Data 609 Assignment #5
Joseph Simone
4/1/2020
Week 9 Homework Assignment
Section 10.2
Page 404 #2
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.
Rose’s Game Descision
Let:
\(x = \textrm{Percent of Time playing: R1}\)
\((1- x) = \textrm{Percent of Time Playing: R2}\)
\(VP = \textrm{Value of Payoff for Rose}\)
Objective:
\(\textrm{Maximize}~VP\)
Subject to:
\(VP \le 10x+5(1-x)\)
\(VP \le 10x\)
\(x \ge 0\)
\(x \le 1\)
Alegebraic Solution
There are six \(_4C_2=6\) possible ways of intersecting four distinct constraints two at a time, with 4 contraints in place.
The constraint boundaries for \(x \ge 0\) & \(x \ge 1\) are parallel, therefore, we have five possible intersection points.
Beginning with the \(1^{st}\) four intersections, they can be found by setting \(x=0\) & \(x=1\) in the constraints that represents Colin’s two strategies:
\(x=0: VP = 10 \times 0 + 5 \times (1- 0) = 5\)
\(x=1: VP = 10 \times 1 + 5 \times (1 - 1) = 10\)
\(x=0: VP = 10 \times 0 = 0\)
\(x=1: VP = 10 \times = 10\)
Then, the \(5^{th}\) intersection can be found by setting these two constraint equations equal to one other:
\(VP = 10x + 5(1-x) = 5x + 5\)
\(VP = 10x\)
Simultaneously solving the two equations produces \(x=1\) & \(VP = 10\).
Summarize potential outcomes & which solutions are probable:
| x | VP | Outcome |
|---|---|---|
| 0 | 5 | N |
| 1 | 10 | Y |
| 0 | 0 | Y |
| 1 | 10 | Y |
| 1 | 10 | Y |
Graphical Solution
ggplot(data.frame(x=c(-1,1)),aes(x)) +
stat_function(fun=function(x) 5*x + 5, geom="line", aes(col='Colins pure C1 = 10x + 5(1-x)')) +
stat_function(fun=function(x) 10*x, geom="line", aes(col='Colins pure C2 = 10x')) +
geom_vline(xintercept=0, aes(col= 'x=0')) +
geom_vline(xintercept=1, aes(col= 'x=1')) +
geom_point(data=ips, aes(x,y)) +
theme_bw() +
labs(title = 'Graphical Solution for Rose', y = "VP") +
annotate('text', x = 0.1, y = 15, label="x = 0", size=3 ) +
annotate('text', x = 1.1, y = 15, label="x = 1", size=3 ) +
annotate('text', x = -0.1, y = 0.2, label="(0, 0)", size=3 ) +
annotate('text', x = -0.1, y = 5.2, label="(0, 5)", size=3 ) +
annotate('text', x = 0.85, y = 10.5, label="(1, 10)", size=3 )
The Value of Power sets a maximum value of 10 when x = 1, when Rose plays R1 100% of the time
Colin’s Game Decision
Let:
\(y = \textrm{percent of time playing C1}\)
\((1- y) = \textrm{percent of time playing C2}\)
\(VP = \textrm{value of payoff for Rose}\)
Objective:
\(\textrm{Minimize}~VP\)
Subject to:
\(VP \ge 10\)
\(VP \ge 5y\)
\(y \ge 0\)
\(y \le 1\)
Alegebraic Solution
There are six \(_4C_2=6\) possible ways of intersecting four distinct constraints two at a time, with 4 contraints in place.
The constraint boundaries for \(x \ge 0\) & \(x \ge 1\) are parallel, therefore, we have five possible intersection points.
Beginning with the \(1^{st}\) four intersections, they can be found by setting \(x=0\) & \(x=1\) in the constraints that represents Rose’s two strategies:
\(x=0: VP = 10 \times 0 + 5 \times (1- 0) = 5\)
\(x=1: VP = 10 \times 1 + 5 \times (1 - 1) = 10\)
\(x=0: VP = 10 \times 0 = 0\)
\(x=1: VP = 10 \times = 10\)
Then, the \(5^{th}\) intersection can be found by setting these two constraint equations equal to one other:
\(VP = 10x + 5(1-x) = 5x + 5\)
\(VP = 10x\)
Simultaneously solving the two equations produces \(x=1\) & \(VP = 10\).
Summarize potential outcomes & which solutions are probable:
| y | VP | Outcome |
|---|---|---|
| 0 | 10 | N |
| 1 | 10 | Y |
| 0 | 0 | Y |
| 1 | 5 | Y |
| 2 | 10 | Y |
Graphical Solution
ggplot(data.frame(x=c(-1,2)),aes(x)) +
stat_function(fun=function(x) 10, geom="line", aes(col='Roses pure R1 = 10')) +
stat_function(fun=function(x) 5*x, geom="line", aes(col='Roses pure R2 = 5x')) +
geom_vline(xintercept=0, aes(col= 'y=0')) +
geom_vline(xintercept=1, aes(col= 'y=1')) +
geom_point(data=ips, aes(x,y)) +
theme_bw() +
labs(title = 'Graphical Solution for Colin', y = "VP", x = 'y') +
annotate('text', x = 0.1, y = 15, label="y = 0", size=3 ) +
annotate('text', x = 1.1, y = 15, label="y = 1", size=3 ) +
annotate('text', x = -0.1, y = 0.2, label="(0, 0)", size=3 ) +
annotate('text', x = -0.1, y = 10.5, label="(0, 10)", size=3 ) +
annotate('text', x = 0.85, y = 10.5, label="(1, 10)", size=3 ) +
annotate('text', x = 0.85, y = 5.5, label="(1, 5)", size=3 ) +
annotate('text', x = 1.85, y = 10.5, label="(2, 10)", size=3 )
The minimum value of VP is still at 10, regardless of which value of y selected. Colin can play C1 any percent of the time between 0% and 100%.
Solution:
Based on the possible outcome solutions, Colin could minimizes Rose’s payoff by playing C1 either none (0%) or all (100%) of the time.
However, if Rose plays R1 all (100%) of the time, Colin can take advantage by playing the pure C2 strategy.
This outcome can be noted when reviewing the graphical solution from Rose’s Game Decision.
Section 10.7
Page 454 #3
Doc Holliday versus Ike Clanton (2-person, 3-strategy game) are used in this problem. On October 26, 1881, the bad blood between the Earps, Clantons, and McLaurys came to a head at the O.K. Corral. Billy Clanton, Frank McLaury, and Tom McLaury were killed. Doc Holliday, Virgil and Morgan Earp were injured. Miraculously, Wyatt Earp was unharmed, while the unarmed Isaac ‘‘Ike’’ Clanton survived by running away. Many people believed that Doc Holliday shotgunned an unarmed Tom McLaury in the back as he was attempting to flee the scene.
Ike Clanton and his friends and associates, known as the ‘‘cowboys,’’ swore to get their revenge on the Earps and Holliday. In the ensuing months, Morgan Earp was mur- dered and Virgil Earp seriously wounded in an ambush. A few days later, Wyatt Earp apparently shot and killed Frank Stillwell, a Clanton associate, and another man believed to be involved in the ambush. Over the next few years, many more of the cowboys were killed.
Although close to death from tuberculosis, in 1887 Doc Holliday decided to look up Ike Clanton and to settle their differences once and for all. On June 1, 1887, Doc Holliday and J. V. Brighton cornered Ike Clanton near Springerville, Arizona. Doc told J.V. to stay out of it for the time being.
Ike and Doc each had a pistol and shotgun. Ike and Doc, spurning their pistols in favor of their shotguns, pressed forward toward each other. At long range, Ike—with his cowboy background—was a better shot than Doc. At middle range, Doc—the seasoned gunfighter—could outgun Ike. Both desperadoes were deadly at close range. The prob- abilities that either rogue would kill the other with a blast from his single-shot shotgun appear in the following table:
In this problem, the payoff to Doc is 10 units if Doc survives and Ike is killed, -10 units if Doc is killed and Ike survives. Doc and Ike’s strategies are as follows:
Compute the payoff matrix for Doc Holiday and solve the game.
Solution
Considering the given payoff to Doc and Ike, Constructing the payoff matrix below:
| Kill Probability | |||
|---|---|---|---|
| Gunslinger’s Shot | Longe Range | Middle Range | Short Range |
| —————- | ————- | ————- | ———– |
| Ike Clanton | 0.5x(-10) | 0.6x(-10) | 1.0x(-10) |
| Doc Holiday | 0.3x(10) | 0.8x(10) | 1.0x(10) |
| Kill Probability | |||
|---|---|---|---|
| Gunslinger’s Shot | Longe Range | Middle Range | Short Range |
| —————- | ————- | ————- | ———– |
| Ike Clanton | -5 | -6 | -1 |
| Doc Holiday | 3 | 8 | 1 |
Just Doc’s Payoff
| Kill Probability | |||
|---|---|---|---|
| Gunslinger’s Shot | Longe Range | Middle Range | Short Range |
| —————- | ————- | ————- | ———– |
| Doc Holiday | 3 | 8 | 1 |
Solving the game for both using MiniMax Method:
| Kill Probability | |||||
|---|---|---|---|---|---|
| Gunslinger’s Shot | Longe Range | Middle Range | Short Range | Row Min | Row Max |
| —————- | ————- | ————- | ———– ———– ——– | ||
| Ike Clanton | -5 | -6 | -1 | -6 | 1 |
| Doc Holiday | 3 | 8 | 1 | 1 | |
| Column Max | 3 | 8 | 1 | ||
| Column Min | 1 |
Here Maximin is the Maximum of the Row Minimum and Minimax is the Minimum of the Comumn Maximum Values and the Long Range, the Middle Range abd Short Range.
The above table is the Value of Minimax is equal to the Maximin which implies that Pure Strategy Saddle Point Exist. Therefore, both Pure Strategy is the best strategy among all possible strategies.
Week 10 Homework Assignment
Section
Page 478, #6
Suggest other phenomena for which the model described in the text might be used.
Solution
Let’s say you want toset-up a no-interest checking account for your son, a college student. In the long term, the account will retain a value between \(L\) and \(H\). However, the account holder continously withdraws from account. Assume a continuous approximation is reasonable, in terms of the amounts that are proportional to total funds are always in the account. Therfore, depositing a flat rate into the account at regularly. Finally you can solve for both of these quantities by using the Drug Dosage Model discussed in the Text.