International Guidance and Controls

International Guidance and Controls

Let S be the cost of project lateness including direct and indirect costs.  According to the case the direct cost of lateness is $0.8m per month of lateness, and the indirect cost of lateness is the cost of lost reputation.  As an example, if the project is two months late and the cost of lost reputation is estimated to be $2m, then S is $3.6m.

Let:

        SW Only = software only;

        HW Now = expand hardware now;

        Delay HW = delay hardware decision;

        OT = project finished on time;

        Late = project finished late;

        FP = favorable software progress in the first five months;

        NP = learning nothing new in the first five months; and

        UP = unfavorable software progress in the first five months.

Consider the decision tree shown in Figure 1 where the nodes have been numbered for easy reference.  The decision choices at the initial decision node (D1) are: SW Only, Delay HW, and HW Now.  Do we need to consider all three decision options at D1?  Note that there is no explicit cost to delaying the hardware expansion decision.  So you will be right to argue that the SW Only decision is dominated by the Delay HW decision, and thus can be disregarded.  Nevertheless, I shall, if only for pedagogical reasons, keep the SW Only option alive.  You might also argue that in reality delaying the hardware decision may impede the software progress (Parkinson’s Law: Work expands to fill the available time.) and thus may very well have some costs.  

Note that in the present problem the objective is to minimize expected cost.  In the prose below, I am going to omit the $ sign.  

At C1, the expected cost is 3 + 0.2S.  

At C3, the expected cost is 3 + 0.1S.

At C4, the expected cost is 3 + 0.2S.

At C5, the expected cost is 3 + 0.6S.

What is the optimal choice at D2?  That depends on which of the two choices at D2 has the lower expected cost, which in turn depends on the value of S.  The two expected costs are 3.75 and 3 + 0.1S.  We equate these two expected costs (3.75 = 3 + 0.1S) to find that for S ≤ 7.5 the right decision would be to stick with SW Only.  Similarly, at D3, we equate the two expected costs (3.75 = 3 + 0.2S) to find that for S ≤ 3.75 the right decision would be to stick with SW Only.  Finally, at D4, we equate the two expected costs (3.75 = 3 + 0.6S) to find that for S ≤ 1.25 the right decision would be to stick with SW Only.