生产运作管理作业第九章:
7.⑴ W1=max{n,N+「2n/5」}=max{7,11+「2×7/5」}=14
周末休息(W1-n)=(14-7)=7(人),余下7人,1—(W1-N)=1—(14-11)=1-3号周一休息,紧接着的(W1-N)=(14-11)=3人周二休息,即3-5号周二休息。具体安排如下表:
工人号 |
周一 |
周二 |
周三 |
周四 |
周五 |
周六 |
周日 |
周一 |
周二 |
周三 |
周四 |
周五 |
周六 |
周日 |
1 |
× |
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× |
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× |
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× |
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2 |
× |
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× |
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× |
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3 |
× |
× |
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× |
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× |
× |
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4 |
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× |
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× |
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× |
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5 |
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× |
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6 |
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7 |
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8 |
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× |
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× |
× |
9 |
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× |
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× |
× |
10 |
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× |
× |
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× |
× |
11 |
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× |
× |
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× |
× |
12 |
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× |
× |
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× |
× |
13 |
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× |
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× |
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14 |
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× |
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× |
× |
⑵ W2=max{n,N+「2n/5」,「(2N+2n)/5」}=max{7,14,12}=14
K==max{0,2N+n-2W2}= max{0,1}=1
1号周五周六休息,2号周日周一休息,接下来6个工人周六周日休息。具体安排如下表:
工人号 |
周一 |
周二 |
周三 |
周四 |
周五 |
周六 |
周日 |
周一 |
周二 |
周三 |
周四 |
周五 |
周六 |
周日 |
1 |
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× |
× |
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× |
× |
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2 |
× |
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× |
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× |
3 |
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× |
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× |
× |
4 |
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× |
× |
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× |
× |
5 |
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× |
× |
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× |
× |
6 |
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× |
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× |
× |
7 |
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× |
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× |
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8 |
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× |
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× |
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9 |
× |
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× |
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10 |
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× |
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× |
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11 |
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× |
× |
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12 |
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× |
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× |
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13 |
× |
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× |
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14 |
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× |
× |
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× |
× |
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⑶ W3=max{2n,N+「2,2n/5」}= 14,(W3-2n)=(14-2×7)=0(人)在周末休息,余下的14人分为A,B两组,每组7人,A,B两组工人分别在第一周周末,第二周周末休息。安排A组第二周休息日,由于5W3﹥5N+2n,所以1-(W3-N)=1-3号周五休息。具体安排如下表:
工人号 |
周一 |
周二 |
周三 |
周四 |
周五 |
周六 |
周日 |
周一 |
周二 |
周三 |
周四 |
周五 |
周六 |
周日 |
1 |
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× |
× |
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× |
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× |
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2 |
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× |
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3 |
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5 |
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6 |
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7 |
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8 |
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9 |
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10 |
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× |
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11 |
× |
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× |
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12 |
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× |
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13 |
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× |
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14 |
× |
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× |
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(4)W4= max{2n,N+「2,2n/5」,「(4N+4n)/5」}=15,将W4名工人分为AB两组,A组「W4/2」=8名工人,安排在第一周周末休息,B组则为7名工人,安排在第二周周末休息,k= max{0,4N+2n-4W4}=0,所以在保证平时有11人上班,周末有7人上班的前提下,具体的安排如下表:
工人号 |
周一 |
周二 |
周三 |
周四 |
周五 |
周六 |
周日 |
周一 |
周二 |
周三 |
周四 |
周五 |
周六 |
周日 |
1 |
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× |
× |
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× |
× |
2 |
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7 |
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8 |
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9 |
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× |
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10 |
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11 |
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12 |
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13 |
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× |
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14 |
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15 |
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