物流培训（德国版）

K. . “ 28 2/5 128 2417 000 “ K. he is . . F It F . is by . D as F it is to F D 0 00 0 00 000 00 M B D & V 3,56 1 U .) atl B (.) 78 C 17 C 104 6 S S C 23 C 204 & S D K. . “ 28 2/5 128 2417 000 “ D K. 28 ( 28 K. 28 ( - 28 - A,B ) + ) | | . A, B + + ( = of AIs . ( K. 28 n 2=a K. B = 28 30 r. 30 1 0 K. 28 - K. 28 - I I K. 28 K. 28 K. = : 00 K. . “ 28 2/5 128 2417 000 “ .? +60620:0931+1+1234567 +:96A: 20+ :20060620:102 Y+23436+ +A :1000 :1 +1 +1234567 nd .? +60620:0931+1+ 1234567+:96A:20+: 20060620:102Y+23436+ +A:1000:1+1 +1234567 As a 28 Z 11 a In it a so of 1 200606200730 is It to 203 is 203 of 203 11:200606200730:203 or by :“ of +“. 1:200606200730:203 ,+ 28 00 01 . . . . . . . . K. “ . I 28 K. 12 K. - 1977 - 1988 / 89 of N 1985 K. - in K. 001 / 2002 K. - D 2 - D 2 3 4 5 6 7 8 9 0 1 2 4 8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 0 0 1 1 0 0 20 21 22 23 r. of 7 8 1 2 4 8 1 1 1 1 1 1 r. of 2 4 7 P 6 7 8 2 / 5 1 1 1 1 1 P = 1 K. 2 4 8 2 4 7 P 0 0 1 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 1 0 1 1 0 1 0 0 0 1 1 0 0 0 0 0 1 1 0 1 2 3 4 5 6 7 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 0 0 1 0 1 0 8 9 1 4C 2/5 5 = 32 = 10 2 P = 1 1 0 0 0 0 = = 1 2 K. - D = of 0 ( 2/5 - D K. 1 0 1 0 0 0 0 1 1 0 1 0 1 0 0 0 0 1 1 0 K. 1 4 2 9 0 3 1 3 1 3 (3 + 4 + 6 +9 + 0 = 21 1 10 = 1 10 - 1 = 9 10X 10X (0V+3) + (1V+2) V V 1) V“s 2) = 2V + 7 or in 1V+2 K. 2 / 5 L = V z ( 10X) F = (N(2V+7) + (4V+6) + + (N+1) + 2R = (N(2V+3) + (2) + + (N+1)X + 2R = ( N/2 (4V+6) + (V+6