Practice Problems

Author: AustinTSchaffer

OMSCS/Courses/GA/06.5 - NP - Graph Problems.md

@@ -321,7 +321,7 @@ $IS \rightarrow VC$
 - [[8.4 - Clique 3]]
 - [[8.10 - NP-Completeness by Generalization]]
 - [[8.14 - Clique + IS (TODO)]]
-- [[8.19 - Kite (TODO)]]
+- [[8.19 - Kite]]
 
 Remember for all of these practice problems, and for future homeworks in this course.
 - Follow this general procedure

OMSCS/Courses/GA/Office Hours/2026-03-15 - TA Office Hours - Exam 2.md

@@ -34,7 +34,7 @@ This OH covers the following strategies, outlined below. It also covers the foll
 
 ## Max-Flow Strategies
 - Convert to flow network
-	- [[7.3 - Cargo Plane (TODO)]]
+	- [[7.3 - Cargo Plane]]
 	- Adding new source and sink
 - Updating capacities
 	- infinite capacity

OMSCS/Courses/GA/Practice Problems/7.1 - Dual LP Optimality (TODO).md

@@ -0,0 +1,29 @@
+---
+tags:
+  - OMSCS
+  - Algorithms
+  - Practice
+---
+# 7.1 - Geometric Solution for an LP
+
+Consider the following LP
+
+- $\max 5x + 3y$
+- $5x-2y\ge0$
+- $x+y \le 7$
+- $x \le 5$
+- $x,y \ge 0$
+
+Plot the feasible region and identify the optimal solution.
+
+## Normal Form
+- $\max 5x + 3y$
+- $-5x+2y\le0$
+- $x+y \le 7$
+- $x \le 5$
+- $0 \le x,y$
+
+## Plot
+The green line is $5x+3y=31$. This line intersects the shaded region at $(x=5,y=2)$. This solution satisfies all constraints and is optimal.
+
+![[Pasted image 20260403203210.png]]

OMSCS/Courses/GA/Practice Problems/7.1 - Geometric Solution for an LP.md

@@ -4,7 +4,7 @@ tags:
   - Algorithms
   - Practice
 ---
-# 7.3 - Cargo Plane (TODO)
+# 7.3 - Cargo Plane
 ![[Pasted image 20260317223151.png]]
 
 The cargo plane has a capacity. Maximum of 100 tons and 60 $m^3$. There are 3 materials which can be transported:
@@ -22,3 +22,23 @@ The cargo plane has a capacity. Maximum of 100 tons and 60 $m^3$. There are 3 ma
 	- 20 cubic meters available
 	- $12,000 per cubic meter
 
+## Standard Form
+- Objective: $\max \space (1000)m_1+(1200)m_2+(12000)m_3$
+- Constraints
+	- $m_1,m_2,m_3 \ge 0$
+	- $m_1 \le 40$
+	- $m_2 \le 30$
+	- $m_2 \le 20$
+	- $2m_1+m_2+3m_3 \le 100$
+	- $m_1+m_2+m_3 \le 60$
+
+## Geometric View
+![[Pasted image 20260403205315.png]]
+
+![[Pasted image 20260403205324.png]]
+
+![[Pasted image 20260403205422.png]]
+
+![[Pasted image 20260403205433.png]]
+
+![[Pasted image 20260403205441.png]]

…ice Problems/7.3 - Cargo Plane (TODO).md …A/Practice Problems/7.3 - Cargo Plane.mdOMSCS/Courses/GA/Practice Problems/7.3 - Cargo Plane (TODO).md renamed to OMSCS/Courses/GA/Practice Problems/7.3 - Cargo Plane.md OMSCS/Courses/GA/Practice Problems/7.3 - Cargo Plane (TODO).md renamed to OMSCS/Courses/GA/Practice Problems/7.3 - Cargo Plane.md

@@ -0,0 +1,34 @@
+---
+tags:
+  - OMSCS
+  - Algorithms
+  - Practice
+---
+# 7.4 - LP for Duff Beer
+> Moe is deciding how much Regular Duff beer and how much Duff Strong beer to order each week. Regular Duff costs Moe $1 per pint and he sells it at $2 per pint; Duff Strong costs Moe $1.50 per pint and he sells it at $3 per pint. However, as part of a complicated marketing scam, the Duff company will only sell a pint of Duff Strong for each two pints or more of Regular Duff that Moe buys. Furthermore, due to past events that are better left untold, Duff will not sell Moe more than 3,000 pints per week. Moe knows that he can sell however much beer he has. Formulate a linear program for deciding how much Regular Duff and how much Duff Strong to buy, so as to maximize Moe’s profit. Solve the program geometrically.
+
+- Variables
+	- $dr:$ Pints of Duff Regular
+	- $ds:$ Pints of Duff Strong
+
+The complicated part of this problem is "the Duff company will only sell a pint of Duff Strong for each two pints or more of Regular Duff that Moe buys". This essentially says:
+
+- $ds \le \frac{1}{2}dr$
+- $ds - 0.5dr \le 0$
+- If $ds-0.5dr = 0$, then Moe bought 2x as much Regular Duff as he did Strong Duff.
+- If $ds-0.5dr < 0$, then Moe bough the more than 2x Regular Duff than he did Strong Duff.
+- If $ds - 0.5dr > 0$, then Moe is in breach of contract.
+- I guess it wasn't that complicated.
+- (Turns out it was a little complicated. I originally had $ds \le 2dr$, which had the opposite effect.)
+
+## Standard Form
+- Objective, maximize weekly profit: $\max \space (2-1)dr + (3-1.50)ds$
+- Constraints
+	- $dr,ds \ge 0$
+	- $dr+ds \le 3000$
+	- $ds - 0.5dr \le 0$
+
+## Geometric Solution
+![[Pasted image 20260403215232.png]]
+	
+The purple line is $dr+1.5ds=3500$, which intersects the region of feasibility at $(dr=2000,ds=1000)$. This meets the specified constraints and maximizes profit.

OMSCS/Courses/GA/Practice Problems/7.4 - LP for Duff Beer (TODO).md

@@ -0,0 +1,63 @@
+---
+tags:
+  - OMSCS
+  - Algorithms
+  - Practice
+---
+# 7.5 - LP for Canine Products
+> The Canine Products company offers two dog foods, Frisky Pup and Husky Hound, that are made from a blend of cereal and meat. A package of Frisky Pup requires 1 pound of cereal and 1.5 pounds of meat, and sells for $7. A package of Husky Hound uses 2 pounds of cereal and 1 pound of meat, and sells for $6. Raw cereal costs $1 per pound and raw meat costs $2 per pound. It also costs $1.40 to package the Frisky Pup and $0.60 to package the Husky Hound. A total of 240,000 pounds of cereal and 180,000 pounds of meat are available each month. The only production bottleneck is that the factory can only package 110,000 bags of Frisky Pup per month. Needless to say, management would like to maximize profit.
+> 
+> (a) Formulate the problem as a linear program in two variables.
+> 
+> (b) Graph the feasible region, give the coordinates of every vertex, and circle the vertex maximizing profit. What is the maximum profit possible?
+
+- Variables
+	- $f$ - Frisky Pup
+	- $h$ - Husky Hound
+- costs
+	- cereal costs $1
+	- meat costs $2
+- Availability
+	- 240,000 cereal
+	- 180,000 meat
+	- 110,000 Frisky Bags
+- $f$
+	- requires 1 cereal ($\$1$)
+	- requires 1.5 meat ($1.5*\$2=\$3$)
+	- sells for $\$7$
+	- ingredient cost: $\$4$
+	- packaging costs: $\$1.40$
+	- profit: $\$7-\$4-\$1.4=\$1.6$
+- $h$
+	- requires 2 cereal ($\$2$)
+	- requires 1 meat ($\$2$)
+	- sells for $\$6$
+	- ingredient cost: $\$4$
+	- packaging cost: $\$0.6$
+	- profit: $\$6-\$4-\$0.6=\$1.4$
+- Timespan: Monthly
+
+## Standard Form
+- Objective: $\max \space 1.6f+1.4h$
+- Constraints:
+	- $f,h \ge 0$
+	- $f \le 1.1*10^5$
+	- $f+2h \le 2.4*10^5$
+	- $1.5f+h \le 1.8*10^5$
+
+## Graph
+$f$ is on the x-axis. $h$ is on the y-axis. Purple is at $p=\$2.22*10^5$, which intersects the feasible region at $(h=9*10^4,f=6*10^4)$.
+
+![[Pasted image 20260403221422.png]]
+
+All of the vertex pairs are:
+
+| f       | h       | Profit       |
+| ------- | ------- | ------------ |
+| 0       | 0       | 0            |
+| 110,000 | 0       | $176,000     |
+| 110,000 | 15,000  | $197,000     |
+| 60,000  | 90,000  | **$222,000** |
+| 0       | 120,000 | $168,000     |
+
+These coordinates can be determined by taking the intersection of pairs of constraint boundaries, and checking whether that point is feasible against the other constraints.