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Mohon bantuannya: jika ada mahasiswa dan mahasiswi yang ingin:
1. mendapatkan uang tambahan
2.mencari pengalaman kerja/ portfolio
3.mencari referensi perusahaan4.menyukai tantangan ini ada kesempatan yang menarik ! Perusahaan temanku lagi butuh tenaga untuk pekerjaan ICT,per project basisuntuk pembuatan Website.
Syarat teknis:
1.Mahir PHP & ASP classic
2.Mahir VB for MS Acces
3.punya keahlian di bidang disain grafis lebih disukai
Syarat non teknis:
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Tahapan seleksi
1. Evaluasi CV
2. Wawancara
3. Test
4. Diterima atau tidak diterima Jika ada yang berminat, mahasiswa atau mahasiswi dapat mengirimkan CV via email ke alamat dibawah iniemail :
titus@bali-antique-art.com jika ibu Anna ada ide atau usul lain ( dimasukan dalam penilaian tugas kuliah ?)
saya siap berkoodinasi terima kasih sebelumnya Ibu Annasemoga berkenan
SalamRully Mardjono
Friday, April 24, 2009
Wednesday, April 22, 2009
Engineering Economic Analysis
The Decision Making Process
Decision making may take place by default, this is, without consciously recognizing that an opportunity for decision making exists. This fact leads us to a first element in a definition of decision-making. To have a decision making situation, there must be at least two alternatives available. If only one course of action is available, there can be no decision making, for there is nothing to decide. We would have no alternative but to proceed with single available course of action. (One might argue that it is a rather unusual situation when there are no alternative courses o action. More frequently, alternatives simply are not recognized.)
At this point we might console that the decision-making process consist of choosing from among alternative courses of action. But this is an inadequate definition. Consider the following:
At a horse race,a bettor was uncertain which of the five horses to bet on in the next race. He closed his eyes and pointed his finger at the list of horses printed in the racing program. Upon opening his eyes, he saw that he as pointing to horse number four. He hurried off to place his bet on that horse.
Does the racehorse selection represent the process of decision making? Yes, it clearly was a process of choosing among alternatives (assuming the bettor had already ruled out the “do-nothing” alternative of placing no bet). But the particular method of deciding seems inadequate and irrational. We want to deal with rational decision making.
Rational Decision Making
Rational decision making is a complex process that contains a number of essential elements; although somewhat arbitrary, we define the rational decision-making process in terms of eight steps:
1. Recognition of a problem;
2. Definition of the goal or objective;
3. Assembly of relevant data;
4. Identification of feasible alternatives;
5. Selection of the criterion for judging which is the best alternative;
6. Construction of the interrelationships between the objective, alternatives, data, and the criterion;
7. Prediction of the outcomes or each alternative; and,
8. Choice of the best alternative to achieve the objective.
The following sections will describe these elements in greater detail.
Decision-Process System
We know that decision making cannot begin until the existence of a problem is recognized. But from that point on, there is no fixed path to choosing the best alterntive. Problems seldom can be solved by the sequential approach of Figure 2-1. This is because it is usually difficult, or impossible, to complete one element in the process without considering the effect on other elements in decision making. The gathering of relevant data may suggest feasible alternatives. But it could just as easily be that in identifying feasible alternatives, one will need additional data not yet assembled. This, decision making cannot be seen as an eight-step process that proceeds sequentially from Step 1 to Step 8.
A somewhat better diagram of the decision process is illustrated in Fig. 2-2. This diagram groups the elements in a more flexible, and therefore more realistic, manner. There is no attempt to dictate which comes first-the objective or goal, the feasible alternatives, or the relevant data. In fact,the implication is that once one has recognized the problem, several elements of the decision-making process may be considered concurrently. We mentioned earlier that the eight elements of Figures 2-1 and 2-2 are somewhat arbitrary and artificial, so it is not too surprising that we have difficulty drawing a diagram that properly represents the interrelationship between the elements.
Even Fig. 2-2 seems to siggest that once the relevant data, for example, are determined, that element of the decision process has been concluded. We objected to that concept in the linier relationship in Fig. 2-1,and so we are equally critical of Fig. 2-2 in this respect. The missing aspect of both Figures 2-1 and 2-2 is feedback. No matter where one is in the decision-making process, there will frequently be a need to go back and redo or extend the work on some other element in the process. In other words, one may pass through a particular element several times while in the decision process system. This feedback, where subsequent elements influence previously determined elements, is difficult to show in a diagram, for there would likely be prospective paths from any element to most other elements. To redraw Fig. 2-2 with these additional pathways would useful schematic diagram of thedecision process.
SUMMARY
For rational decision making to take place,there must be an effort to select,by a logical method of analysis,the best alternative from among the feasible alternatives. While difficult to isolate into discrete items,the analysis can be thought of as including eight elements.
1. Recognition of a problem: the realization that a problem exists is the first step in problem solving.
2. Definition of the goal or objective to be accomplished: what is the task?
3. Assembly of relevant data: What are the facts? Do we need to gather additional data? Is the additional information worth at least what it costs us to obtain it?
4. identification of feasible alternatives: what are the practical alternative ways of accomplishing our objective or task?
5. Selection of the criterion for judging the best alternative: what is the single criterion most important to the solution of the problem? There are many possible criteria from which to choose. They may be political, economic, ecological, humanitarian, or whatever. A single criterion may be selected, or it may be a composite of several different criteria.
6. Construction of the various interrelationships: this phase is frequently called mathematical modeling .
7. Prediction of outcomes for each alternative.
8. choice of the best alternative to achieve the odjective.
The decision process system is not a matter of proceeding from the first element to the last one, for there is no mandatory sequence that must be followed. In fact, as one proceeds it is often necessary to go back and re-examine maker is more likely to be the person who performs the analysis than the person who selects the resulting alternative to be adopted.
Decision making may take place by default, this is, without consciously recognizing that an opportunity for decision making exists. This fact leads us to a first element in a definition of decision-making. To have a decision making situation, there must be at least two alternatives available. If only one course of action is available, there can be no decision making, for there is nothing to decide. We would have no alternative but to proceed with single available course of action. (One might argue that it is a rather unusual situation when there are no alternative courses o action. More frequently, alternatives simply are not recognized.)
At this point we might console that the decision-making process consist of choosing from among alternative courses of action. But this is an inadequate definition. Consider the following:
At a horse race,a bettor was uncertain which of the five horses to bet on in the next race. He closed his eyes and pointed his finger at the list of horses printed in the racing program. Upon opening his eyes, he saw that he as pointing to horse number four. He hurried off to place his bet on that horse.
Does the racehorse selection represent the process of decision making? Yes, it clearly was a process of choosing among alternatives (assuming the bettor had already ruled out the “do-nothing” alternative of placing no bet). But the particular method of deciding seems inadequate and irrational. We want to deal with rational decision making.
Rational Decision Making
Rational decision making is a complex process that contains a number of essential elements; although somewhat arbitrary, we define the rational decision-making process in terms of eight steps:
1. Recognition of a problem;
2. Definition of the goal or objective;
3. Assembly of relevant data;
4. Identification of feasible alternatives;
5. Selection of the criterion for judging which is the best alternative;
6. Construction of the interrelationships between the objective, alternatives, data, and the criterion;
7. Prediction of the outcomes or each alternative; and,
8. Choice of the best alternative to achieve the objective.
The following sections will describe these elements in greater detail.
Decision-Process System
We know that decision making cannot begin until the existence of a problem is recognized. But from that point on, there is no fixed path to choosing the best alterntive. Problems seldom can be solved by the sequential approach of Figure 2-1. This is because it is usually difficult, or impossible, to complete one element in the process without considering the effect on other elements in decision making. The gathering of relevant data may suggest feasible alternatives. But it could just as easily be that in identifying feasible alternatives, one will need additional data not yet assembled. This, decision making cannot be seen as an eight-step process that proceeds sequentially from Step 1 to Step 8.
A somewhat better diagram of the decision process is illustrated in Fig. 2-2. This diagram groups the elements in a more flexible, and therefore more realistic, manner. There is no attempt to dictate which comes first-the objective or goal, the feasible alternatives, or the relevant data. In fact,the implication is that once one has recognized the problem, several elements of the decision-making process may be considered concurrently. We mentioned earlier that the eight elements of Figures 2-1 and 2-2 are somewhat arbitrary and artificial, so it is not too surprising that we have difficulty drawing a diagram that properly represents the interrelationship between the elements.
Even Fig. 2-2 seems to siggest that once the relevant data, for example, are determined, that element of the decision process has been concluded. We objected to that concept in the linier relationship in Fig. 2-1,and so we are equally critical of Fig. 2-2 in this respect. The missing aspect of both Figures 2-1 and 2-2 is feedback. No matter where one is in the decision-making process, there will frequently be a need to go back and redo or extend the work on some other element in the process. In other words, one may pass through a particular element several times while in the decision process system. This feedback, where subsequent elements influence previously determined elements, is difficult to show in a diagram, for there would likely be prospective paths from any element to most other elements. To redraw Fig. 2-2 with these additional pathways would useful schematic diagram of thedecision process.
SUMMARY
For rational decision making to take place,there must be an effort to select,by a logical method of analysis,the best alternative from among the feasible alternatives. While difficult to isolate into discrete items,the analysis can be thought of as including eight elements.
1. Recognition of a problem: the realization that a problem exists is the first step in problem solving.
2. Definition of the goal or objective to be accomplished: what is the task?
3. Assembly of relevant data: What are the facts? Do we need to gather additional data? Is the additional information worth at least what it costs us to obtain it?
4. identification of feasible alternatives: what are the practical alternative ways of accomplishing our objective or task?
5. Selection of the criterion for judging the best alternative: what is the single criterion most important to the solution of the problem? There are many possible criteria from which to choose. They may be political, economic, ecological, humanitarian, or whatever. A single criterion may be selected, or it may be a composite of several different criteria.
6. Construction of the various interrelationships: this phase is frequently called mathematical modeling .
7. Prediction of outcomes for each alternative.
8. choice of the best alternative to achieve the odjective.
The decision process system is not a matter of proceeding from the first element to the last one, for there is no mandatory sequence that must be followed. In fact, as one proceeds it is often necessary to go back and re-examine maker is more likely to be the person who performs the analysis than the person who selects the resulting alternative to be adopted.
Tuesday, April 21, 2009
UTS M Gama I
NIM : 06407017
Gama sell key chain. And Gama has an opportunity to receive Rp.400.000 one month from now. If he can earn 10% on his sale in the normal course of events, what is the most he should pay now for this opportunity?
We can know :
FV1 = Rp.400.000, n = 1 and i = 0,1
PV = FVn/(1+i)n
Answer :
PV = Rp.400.000/(1+ 0,1)
= Rp.363.600
The value today (“present value”) of Rp.400.000 received one month from today, given an opportunity cost of 10% is Rp.363.600 .
Gama sell key chain. And Gama has an opportunity to receive Rp.400.000 one month from now. If he can earn 10% on his sale in the normal course of events, what is the most he should pay now for this opportunity?
We can know :
FV1 = Rp.400.000, n = 1 and i = 0,1
PV = FVn/(1+i)n
Answer :
PV = Rp.400.000/(1+ 0,1)
= Rp.363.600
The value today (“present value”) of Rp.400.000 received one month from today, given an opportunity cost of 10% is Rp.363.600 .
UTS Yuli Ecko P
NIM 06407029
Ecko sell hard laptop case. And Ecko has an opportunity to receive Rp.500.000 one month from now. If she can earn 10% on her sale in the normal course of events, what is the most she should pay now for this opportunity?
We can know :
FV1 = Rp.500.000, n = 1 and i = 0,1
PV = FVn/(1+i)
nAnswer :
PV = Rp.500.000/(1+ 0,1)
= Rp.454.500The value today (“present value”) of Rp.500.000 received one month from today, given an opportunity cost of 10% is Rp.454.500 .
Ecko sell hard laptop case. And Ecko has an opportunity to receive Rp.500.000 one month from now. If she can earn 10% on her sale in the normal course of events, what is the most she should pay now for this opportunity?
We can know :
FV1 = Rp.500.000, n = 1 and i = 0,1
PV = FVn/(1+i)
nAnswer :
PV = Rp.500.000/(1+ 0,1)
= Rp.454.500The value today (“present value”) of Rp.500.000 received one month from today, given an opportunity cost of 10% is Rp.454.500 .
UTS Andhika W
NIM : 06407009
Class of 601 sells and promotion product. sandals, bag laptop, t-shirt, and laptop case or hard disk external case.
exam : product bag laptop. bag laptop can be offered with two alternatives payment. which can be total payment and credit payment, If costumer choose The first payment is totally payment with price @Rp 100,000,- or and the another payment is everyweeks must paid with price @ Rp 25000,- in 4 weeks. What the payment will be choose if costumer have a analysis with interest 10%?
Sollution :
~ first alternative : p1 = Rp 100.000,-
~ second alternative with p2 :
p2 = RP 25.000,- (P/A, 10%, 4)
= Rp 25.000,- (6,1446)
= Rp 153.600,-
Because the P value from the first alternatives payment has a big payment then the second payment so the costumer will choose second payment.
Class of 601 sells and promotion product. sandals, bag laptop, t-shirt, and laptop case or hard disk external case.
exam : product bag laptop. bag laptop can be offered with two alternatives payment. which can be total payment and credit payment, If costumer choose The first payment is totally payment with price @Rp 100,000,- or and the another payment is everyweeks must paid with price @ Rp 25000,- in 4 weeks. What the payment will be choose if costumer have a analysis with interest 10%?
Sollution :
~ first alternative : p1 = Rp 100.000,-
~ second alternative with p2 :
p2 = RP 25.000,- (P/A, 10%, 4)
= Rp 25.000,- (6,1446)
= Rp 153.600,-
Because the P value from the first alternatives payment has a big payment then the second payment so the costumer will choose second payment.
UTS Endah Ardini
Name : Endah Ardini
NIM : 06403070
Mata Hati foundation is a foundation that move on the social major for help those “street child”. Mata Hati foundation has a plan to donate a art building especially for the music things with the maintenance cost for a long term. That foundation decide to put their charity money into bank that offer a 12 % of interest a year. Maintenance cost for the building is 4 million a year and every 10 year has to repair with the repairing cost of 20 million each. If the saving money for the maintenance cost are 150 million, what the maximum cost of build the building then they have enough money for the maintenance forever time?
Knowing :
CC = 150 million
i = 12 %
A = 4 million + 20 million (A/F,12%,10)
= 4 million + { 20 million (0,05698)}
= 5,139,600 million
What been asking was the build cost (first investment = P) are ???
CC = P + A/i
P = CC – A/i
= 150 million – ( 5,139,600 / 0,12 )
= 150 million – 42,830 million
= 107,170 million
NIM : 06403070
Mata Hati foundation is a foundation that move on the social major for help those “street child”. Mata Hati foundation has a plan to donate a art building especially for the music things with the maintenance cost for a long term. That foundation decide to put their charity money into bank that offer a 12 % of interest a year. Maintenance cost for the building is 4 million a year and every 10 year has to repair with the repairing cost of 20 million each. If the saving money for the maintenance cost are 150 million, what the maximum cost of build the building then they have enough money for the maintenance forever time?
Knowing :
CC = 150 million
i = 12 %
A = 4 million + 20 million (A/F,12%,10)
= 4 million + { 20 million (0,05698)}
= 5,139,600 million
What been asking was the build cost (first investment = P) are ???
CC = P + A/i
P = CC – A/i
= 150 million – ( 5,139,600 / 0,12 )
= 150 million – 42,830 million
= 107,170 million
UTS Haris Abidin
Name : Haris Abidin
NIM : 06407055
A pump that is driven by a 20HP motor strength is needed to drain water from a tunnel. Number of pump hours operation for 1 year depending on uncertain. Pump unit that is required is estimated to haverain fall a period of 4 years. 2 alternatives considered:
Proposal A: propose to build a power line equipped with the electric motor 1 with a total cost of $ 14,000. Value remaining at the end of the year - 4 = $ 2000. Cost-per-hour flow of $ 8.4 operations; automaticmaintenance cost $ 1,200 per year. No tools required bantu equipment.
Proposal B: proposed to buy a gas motor for $ 5500. This motor does not have any residual value at the end of the year to-4. Costs for fuel and oil per hour = $ 4.2 operation. Maintenance cost of $ 1.5 per hour operation. And the wages paid when the motor runs $ 8 per hour.
Interest rate of 10%
Answer :
Solutions:
Proposal A:
TCA = total annual cost of the proposal ekivalen A
CRA = annual cost recovery ekivalen capital (capital recovery)
= (P - S) (A / P, 10%, 4) + S (0.10)
= ($ 14,000 - 2000) (0.3155) + 2000 (0.10)
= $ 3986
M = annual maintenance cost = $ 1200
C = cost per hour operating flow = $ 8.4
t = number of operating hours per year
TCA = CRA + M + C.t
= $ 3986 + 1200 + 8.4 t
= 5186 + 8.4 t
Proposal B:
TCB = total annual cost of the proposal ekivalen B
CRB fee = annual capital recovery ekivalen = (P - S) (A / P, 10%, 4) + S (0.10)
= ($ 5500) (0.3155)
= $ 1735.25
H = the cost of fuel, maintenance, wages per hour
= $ 4.2 + $ 1.5 + $ 8 = $ 13.7 per hour
t = number of operating hours per year
TCB = CRB + H.t
= $ 1735.25 + 13.7 t
TCA TCB =
5186 + 8.4 t = $ 1735.25 + 13.7 t
t = 651.08 hours ~ 651 hours
TCA = 5186 + 8.4 (651) = $ 10,654.4
NIM : 06407055
A pump that is driven by a 20HP motor strength is needed to drain water from a tunnel. Number of pump hours operation for 1 year depending on uncertain. Pump unit that is required is estimated to haverain fall a period of 4 years. 2 alternatives considered:
Proposal A: propose to build a power line equipped with the electric motor 1 with a total cost of $ 14,000. Value remaining at the end of the year - 4 = $ 2000. Cost-per-hour flow of $ 8.4 operations; automaticmaintenance cost $ 1,200 per year. No tools required bantu equipment.
Proposal B: proposed to buy a gas motor for $ 5500. This motor does not have any residual value at the end of the year to-4. Costs for fuel and oil per hour = $ 4.2 operation. Maintenance cost of $ 1.5 per hour operation. And the wages paid when the motor runs $ 8 per hour.
Interest rate of 10%
Answer :
Solutions:
Proposal A:
TCA = total annual cost of the proposal ekivalen A
CRA = annual cost recovery ekivalen capital (capital recovery)
= (P - S) (A / P, 10%, 4) + S (0.10)
= ($ 14,000 - 2000) (0.3155) + 2000 (0.10)
= $ 3986
M = annual maintenance cost = $ 1200
C = cost per hour operating flow = $ 8.4
t = number of operating hours per year
TCA = CRA + M + C.t
= $ 3986 + 1200 + 8.4 t
= 5186 + 8.4 t
Proposal B:
TCB = total annual cost of the proposal ekivalen B
CRB fee = annual capital recovery ekivalen = (P - S) (A / P, 10%, 4) + S (0.10)
= ($ 5500) (0.3155)
= $ 1735.25
H = the cost of fuel, maintenance, wages per hour
= $ 4.2 + $ 1.5 + $ 8 = $ 13.7 per hour
t = number of operating hours per year
TCB = CRB + H.t
= $ 1735.25 + 13.7 t
TCA TCB =
5186 + 8.4 t = $ 1735.25 + 13.7 t
t = 651.08 hours ~ 651 hours
TCA = 5186 + 8.4 (651) = $ 10,654.4
UTS Nizar Hadi
Name : Nizar Hadi
NIM : 06407047
An investor buy land properties for $100.000, where above it will build an office building that will be rent to other customer. There’s 3 alternatife which can do in the term of build that building :
1. Build a 2 floor office with 2000 m2 width
2. Build a 4 floor office with 3500 m2 width
3. Build a 8 floor office with 8000 m2 width
Cost estimated and profit from that building for over 20 years is :
a) Which alternative that best for the investor ?
b) What is the minnimal rent cost must be decrease from A and C alternative, so the investor will choose B alternative?
c) What is the max build cost from an alternative that choose from option a), so investor consistens with his choise?
Answer :
a) The best alternative for investor are the alternative C
b) The alternative must be decreased because of the average of all the cost is too high.
c) Alternative C :
8 X 1 rent 8000 m2 width
Cost : $ 2.100.000 X 1 = $ 2.100.000
Rent : $ 32 X 8000 = $ 256.000
Value: $ 400.000 X 1 = $ 400.000
TOTAL = $ 2.756.000
NIM : 06407047
An investor buy land properties for $100.000, where above it will build an office building that will be rent to other customer. There’s 3 alternatife which can do in the term of build that building :
1. Build a 2 floor office with 2000 m2 width
2. Build a 4 floor office with 3500 m2 width
3. Build a 8 floor office with 8000 m2 width
Cost estimated and profit from that building for over 20 years is :
a) Which alternative that best for the investor ?
b) What is the minnimal rent cost must be decrease from A and C alternative, so the investor will choose B alternative?
c) What is the max build cost from an alternative that choose from option a), so investor consistens with his choise?
Answer :
a) The best alternative for investor are the alternative C
b) The alternative must be decreased because of the average of all the cost is too high.
c) Alternative C :
8 X 1 rent 8000 m2 width
Cost : $ 2.100.000 X 1 = $ 2.100.000
Rent : $ 32 X 8000 = $ 256.000
Value: $ 400.000 X 1 = $ 400.000
TOTAL = $ 2.756.000
UTS Surya Halim
Name : Surya Halim
Nim : 06507002
A company will decide to have an equipment which can saving an operation cost. Two optional are, each cost $1000 with usage time period of 5 years without unleft. A equipment will hope to saving $300 for annualy. B equipment will save $400 on the first year, decrease $50 every year. On interest of 7%, which equipment to be choose?
Answer :
Equipment A :
PW of cost = $1000
PW of benefit = $300 (P/A,7%,5) = 300(4.100) = $1230
B/C = 1230 / 1000 = 1.23
Equipment B :
PW of cost = $ 1000
PW of benefit = 400(P/A,7%,5) – 50 (P/G,7%,5) = 400(4.100) – 50(7.647) =$1258
B/C = 1258 / 1000 = 1.26
CHOOSE EQUIPMENT B
Machine X:
EUAB = $95
EUAC = 200(A/P,10%,6) – 50(A/F,10%,6) = 200(0.2296) – 50(0.1296) = $40
Machine Y:
EUAB = $120
EUAC = 700(A/P,10%,12) – 150(A/F,10%,12)
= 700(0.1468) – 150(0.0468) = $96
Machine Y – machine X:
ΔB / ΔC = (120 – 95) / (96 – 40) = 25/56 = 0.45 à Pilih machine X
CHOOSE EQUIPMENT A.
Nim : 06507002
A company will decide to have an equipment which can saving an operation cost. Two optional are, each cost $1000 with usage time period of 5 years without unleft. A equipment will hope to saving $300 for annualy. B equipment will save $400 on the first year, decrease $50 every year. On interest of 7%, which equipment to be choose?
Answer :
Equipment A :
PW of cost = $1000
PW of benefit = $300 (P/A,7%,5) = 300(4.100) = $1230
B/C = 1230 / 1000 = 1.23
Equipment B :
PW of cost = $ 1000
PW of benefit = 400(P/A,7%,5) – 50 (P/G,7%,5) = 400(4.100) – 50(7.647) =$1258
B/C = 1258 / 1000 = 1.26
CHOOSE EQUIPMENT B
Machine X:
EUAB = $95
EUAC = 200(A/P,10%,6) – 50(A/F,10%,6) = 200(0.2296) – 50(0.1296) = $40
Machine Y:
EUAB = $120
EUAC = 700(A/P,10%,12) – 150(A/F,10%,12)
= 700(0.1468) – 150(0.0468) = $96
Machine Y – machine X:
ΔB / ΔC = (120 – 95) / (96 – 40) = 25/56 = 0.45 à Pilih machine X
CHOOSE EQUIPMENT A.
UTS Fadilah Oki
Name : Fadilah Oki
NIM : 065.07.007
An automatic essembly machine can bought for 20 milion with left of 2 million. This machine can estimate to afford income of 2 million a year.
If the company have an estimate of that machine was 15 years. Did the company should buy that machine?
(use the pay back periode)
a. With assumption of buy interest = 0
Answer :
a. If the interest buying method was = 0, so with module (4.24) will have :
0 = -20 million + N' (2000) + 2000
20000 - 2000 = N' (2000)
18000 = N' (2000)
N' = 18000 / 2000
N' = 9 year
NIM : 065.07.007
An automatic essembly machine can bought for 20 milion with left of 2 million. This machine can estimate to afford income of 2 million a year.
If the company have an estimate of that machine was 15 years. Did the company should buy that machine?
(use the pay back periode)
a. With assumption of buy interest = 0
Answer :
a. If the interest buying method was = 0, so with module (4.24) will have :
0 = -20 million + N' (2000) + 2000
20000 - 2000 = N' (2000)
18000 = N' (2000)
N' = 18000 / 2000
N' = 9 year
UTS Raden Dani Januardi
Name : Raden Dani Januardi
NIM : 065.07.005
What is the value of S P Rp 10,000,000 if j12 = 12% for:
a. 5 years
b. 25 years
Answer:
a. P = Rp 10,000,000
i = 12% / 12 = 1% = 0.01
n = 5 years x 12 months = 60
S = P (1 + i) ⁿ
= Rp 10,000,000 (1 +0.01) 60
= Rp 18,166,967
b. P = Rp 10,000,000
i = 1% = 0.01
n = 25 years x 12 months = 300
S = P (1 + i) n
= Rp 10,000,000 (1 +0.01) 300
Rp = 197884662.6
Description:
Compound interest calculation
S = P (1 + i) n with i = jm / m
With
P = initial principal value (principal)
S = value end
n = Number of interest calculation period
m = number of times a year in interest rates, namely 2 to semesteran, 4 to
dst quarter.
Jm = nominal annual interest rate calculation to the period m times per year.
i = interest rate interest rates periodically
NIM : 065.07.005
What is the value of S P Rp 10,000,000 if j12 = 12% for:
a. 5 years
b. 25 years
Answer:
a. P = Rp 10,000,000
i = 12% / 12 = 1% = 0.01
n = 5 years x 12 months = 60
S = P (1 + i) ⁿ
= Rp 10,000,000 (1 +0.01) 60
= Rp 18,166,967
b. P = Rp 10,000,000
i = 1% = 0.01
n = 25 years x 12 months = 300
S = P (1 + i) n
= Rp 10,000,000 (1 +0.01) 300
Rp = 197884662.6
Description:
Compound interest calculation
S = P (1 + i) n with i = jm / m
With
P = initial principal value (principal)
S = value end
n = Number of interest calculation period
m = number of times a year in interest rates, namely 2 to semesteran, 4 to
dst quarter.
Jm = nominal annual interest rate calculation to the period m times per year.
i = interest rate interest rates periodically
UTS Sinarino Wicaksono
Name : Sinarino Wicaksono
NIM : 064.07.026
A businessman sell wallet. And the businessman has an opportunity to receive $300 two year from now. If he can earn 6% on his sale in the normal course of events, what is the most she should pay now for this opportunity?
We can know :
FV1 = $300, n = 2 and i = 0.06
PV = FVn/( 1 + i )^n
Answer :
PV = $300/(( 2 + 0.06 ))^2
= $70.7
The value today (“present value”) of $300 received two year from today, given an opportunity cost of 5% is $70.7
NIM : 064.07.026
A businessman sell wallet. And the businessman has an opportunity to receive $300 two year from now. If he can earn 6% on his sale in the normal course of events, what is the most she should pay now for this opportunity?
We can know :
FV1 = $300, n = 2 and i = 0.06
PV = FVn/( 1 + i )^n
Answer :
PV = $300/(( 2 + 0.06 ))^2
= $70.7
The value today (“present value”) of $300 received two year from today, given an opportunity cost of 5% is $70.7
UTS Cynthia Dewi
Name : Cynthia Dewi
NIM : 064.07.019
A Bussinesman makes a product like keychain. The keychain has offered by Manufacture Company, with two alternatives payment. The first payment is totally payment with price @Rp, 10.000.000,- and the another payment is every year must paid with price @Rp, 2.000.000,- in 6 years. The first and the oddment @Rp, 1.000.000,- in 7 years. What the payment will be choose, if the Bussinessman have a analisis with interest 15% ?
Sollution :
1). First alternative : P1 = Rp 10.000.000,-
2). Second alternative :
P2 = Rp 2 million (P/A,15%, 6) + Rp 1 million ((P/A, 15%, 13) – (P/A, 15%, 6))
= Rp 2 million (3,785) + Rp 1 million (5,583 – 3,785)
= Rp 7,57 million + Rp 1,80 million
= Rp 9,370,000,-
- The Bussinessman will choose the first payment, because the first payment has a big payment than the second payment
NIM : 064.07.019
A Bussinesman makes a product like keychain. The keychain has offered by Manufacture Company, with two alternatives payment. The first payment is totally payment with price @Rp, 10.000.000,- and the another payment is every year must paid with price @Rp, 2.000.000,- in 6 years. The first and the oddment @Rp, 1.000.000,- in 7 years. What the payment will be choose, if the Bussinessman have a analisis with interest 15% ?
Sollution :
1). First alternative : P1 = Rp 10.000.000,-
2). Second alternative :
P2 = Rp 2 million (P/A,15%, 6) + Rp 1 million ((P/A, 15%, 13) – (P/A, 15%, 6))
= Rp 2 million (3,785) + Rp 1 million (5,583 – 3,785)
= Rp 7,57 million + Rp 1,80 million
= Rp 9,370,000,-
- The Bussinessman will choose the first payment, because the first payment has a big payment than the second payment
UTS Oktarina Maharani
Name : Oktarina Maharani
NIM : 064.07.007
Class F 601 sells a product like sandals. The sandals can be offered with two alternatives payment. Which can be total payment and credit payment, If costumer choose The first payment is totally payment with price @Rp 25,000,- or and the another payment is everyday must paid with price @Rp 3000,- in 10 days. What the payment will be choose if costumer have a analysis with interest 10%?
Sollution :
~ First alternative : P1 = Rp 25.000,-
~ Second Alternative with P2 :
P2 = Rp 3000 ( P/A, 10%, 10 )
= Rp 3000 ( 6,1446)
= Rp 18500
Because the P value from the first alternatives payment has a big payment then the second payment so the costumer will choose second payment.
NIM : 064.07.007
Class F 601 sells a product like sandals. The sandals can be offered with two alternatives payment. Which can be total payment and credit payment, If costumer choose The first payment is totally payment with price @Rp 25,000,- or and the another payment is everyday must paid with price @Rp 3000,- in 10 days. What the payment will be choose if costumer have a analysis with interest 10%?
Sollution :
~ First alternative : P1 = Rp 25.000,-
~ Second Alternative with P2 :
P2 = Rp 3000 ( P/A, 10%, 10 )
= Rp 3000 ( 6,1446)
= Rp 18500
Because the P value from the first alternatives payment has a big payment then the second payment so the costumer will choose second payment.
UTS Dinda Maharani A. Loebis
Name : Dinda Maharani A. Loebis
NIM : 064.07.041
1. Rani sell hard disk external case. And Rani has an opportunity to receive $200 one year from now. If she can earn 5% on her sale in the normal course of events, what is the most she should pay now for this opportunity?
We can know :
FV1 = $200, n = 1 and i = 0.05
PV = FVn/( 1 + i )^n
Answer :
PV = $200/(( 1 + 0.05 ))
= $190.47
The value today (“present value”) of $200 received one year from today, given an opportunity cost of 5% is $190.47.
2. Rani sells a product like hard disk external case. The hard disk external case has offered with two alternatives payment. The first payment is totally payment with price Rp 50.000,- and the another payment is everyday must paid with price Rp 10.000,- in 7 days. What the payment will be choose if Rani have a analysis with interest 10%?
Sollution :
~ First alternative : P1 = Rp 50.000,-
~ Second Alternative :
P2 = Rp 10.000,- ( P/A, 10%, 7 )
= Rp 10.000,- ( 4,8684)
= Rp 48684
Rani will choose the first payment, because the first payment has a big payment then the second payment.
NIM : 064.07.041
1. Rani sell hard disk external case. And Rani has an opportunity to receive $200 one year from now. If she can earn 5% on her sale in the normal course of events, what is the most she should pay now for this opportunity?
We can know :
FV1 = $200, n = 1 and i = 0.05
PV = FVn/( 1 + i )^n
Answer :
PV = $200/(( 1 + 0.05 ))
= $190.47
The value today (“present value”) of $200 received one year from today, given an opportunity cost of 5% is $190.47.
2. Rani sells a product like hard disk external case. The hard disk external case has offered with two alternatives payment. The first payment is totally payment with price Rp 50.000,- and the another payment is everyday must paid with price Rp 10.000,- in 7 days. What the payment will be choose if Rani have a analysis with interest 10%?
Sollution :
~ First alternative : P1 = Rp 50.000,-
~ Second Alternative :
P2 = Rp 10.000,- ( P/A, 10%, 7 )
= Rp 10.000,- ( 4,8684)
= Rp 48684
Rani will choose the first payment, because the first payment has a big payment then the second payment.
UTS Dwi Indah Fitriyanti
Name : Dwi Indah Fitriyanti
NIM : 064.07.038
An enterpreuner sells a product like t-shirt. The t-shirt can be offered with two alternatives payment. Which can be total payment and credit payment,If you choose The first payment is totally payment with price @Rp 50.000,- and the another payment is everyday must paid with price @Rp 13.000,- in 5 days. What the payment will be choose if you have a analysis with interest 5%?
Sollution :
~ First alternative : P1 = Rp 50.000,-
~ Second Alternative :
P2 = Rp 13.000 ( P/A, 5%, 5 )
= Rp 13.000 ( 4,3295)
= Rp 56.300
Because the second payment has a big payment then the first payment so the enterpreuner will choose second payment.
NIM : 064.07.038
An enterpreuner sells a product like t-shirt. The t-shirt can be offered with two alternatives payment. Which can be total payment and credit payment,If you choose The first payment is totally payment with price @Rp 50.000,- and the another payment is everyday must paid with price @Rp 13.000,- in 5 days. What the payment will be choose if you have a analysis with interest 5%?
Sollution :
~ First alternative : P1 = Rp 50.000,-
~ Second Alternative :
P2 = Rp 13.000 ( P/A, 5%, 5 )
= Rp 13.000 ( 4,3295)
= Rp 56.300
Because the second payment has a big payment then the first payment so the enterpreuner will choose second payment.
UTS Alqhadafi
My name is Alqhadafi
My NIM are 06407033
The question are...
An web developer has found a bug on one of URL on the web that he assume he can fix it to make more better 25 % on that URL to go much better than it was. The URL company offer the web developer to try fix it with 2 way of payment, it is pay all of it now for 100 million or pay every year 20 million for the first 7 years and the rest 6 million for the next 8 years. Which payment that the web developer choose if he analize it with 15 % of interest ?
Answer :
*> The first method was P1 = 100 million
*> The second method was,,
P2 = 20 million (P/A, 15%, 7) + 6 million (P/A, 15%, 8) (P/F, 15%, 7)
= 20 million (4,160) + 6 million (4,487) (0,3759)
= 83,20 million + 10,120 million
= 93,320 million
So, because the first alternative was bigger then the second one,, the web developer choose the first one.
My NIM are 06407033
The question are...
An web developer has found a bug on one of URL on the web that he assume he can fix it to make more better 25 % on that URL to go much better than it was. The URL company offer the web developer to try fix it with 2 way of payment, it is pay all of it now for 100 million or pay every year 20 million for the first 7 years and the rest 6 million for the next 8 years. Which payment that the web developer choose if he analize it with 15 % of interest ?
Answer :
*> The first method was P1 = 100 million
*> The second method was,,
P2 = 20 million (P/A, 15%, 7) + 6 million (P/A, 15%, 8) (P/F, 15%, 7)
= 20 million (4,160) + 6 million (4,487) (0,3759)
= 83,20 million + 10,120 million
= 93,320 million
So, because the first alternative was bigger then the second one,, the web developer choose the first one.
UTS Fenno Steginga Assignment
Name : Fenno Steginga
NIM : 064.06.036
Lecture : Mam Anna Tohir
This is my mid test assignment.
Question,,
A concrete aggregate mix is required to contain at least 31% sand by volume for proper batching. One source of material, which has 25% sand and 75% coarse aggregate, sells for $3 per cubic meter. Another source, which has 40% sand and 60% coarse aggregate, sells for $4.40 per cubic meter. Determine the least cost per cubic meter of blended aggregates.
Answer :
The least cost of blended aggregates will result from maximum use of the lower cost material. The higher cost material will be used to increase the proportion of sand up to the minimum level (31%) specified.
Let x = portion of blended aggregates from $3.00/m cubic source
1 – x = portion of blended aggregates from $4.00/m cubic source
Sand balance :
X(0.25) + (1 – x)(0.40) = 0.31
0.25x + 0.40 – 0.40x = 0.31
X = 0.31 – 0.40 = -0.09
0.25 – 0.40 -0.15
= 0.60
Thus the blended aggregates will contain :
60% of $3.00/m cubic material
40% of $4.40/m cubic material
The least cost per cubic yard of blended aggregates :
= 0.60($3.00) + 0.40($4.40) = 1.80 + 1.76
= $3.56/m cubic
NIM : 064.06.036
Lecture : Mam Anna Tohir
This is my mid test assignment.
Question,,
A concrete aggregate mix is required to contain at least 31% sand by volume for proper batching. One source of material, which has 25% sand and 75% coarse aggregate, sells for $3 per cubic meter. Another source, which has 40% sand and 60% coarse aggregate, sells for $4.40 per cubic meter. Determine the least cost per cubic meter of blended aggregates.
Answer :
The least cost of blended aggregates will result from maximum use of the lower cost material. The higher cost material will be used to increase the proportion of sand up to the minimum level (31%) specified.
Let x = portion of blended aggregates from $3.00/m cubic source
1 – x = portion of blended aggregates from $4.00/m cubic source
Sand balance :
X(0.25) + (1 – x)(0.40) = 0.31
0.25x + 0.40 – 0.40x = 0.31
X = 0.31 – 0.40 = -0.09
0.25 – 0.40 -0.15
= 0.60
Thus the blended aggregates will contain :
60% of $3.00/m cubic material
40% of $4.40/m cubic material
The least cost per cubic yard of blended aggregates :
= 0.60($3.00) + 0.40($4.40) = 1.80 + 1.76
= $3.56/m cubic
Announcement
Dear Students,
There are 2 things that I have to announce:
1. Please, attending the class tomorrow (April 22, 2009) for midtest at 10.00 AM in room AE 101B
2. There will be an opportunity for exchange students with one of Taiwan's University, please take a look here
If there are any question and unclear statement, please posting the comment.
Regards,
Anna
There are 2 things that I have to announce:
1. Please, attending the class tomorrow (April 22, 2009) for midtest at 10.00 AM in room AE 101B
2. There will be an opportunity for exchange students with one of Taiwan's University, please take a look here
If there are any question and unclear statement, please posting the comment.
Regards,
Anna
Sunday, April 19, 2009
UTS 601
A man sell hard disk external case. And The Man has an opportunity to receive $300 two year from now. If she can earn 6% on her sale in the normal course of events, what is the most she should pay now for this opportunity?
We can know :
FV2 = $300, n = 2 and i = 0.06
PV = FVn/( 1 + i )^n
Answer :
PV = $300/〖( 1 + 0.06 )〗^2
= $266.99
The value today (“present value”) of $300 received two year from today, given an opportunity cost of 6% is $266.99.
We can know :
FV2 = $300, n = 2 and i = 0.06
PV = FVn/( 1 + i )^n
Answer :
PV = $300/〖( 1 + 0.06 )〗^2
= $266.99
The value today (“present value”) of $300 received two year from today, given an opportunity cost of 6% is $266.99.
Thursday, April 16, 2009
Class 601 Activities
We study a lot from our lecture, mam Anna Tohir, Mr. Ahmad Zuhdi, and all the guest lecture ( like, Mr. Wirawan, and Mr. Ruli), that gave us their precious time to come and teach us on our "moving" class.
:-) :-)
This is some of our pic.
Tuesday, April 7, 2009
About Class 601
Class of 601 consist of active and bright students from Information Engineering Faculty who join Information Economic subject, they are Fenno Steginga as Head of the class, Fadilah Oky, Haris Abidin, Sandy Kurniawan, Cynthia Dewi, Dwi Indah, Oktarina Maharani, Dinda Maharani A Loebis, Nizar Hadi, Alqhadafi, Yuli Eko Purwanto, Adhitya Setiadi, Dany, M Gama Iffahindra, Andhika Wijanarko, Sinarino Wicaksono, Surya Halim.
Information Economic subject is a part of Micro Economic, a studies how information affects an economy and economic decission. The aim of this subject for the students, they are willing to include this topic in the curriculum to be implemented as an actual and real, the study shows that a deep learning strategy to be implemented both in economic or information side. The teaching method by using active learning courage and dig potential know-how of each students. Team work is the key to solve the study cases from the exam were given. Visiting study and visiting Professional Person in Industrial Engineering and Information also provided to rich the knowledge for students.
We hope the students can absorb, grab and implement the Information Economic knowledge with the aim to be use for them in the future.
Regards,
Anna Tohir, BSc, M.Eng, DBA (cand)
Lecture
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