Discrete Mathematics with Graph Theory (3rd Edition) 314

# Discrete Mathematics with Graph Theory (3rd Edition) 314 -...

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312 Solutions to Exercises 2. (a) [BB] Since the times required for the tasks depend on which other tasks have already been com- pleted and since tasks cannot take place at the same time, this is a Type I scheduling problem. s C{A,14) 2 E(C,17) CD 3@ T(E,19) (b) [BB] The critical path is SACET; that is, pour the mold, polish, calibrate, inspect. The shortest time for the project is the length of this path: 19 units of time. 3. The weight on arc AC becomes 9, changing the label on C to (A, 17). Thus, the label on E changes to (B, 19) and the label at the finish changes to (E, 21). The critical path is now SABET; that is, first pour the mold, then calibrate, then polish and finally inspect. Now 21 time units are required.
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Unformatted text preview: 4. The critical path is SABEGT: Pour the mold, calibrate, have a coffee break, polish and inspect. The shortest time is 23 units. 5. [BB] The new digraph is shown. The project now requires 18 units of time. The critical path is SF EW PaDV CT. The slack in Pl is now 1; the slack in the installation of kitchen cabinets is 3. All other tasks have slack 0 since they lie on the critical path. E(F,9) 1 V(D,16) 2 C(V,18) 6. (a) Since the time for each task depends on the tasks which have already been completed, this is a type I scheduling problem....
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## This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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