agridat (version 1.23)

hazell.vegetables: Gross profit for 4 vegetable crops in 6 years

Description

Gross profit for 4 vegetable crops in 6 years

Usage

data("hazell.vegetables")

Format

A data frame with 6 observations on the following 5 variables.

year

year factor, 6 levels

carrot

Carrot profit, dollars/acre

celery

Celery profit, dollars/acre

cucumber

Cucumber profit, dollars/acre

pepper

Pepper profit, dollars/acre

Details

The values in the table are gross profits (loss) in dollars per acre. The criteria in the example below are (1) total acres < 200, (2) total labor < 10000, (3) crop rotation.

The example shows how to use linear programming to maximize expected profit.

References

Carlos Romero, Tahir Rehman. (2003). Multiple Criteria Analysis for Agricultural Decisions. Elsevier.

Examples

if (FALSE) {
  
  library(agridat)
  data(hazell.vegetables)
  dat <- hazell.vegetables
  
  libs(lattice)
  xyplot(carrot+celery+cucumber+pepper ~ year,dat,
         ylab="yearly profit by crop",
         type='b', auto.key=list(columns=4),
         panel.hline=0)

  # optimal strategy for planting crops (calculated below)
  dat2 <- apply(dat[,-1], 1, function(x) x*c(0, 27.5, 100, 72.5))/1000
  colnames(dat2) <- rownames(dat)
  barplot(dat2, legend.text=c("     0 carrot", "27.5 celery", " 100 cucumber", "72.5 pepper"),
          xlim=c(0,7), ylim=c(-5,120),
          col=c('orange','green','forestgreen','red'),
          xlab="year", ylab="Gross profit, $1000",
          main="hazell.vegetables - retrospective profit from optimal strategy",
          args.legend=list(title="acres, crop"))

  libs(linprog)
  # colMeans(dat[ , -1])
  # 252.8333 442.6667 283.8333 515.8333
 
  # cvec = avg across-years profit per acre for each crop
  cvec <- c(253, 443, 284, 516)
  
  # Maximize c'x for Ax=b
  A <- rbind(c(1,1,1,1), c(25,36,27,87), c(-1,1,-1,1))
  colnames(A) <- names(cvec) <- c("carrot","celery","cucumber","pepper")
  rownames(A) <- c('land','labor','rotation')

  # bvec criteria = (1) total acres < 200, (2) total labor < 10000,
  # (3) crop rotation.

  bvec <- c(200,10000,0)
  const.dir <- c("<=","<=","<=")

  m1 <- solveLP(cvec, bvec, A, maximum=TRUE, const.dir=const.dir, lpSolve=TRUE)
  # m1$solution # optimal number of acres for each crop
  #   carrot    celery  cucumber    pepper
  #  0.00000  27.45098 100.00000  72.54902
  
  # Average income for this plan
  ## sum(cvec * m1$solution)
  ## [1] 77996.08

  # Year-to-year income for this plan
  ## as.matrix(dat[,-1]) 
  ##           [,1]
  ## [1,]  80492.16
  ## [2,]  80431.37
  ## [3,]  81884.31
  ## [4,] 106868.63
  ## [5,]  37558.82
  ## [6,]  80513.73

  # optimum allocation that minimizes year-to-year income variability.
  # brute-force search

  # For generality, assume we have unequal probabilities for each year.
  probs <- c(.15, .20, .20, .15, .15, .15)
  # Randomly allocate crops to 200 acres, 100,000 times
  #set.seed(1)
  mat <- matrix(runif(4*100000), ncol=4)
  mat <- 200*sweep(mat, 1, rowSums(mat), "/")
  # each row is one strategy, showing profit for each of the six years
  # profit <- mat 
  profit <- tcrossprod(mat, as.matrix(dat[,-1])) # Each row is profit, columns are years
  # calculate weighted variance using year probabilities
  wtvar <- apply(profit, 1, function(x) cov.wt(as.data.frame(x), wt=probs)$cov)
  # five best planting allocations that minimizes the weighted variance
  ix <- order(wtvar)[1:5]
  mat[ix,]
  ## carrot celery cucumber pepper
  ##          [,1]     [,2]     [,3]     [,4]
  ## [1,] 71.26439 28.09259 85.04644 15.59657
  ## [2,] 72.04428 27.53299 84.29760 16.12512
  ## [3,] 72.16332 27.35147 84.16669 16.31853
  ## [4,] 72.14622 29.24590 84.12452 14.48335
  ## [5,] 68.95226 27.39246 88.61828 15.03700

}