Suppose you\'re managing a consulting team of expert computer backers, and each

Question

Suppose you're managing a consulting team of expert computer backers, and each week you have to choose a job for them to undertake. Now, as you can well imagine, the set of possible jobs is divided into those that are low-stress (e-g, setting up a Web site for a class at the local elementary school) and those that are high-stress (e.g., protecting the nation's most valuable secrets, or helping a desperate group of Cornell students finish a project that has something to do with compilers). The basic question, each week, is whether to take on a low-stress job or a high-stress job If you select a low-stress job for your team in week i, then you get a revenue of 4>0 dollars; if you select a high-stress job, you get a revenue of hy>0 dollars. The catch, however, is that in order for the team to take on a high-stress job in week i, it's required that they do no job (of either type) in week -1;they need a full week of prep time to get ready for the crushing stress level. On the other hand, it's okay for them to take a low- stress job in week i even if they have done a job (of either type) in week So, given a sequence of n weeks, a plan is specified by a choice of "Sow-stress," "high-stress," or "none" for each of the n weeks, with the property that if "high-stress" is chosen for week i>1, then "none" has to be chosen for week i-1. at's okay to choose a high-stress job in week 1.) The value of the plan is determined in the natural way: for each i, you add 4 to the value if you choose "low-stress" in week i, and you add h, to the value if you choose "high-stress" in week i. (You add 0 if you choose none" in week i) The problem. Given sets of values and hy,h. .h, finda plan of maximum value. (Such a plan will be called optimal.) Example. Suppose n-4, and the values of and hy are given by the following table. Then the plan of maximum value would be to choose none" in week 1, a high-stress job in week 2, and low-stress jobs in weeks 3 and 4. The value of this plan would be 0+50+10+10-70 Week 1 Week 2 Week 3 Week 4 10 10 50

Explanation / Answer

Solution:

Algorithm:

if number == 1
   then
   tempSolution= max(l[number],h[number])
else if number == 2 then
   tempSolution= max(optimalPlan(1, l, h)+ l[2], h[2])
else
   tempSolution= max(optimalPlan(number 1, l, h) + l[number], optimalPlan(number 2, l, h) + h[number])
end if
return Value

FindOptimalValue(number, l, h)

//Initialisation
for itterator = 1 ! number do
   tempSolution[itterator] = 0
end for

for itterator = 1 ! number do
   if itterator == 1 then
       tempSolution[itterator] max(l[itterator], h[itterator])
   else if itterator == 2 then
       tempSolution[itterator] max(tempSolution[1] + l[2], h[2])
   else
       tempSolution[itterator] max(tempSolution[itterator 1] + l[itterator], tempSolution[itterator 2] + h[itterator])
   end if
end for

return Value[number]
//Recurrsive function to find the optimal plan
OPtimalPlan(number, l, h, Value)

for itterator = 1 ! number do
   WeekVal[itterator]
end for
if tempSolution[number] l[number] = tempSolution[number 1] then
   WeekVal[number] ”Low stress”
   OPtimalPlan(number-1, l, h, Value)
else
   WeekVal[number] ”High stress”
   OPtimalPlan(number-2, l, h, Value)
end if
return WeekVal

I hope this helps if you find any problem. Please comment below. Don't forget to give a thumbs up if you liked it. :)

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