1er Parcial Muestra-erika
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UNIVERSIDAD DE MONTERREY DIVISION DE ARQUITECTURA, DISEÑO E INGENIERIA DEPARTAMENTO DE INGENIERIA
PRIMER PARCIAL MUESTRA
INGENIERÍA ECONÓMICA MAESTRO :JUAN SILLERO PÉREZ
ERIKA LOZANO CHAVEZ 58519
SAN PEDRO, GARZA GARCÍA A 13 DE FEBRERO DEL 2006
EJEMPLOS DE REACTIVOS PARA EL PRIMER EXAMEN PARCIAL PROBLEM(2A2). A company produces and sells a consumer product and thus far has been able to control the volume of the product by varying the selling price. The company is seeking to maximize its net profit. It has been concluded that the relationship between price and demand, per month, is approximately D = 500 − 5 p, where p is the price per unit in dollars. The fixed cost is $1,000 per month, and the variable cost is $20 per unit. Obtain the answer, to the following questions: a. What is the optimal number of units that should be produced and sold per month? D= 500 –5p p= (D500 )/5 UT =((D500 )/5)D – (CF+CV) UT =D2/5 + 80D – 1000 dUT = 2D/5 + 80 D= 80/(2/5) = 200 b.
What is the maximum profit per month?
UT =(200)2/5 + 80(200) – 1000 UT = 7,000 c. What are the breakeven sales quantities (range of profitable demand volume)? D= 80 ±√ (80 2 –4(1/5)(1000)) 2(1/5) D1= 12.91 D2= 387.08 PROBLEM( 2A3). a.) Compare the probable part cost from Machine A and Machine B, assuming that each will make the part to the same specification. Which machine yields the lowest part cost? Assume that the interest rate is negligible. b.) If the cost of labor can be cut in half by using parttime employees, which machine should be recommended? Machine A Machine B
Initial capital investment Life Market (salvage) value Parts required per year Labor cost per hour Time to make one part Maintenance cost per year a) Tiempo de producción Costo de op 1 pieza Mantenimiento /pieza Inversión maq /pieza CT
$35,000 10 years $3,500 10,000 $16 20 minutes $1,000
Maq A 20 min (1/3)(16)= $5.33 $0.01 $0.315 $5.745
$150,000 8 years $15,000 10,000 $20 10 minutes $3,000 Maq B 20 min (1/6)(20)= $3.33 $0.3 $1.6875 $5.32
a) es mejor la máquina B ya que tiene un costo de $5.32 b) Tiempo de producción Costo de op 1 pieza Mantenimiento /pieza Inversión maq /pieza CT
Maq A 20 min (1/3)(8)= $2.66 $0.01 $0.315 $3.08
Maq B 20 min (1/6)(10)= $1.66 $0.3 $1.6875 $3.65
a) es mejor la máquina A, teniendo medio turno ya que tiene un costo de $3.08 PROBLEM (Question 2) A large work area can be illuminated satisfactorily by either one of two methods. By using incandescent lighting, the initial cost is low ($10,000) but the operating costs are relatively high ($12.50 per hour). The other lighting system utilizes fluorescent lamps having a high initial cost $25,000) but relatively low operating costs ($4.25 per hour). Only one year (8,760 hours) is to be used in deciding between these two types of lighting. Furthermore, the number of hours of operation during the year (H) must be considered in making the decision. (a) What value of H makes the costs of both types of lighting the same? 10,000 + 12.50 (H) = 25,000 + 4.25(H) 10,00025,000=4.25(H) – 12.50(H) 15,000= 8.25(H) H= 1818.18
(b) Which lighting system would you recommend and why?
Costo inicial Costo de operación
Incandescent $10,000 12.50/h
Fluorescent $25,000 4.25 /h
Incandescent $10,000 + $12.50(8,760 hours) = $119.500 Fluorescent $25,000 + 4.25 (8,760 hours) = $62,230 Es mejor el sistema de Fluorescente ya que tiene un costo de $62,230
PROBLEM (Question 3) In the design of an automobile radiator, an engineer has a choice of using either a brass copper alloy casting or a plastic molding. Either material provides the same service. However, the brasscopper alloy casting weighs 25 pounds, compared with 16 pounds for the plastic molding. Every pound of extra weight in the automobile has been assigned a penalty of $4 to account for increased fuel consumption during the life cycle of the car. The brasscopper alloy casting costs $3.35 per pound while the plastic molding costs $7.40 per pound (no machining is required for plastic). Machining costs per casting are $6.00 for the brasscopper alloy. Which material should the engineer select, and what is the difference in unit costs?
Peso Peso estra Costo
B 25 pounds $4 $3.35 /pound
C 16 pounds $4 $7.40 /pound
Costo de máquina Costo Extra CT
$6 $83.75 $100 $189.75
0 $118.4 $64 $182.4
Es mejor el radiador de plástico $182.4 Hay diferencia de $7.35 más económico PROBLEM (Question 4 ) You plan to provide a retirement fund for yourself by making endofyear deposits of $1,000 for each of the next 35 years into a fund which earns 5% interest (i = 5% per year). After 35 years you will withdraw $5,082 per year until the fund is exhausted. (a) Draw a cash flow diagram from your perspective (viewpoint). A = 5,082 F
35 P= 1,000
N =? P=90,320.3
F=A(F/A, .05, 35)=90,320.3 (b) How long (to the nearest year) will it take to exhaust the fund that you have built up for 35 years?
P = A(P/A,.15,N) 90,320 = 5,082(P/A,.15,N) 90,320/ 5,082 = 17.77 busco 17.77 en las tables con un 5% P/A N= 45 años aprox.
(c) List at least 2 key assumptions that are necessary in your answer to (b). La tasa no cambia
PROBLEM
a) P= G(P/G, i, N) P= G(P/G, 0.15, 7) ( 1.15)1 P = G(8.862) b) P=z –z(1.12)1–z(1.12)2–z(1.12)3–z(1.12)4 = 1,000 (1.12)2–5,000(1.12)6 z(1+ (1.12)1+(1.12)2+(1.12)3+(1.12)4) = 1735.96 z= 1735.96/4.03 Z = 429.97
PROBLEM.
P1 = 1,000 P2
P2= A(P/A, .12,7) P2= 800(4.5638)(1.12)1 P2= 3259.85
A = 800
0 1
2
3 4
5
6
7
8
1800 1600 1400 1200 1000 800
0 1
2
3 4
5
6
7
P3= G(P/G, .12,6) P3= 200(8.930)(1.12)1 P3= 1594.64
P1+ P2+ P3 = 3854.49 P= Z/2(1.12)2+z(1.12)3+z(1.12)4+z(1.12)5+ 2000(1.12)6 P= z(2.31)+ 1013.26 z(2.31)+ 1013.26 = 3854.49 z= 2841.22/ 2.31 z= 1229.96
PROBLEM Based in the following CF diagrams, calculate the unknown values (?) frown the table below using the corresponding factors .
P
0
F
A/year
1
2
3
4
5
n-1
n
i = % annual
Inciso F P a) $18,000 b) $22,000 c) ? d) $10,500 e) ? Note: Extrapolate if necessary a)18,000/3,000 = 6 Busco 6 en la tabla P/A con i =6% N= 8 años aprox. b) A=P(A/P, 8%, 5) = 22,000(0.2505) = 5511 c) P= A(P/A, 5%, 8) = 5,000(6.4632) = 32,316 d)10,500/500 = 21 busco 21 en cada tabla de F/A con n = 10 i = 15.5% aprox e)F=A((1.09)81)/0.09) = 55,142.36
A $3000 ? $5000 $500 $5000
i 6% 8% 5% ? 9%
n ? 5 8 10 8
PROBLEM Hoy puedes comprar una máquina con valor de $100,000 que genera ahorros anuales en la operación de la planta. Si tienes planeado mantener la máquina por cuatro años y sabes que el valor de salvamento en ese año será sólo el 40% de su costo. ¿De cuánto deben ser los ahorros anuales que genere para que la inversión se justifique? (Asumir un 15% de intereses sobre la inversión ). (Elaborar diagrama de flujo y mostrar los cálculos realizados) A =?
N =4 P=100,000
P = F (P/F, .15,4) = 40,000(1.15)4 = 22,870.12 100,000 22,870.12= 77,129.87 A = P(A/P, .15, 4) = 77,129.87(0.3503, .15, 4)= 27,018.59 Se deberá de ahorrar 27,018.59 por año
PROBLEM Suponiendo que tienes un tío rico que desea repartirlo entre sus sobrinos $2,000,000 en una cantidad de $200,000 cada año. Si deposita esa cantidad en un banco que le da el 7% anual, a) ¿Cuántos años le durará esa cantidad ? 2000000/200000 = 10 busco 10 en P/A con i=7% N = 18 años aprox b) Si consigue un 9% en vez del 6%. ¿Cuánto le durará? 2000000/200000 = 10 busco 10 en P/A con i=9% N = 25 años aprox (Elaborar diagrama de flujo en cada caso) A =200,000
N =? P=2000,000
PRIMER PARCIAL MUESTRA
INGENIERÍA ECONÓMICA MAESTRO :JUAN SILLERO PÉREZ
ERIKA LOZANO CHAVEZ 58519
SAN PEDRO, GARZA GARCÍA A 13 DE FEBRERO DEL 2006
EJEMPLOS DE REACTIVOS PARA EL PRIMER EXAMEN PARCIAL PROBLEM(2A2). A company produces and sells a consumer product and thus far has been able to control the volume of the product by varying the selling price. The company is seeking to maximize its net profit. It has been concluded that the relationship between price and demand, per month, is approximately D = 500 − 5 p, where p is the price per unit in dollars. The fixed cost is $1,000 per month, and the variable cost is $20 per unit. Obtain the answer, to the following questions: a. What is the optimal number of units that should be produced and sold per month? D= 500 –5p p= (D500 )/5 UT =((D500 )/5)D – (CF+CV) UT =D2/5 + 80D – 1000 dUT = 2D/5 + 80 D= 80/(2/5) = 200 b.
What is the maximum profit per month?
UT =(200)2/5 + 80(200) – 1000 UT = 7,000 c. What are the breakeven sales quantities (range of profitable demand volume)? D= 80 ±√ (80 2 –4(1/5)(1000)) 2(1/5) D1= 12.91 D2= 387.08 PROBLEM( 2A3). a.) Compare the probable part cost from Machine A and Machine B, assuming that each will make the part to the same specification. Which machine yields the lowest part cost? Assume that the interest rate is negligible. b.) If the cost of labor can be cut in half by using parttime employees, which machine should be recommended? Machine A Machine B
Initial capital investment Life Market (salvage) value Parts required per year Labor cost per hour Time to make one part Maintenance cost per year a) Tiempo de producción Costo de op 1 pieza Mantenimiento /pieza Inversión maq /pieza CT
$35,000 10 years $3,500 10,000 $16 20 minutes $1,000
Maq A 20 min (1/3)(16)= $5.33 $0.01 $0.315 $5.745
$150,000 8 years $15,000 10,000 $20 10 minutes $3,000 Maq B 20 min (1/6)(20)= $3.33 $0.3 $1.6875 $5.32
a) es mejor la máquina B ya que tiene un costo de $5.32 b) Tiempo de producción Costo de op 1 pieza Mantenimiento /pieza Inversión maq /pieza CT
Maq A 20 min (1/3)(8)= $2.66 $0.01 $0.315 $3.08
Maq B 20 min (1/6)(10)= $1.66 $0.3 $1.6875 $3.65
a) es mejor la máquina A, teniendo medio turno ya que tiene un costo de $3.08 PROBLEM (Question 2) A large work area can be illuminated satisfactorily by either one of two methods. By using incandescent lighting, the initial cost is low ($10,000) but the operating costs are relatively high ($12.50 per hour). The other lighting system utilizes fluorescent lamps having a high initial cost $25,000) but relatively low operating costs ($4.25 per hour). Only one year (8,760 hours) is to be used in deciding between these two types of lighting. Furthermore, the number of hours of operation during the year (H) must be considered in making the decision. (a) What value of H makes the costs of both types of lighting the same? 10,000 + 12.50 (H) = 25,000 + 4.25(H) 10,00025,000=4.25(H) – 12.50(H) 15,000= 8.25(H) H= 1818.18
(b) Which lighting system would you recommend and why?
Costo inicial Costo de operación
Incandescent $10,000 12.50/h
Fluorescent $25,000 4.25 /h
Incandescent $10,000 + $12.50(8,760 hours) = $119.500 Fluorescent $25,000 + 4.25 (8,760 hours) = $62,230 Es mejor el sistema de Fluorescente ya que tiene un costo de $62,230
PROBLEM (Question 3) In the design of an automobile radiator, an engineer has a choice of using either a brass copper alloy casting or a plastic molding. Either material provides the same service. However, the brasscopper alloy casting weighs 25 pounds, compared with 16 pounds for the plastic molding. Every pound of extra weight in the automobile has been assigned a penalty of $4 to account for increased fuel consumption during the life cycle of the car. The brasscopper alloy casting costs $3.35 per pound while the plastic molding costs $7.40 per pound (no machining is required for plastic). Machining costs per casting are $6.00 for the brasscopper alloy. Which material should the engineer select, and what is the difference in unit costs?
Peso Peso estra Costo
B 25 pounds $4 $3.35 /pound
C 16 pounds $4 $7.40 /pound
Costo de máquina Costo Extra CT
$6 $83.75 $100 $189.75
0 $118.4 $64 $182.4
Es mejor el radiador de plástico $182.4 Hay diferencia de $7.35 más económico PROBLEM (Question 4 ) You plan to provide a retirement fund for yourself by making endofyear deposits of $1,000 for each of the next 35 years into a fund which earns 5% interest (i = 5% per year). After 35 years you will withdraw $5,082 per year until the fund is exhausted. (a) Draw a cash flow diagram from your perspective (viewpoint). A = 5,082 F
35 P= 1,000
N =? P=90,320.3
F=A(F/A, .05, 35)=90,320.3 (b) How long (to the nearest year) will it take to exhaust the fund that you have built up for 35 years?
P = A(P/A,.15,N) 90,320 = 5,082(P/A,.15,N) 90,320/ 5,082 = 17.77 busco 17.77 en las tables con un 5% P/A N= 45 años aprox.
(c) List at least 2 key assumptions that are necessary in your answer to (b). La tasa no cambia
PROBLEM
a) P= G(P/G, i, N) P= G(P/G, 0.15, 7) ( 1.15)1 P = G(8.862) b) P=z –z(1.12)1–z(1.12)2–z(1.12)3–z(1.12)4 = 1,000 (1.12)2–5,000(1.12)6 z(1+ (1.12)1+(1.12)2+(1.12)3+(1.12)4) = 1735.96 z= 1735.96/4.03 Z = 429.97
PROBLEM.
P1 = 1,000 P2
P2= A(P/A, .12,7) P2= 800(4.5638)(1.12)1 P2= 3259.85
A = 800
0 1
2
3 4
5
6
7
8
1800 1600 1400 1200 1000 800
0 1
2
3 4
5
6
7
P3= G(P/G, .12,6) P3= 200(8.930)(1.12)1 P3= 1594.64
P1+ P2+ P3 = 3854.49 P= Z/2(1.12)2+z(1.12)3+z(1.12)4+z(1.12)5+ 2000(1.12)6 P= z(2.31)+ 1013.26 z(2.31)+ 1013.26 = 3854.49 z= 2841.22/ 2.31 z= 1229.96
PROBLEM Based in the following CF diagrams, calculate the unknown values (?) frown the table below using the corresponding factors .
P
0
F
A/year
1
2
3
4
5
n-1
n
i = % annual
Inciso F P a) $18,000 b) $22,000 c) ? d) $10,500 e) ? Note: Extrapolate if necessary a)18,000/3,000 = 6 Busco 6 en la tabla P/A con i =6% N= 8 años aprox. b) A=P(A/P, 8%, 5) = 22,000(0.2505) = 5511 c) P= A(P/A, 5%, 8) = 5,000(6.4632) = 32,316 d)10,500/500 = 21 busco 21 en cada tabla de F/A con n = 10 i = 15.5% aprox e)F=A((1.09)81)/0.09) = 55,142.36
A $3000 ? $5000 $500 $5000
i 6% 8% 5% ? 9%
n ? 5 8 10 8
PROBLEM Hoy puedes comprar una máquina con valor de $100,000 que genera ahorros anuales en la operación de la planta. Si tienes planeado mantener la máquina por cuatro años y sabes que el valor de salvamento en ese año será sólo el 40% de su costo. ¿De cuánto deben ser los ahorros anuales que genere para que la inversión se justifique? (Asumir un 15% de intereses sobre la inversión ). (Elaborar diagrama de flujo y mostrar los cálculos realizados) A =?
N =4 P=100,000
P = F (P/F, .15,4) = 40,000(1.15)4 = 22,870.12 100,000 22,870.12= 77,129.87 A = P(A/P, .15, 4) = 77,129.87(0.3503, .15, 4)= 27,018.59 Se deberá de ahorrar 27,018.59 por año
PROBLEM Suponiendo que tienes un tío rico que desea repartirlo entre sus sobrinos $2,000,000 en una cantidad de $200,000 cada año. Si deposita esa cantidad en un banco que le da el 7% anual, a) ¿Cuántos años le durará esa cantidad ? 2000000/200000 = 10 busco 10 en P/A con i=7% N = 18 años aprox b) Si consigue un 9% en vez del 6%. ¿Cuánto le durará? 2000000/200000 = 10 busco 10 en P/A con i=9% N = 25 años aprox (Elaborar diagrama de flujo en cada caso) A =200,000
N =? P=2000,000
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