客到达(不可能有顾客接受完服务后离开!)} =p(Y=0) (2△n)-N 1-△t+o(-A△)=1-Mt+o(△n),△t→0 由全概率公式,有 P0(t+△)=p(A)=p(B1)p(AB1)+p(B2)P(A|B2)+p(B3)p(A|B3) =0·AM+P0(1)[1-AM+o(△)+p(1)[A△+o(△) =P0(1)(1-A△)+P1(1)M+[p0()+P1()]o(△r) =P0(D)·(1-A)+P1(1)·At+o(△) 于是 p0(t+△)-p0(1)P0(D)·(1-2△1)+Hp1()·M+o(△)-P0(t) p0()△t+HP1(D)△t+o(△) n po(+up,O O(△) dpo()= lim Po(+A)-Po(0) =lm(-p0()+P()+(4 =lm[-p0()+P1()+m4) -元P0()+4P1()+0=-P0(1)+P(D)(3) dp () 0.n≥1 在顾客流平稳状态时,有 dt p0( 即 0 dt AλPn1()-(2+p)P2()+HPn1()=0,n≥1 (4) Ap0(1)+H(1)=0 令p=二<1,代入(4),得运 筹 学 讲 义 5 = (   ) =1− =1−[1−  + (−  )] =  − (−  ) =  + ( ), → 0 −  p Z t e t o t t o t t o t t t       ； ( | ) { p A B2 = p 在时刻 t + t 时，排队系统内有 0 个顾客|在时刻 t 时，排队系统内有 0 个顾客 } = p{ 在时间 [t,t + t] 内，无顾客到达（不可能有顾客接受完服务后离开！） } 1 ( ) 1 ( ), 0 0! ( ) ( 0) 0 = = −  + −  = −  +   →  = = = −  −  e e t o t t o t t t p Y t t       . 由全概率公式，有 ( ) (1 ) ( ) ( ). ( ) (1 ) ( ) [ ( ) ( )] ( ) 0 ( ) [1 ( )] ( ) [ ( )] ( ) ( ) ( ) ( | ) ( ) ( | ) ( ) ( | ) 0 1 0 1 0 1 0 1 0 1 1 2 2 3 3 p t t p t t o t p t t p t t p t p t o t t p t t o t p t t o t p t t p A p B p A B p B p A B p B p A B =  −  +   +  =  −  +   + +   =   +  −  +  +   +  +  = = + +        于是， . ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (1 ) ( ) ( ) ( ) 0 1 0 1 0 0 0 1 0 t o t p t p t t p t t p t t o t t p t t p t t o t p t t p t t p t   = − + +  −  +  +  =   −  +   +  − =  +  −       ( ) ( ) 0 ( ) ( ). (3) ( ) lim [ ( ) ( )] lim ] ( ) lim [ ( ) ( ) ( ) ( ) lim ( ) 0 1 0 1 0 0 1 0 0 1 0 0 0 0 0 p t p t p t p t t o t p t p t t o t p t p t t p t t p t dt dp t t t t t         = − + + = − +   = − + +   = − + +  +  − =  →  →  →  → 在顾客流平稳状态时，有      = =  0 ( ) 0, 1 ( ) 0 dt dp t n dt dp t n ，即 (4) ( ) ( ) 0 ( ) ( ) ( ) ( ) 0, 1 0 1 1 1    − + = − − + + + =  p t p t pn t pn t pn t n       令 =  1    ，代入（4），得