题目

Problem

The network in Figure 3 shows the activities that need to be undertaken by a company to complete a project. Each activity is represented by an arc. The duration of the activity, in days, is shown in brackets. Each activity requires exactly one worker. The early event times and the late event times are shown at the vertices.

It is given that the total float on activity F is twice the total float on activity D.

It is also given that the total duration of the activities on the path BDFM is 10 days less than the duration of the critical path.

(a) Determine the value of $x$ and the value of $y$ . You must make your method and working clear.

(4)

(b) Draw a cascade chart for this project on Grid 1 in the answer book.

(4)

(c) Use your cascade chart to determine the minimum number of workers needed to complete the project in the shortest possible time. You must make specific reference to time and activities. (You do not need to provide a schedule of the activities.)

(2)

题目中文翻译

图 3 中的网络显示了公司完成项目需要进行的活动。每项活动由弧表示。活动的持续时间（单位：天）显示在括号中。每项活动恰好需要一名工人。最早事件时间和最迟事件时间显示在顶点处。

已知活动 F 的总时差是活动 D 总时差的两倍。

还已知路径 BDFM 上活动的总持续时间比关键路径的持续时间少 10 天。

(a) 确定 $x$ 的值和 $y$ 的值。必须清楚说明方法和计算过程。

(b) 在答案本的网格 1 上画出此项目的级联图。

解答

(a)

思路

活动的总时差等于其终点的最迟事件时间减去起点的最早事件时间，再减去活动持续时间。先利用 F 的总时差是 D 的两倍列出一个方程；关键路径长为 26 天，而路径 BDFM 比关键路径短 10 天，再列出第二个方程，联立即可求出 $x,y$ 。

答题过程

The total floats of activities F and D are

\begin{align*} \operatorname{TF}(F)=&\,22-8-y=14-y,\\ \operatorname{TF}(D)=&\,8-3-x=5-x. \end{align*}

Hence

14-y=2(5-x),

-2x+y=4. \tag{1}

The critical path has duration $26$ days. Since path BDFM is $10$ days shorter,

3+x+y+3=26-10,

giving

x+y=10. \tag{2}

Solving (1) and (2),

\boxed{x=2,\quad y=8}.

(b)

思路

把每项活动画在其最早可开始的时间，并在活动结束后以虚线延伸至最迟结束时间。关键活动 A、G、I、L 没有时差；其余活动的实线与时差区间如下表所示，据此画出级联图。

答题过程

Using $x=2$ and $y=8$ , the activity and float intervals are:

Activity	Activity interval	Float interval
A	$0$ to $8$	none
B	$0$ to $3$	$3$ to $5$
C	$0$ to $4$	$4$ to $6$
D	$3$ to $5$	$5$ to $8$
E	$4$ to $11$	$11$ to $13$
F	$8$ to $16$	$16$ to $22$
G	$8$ to $13$	none
H	$3$ to $11$	$11$ to $13$
I	$13$ to $22$	none
J	$13$ to $21$	$21$ to $23$
K	$13$ to $21$	$21$ to $23$
L	$22$ to $26$	none
M	$22$ to $25$	$25$ to $26$

The corresponding cascade chart is:

(c)

思路

要在 26 天内完成项目，级联图中的活动只能在各自允许的时段内安排。观察第 15 天至第 16 天之间，F、I、J、K 四项活动必定同时进行，因此至少需要四名工人；而级联图也能以四名工人完成全部活动，所以最少人数为 4。

答题过程

For any time $t$ such that

15<t<16,

activities F, I, J and K must all be taking place. Therefore at least four workers are required. The cascade chart shows that the project can be completed in $26$ days using four workers.

Hence the minimum number of workers is

\boxed{4}.

IAL 2022 Jan D1 Q4

题目

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(a)

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(b)

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(c)

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