### Engineering Management (Solved Numerical Problems)

The following table gives the activities in a construction project and other relevant information:

Indirect costs vary as follows:

(a) Draw an arrow diagram for the project.
(b) Determine the project duration which will result in minimum total project cost.

(AMIE Summer 2021, 15 marks)

The network for normal activity times indicates a project duration of 13 months with critical path: A – D – G (1 – 2 – 4 – 5) as shown in following figure.

The crash cost slope for various critical activities of the project is given in following table.

Among the critical activities A, D and G, the least expensive activity is A. This means that the duration of this activity should be crashed by maximum of one month. Thus, project duration reduces to 12 months and new total project cost is as follows:

Total crashed cost = Total direct cost + Indirect cost due to crashing of activity A by one week + Indirect cost for 12 months = 713 + 1 × 30 + 250 = ₹ 993

Since existing critical path remains same even after crashing critical activity A by one month, therefore next lowest cost slope activity D is crashed by 2 months. Thus, project duration reduces to 10 months as shown in following figure.

In this figure, there are three critical paths: 1 – 2 – 5; 1 – 2 – 4 – 5 and 1 – 4 – 5, each having a duration of 10 months.

The new total project cost becomes:

Total crashed cost = Total direct cost + Indirect cost due to crashing of activity D by 2 months
+ Indirect cost for 10 months = 713 + (30 + 2 × 50) + 100 = ₹ 943

Since the critical activities A and D cannot be further crashed, therefore activities F and G may be
crashed by 2 months each. The revised project network is shown in following figure with total project duration of 8 months.

The new total project cost then becomes:

Total crashed cost = Total direct cost + Increased direct cost due to crashing of activities F and G
+ Indirect cost for 8 months = 713 + (30 + 100 + 2 × 30 + 2 × 70) + 50 = ₹ 1093

Since the total project cost for 8 months is more than the cost for 10 months, therefore, further crashing is not desirable.

Hence, optimum pair of time and cost associated with the project is 10 months and Rs 993, as shown in following table.

An operation manager narrowed the search for a new facility location to four communities. The annual fixed costs (land, property taxes, insurance, equipment, and buildings) and the variable costs (labour, materials, transportation, and variable overhead) are as follows:

(i) If the expected demand is 15,000 units per year, which is option is the best?
(ii) Using break-even analysis, calculate the breakeven quantities over the relevant ranges.

(AMIE Summer 2021, 12 marks)

Solution

To plot a community’s total cost line, let us first compute the total cost for two output levels: Q = 0 and 0 = 20,000 units per year. For the Q = 0 level, the total cost is simply the fixed costs. For the Q = 20,000 level, the total cost (fixed plus variable costs) is as follows:

The following figure shows the graph of the total cost lines. The line for community A goes from (0, 150) to (20, 1,390). The graph indicates that community A is best for low volumes, B for intermediate volumes, and C for high volumes. We should no longer consider community D, because both its fixed and its variable costs are higher than community C’s.
The break-even quantity between A and B lies at the end of the first range, where A is best, and the beginning of the second range, where B is best. We find it by setting both communities’ total cost equations equal to each other and solving:

The break-even quantity between B and C lies at the end of the range over which B is best and the beginning of the final range where C is best. It is,

No other break-even quantities are needed. The break-even point between A and C lies above the shaded area, which does not mark either the start or the end of one of the three relevant ranges.
Management located the new facility at community C, because the 15,000 units-per-year demand forecast lies in the high-volume range. These results can also be used as an input for a final decision using a preference matrix, where other non quantitative factors could also be incorporated into the decision-making process.

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