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财务报表分析 Financial Statement Analysis BMAN20081

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财务报表分析 Financial Statement Analysis BMAN20081

For KDL, the weighted average cost would be
$$
\$ 321,600 / 7,600 \text { units }=\$ 42.3158 \text { per unit }
$$
Cost of goods sold using the weighted average cost method would be
$$
5,600 \text { units at } \$ 42.3158=\$ 236,968
$$
Ending inventory using the weighted average cost method would be
$$
2,000 \text { units at } \$ 42.3158=\$ 84,632
$$

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BMAN20081 COURSE NOTES ：

is sold, it is sold from the top of the stack, so the last lumber in is the first lumber out. Theoretically, a company should choose this method under U.S. GAAP if the physical inventory flows in this manner. ${ }^{23}$ Under the LIFO method, in the KDL example, it would be assumed that the 2,000 units remaining in ending inventory would have come from the first quarter’s purchases: 26
Ending inventory 2,000 units at $\$ 40$ per unit $=\$ 80,000$
The remaining costs would be allocated to cost of goods sold under LIFO:
Total costs of $\$ 321,600$ less $\$ 80,000$ remaining in ending inventory $=\$ 241,600$
Alternatively, the cost of the last 5,600 units purchased is allocated to cost of goods sold under LIFO:
1,900 units at $\$ 45$ per unit $+2,200$ units at $\$ 43$ per unit $+1,500$ units at $\$ 41$ per unit $=\$ 241,600$








投资分析 Investment Analysis BMAN20072

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投资分析 Investment Analysis BMAN20072

Divide both sides by capital
$$
V / C=1+(\mathrm{NPV} / C)
$$
where $V / C=$ Market-to-book ratio of a firm.
But
$$
\mathrm{NPV}=\text { MVA },
$$
$=$ Present value of expected EVA
$=$ EVA/WACC if the EVA stream is assumed to be perpetual
Therefore,
$$
\begin{aligned}
V / C &=1+[(\mathrm{EVA} / \mathrm{WACC}) / C] \
&=1+[\mathrm{ROC}-\mathrm{WACC}] / \text { WACC}
\end{aligned}
$$
That is, the $V / C$ ratio is greater than 1 if the firm has discounted positive EVA.

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BMAN20072 COURSE NOTES ：

FCF is the cash flow available to all investors in the company – both shareholders and bondholders after consideration for taxes, capital expenditure, and working capital investment.
Free cash flow $=$ NOPAT $+$ depreciation $-$ capital expenditure $-(+)$ increases (decreases) in working capital investment,
where
NOPAT $=$ Net Operating Profit after tax $=$ Earnings before Interest but after Taxes, $=\operatorname{EBIT}(1-$ Tax rate $)$,
EBIT $=$ Revenue $-$ cost of goods sold – operating expenses $-$ depreciation.
Estimation of cash flows requires NOPAT, capital expenditure, and networking capital. In calculating NOPAT, interest is not deducted because the discount rate, weighted average cost of capital (WACC), incorporates after-tax cost of debt.








管理会计的基础知识 Fundamentals of Management Accounting BMAN10632

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管理会计的基础知识 Fundamentals of Management Accounting BMAN10632

The first step in the preparation of a flexible budget is the determination of cost behaviour patterns, which means deciding whether costs are fixed, variable or semi-variable.

  • Fixed costs are easy to spot. They remain constant as activity levels change.
  • For non-fixed costs, divide each cost figure by the related activity level. If the cost is a variable cost, the cost per unit will remain constant. If the cost is a semi-variable cost, the unit rate will reduce as activity levels increase.

The second step in the preparation of a flexible budget is to calculate the budget cost allowance for each cost item.
Budget cost allowance $=$ budgeted fixed cost ${ }^{*}+\left(\right.$ number of units $x$ variable cost per unit) ${ }^{\star *}$

  • nil for variable cost
    ** nil for fixed cost
    Semi-variable costs therefore need splitting into their fixed and variable components so that the budget cost allowance can be calculated.
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BMAN10632 COURSE NOTES ：

A fixed budget is a budget which is set for a single activity level.
A flexible budget is a budget which recognises different cost behaviour patterns and is designed to change as volume of activity changes.
Control involves comparing a flexible budget (based on the actual activity level) with actual results. The differences between the flexible budget figures and the actual results are budget variances.

Budgets can be used in a system of reward strategies for managers. They act as targets or benchmarks for managerial performance.








高级数学 Advanced Mathematics ECON10071

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高级数学 Advanced Mathematics ECON10071
问题 1.

Let $D_{n}:=A \backslash E_{n} \downarrow D:=A \backslash E$. We want to show $v\left(D_{n}\right) \downarrow v(D)$. Clearly $v\left(D_{n}\right) \downarrow \alpha$ for some $\alpha \geq v(D)$. If $\alpha>v(D)$, take $0<\varepsilon<\alpha-v(D)$. For each $n=1,2, \ldots$, take a compact $C_{n} \subset D_{n}$ with $\mu^{*}\left(C_{n}\right)>v\left(D_{n}\right)-\varepsilon / 3^{n}$. Let $F_{n}:=\bigcap_{1 \leq j \leq n} C_{j}$. Then for each $n$,
$$
C_{n} \subset F_{n} \cup \bigcup_{1 \leqq i \leq n} C_{n} \backslash C_{i} .
$$

证明 .

Since all compact sets are in the algebra $\mathcal{M}(v)$,
$$
v\left(C_{n}\right) \leq v\left(F_{n}\right)+\sum_{1 \leq i \leq n} v\left(C_{n} \backslash C_{i}\right) .
$$
Since $C_{n} \subset D_{n} \subset D_{i}$ for $i \leq n$
$$
v\left(C_{n}\right) \leq v\left(F_{n}\right)+\sum_{1 \leq i \leq n} \varepsilon / 3^{i}
$$


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ECON10071 COURSE NOTES ：

$$
\int_{-\pi}^{\pi}\left|\left(e^{i(n+1) x}-e^{-i n x}\right) /\left(e^{i x}-1\right)\right| d x \rightarrow \infty
$$
or equivalently, dividing numerator and denominator by $e^{i x / 2}$,
$$
\int_{0}^{\pi}\left|\sin \left(\left(n+\frac{1}{2}\right) x\right)\right| / \sin \left(\frac{x}{2}\right) d x \rightarrow \infty
$$
Now $\sin (n x+x / 2)=\sin (n x) \cos (x / 2)+\cos (n x) \sin (x / 2)$. For $\theta:=x / 2$, we have $0 \leq \theta \leq \pi / 2$, so $\cos ^{2} \theta \leq \cos \theta, 1-2 \cos \theta+\cos ^{2} \theta \leq \sin ^{2} \theta$, and $|1-\cos \theta| \leq \sin \theta$. So it will be enough to show that
$$
\int_{0}^{\pi}|\sin (n x)| / \sin \left(\frac{x}{2}\right) d x \rightarrow \infty
$$








数学概论 Introductory Mathematics ECON10061

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数学概论 Introductory Mathematics ECON10061
问题 1.

An elliptical platform is designed to have a surface area of 725 square feet. To give the desired form, the major axis must be $1 \frac{1}{2}$ times as long as the minor axis. Determine the dimensions of the major and minor axes.
Let $x=$ minor axis
$$
\begin{aligned}
&b=0.5 x \
&a=0.75 x
\end{aligned}
$$
$$
1.5 x=\text { major axis }
$$

证明 .

\begin{aligned}
725 \mathrm{sq} \mathrm{} \mathrm{ft} &=3.1416(0.75 x)(0.5 x) \
725 \mathrm{sq} \mathrm{} \mathrm{ft} &=1.1781 x^{2} \
x^{2} & \approx 615.3977 \mathrm{sq} \mathrm{ft} \
x &=24.807 \mathrm{ft} \
\text { Minor axis } & \approx 24.8 \mathrm{ft} \text { Ans } \
\text { Major axis } & \approx 1.5(24.807 \mathrm{ft})=37.2 \mathrm{ft} \text { Ans }
\end{aligned}


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ECON10061 COURSE NOTES ：

Find the area of the outside circle.
$A_{B}=0.7854 d^{2}$
$A_{B} \Rightarrow(0.7854)(4.00 \mathrm{in})^{2} \approx 12.5664 \mathrm{sq}$ in
Find the area of the hole.
$A_{B} \approx(0.7854)(3.40 \mathrm{in})^{2} \approx 9.079224 \mathrm{sq}$ in
Find the cross-sectional area.
$12.5664 \mathrm{sq}$ in $-9.079224 \mathrm{sq}$ in $\approx 3.487176 \mathrm{sq}$ in
Find the volume.
$V=3.487176 \mathrm{sq}$ in $\times 50.0$ in $\approx 174 \mathrm{cu}$ in Ans








金融基础知识 Fundamentals of Finance BMAN10552/BMAN10621A/BMAN23000

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金融基础知识 Fundamentals of Finance BMAN10552
问题 1.

Sure! Take the case of Blue Skies. Suppose that it has 2 million shares outstanding. It plans to pay a dividend of $\mathrm{DIV}_{1}=\$ 3$ a share. So the total dividend payment is 2 million $\times \$ 3=\$ 6$ million. Investors expect a steady dividend growth of 8 percent a year and require a return of 12 percent. So the total value of Blue Skies is
$$
\mathrm{PV}=\frac{\$ 6 \text { million }}{.12-.08}=\$ 150 \text { million }
$$

证明 .

Alternatively, we could say that the total value of the company is the number of shares times the value per share:
$$
\mathrm{PV}=2 \text { million } \times \$ 75=\$ 150 \text { million }
$$
Of course things are always harder in practice than in principle. Forecasting cash flows and settling on an appropriate discount rate require skill and judgment. As the nearby box shows, there can be plenty of room for disagreement.


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BMAN10552 COURSE NOTES ：

$$
P_{0}=\frac{\mathrm{DIV}{1}+P{1}}{1+r}=\frac{\$ 5+\$ 105}{1.10}=\$ 100
$$
Since dividends and share price grow at 5 percent,







宏观经济学Macroeconomics ECON10241/ECON10262

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宏观经济学1 Macroeconomics 1 ECON10241
问题 1.

Consider the $2 \times 2$ covariance matrix
$$
\Sigma=\left(\begin{array}{cc}
\sigma_{1}^{2} & \rho \sigma_{1} \sigma_{2} \
\rho \sigma_{1} \sigma_{2} & \sigma_{2}^{2}
\end{array}\right)
$$
The Cholesky factor

证明 .

Suppose we posit a factor structure for $x$ :
$$
\left(\begin{array}{l}
x_{1 t} \
x_{2 t}
\end{array}\right)=\left(\begin{array}{cc}
\sigma_{1} & 0 \
\rho \sigma_{2} & \sqrt{1-\rho^{2}} \sigma_{2}
\end{array}\right)\left(\begin{array}{c}
z_{1 t} \
z_{2 t}
\end{array}\right) \text { where }\left(\begin{array}{l}
z_{1 t} \
z_{2 t}
\end{array}\right) \sim N(0, I)
$$


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ECON10241 COURSE NOTES ：

$$
\Delta p_{t}=u_{t}+\left(\lambda+c-c(1-\beta) L-c \beta L^{2}\right) \epsilon_{q t}
$$
Then recall that since $q_{t}=(1+\beta L) \epsilon_{q, t}$,
$$
\epsilon_{q, t}=(1+\beta L)^{-1} q_{t}
$$
Substituting into the $\Delta p_{t}$ equation gives:
$$
\Delta p_{t}=u_{t}+\left(\lambda+c-c(1-\beta) L-c \beta L^{2}\right)(1+\beta L)^{-1} q_{t}
$$







微观经济学 Microeconomics ECON10232/ECON10331

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微观经济学2 Microeconomics 2 ECON10232
问题 1.

Consider farmer $i$ in isolation. Each season she chooses an input level $\left(k_{i /}\right)$ and observes output $\left(q_{i}\right)$. Thus, she can infer $x_{i \rho}$ and she uses this information to update her beliefs about $x^{}$. Suppose that the variance of her beliefs about $x^{}$ in period $t-1$ is $\sigma_{k, t-1}^{2}$. Then after observing $x_{u}$ and applying Bayes’s rule to update her beliefs about $x^{*}$,
$$
\sigma_{K i t}^{2}=\frac{1}{\frac{1}{\sigma_{K i, t-1}^{2}}+\frac{1}{\sigma_{u}^{2}}}
$$

证明 .

where $I_{t-1}$ is the number of trials of the new technology that $i$ has observed on her own farm from period 0 to period $t$ – 1. So we can write expected profits as a function of the number of trials:
$$
\mathrm{Eq}{i t}\left(I{t-1}\right)=1-\frac{1}{\rho_{i 0}+I_{t-1} \rho_{0}}-\sigma_{u}^{2} .
$$
Additional experiments with the new technology increase future expected profits:
$$
\frac{\partial \mathrm{Eq}{t}\left(I{t-1}\right)}{\partial I_{t-1}}=\frac{\rho_{0}}{\left(\rho_{i 0}+I_{t-1} \rho_{0}\right)^{2}} .
$$


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ECON10232 COURSE NOTES ：

In an efficient allocation, the suplus generated by training and technology adoption is maximized
$$
\operatorname{Max}_{\gamma \in{1,0}, \tau} y \alpha \tau-(1+r)(c(\tau)+\delta y)
$$
The first term is the additional output generated by training and adoption of technology, and the second term is their cost (which is incurred in the first period), if $\gamma=0$, the optimal level of training is obviously $\tau=0$. If $\gamma=1$, then the optimal level of training is $t$ such that
$$
\alpha=(1+r) c^{\prime}\left(T^{h}\right)
$$
If $\delta>0$, it is clearly not profitable to innovate unless workers are trained. Suppose in addition that
$$
\alpha T^{h}-(1+r)\left(c\left(T^{h}\right)+\delta\right)>0
$$







双曲几何学 Hyperbolic Geometry MATH32051

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编码理论 Coding Theory MATH32032
问题 1.

If $m$ is either elliptic or loxodromic, we may write $n(z)=p \circ m \circ p^{-1}(z)=\alpha^{2} z$, where $\alpha^{2} \in \mathbb{C}-{0,1}$. When we normalize so that the determinant of $n$ (or equivalently of $m$ ) is 1 , we need to write
$$
n(z)=\frac{\alpha z}{\alpha^{-1}},
$$
and so
$$
\tau(n)=\tau\left(p \circ m \circ p^{-1}\right)=\tau(m)=\left(\alpha+\alpha^{-1}\right)^{2}
$$

证明 .

In the case in which $m$ is elliptic, so that $|\alpha|=1$, write $\alpha=e^{i \theta}$ for some $\theta$ in $(0, \pi)$. Calculating, we see that
$$
\tau(m)=\left(\alpha+\alpha^{-1}\right)^{2}=\left(e^{i \theta}+e^{-i \theta}\right)^{2}=4 \cos ^{2}(\theta)
$$
In particular, we have that $\tau(m)$ is real and lies in the interval $[0,4)$.
In the case in which $m$ is loxodromic, so that $|\alpha| \neq 1$, we write $\alpha=\rho e^{i \theta}$ for some $\rho>0, \rho \neq 1$, and some $\theta$ in $[0, \pi)$. Calculating, we see that
$$
\alpha+\alpha^{-1}=\rho e^{i \theta}+\rho^{-1} e^{-i \theta}
$$


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MATH32051 COURSE NOTES ：

Note that in the latter case, because $C(\overline{\mathbb{R}})=\overline{\mathbb{R}}$, we may instead consider the composition
$$
m \circ C(z)=m(\bar{z})=\frac{a z+b}{c z+d},
$$
and so reduce ourselves to considering just the former case in which
$$
m(z)=\frac{a z+b}{c z+d},
$$
where $a, b, c, d \in \mathbb{C}$ and $a d-b c=1$.







编码理论 Coding Theory MATH32032

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编码理论 Coding Theory MATH32032
问题 1.

A $q$-nary block code $\mathrm{B}(n, k, d)$ with code word length $n$ can be illustrated as a subset of the so-called code space $\mathbb{F}_{q}^{n}$. Such a code space is a graphical illustration of all possible $q$-nary words or vectors. 4 The total number of vectors of length $n$ with weight $w$ and $q$-nary components is given by
$$
\left(\begin{array}{l}
n \
w
\end{array}\right)(q-1)^{w}=\frac{n !}{w !(n-w) !}(q-1)^{w}
$$
with the binomial coefficients
$$
\left(\begin{array}{l}
n \
w
\end{array}\right)=\frac{n !}{w !(n-w) !}
$$

证明 .

The total number of vectors within $\mathbb{F}{q}^{n}$ is then obtained from $$ \left|\mathbb{F}{q}^{n}\right|=\sum_{w=0}^{n}\left(\begin{array}{c}
n \
w
\end{array}\right)(q-1)^{w}=q^{n} .
$$


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MATH32032 COURSE NOTES ：

of choosing $w$ out of $n$ positions, the probability of $w$ errors at arbitrary positions within an $n$-dimensional binary received word follows the binomial distribution
$$
\operatorname{Pr}{w \text { errors }}=\left(\begin{array}{c}
n \
w
\end{array}\right) \varepsilon^{w}(1-\varepsilon)^{n-w}
$$
with mean $n \varepsilon$. Because of the condition $\varepsilon<\frac{1}{2}$, the probability $\operatorname{Pr}{w$ errors $}$ decreases with increasing number of errors $w$, i.e. few errors are more likely than many errors.

The probability of error-free transmission is $\operatorname{Pr}{0$ errors $}=(1-\varepsilon)^{n}$, whereas the probability of a disturbed transmission with $\mathbf{r} \neq \mathbf{b}$ is given by
$$
\operatorname{Pr}{\mathbf{r} \neq \mathbf{b}}=\sum_{w=1}^{n}\left(\begin{array}{l}
n \
w
\end{array}\right) \varepsilon^{w}(1-\varepsilon)^{n-w}=1-(1-\varepsilon)^{n} .
$$