Updated the codes

Author: analyticswithali

Integer programming model.Rmd

@@ -1,19 +1,21 @@
 ---
 output:
  html_document:
+ toc: true
+ toc_depth: '4'
  df_print: paged
 ---
 
-#### **To:** Seattle Museum of Modern Art: Board of directors 
+##### **To:** Seattle Museum of Modern Art: Board of directors 
 
-#### **From:** Afsar Ali 
+##### **From:** Afsar Ali 
 
-#### **Date:** `r format(Sys.time(), '%B %d, %Y')` 
+##### **Date:** `r format(Sys.time(), '%B %d, %Y')` 
 
-#### **RE:** Analysis of 3 Proposal: Artie's Dream 
+##### **RE:** Analysis of 3 Proposal: Artie's Dream 
 
 ***
-```{r message=FALSE, warning=FALSE, include=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE}
 
 # Title: Artie's Dream
 # Purpose: Final Project 
@@ -35,7 +37,7 @@ library(kableExtra)
 library(formattable)
 ```
 
-```{r message=FALSE, warning=FALSE, include=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE}
 
 # Import Data 
 data <- read.csv("5260_S18_Arties_Dream.csv", header = TRUE)
@@ -46,7 +48,7 @@ data <- read.csv("5260_S18_Arties_Dream.csv", header = TRUE)
 Artie Fophartie has recently caught our attention with his willingness to fund an entire exhibit and generously donating $1 million to the museum from his lottery ticket winnings. Since the exhibit pieces would remain on display for two months, a thorough analysis is needed to calculate the cost and determine which art pieces should be displayed. This analysis is based on the 34 art pieces that Artie selected from the modern art community. Celeste McKenzie and Artie made specific requirements for the exhibit and based on these requirements there are three proposals for you to review. The first proposal reflects the budget of $4 million set by Artie. The second proposal reflects Celeste's decision to include at least 20 art pieces to ensure that the exhibit draws the attention of the public. The third proposal reflects cost uncertainty.
 
 
-```{r message=FALSE, warning=FALSE, include=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE}
 # This model has choice variables that are binary
 p1 <- make.lp(0, 35)
 lp.control(p1, sense = "max")
@@ -130,7 +132,7 @@ With $4 million to fund the exhibit and the specific requirements from Artie and
 
 *Table 1: List of 15 art pieces displayed without exceeding the budget*
 
-```{r echo=FALSE, message=FALSE, warning=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 #Stracture Table
 data$Solution <- get.variables(p1)[1:34]
 data1 <- subset(data, Solution == 1)
@@ -149,7 +151,7 @@ kable("html", row.names = TRUE) %>%
 ```
 
 
-```{r include=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 
 # This model has choice variables that are binary
 p1 <- make.lp(0, 35)
@@ -233,7 +235,7 @@ In proposal 2, we are considering at least 20 pieces to ensure that the exhibit
 
 *Table 2: Minimum cost is $5.45 Million for displaying at least 20 art pieces*
 
-```{r echo=FALSE, message=FALSE, warning=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 #Stracture Table
 data$Solution2 <- get.variables(p1)[1:34]
 data2 <- subset(data, Solution2 == 1)
@@ -252,7 +254,7 @@ data2 %>%
 ```
 
 
-```{r include=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 # Number of simulations
 set.seed(1234)
 n <- 1000
@@ -348,7 +350,7 @@ We know from our previous experiences, cost associated with each art piece varie
 
 
 *Table 3: Impact on Proposal 2 with cost uncertainty ranges from $4.9 Million to $5.9 Million*
-```{r echo=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 # Make table of output
 out_rp3 <- round(rbind(apply(rp3, 2, mean), apply(rp3, 2, sd), 
  apply(rp3, 2, max), apply(rp3, 2, min)), 2)

Integer programming model.html Integer_programming_model.htmlInteger programming model.html renamed to Integer_programming_model.html

@@ -1,17 +1,21 @@
 ---
-output: html_document
+output:
+ html_document:
+ toc: true
+ toc_depth: '4'
+ df_print: paged
 ---
 
-#### **To:** Seymore Butts
+##### **To:** Seymore Butts
 
-#### **From:** Afsar Ali 
+##### **From:** Afsar Ali 
 
-#### **Date:** `r format(Sys.time(), '%B %d, %Y')` 
+##### **Date:** `r format(Sys.time(), '%B %d, %Y')` 
 
-#### **RE:** Analysis of the Proposals to reduce average level of in-process inventory 
+##### **RE:** Analysis of the Proposals to reduce average level of in-process inventory 
 
 ***
-```{r message=FALSE, warning=FALSE, include=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 
 # Title: Artie's Dream
 # Purpose: Final Project 
@@ -39,7 +43,7 @@ library(formattable)
 How can we reduce the average number of poster sheets waiting to complete inspection? I was tasked with analyzing and producing solutions to reduce the backlog by taking into account the cost associated with each task and cost of in-process inventory. According to my findings, we should only use 8 printing press, increase the print time by 15 mins and hire an additional inspector. This will reduce our **Total cost by $50.63 per hour** from current the total cost of $203.14 per hour to $152.51 per hour and optimizes our overall operation. With the task of analyzing the two initial proposals we discussed, I was able to create a third solution that maximizes the poster sheet flow while minimizing our cost. 
 
 
-```{r message=FALSE, warning=FALSE, include=FALSE, paged.print=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 #Current Status:
 
 # Evaluating the status quo print reproductions of the paintings and photographs
@@ -102,7 +106,7 @@ Table 1 breaks down our status quo. Our current **total cost is $203.14 per hour
 
 *Table 1: Currently the Total Cost is $203.14 per hour *
 
-```{r echo=FALSE, message=FALSE, warning=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 now <- as.data.frame(cbind(p4_p[3,],#using 10 presses and 1 inspector
  p4_i[1,])) #using 10 presses and 1 inspector
 #L is the all inventory (work in process) of the queue system while Lq (ork To Process)is the portion of L that is idle (queue waiting)
@@ -128,7 +132,7 @@ now %>%
  column_spec(4, bold = T) 
 ```
 
-```{r message=FALSE, warning=FALSE, include=FALSE, paged.print=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 ##Proposal 4: Take slightly longer to make the prints (which would increase their average time to make a print to 1.2 hours), so that the inspector can keep up with his output better. This also would reduce the cost of the power for running each press from $7.00 to $6.50 per hour. (By contrast, decreasing the time would increase this cost to $7.50 per hour while decreasing the average time to make a print to 0.8 hour.):
 
 # Evaluating the status quo print reproductions of the paintings and photographs
@@ -191,7 +195,7 @@ If we were to take slightly longer time to make the prints, so that the inspecto
 
 *Table 2: Decreasing print time by 10 mins would increase Total Cost by $23.24 per hour *
 
-```{r echo=FALSE, message=FALSE, warning=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 p4a <- as.data.frame(cbind(p4_p[2,],#using 10 presses and 1 inspector
  p4_i[1,])) #using 10 presses and 1 inspector
 #L is the all inventory (work in process) of the queue system while Lq (ork To Process)is the portion of L that is idle (queue waiting)
@@ -221,7 +225,7 @@ now1 %>%
  add_header_above(c(" ", "Current" = 3, "Proposed: Decreasing print time" = 3))
 ```
 
-```{r message=FALSE, warning=FALSE, include=FALSE, paged.print=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 ##Proposal 4: Take slightly longer to make the prints (which would increase their average time to make a print to 1.2 hours), so that the inspector can keep up with his output better. This also would reduce the cost of the power for running each press from $7.00 to $6.50 per hour. (By contrast, decreasing the time would increase this cost to $7.50 per hour while decreasing the average time to make a print to 0.8 hour.):
 
 # Evaluating the status quo print reproductions of the paintings and photographs
@@ -284,7 +288,7 @@ By contrast, if we were increasing the print time, it would increase the cost of
 
 *Table 3: Increasing print time by 15 mins would decrease the Total Cost by $9.63 per hour *
 
-```{r echo=FALSE, message=FALSE, warning=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 p4b <- as.data.frame(cbind(p4_p[5,],#using 10 presses and 1 inspector
  p4_i[1,])) #using 10 presses and 1 inspector
 #L is the all inventory (work in process) of the queue system while Lq (ork To Process)is the portion of L that is idle (queue waiting)
@@ -314,7 +318,7 @@ now2 %>%
  add_header_above(c(" ", "Current" = 3, "Proposed: Increasing print time" = 3))
 ``` 
 
-```{r message=FALSE, warning=FALSE, include=FALSE, paged.print=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 #Proposal 5: Substitute a more experienced inspector for this task. She is somewhat faster at 7 minutes per poster, so she should keep up better. However, this inspector is in a job classification that calls for a total compensation (including benefits) of $19 per hour, whereas the current inspector is in a lower job classification where the compensation is $17 per hour:
 
 # Evaluating the status quo print reproductions of the paintings and photographs
@@ -377,7 +381,7 @@ If we were to substitute a more experienced inspector, the **total cost would de
 
 *Table 4: Experienced inspector would decrease the Total Cost by $18.36 per hour *
 
-```{r echo=FALSE, message=FALSE, warning=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 p5a <- as.data.frame(cbind(p4_p[3,],#using 10 presses and 1 inspector
  p4_i[1,])) #using 10 presses and 1 inspector
 #L is the all inventory (work in process) of the queue system while Lq (ork To Process)is the portion of L that is idle (queue waiting)
@@ -409,7 +413,7 @@ now3 %>%
  add_header_above(c(" ", "Current" = 3, "Proposed: Increasing print time" = 3, "Proposed: Experienced inspector" = 3))
 ``` 
 
-```{r message=FALSE, warning=FALSE, include=FALSE, paged.print=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 # My Proposal:
 
 # Evaluating the status quo print reproductions of the paintings and photographs
@@ -472,7 +476,7 @@ The recommended proposal is a combination of both of the initial proposal. In Ta
 
 *Table 5: 2 Inspector, 8 Presses at increased rate would decrease the Total Cost by $50.63 per hour*
 
-```{r echo=FALSE, message=FALSE, warning=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 
 p6 <- as.data.frame(cbind(p4_p[3,],#using 8 presses and 2 inspector
  p4_i[2,])) #using 8 presses and 1 inspector

Inventory optimization model.Rmd

@@ -1,5 +1,5 @@
 ---
-title: "Ali_BUAN Midterm Project"
+title: "Network optimization Project"
 author: "Afsar Ali"
 date: '`r format(Sys.time(), "%B %d, %Y")`'
 output:
@@ -9,7 +9,7 @@ output:
 ---
 
 # Objective
-```{r}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 # Course: BUAN 5260
 # Title: Week 4-Network optimization
 # Purpose: USE IFRC Information information to model scenarios 
@@ -18,7 +18,7 @@ output:
 
 ```
 
-```{r, echo=FALSE, message=FALSE, warning=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 # Clear packages 
 if(is.null(sessionInfo()$otherPkgs) == FALSE)lapply(
  paste("package:", names(sessionInfo()$otherPkgs), sep=""), 
@@ -39,7 +39,7 @@ set.seed(123)
 ```
 #Loading and cleaning the data
 
-```{r, message=FALSE, warning=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 #load Data 
 ifrc <- read.csv("5260_S18_Aiding_Africa_Data.csv", skip = 1)
 #Naming and Creating each table
@@ -80,7 +80,7 @@ edges$ID <- paste(edges$From, edges$To, sep = ' > ')
 
 #1 Netowrk Map
 - Make sense all the Air Routes are the Fastest 
-```{r, message=FALSE, warning=FALSE}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 net <- graph_from_data_frame(d=edges, vertices=nodes, directed=T)
 
 net$layout <- matrix(c(-800, -800, 
@@ -123,7 +123,7 @@ plot(net, edge.arrow.size=.3, edge.label = round(E(net)$Time, 2),
 +Jacksonville, FL > Niamey, Niger = 22.87 
 
 
-```{r}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 #Create Table for shortest time
 distMatrix <- shortest.paths(net, v=V(net), to=V(net))
 
@@ -152,7 +152,7 @@ plot(shortMatrix, edge.arrow.size=.2, edge.label = round(E(net)$Time, 2))
 #3 - 2nd Plan considering cost and constraints 
 
 
-```{r}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 #Part 3 - Min Cost
 # Set up model
 min_cost <- make.lp(0, 30)
@@ -251,7 +251,7 @@ tm1
 
 ##Graph the Min Cost Solution
 
-```{r}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 
 # Include solution in edges dataframe
 edges$flow <- get.variables(min_cost)
@@ -292,7 +292,7 @@ plot(g, edge.width = 15*E(g)$Mincost/max(E(g)$Mincost),
 
 #4 - Last Plan Max Flow with many constraints 
 
-```{r}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 #Maximum Flow
 
 # Set up model
@@ -406,7 +406,7 @@ tm3
 
 ##Graph Max Cost Solution
 
-```{r}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 # Include solution in edges dataframe
 edges$Mflow <- get.variables(max_flow)[3:32]
 edges$TotalAid <- edges$Mflow * edges$Capacity
@@ -452,7 +452,7 @@ E(g1)$TotalAid
 
 #4 - Last Plan Testing Max Flow by relaxing some constraints to congos
 
-```{r}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 #Maximum Flow
 
 # Set up model
@@ -567,7 +567,7 @@ tm3
 
 ##Graph Max Cost Solution
 
-```{r}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 # Include solution in edges dataframe
 edges$Mflow <- get.variables(max_flow)[3:32]
 edges$TotalAid <- edges$Mflow * edges$Capacity
@@ -615,7 +615,7 @@ E(g1)$TotalAid
 
 ##Fuction to run Matrix
 
-```{r}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 
 
 
@@ -744,7 +744,7 @@ write.csv(CM, file = "MyData.csv")
 
 #Using I graph to calculate min and max (didnt work)
 
-```{r}
+```{r echo=TRUE, message=FALSE, warning=FALSE, paged.print=TRUE}
 #Addmore attributes to the Graph
 
 E(net)$Group <- E(net)$Type