If you are a student, ask your professor if they can share the Official Instructor’s Solution Manual. McGraw-Hill provides this only to faculty, but some professors make chapter-by-chapter solutions available on the university’s LMS (Moodle, Blackboard, Canvas).
Solutions typically solve for one of three metrics to determine the "best" alternative:
Las primeras versiones de la 4ta edición tenían erratas en el libro de texto. El mejor solucionario incluye notas aclaratorias y corrige esas discrepancias, algo que ni siquiera el apéndice del libro oficial hace.
To effectively use the solutions, one must understand the textbook's organization. The 4th Edition is typically divided into three main parts: If you are a student, ask your professor
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Usa el solucionario para verificar tu respuesta, no para obtenerla. Si tu resultado difiere, revisa tu proceso antes de ver el procedimiento del libro.
Below is a sample solution demonstrating how the Solucionario typically approaches a Chapter 4 or 5 problem. To effectively use the solutions, one must understand
Problem Statement: An engineer proposes a new machine that costs $20,000. It is expected to have a salvage value of $4,000 after 5 years. Operating costs are $3,000 per year. At an interest rate (MARR) of 10% per year, what is the Present Worth (PW) of ownership?
Solution Manual Approach:
Calculation: $$PW = \textInitial Cost + \textPW of Operating Costs - \textPW of Salvage$$ $$PW = -20,000 - 3,000(P/A, 10%, 5) + 4,000(P/F, 10%, 5)$$ Part 2: Evaluation of Projects and Alternatives (Chapters
Look up factors in the 4th Edition tables: $(P/A, 10%, 5) = 3.7908$ $(P/F, 10%, 5) = 0.6209$
$$PW = -20,000 - 3,000(3.7908) + 4,000(0.6209)$$ $$PW = -20,000 - 11,372.40 + 2,483.60$$ $$PW = -28,888.80$$
Conclusion: The project has a negative Present Worth of $-28,888.80. If this were a cost-only project, this value is acceptable for comparison. If it were a revenue project, it would be rejected unless it provided intangible benefits.