Choose 2 problems from the following list: Page 385: #1 a, #1 c Page 404: #2 a Page 413: #3 Page 420: #1 Page 428: #3 Page 440: #2 Page 454: #3

Choose 2 problems from the following list: Page 469: #3 Page 478: #6 Page 481: #1 Page 522: #21 and #22

Chapter 10: Game theory

10.2 Total Conflict as a Linear Program Model: Pure and Mix strategy

Page 404: #2 a

For problems a-g build a linear programming model for each player’s decisions and solve it both geometrically and algebraically. Assume the row player is maximizing his payoffs which are shown in the matrices below

a <- data.frame('name'=rep('Rose',2),'strategy'=c('R1','R2'),'C1'=c(10,5),'C2'=c(10,0))
kable(a, "html", align = "c") %>%
  kable_styling(full_width = F) %>%
  column_spec(1, bold = T) %>%
  collapse_rows(columns = 1) %>% 
  add_header_above(c(" "," ", "Colin" = 2))

		Colin
name	strategy	C1	C2
Rose	R1	10	10
Rose	R2	5	0

Solution

Let

S: average payoffs

x: Portion of the time to play R1 strategy

1- x: Portion of the time to play R2 strategy

If Colin plays a pure C1 strategy, the expected value is

EV(C1) = 10x+5(1-x)

Similarly, if Colin plays C2 strategy, the expected value is

EV(C2) = 10x

The objective expression:

maximize S

Constraints

S 10x+5(1-x)

S 10x

0 x 1

These inequations could be rewritten as follows:

S - 5x <= 5

S - 10x <=0

x <= 1

x >= 0

The constraints coule be write as:

\[ \begin{equation} \begin{bmatrix} 1 & -5 \\ 1 & -10 \end{bmatrix} \begin{bmatrix} S \\ x \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \end{bmatrix} \end{equation} \] let: \[ \begin{equation} {f} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{equation} \] and \[ \begin{equation} {x} = \begin{pmatrix} S \\ x \end{pmatrix} \end{equation} \]

The objective function:$max \ f^Tx$

The constraints: Ax <= c

suppressMessages(suppressWarnings(library(linprog)))  

# form c vector (f in object fucction)
cvec <- c(1, 0)

# form b vector (c in constraints expression)
bvec <- c(5,0) 

# form matrix A
A <- matrix(c(1,1,-5,-10),nrow=2,byrow=F)

res <- solveLP(cvec,bvec,A,maximum=TRUE)

## Warning in solveLP(cvec, bvec, A, maximum = TRUE): Simplex algorithm (phase
## 2) did not reach optimum.

res

## 
## 
## Results of Linear Programming / Linear Optimization
## 
## Objective function (Maximum): 10 
## 
## Iterations in phase 1: 0
## Iterations in phase 2: 1000 (equals 'maxiter' !!!)
## Solution
##   opt
## 1  10
## 2   1
## 
## Basic Variables
##   opt
## 1  10
## 2   1
## 
## Constraints
##   actual dir bvec free dual dual.reg
## 1      5  <=    5    0    2        5
## 2      0  <=    0    0   -1      Inf
## 
## All Variables (including slack variables)
##     opt cvec min.c max.c marg marg.reg
## 1    10    1     0   Inf   NA       NA
## 2     1    0    -5   Inf   NA       NA
## S 1   0    0  -Inf     2   -2        5
## S 2   0    0  -Inf    -1    1      Inf
## Simplex algorithm phase 2 did not find the optimal solution within the number of iterations specified by argument 'maxiter'

The maximum payoffs for the row player is 10.

Or we can solive the linear programming by the interior point method and graphically

suppressMessages(suppressWarnings(library(intpoint)))
t <- 1
c <- t(cvec)
bm = c(5,0)
m = A
interior_point(t, c, bm , m, e = 1e-04, a1 = 1, a2 = 0.97)

## $`The optimum value of Z is`
## [1] 10
## 
## $`The optimal solution X`
##      [,1]
## [1,]   10
## [2,]    1
## 
## $`Number of iterations`
## [1] 5

** The result is the same: the maximum payoffs for the row player is 10.**

10.3 Desition Theory Revisits: Game Theory Against Nature

Page 413: #3

We are considering three alternatives A, B, or C or a mix of the three alternatives under uncertain conditions of the economy. The payoff matrix is as follows:

tbl.2<- data.frame('Alternative'=c('A','B','C'),'No.1'=c(3000,1000,4500),'No.2'=c(4500,9000,4000),'No.3'=c(6000,2000,3500))
kable(tbl.2, "html", align = "c") %>%
  kable_styling(full_width = F) %>%
  column_spec(1, bold = T) %>%
  collapse_rows(columns = 1) %>% 
  add_header_above(c(" ", "Condistions" = 3))

	Condistions
Alternative	No.1	No.2	No.3
A	3000	4500	6000
B	1000	9000	2000
C	4500	4000	3500

Set up and solve both the investor’s and the economy’s game.

Solution

V: Net profit in hundreds of thousands of dollars x1: Portion of the time that the investor should play A X2: Portion of the time that the investor should play B 1-x1-x2: Portion of the time that the investor should play C

objective function:

Maximize V

constraints:

$V <= 3000 \times x1+4500\times X2+6000\times (1-X1-X2) Economy condition A$ $V <= 1000 \times x1+9000\times X2+2000\times (1-X1-X2) Economy condition B$ $V <= 4500 \time sx1+4000\times X2+3500\times (1-X1-X2) Economy condition C$ $x1 \leqslant 1$ $x1 \geqslant 0$ $x2 \leqslant 1$ $x2 \geqslant 0$

These inequations could be rewritten as follows:

S - 5x <= 5

S - 10x <=0

x <= 1

x >= 0

The constraints coule be write as:

\[ \begin{equation} \begin{bmatrix} 1 & 3000 & 1500 \\ 1 & -1000 & 7000 \\ 1 & 1000 & 500 \end{bmatrix} \begin{bmatrix} V \\ x1 \\ x2 \end{bmatrix} = \begin{bmatrix} 6000 \\ 2000 \\ 3500 \end{bmatrix} \end{equation} \] let: \[ \begin{equation} {f} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \end{equation} \] and \[ \begin{equation} {x} = \begin{pmatrix} V \\ x1 \\ x2 \end{pmatrix} \end{equation} \]

The objective function:$max \ f^Tx$

The constraints: Ax <= c

#suppressMessages(suppressWarnings(library(linprog)))  

# form c vector (f in object fucction)
cvec <- c(1, 0,0)

# form b vector (c in constraints expression)
bvec <- c(6000,2000,3500) 

# form matrix A
A <- matrix(c(1,1,1,3000,1000,-1000,1500,-7000,-500),nrow=3,byrow=F)

res <- solveLP(cvec,bvec,A,maximum=TRUE)
res

## 
## 
## Results of Linear Programming / Linear Optimization
## 
## Objective function (Maximum): 4125 
## 
## Iterations in phase 1: 0
## Iterations in phase 2: 3
## Solution
##           opt
## 1 4125.000000
## 2    0.441667
## 3    0.366667
## 
## Basic Variables
##           opt
## 1 4125.000000
## 2    0.441667
## 3    0.366667
## 
## Constraints
##   actual dir bvec free        dual dual.reg
## 1   6000  <= 6000    0 2.50000e-01  2038.46
## 2   2000  <= 2000    0 5.55112e-17  6625.00
## 3   3500  <= 3500    0 7.50000e-01  5500.00
## 
## All Variables (including slack variables)
##             opt cvec min.c   max.c  marg marg.reg
## 1   4125.000000    1     0     Inf    NA       NA
## 2      0.441667    0     0 2647.06    NA       NA
## 3      0.366667    0 -3750    0.00    NA       NA
## S 1    0.000000    0  -Inf    0.25 -0.25  2038.46
## S 2    0.000000    0  -Inf    0.00  0.00  6625.00
## S 3    0.000000    0  -Inf    0.75 -0.75  5500.00

The geometric solution:

suppressMessages(suppressWarnings(library(intpoint)))
t <- 1
c <- t(cvec)
bm = c(6000,2000,3500)
m = A
interior_point(t, c, bm , m, e = 1e-04, a1 = 1, a2 = 0.97)

## $`The optimum value of Z is`
## [1] 4125
## 
## $`The optimal solution X`
##              [,1]
## [1,] 4125.0000000
## [2,]    0.4416667
## [3,]    0.3666667
## 
## $`Number of iterations`
## [1] 14

The solution as x1=0.44,x2=0.36,V=$4125, giving the strategy for the investor as 44% A, 36% B and 20% C. Thus, if the investor plays a mixed strategy of 44% A, 36% B and 20% C its net profit will be $4125 regardless of whether the economy is in condition 1,2 or 3 or any mix of the condistions. At this point, theinvestor has a conservative strategy thatguarantees $4125.

Economy’s game:

V: Net profit in hundreds of thousands of dollars y1: Portion of the time that the economy in condition No.1 y2: Portion of the time that the economy in condition No.2 1-x1-x2: Portion of the time that the economy in condition No.3

objective function:

Maximize V

constraints:

$V <= 3000 \times x1+1000\times X2+4500\times (1-X1-X2) Economy condition A$ $V <= 4500 \times x1+9000\times X2+4000\times (1-X1-X2) Economy condition B$ $V <= 6000 \time sx1+2000\times X2+3500\times (1-X1-X2) Economy condition C$ $y1 \leqslant 1$ $y1 \geqslant 0$ $y2 \leqslant 1$ $y2 \geqslant 0$

These inequations could be rewritten as follows:

S - 5x <= 5

S - 10x <=0

x <= 1

x >= 0

The constraints coule be write as:

\[ \begin{equation} \begin{bmatrix} 1 & 1500 & 3500 \\ 1 & -500 & -5000 \\ 1 & -2500 & 1500 \end{bmatrix} \begin{bmatrix} V \\ x1 \\ x2 \end{bmatrix} = \begin{bmatrix} 4500 \\ 4000 \\ 3500 \end{bmatrix} \end{equation} \] let: \[ \begin{equation} {f} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \end{equation} \] and \[ \begin{equation} {x} = \begin{pmatrix} v \\ x1 \\ x2 \end{pmatrix} \end{equation} \]

The objective function:$max \ f^Tx$

The constraints: Ax <= c

#suppressMessages(suppressWarnings(library(linprog)))  

# form c vector (f in object fucction)
cvec <- c(1, 0,0)

# form b vector (c in constraints expression)
bvec <- c(4500,4000,3500) 

# form matrix A
A <- matrix(c(1,1,1,1500,-500,-2500,3500,-5000,1500),nrow=3,byrow=F)

res <- solveLP(cvec,bvec,A,maximum=TRUE)
res

## 
## 
## Results of Linear Programming / Linear Optimization
## 
## Objective function (Maximum): 4125 
## 
## Iterations in phase 1: 0
## Iterations in phase 2: 2
## Solution
##       opt
## 1 4125.00
## 2    0.25
## 3    0.00
## 
## Basic Variables
##         opt
## 1   4125.00
## 2      0.25
## S 2    0.00
## 
## Constraints
##   actual dir bvec free  dual dual.reg
## 1   4500  <= 4500    0 0.625     1000
## 2   4000  <= 4000    0 0.000       NA
## 3   3500  <= 3500    0 0.375    11000
## 
## All Variables (including slack variables)
##         opt cvec min.c       max.c      marg marg.reg
## 1   4125.00    1     0         Inf        NA       NA
## 2      0.25    0 -2500 1500.000000        NA       NA
## 3      0.00    0  -Inf 2750.000000 -2750.000      0.5
## S 1    0.00    0  -Inf    0.625000    -0.625   1000.0
## S 2    0.00    0    NA    0.366667     0.000       NA
## S 3    0.00    0  -Inf    0.375000    -0.375  11000.0

The geometric solution:

suppressMessages(suppressWarnings(library(intpoint)))
t <- 0
c <- t(cvec)
bm = c(4500,4000,3500)
m = A
interior_point(t, c, bm , m, e = 1e-04, a1 = 1, a2 = 0.97)

## $`The optimum value of Z is`
## [1] 0
## 
## $`The optimal solution X`
##         [,1]
## [1,] 4125.00
## [2,]    0.25
## [3,]    0.00
## 
## $`Number of iterations`
## [1] 14

Thr solution is economy plays 25% condition No.1, 0% condition No.2 and 75% condition No.3. To minimize profit is $4125. No matter in what economy condition, could limit the investor’s profit to $4125.

Chapter 11

11.1 Population Growth

Page 469: #3

The following data were obtained for the growth of a sheep population introduced into a new environment on the island of Tasmania (adapted from J. Davidson, ``On the Growth of the Sheep Population in Tasmania,’’ Trans. R. Soc. S. Australia 62(1938): 342-346).

t..year.	1814	1824	1834	1844	1854	1864
P.t.	125	275	830	1200	1750	1650

Make an estimate of M by graphing P(t).
Plot lnP=[P/(M-P)]against t . If a logistic curve seems reasonable, estimate rM and t*.

Solution

t <- c(1814, 1824, 1834, 1844, 1854, 1864)
p <- c(125, 275, 830, 1200, 1750, 1650)

ggplot(data.frame(x=t,y=p),aes(x=x,y=y, fill = p)) + 
  geom_line() +
  geom_point() + 
  ggtitle("Sheep Population Growth") + 
  theme(plot.title = element_text(lineheight=.8, face="bold"))+
  xlab("Years")+
  ylab("Population")

The sheep population appears to reach a maximum level, and the last data point suggests that the data never exceeded 1750. I estimate M = 1750.

Then plot Years elapsed and Change p(t+1) - p(t)

yearlapsed<- t-t[1]
p_changes<-0
for (i in 1:length(p)-1) {
  p_changes<- c(p_changes, p[i+1] -p[i]) 
  }
p_changes

## [1]    0  150  555  370  550 -100

yearlapsed

## [1]  0 10 20 30 40 50

ggplot(data.frame(x=yearlapsed,y=p_changes),aes(x=x,y=y, fill = p_changes)) + 
  geom_line() +
  geom_point() + 
  ggtitle("Sheep Population Growth Change") + 
  theme(plot.title = element_text(lineheight=.8, face="bold"))+
  xlab("Elapsed Years")+
  ylab("Growth Change")

The plot suggests that the logistic curve is reasonable. Then estimate rM and t???.

M <- 1800

#ln [P/ M - P]
ln_pMp <- log(p/(M-p))

suppressMessages(suppressWarnings(library(Laurae)))

# plot ln [P/ M - P] versus t
ggplot(data.frame(x=t,y=ln_pMp),aes(x=x,y=y, fill = ln_pMp)) + 
  geom_line() +
  geom_point() + 
  stat_smooth(method="lm")+
  stat_smooth_func(geom="text",method="lm",hjust=0,parse=TRUE)+
  ggtitle("ln [P/ M - P] versus t") + 
  theme(plot.title = element_text(lineheight=.8, face="bold"))+
  xlab("Years (t)")+
  ylab("ln [P/ M - P]")

The above graph does approximate to a straight line. Thus we accept the assumptions of logistic growth for the sheep population. rM = 0.119.

Then estimate the slope (rM) of the graph.

rM <- lm(ln_pMp ~ t)
(rM <- rM$coefficients["t"])

##         t 
## 0.1189136

Then estimate t* with the fomula as follows:

$t*=t_0 -\frac{1}{rM}+\frac{P_0}{M-P_0}$

(t_star <- t[1] - (1/rM)*log(p[1]/(M-p[1])))

##        t 
## 1835.825

$P(t)= \frac{M}{1+e^{-rM(t-t*)}}$

#p_t <- M/(1+exp(rM*t_star)*exp(-rM*t))
exp(rM*t_star)

##            t 
## 6.433129e+94

$P(t)= \frac{1800}{1+6.433129^94 \times e^{(0.1189136 \times t)}}$

Let validate:

t1 <- 1814
(p_t1 <- M/(1+exp(rM*t_star)*exp(-rM*t1)))

##   t 
## 125

t2 <- 1824
(p_t2 <- M/(1+exp(rM*t_star)*exp(-rM*t2)))

##        t 
## 354.3244

t3 <- 1836
(p_t3 <- M/(1+exp(rM*t_star)*exp(-rM*t3)))

##        t 
## 909.3793

when t = 1814, p = 125 which is the same to the real data.

when t = 1824,p = 355 which i higher than the real population 275.

when t = t* = 1835.825 = 1836, p = 910 which is close to p = M/2 = 1800/2= 900.

In general, the logistical model roughly agrees with the population.

(t_star <- t[1] - (1/rM)*log(p[1]/(M-p[1])))

##        t 
## 1835.825

11.7 Linear Equations

Page 522: #21

Oxygen flows through one tube into a liter flask falled with air, and the mixture of oxygen and air (considered well stirred) escapes through another tube. Assuming that air contains 21% oxygen, what percentage of oxygen will the flask contain after 5 L have passed through the intake tube?

Solution

the problem is ‘proportion of oxygen’ vs ‘volume passed through’.

If P(v) is the proportion of Oxygen when a volume v has passed through, then the initial condition is when v=0, P=0.21.

The differential equation would be: dP = dv - dPv x 1, therefore, dP/dv = 1-P, when if an extra dv of oxygen pass into the flask, then an amount dPv comes out.

The Deferential Equation: dP/dv = 1-P.

The initial condition is: P(0) = 0.21

derivs <- function( r,y, parms)
  list(r-y)

suppressMessages(suppressWarnings(library(deSolve)))

yini <- 0.21
volums <- seq(from = 0, to = 5, by = 0.2)
(out <- ode(y = yini, times = volums, func = derivs, parms = NULL))

##    time         1
## 1   0.0 0.2100000
## 2   0.2 0.1906654
## 3   0.4 0.2110880
## 4   0.6 0.2640625
## 5   0.8 0.3436873
## 6   1.0 0.4451348
## 7   1.2 0.5644459
## 8   1.4 0.6983833
## 9   1.6 0.8442956
## 10  1.8 1.0000124
## 11  2.0 1.1637564
## 12  2.2 1.3340726
## 13  2.4 1.5097695
## 14  2.6 1.6898717
## 15  2.8 1.8735808
## 16  3.0 2.0602429
## 17  3.2 2.2493228
## 18  3.4 2.4403821
## 19  3.6 2.6330621
## 20  3.8 2.8270690
## 21  4.0 3.0221622
## 22  4.2 3.2181449
## 23  4.4 3.4148558
## 24  4.6 3.6121629
## 25  4.8 3.8099582
## 26  5.0 4.0081531

Therefore, the precentage of oxygen that flask will contain after 5 L is 4.0081531 Liters.

YunMai_hw9_10

Yun Mai

April 10, 2018

Chapter 10: Game theory

10.2 Total Conflict as a Linear Program Model: Pure and Mix strategy

10.3 Desition Theory Revisits: Game Theory Against Nature

Chapter 11

11.1 Population Growth

11.7 Linear Equations