Future value for Simple interest: 𝐅𝐕 = 𝐏 + 𝐈 = 𝐏(𝟏 + 𝐫𝐭) Power rule:

𝐝
𝐱𝐧 = 𝐧𝐱𝐧−𝟏

/compound linsu
𝐝𝐱
𝐅𝐕
Present value for Compound interest: 𝐏 = 𝐝
(𝟏 + 𝐢)𝐧 it Constant multiple property: (𝐤𝐟(𝐱)) = 𝐤𝐟 ′(𝐱)
𝐝𝐱
(𝟏 + 𝐢)𝐧 − 𝟏
FV = Pxe
Future value of an Annuity: 𝐅𝐕 = 𝐑[ ] Sum and difference property:
𝐝
[𝐟(𝐱) ± 𝐠(𝐱)] = 𝐟 ′(𝐱) ± 𝐠′(𝐱)
𝐢
𝐝𝐱
𝐧
(𝟏 + 𝐢) − 𝟏
𝐅𝐕(𝐥𝐮𝐦𝐩 𝐬𝐮𝐦) = 𝐅𝐕(𝐚𝐧𝐧𝐮𝐢𝐭𝐲) → 𝐏(𝟏 + 𝐢)𝐧 = 𝐑[ ] Total cost = 𝐶(𝑥) → marginal cost = 𝐶′(𝑥)
𝐢
Total revenue = 𝑅(𝑥) → marginal revenue = 𝑅′(𝑥)
Prepaying a loan: Unpaid balance = current value of loan amount – current value of annuity
(𝟏 + 𝒊)𝒎 − 𝟏
Total profit = 𝐏(𝐱) = 𝐑(𝐱) − 𝐂(𝐱) → marginal cost 𝐏′(𝐱) = 𝐑′(𝐱) − 𝐂′(𝐱)]
= 𝑷(𝟏 + 𝒊)𝒎 − 𝑹[ ]
𝒊
__________________________________
__________________________________
𝑑 𝑥
𝐀 𝑒 = 𝑒𝑥
Perpetuities: 𝐏 = 𝑑𝑥
𝐫
salvage investment/ years 𝑑 1
&
earR +
-

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑎𝑛𝑛𝑢𝑎𝑙 𝑟𝑒𝑡𝑢𝑟𝑛 𝑙𝑛 𝑥 =

𝐴𝑅𝑅 = - 𝑑𝑥 𝑥
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡𝑠 investment
𝑑
𝑏 𝑥 = 𝑏 𝑥 𝑙𝑛𝑏 for b > 0, b ≠ 1
- Advantages of ARR: 𝑑𝑥

of 𝑑 1 1
+ This method is easy to calculate time value
ignores money 𝑑𝑥
𝑙𝑜𝑔𝑏 𝑥 = .
𝑙𝑛 𝑏 𝑥
+ It recognizes the profitability factor of investment
𝑑
𝑠𝑖𝑛 𝑥 = 𝑐𝑜𝑠 𝑥
- Disadvantages of ARR: 𝑑𝑥

+ It ignores time value of money. Suppose, if we use ARR to compare two projects having 𝑑
𝑐𝑜𝑠 𝑥 = − 𝑠𝑖𝑛 𝑥
equal initial investments. The project which has higher annual income in the latter years 𝑑𝑥
may rank higher than the one having higher annual income in the beginning years, even if
𝑑
the present value of the income generated by the latter project is higher 𝑓(𝑥)𝑔(𝑥) = 𝑓 ′(𝑥)𝑔(𝑥)
𝑑𝑥
+ It uses accounting income rather than cash flow information. Thus it is suitable for + 𝑓(𝑥)𝑔′(𝑥)
projects having high maintenance costs because their viability also depends on timely cash
inflows. 𝑑 𝑓(𝑥) 𝑓 ′(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥)
=
𝑑𝑥 𝑔(𝑥) 𝑔2 (𝑥)
__________________________________
𝑑
𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐩𝐚𝐲𝐦𝐞𝐧𝐭 𝐁 𝑓[𝑔(𝑥)] = 𝑓 ′[𝑔(𝑥)]𝑔′(𝑥)
𝐏𝐚𝐲𝐛𝐚𝐜𝐤 𝐩𝐞𝐫𝐢𝐨𝐝 = = 𝐀 + 𝑑𝑥
𝐀𝐧𝐧𝐮𝐚𝐥 𝐜𝐚𝐬𝐡 𝐢𝐧𝐟𝐥𝐨𝐰 𝐂
𝑑
[𝑓(𝑥)]𝑛 = 𝑛[𝑓(𝑥)]𝑛−1𝑓 ′(𝑥)
- Advantages of payback period: 𝑑𝑥
of
time value
+ Payback period is very simple to calculate
ignores money __________________________________

+ It can be measure of risk inherent in a project. Since cash flows that occur later in a The relative rate of change of a function 𝑓(𝑥) is
𝑓 ′(𝑥)
𝑓(𝑥)
project’s life are considered more uncertain, payback perios provides an indication of how
certain the project cash inflows are Let the price p and demand x for a product be related by a price-demand equation of the
+ For companies facing liquidity problems, it provides a good ranking of projects that would form x = f(p). Then the elasticity of demand at price p, denoted by E(p), is:
return money early. 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑒𝑚𝑎𝑛𝑑 𝑝𝑓 ′(𝑝)
E(p) = - =
𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑝𝑟𝑖𝑐𝑒 𝑓(𝑝)
- Disadvantages of payback period: It does not take into account the time value of money,
and the cash flows that occur after the payback period. E(p) Demand Interpretation Revenue
<1 Ineslatic Demand is not sensitive to changes in price A price increase will increase
__________________________________ revenue
>1 Elastic Demand is sensitive to changes in price A price increase will decrease
Decision rule: Accept the project only if its NPV is positive or zero. Reject the project revenue
having negative NPV. While comparing two or more exclusive projects having positive =1 Unit A percentage change in price produces the
NPVs, accept the one with highest NPV. same percentage change in demand
As 𝐑(𝐱) = 𝐱𝐩 = 𝐟(𝐩)𝐩, 𝐭𝐡𝐞𝐧 𝐑′(𝐩) = 𝐟(𝐩)[𝟏 − 𝐞 (𝐩)]
𝟏 − (𝟏 + 𝐢)−𝐧
NPV formula when cash inflows are even: 𝐍𝐏𝐕 = 𝐑[ ] − 𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐈𝐧𝐯𝐞𝐬𝐭𝐦𝐞𝐧𝐭
𝐢 𝟏 𝒏
The Number e: 𝒆 = 𝐥𝐢𝐦 (𝟏 + ) = 𝐥𝐢𝐦(𝟏 + 𝒔)𝟏/𝒔
𝒏→∞ 𝒏 𝒔→∞
NPV formula when cash inflows are uneven: 𝐍𝐏𝐕 = [(𝟏 + 𝐢) 𝟏 𝐑𝟏
+
𝐑𝟐
(𝟏 + 𝐢) 𝟐
+. . . ] − 𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐈𝐧𝐯𝐞𝐬𝐭𝐦𝐞𝐧𝐭
Continous compound interest formula: 𝐀 = 𝐏𝐞𝐫𝐭
- Advatage of NPV: Net present value accounts for time value of money. Thus, it is more
reliable than other investment appraisal techniques which do not discount future cash flows, _________________________________________________________________________
such as payback period and accounting rate of return.
𝐛
Total Income for a Continous Income Stream: Total Income = ∫𝐚 𝐟(𝐭)𝐝𝐭
- Disadvantage of NPV: It is based on estimated future cash flows of the project, and
estimates may be far from actual results. 𝐓
Future Value of a Continous Income Stream: FV = ∫𝟎 𝐟(𝐭)𝐞𝐫(𝐓 − 𝐭)𝐝𝐭
__________________________________ 𝐱̅
Consumers’ Surplus: CS = ∫𝟎 [𝐃(𝐱) − 𝐩
̅ ]𝐝𝐭
𝐀
IRR Calculation: 𝐈𝐑𝐑 = 𝐚 + (𝐛 − 𝐚) , where:
𝐀−𝐁 ̅
𝒙
Producers’ Surplus: PS = ∫𝟎 [𝒑
̅ − 𝑺(𝒙)]𝒅𝒕
a: interest rate so that NPV at a is positive, while NPV at b = a + 1% is negative
In a free competitive market, the price of a product is determined by the relationship
A is the NPV at rate a between supply and demand. If 𝐩 = 𝐃(𝐱) and 𝑝 = 𝑆(𝑥) are the price–demand and price–
supply equations, respectively, for a product and if (𝑥, ̅ 𝑝̅)is the point of intersection of these
B is the NPV at rate b = a + 1% equations, then 𝑝̅ is called the equilibrium price and 𝑥̅ is called the equilibrium quantity. If
the price stabilizes at the equilibrium price 𝑝̅ , then this is the price level that will determine
_________________________________________________________________________ both the consumers’ surplus and the producers’ surplus.
Constant function rule: If 𝑦 = 𝑓(𝑥) = 𝐶, then 𝑓 ′ (𝑥) = 0
Thứ tự xét:NPV + IRR > ARR + PP Company Shoes For You’s who is determining whether they should invest in a new project.
Shoes For You’s will expect to invest $500,000 for the development of their new product.
ARR nhớ xét scrap value và depreciation The company estimates that the first year cash flow will be $220,000, the second year cash
flow will be $300,000, and the third year cash flow to be $210,000. The expected return of
NPV, IRR đừng làm tròn, nhớ tính scrap value 15% is used as the discount rate. Is this project acceptable?
Máy cũ -> máy mới: trừ vô initial investment

Scrap value: ko trừ vô initial investment

̅̅̅̅̅̅ = 𝐶(𝑥)
𝑨𝒗𝒈 𝒄𝒐𝒔𝒕 = 𝐶(𝑥) : this is the avg cost to produce x products
𝑥
Company XYZ must decide whether to purchase a piece of factory equipment for $300,000.
𝑴𝒂𝒓𝒈𝒊𝒏𝒂𝒍 𝒂𝒗𝒈 𝒄𝒐𝒔𝒕 = 𝐶(𝑥) ̅̅̅̅̅̅′: after x bits are produced, if we produce 1 more product, The equipment would only last three years, but it is expected to generate $145,000 of
the avg cost is increased/ reduced by … additional annual profit during those years. Company XYZ also thinks it can sell the
equipment for scrap at the end of the third year for about $10,000. The minimum required
rate of return is 15%. Should the company purchase the equipment? Use internal rate of
return (IRR) method for your conclusion.
Max, min

P(x)′ = 0 ,P(x)′′ < 0 => max

P(x)′ = 0 ,P(x)′′ > 0 => min

Rachel owes $7200 to a store. She has agreed to pay the amount in seven months at an
interest rate of 8%. Two months before the loan is due, the store needs $7550 to pay a
wholesaler’s bill and wants to sell Rachel’s note (loan) to a bank. The bank will buy the
note, provided that its return on the investment is 10%. How much will the store receive? Is A city has a wet season and a dry season. A
it enough to pay the bill?
manufacturer in this city produces and sells umbrellas. Based
Rahel has to pay in 7months:
on sales data, the rate at which the umbrellas are sold is
7
FV = 7200 (1 + 0.008 x ) = 7536 𝜋 𝜋
12
𝑓(𝑡) = 750 cos ( 𝑡) + 150 cos ( (𝑡 − 3)) + 1000
6 3
Amount bank pays the store:
7536 where t represents the time, in months, since 1 January 2022. The manufacturer produces
P = 2 = 7412.46 1000 umbrellas per month. When the manufacturer produces more umbrellas than they sell,
1+0.1x
12
the surplus umbrellas are stored in a warehouse. When they produces fewer umbrellas than
they sell, the umbrellas that are stored in the warehouse are sold. At t = 0, the warehouse
contained 5250 umbrellas. During 2022, what was the greatest number of umbrellas stored
An oil spill has fouled 200 miles of Pacific shoreline. The oil company responsible has been in the warehouse at one time?
given 14 days to clean up the shoreline, after which a fine will be levied in the amount of
$10,000/day. The local cleanup crew can scrub five miles of beach per week at a cost of
$500/day. Additional crews, which can also scrub five miles of beach per week, can be
brought in at a cost of $18,000 plus $800/day for each crew. How many additional crews
should be brought in to minimize the total cost to the company?

Smart Manufacturing Company is planning to reduce its labor costs by automating a critical
task that is currently performed manually. The cost to purchase and install a new machine is
$15,000. The installation of machine can reduce annual labor cost by $4,200. The life of the
machine is 15 years. The salvage value of the machine after fifteen years will be zero. The x = C + I = ay + b+ cr + d = b + d = Y(1 -

a) -
c

required rate of return of Smart Manufacturing Company is 25%. Should Smart

Manufacturing Company purchase the machine? Ms = MD = >
Mg* = KeY + 120 +hy
*
Mg -
kg =
Key ther

[iii][] [ii]A x
=

D
+

[]
-

A
xbd b
=
-

=* [ii] [a]
Domain : TXA
Range : tapgiatr !
cofactor :

f(x ,y) 15x2 + 7y2

= Ans =
3x3-7x1 = 2
2
Anz = -(Mx3-7x2)
=

-2
Alz =
441-3x2 =

fx = 30x f((1 , 6)= 30x1 =30 detA anXAestaezXAnz+ 13 xArs

=

nx2 + 1x( 2) = 10

fy
2x2 +

fy(1 2) 28
=

=
14y ,
= 14x2 =

Cramer
=> At (1,2) ,
if we + by 1 ,
862)T by 30 C
A
=
detA
My "
28

Tim die Wi [0] [T

Age (Fath
Be : Tinh
disty dicsdyy facy-fiy
, ,
fac fo

= Sy:minmana
>6

B2 : D ,

= 0 : no info

3x2
,
M
MT : 2 +3 [M Me

At
=
lAl =
ad-ba

singular : no A-1

F=
[r[]
A hospital wants to make a weekly night shift
(12pm-8am)
schedule for its nurses. The demand for nurses for
the night
shift on day j is an integer dj , j= 1, . . . , 7. Every
nurse works
5 days in a row on the night shift. The problem is to
find the
minimal number of nurses the hospital needs to hire.
Formulate the linear programming problem.
Basis feasible solution corresponding to

J= GA' A? ASY can be found by setting Is = clu x = 0

& (c)
=
,

>
-
min
= xe 2
x2 12 9
xy
= = =
, ,

c = (2
,
12 0 , 0, 9
, ,
0) is the basis feasible solution subject to - =

Form the simplex tablue DK0

1 12 -
20 -
> 3 hang f(x)
2 A X De Cly
Min >O
32 Cy <y 26

1 A 2 1 01 10 -

-1 A 120101 01

g A59004213
DK -
10 00 - 1 201

D 1 = (1x1 -

1x0 + 0x0) -
1 = 0

D
n>0, 160 ,
xi = (2 12 0 , 0, 9, 0) is not
, , optimal solution

Only optimal if all DKO

A "enters basis
As max = GDH D63 ,
= 2 = DH ,

DKC0
man
Minh , = exists (positive denominator
Next tablue : A", A", As

A "replaces At
C Axiit isis
M
-
2 A 2 1 O 1 1 O -

I Row 1

-
1 Al 1 -
11 0 o
② 1/
2

O A5 ⑤2 O G g + ⑤ 13

DK 14 -
20 -
3003

New RA = OldR _ /Ra = Old Ra -

New R1/Rz =
Old Rz-2 New R1

A "enters (ASTO)
5/5 :A A replaces A 5
Min =
(10 % 5
1
= exists >
-

12 -
20 -
3
C AX 2) 32 ly Cy <y 26

M
2
-
A 3 315 O
715145 ↓

-
1 Al 8-1151 -

758
-

2/5 O
6
-

3 A 1 -

2 O 2 g + 1
5
DK -

17 0 0 0

Rs Old
New =
Rs/5
Re =
old Re + New R3
Ra = old R2-2 New R3

Smin = -17- >

optimal solution

x* = 10 8 0 3 , 0 1)
, , , ,