2016 Update
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Tina Müller
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7965 changed files with 139854 additions and 31002 deletions
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@ -1,10 +1,13 @@
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See also: [[Knapsack problem]] and [[wp:Continuous_knapsack_problem|Wikipedia]].
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<!-- a thief (or burglar) steals, a robber robs (confronts a person while stealing). Not exactly a perfect definition, but close enough. -- Gerard Schildberger. -->
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A robber burgles a butcher's shop, where he can select from some items. He knows the weights and prices of each items. Because he has a knapsack with 15 kg maximal capacity, he wants to select the items such that he would have his profit maximized. He may cut the items; the item has a reduced price after cutting that is proportional to the original price by the ratio of masses. That means: half of an item has half the price of the original.
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A thief burgles a butcher's shop, where he can select from some items.
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This is the item list in the butcher's:
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The thief knows the weights and prices of each items. Because he has a knapsack with 15 kg maximal capacity, he wants to select the items such that he would have his profit maximized. He may cut the items; the item has a reduced price after cutting that is proportional to the original price by the ratio of masses. That means: half of an item has half the price of the original.
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{| style="text-align: left; width: 80%;" border="4" cellpadding="2" cellspacing="2"
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This is the item list in the butcher's shop:
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{| style="text-align: left; width: 50%;" border="4" cellpadding="2" cellspacing="2"
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|+ Table of potential knapsack items
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|- style="background-color: rgb(255, 204, 255);"
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! Item !! Weight (kg) !! Price (Value)
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@ -30,4 +33,16 @@ This is the item list in the butcher's:
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| Knapsack || <=15 kg || ?
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|}
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'''Which items does the robber carry in his knapsack so that their total weight does not exceed 15 kg, and their total value is maximised?'''
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<br>
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;Task:
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Show which items the thief carries in his knapsack so that their total weight does not exceed 15 kg, and their total value is maximized.
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;Related task:
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* [[Knapsack problem]]
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;See also:
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* Wikipedia article: [[wp:Continuous_knapsack_problem|continuous knapsack]].
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<br><br>
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@ -1,4 +1,4 @@
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; Solve Continuous Knapsdack Problem
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; Solve Continuous Knapsack Problem
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; Nicolas Modrzyk
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; January 2015
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@ -1,14 +1,16 @@
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defmodule KnapsackProblem do
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def price_per_weight( items ), do: (for {name, weight, price} <-items, do: {name, weight, price / weight} )
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def select( max_weight, items ) do
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{_remains, selected_items} = List.foldr( List.keysort(items, 2), {max_weight, []}, &select_until/2 )
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selected_items
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Enum.sort_by( items, fn {_name, weight, price} -> - price / weight end )
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|> Enum.reduce( {max_weight, []}, &select_until/2 )
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|> elem(1)
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|> Enum.reverse
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end
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def task( max_weight, items ) do
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def task( items, max_weight ) do
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IO.puts "The robber takes the following to maximize the value"
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for {name, weight} <- select( max_weight, price_per_weight(items) ), do: :io.fwrite("~.2f of ~s~n", [weight, name])
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Enum.each( select( max_weight, items ), fn {name, weight} ->
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:io.fwrite("~.2f of ~s~n", [weight, name])
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end )
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end
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defp select_until( {name, weight, _price}, {remains, acc} ) when remains > 0 do
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@ -31,4 +33,4 @@ items = [ {"beef", 3.8, 36},
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{"salami", 3.0, 95},
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{"sausage", 5.9, 98} ]
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KnapsackProblem.task( 15, items )
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KnapsackProblem.task( items, 15 )
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@ -0,0 +1,32 @@
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defmodule KnapsackProblem do
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def continuous(items, max_weight) do
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Enum.sort_by(items, fn {_item, {weight, price}} -> -price / weight end)
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|> Enum.reduce_while({max_weight,0}, fn {item, {weight, price}}, {rest, value} ->
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if rest > weight do
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IO.puts "Take all #{item}"
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{:cont, {rest - weight, value + price}}
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else
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:io.format "Take ~.3fkg of ~s~n~n", [rest, item]
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:io.format "Total value of swag is ~.2f~n", [value + rest*price/weight]
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{:halt, :ok}
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end
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end)
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|> case do
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{weight, value} ->
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:io.format "Total: weight ~.3fkg, value ~p~n", [max_weight-weight, value]
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x -> x
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end
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end
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end
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items = [ beef: {3.8, 36},
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pork: {5.4, 43},
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ham: {3.6, 90},
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greaves: {2.4, 45},
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flitch: {4.0, 30},
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brawn: {2.5, 56},
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welt: {3.7, 67},
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salami: {3.0, 95},
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sausage: {5.9, 98} ]
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KnapsackProblem.continuous( items, 15 )
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@ -4,8 +4,8 @@ class KnapsackItem {
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has $.price is rw;
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has $.ppw;
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method new (Str $n, $w, $p) {
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KnapsackItem.bless(*, :name($n), :weight($w), :price($p), :ppw($w/$p))
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method new (Str $n, Rat $w, Int $p) {
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self.bless(:name($n), :weight($w), :price($p), :ppw($w/$p))
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}
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method cut-maybe ($max-weight) {
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@ -1,45 +1,45 @@
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/*REXX program solves the (continuous) burglar's knapsack problem. */
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@.= /*═══════ name weight value ══════*/
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@.1 = 'flitch 4 30 '
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@.2 = 'beef 3.8 36 '
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@.3 = 'pork 5.4 43 '
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@.4 = 'greaves 2.4 45 '
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@.5 = 'brawn 2.5 56 '
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@.6 = 'welt 3.7 67 '
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@.7 = 'ham 3.6 90 '
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@.8 = 'salami 3 95 '
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@.9 = 'sausage 5.9 98 '
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parse arg maxW d . /*get possible arguments from the C.L. */
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if maxW=='' | maxW==',' then maxW=15 /*the burglar's knapsack maximum weight*/
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if d=='' | d==',' then d= 3 /*number of decimal digits in FORMAT. */
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wL=d+length('weight'); nL=d+length('total weight'); vL=d+length('value')
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/*REXX pgm solves the continuous burglar's knapsack problem; items with weight and value*/
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@.= /*═══════ name weight value ══════*/
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@.1 = 'flitch 4 30 '
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@.2 = 'beef 3.8 36 '
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@.3 = 'pork 5.4 43 '
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@.4 = 'greaves 2.4 45 '
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@.5 = 'brawn 2.5 56 '
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@.6 = 'welt 3.7 67 '
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@.7 = 'ham 3.6 90 '
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@.8 = 'salami 3 95 '
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@.9 = 'sausage 5.9 98 '
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parse arg maxW d . /*get possible arguments from the C.L. */
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if maxW=='' | maxW=="," then maxW=15 /*the burglar's knapsack maximum weight*/
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if d=='' | d=="," then d= 3 /*number of decimal digits in FORMAT. */
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wL=d+length('weight'); nL=d+length("total weight"); vL=d+length('value')
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totW=0; totV=0
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do #=1 while @.#\==''; parse var @.# n.# w.# v.# .
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end /*#*/ /* [↑] assign item to separate lists. */
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#=#-1 /*#: is the number of items in @ list.*/
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call show 'unsorted item list' /*display the header and the @ list.*/
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call sortD /*invoke sort (which sorts descending).*/
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do #=1 while @.#\==''; parse var @.# n.# w.# v.# .
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end /*#*/ /* [↑] assign item to separate lists. */
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#=#-1 /*#: is the number of items in @ list.*/
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call show 'unsorted item list' /*display the header and the @ list.*/
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call sortD /*invoke sort (which sorts descending).*/
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call hdr "burglar's knapsack contents"
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do j=1 for # while totW<maxW; f=1 /*process items. */
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if totW+w.j>=maxW then f=(maxW-totW)/w.j /*calculate fract.*/
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totW=totW+w.j*f; totV=totV+v.j*f /*add ───► totals.*/
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call syf left(word('{all}',1+(f\==1)),5) n.j, w.j*f, v.j*f
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end /*j*/ /* [↑] show item.*/
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call sep; say; t='t' /* [↓] $ suppresses trailing zeroes.*/
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call sy left('total weight',nL,'─'), $(format(totW,,d))
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call sy left('total value',nL,'─'), , $(format(totV,,d))
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exit /*stick a fork in it, we're all done. */
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/*────────────────────────────────────────────────────────────────────────────*/
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sortD: do s=2 to #; a=n.s; !=w.s; u=v.s /* [↓] this is a descending sort.*/
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do k=s-1 by -1 to 1 while v.k/w.k<u/!;?=k+1;n.?=n.k;w.?=w.k;v.?=v.k;end
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?=k+1; n.?=a; w.?=!; v.?=u
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end /*s*/
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return /* ↑↑↑ sort algorithm is OK for small arrays.*/
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/*──────────────────────────────────one─liner subroutines─────────────────────*/
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hdr: say; say; say center(arg(1),50,'─'); say; call title; call sep; return
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sep: call sy copies('═',nL), copies("═",wL), copies('═',vL); return
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show: call hdr arg(1); do j=1 for #; call syf n.j,w.j,v.j; end; return
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sy: say left('',9) left(arg(1),nL) right(arg(2),wL) right(arg(3),vL); return
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syf: call sy arg(1), $(format(arg(2),,d)), $(format(arg(3),,d)); return
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title: call sy center('item',nL), center("weight",wL), center('value',vL);return
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$: arg x; if pos(.,x)>1 then x=left(strip(strip(x,'T',0),,.),length(x));return x
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do j=1 for # while totW<maxW; f=1 /*process the items. */
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if totW+w.j>=maxW then f=(maxW-totW)/w.j /*calculate fraction. */
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totW=totW+w.j*f; totV=totV+v.j*f /*add it ───► totals. */
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call syf left(word('{all}',1+(f\==1)),5) n.j, w.j*f, v.j*f
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end /*j*/ /* [↑] display item. */
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call sep; say ' /* [↓] $ suppresses trailing zeroes.*/
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call sy left('total weight', nL, "─"), $(format(totW,,d))
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call sy left('total value', nL, "─"), , $(format(totV,,d))
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exit /*stick a fork in it, we're all done. */
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/*──────────────────────────────────────────────────────────────────────────────────────*/
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sortD: do s=2 to #; a=n.s; !=w.s; u=v.s /* [↓] this is a descending sort. */
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do k=s-1 by -1 to 1 while v.k/w.k<u/!; ?=k+1; n.?=n.k; w.?=w.k;v.?=v.k;end
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?=k+1; n.?=a; w.?=!; v.?=u
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end /*s*/
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return /* ↑↑↑ sort algorithm is OK for small arrays*/
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/*──────────────────────────────────────────────────────────────────────────────────────*/
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hdr: say; say; say center(arg(1),50,'─'); say; call title; call sep; return
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sep: call sy copies('═', nL), copies("═", wL), copies('═', vL); return
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show: call hdr arg(1); do j=1 for #; call syf n.j, w.j, v.j; end; return
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sy: say left('',9) left(arg(1),nL) right(arg(2),wL) right(arg(3),vL); return
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syf: call sy arg(1), $(format(arg(2), , d)), $(format(arg(3), , d)); return
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title: call sy center('item',nL), center("weight", wL), center('value', vL); return
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$: parse arg x;if pos(.,x)>1 then x=left(strip(strip(x,'T',0),,.),length(x)); return x
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@ -0,0 +1,36 @@
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/* create SAS data set */
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data mydata;
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input item $ weight value;
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datalines;
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beef 3.8 36
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pork 5.4 43
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ham 3.6 90
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greaves 2.4 45
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flitch 4.0 30
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brawn 2.5 56
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welt 3.7 67
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salami 3.0 95
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sausage 5.9 98
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;
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/* call OPTMODEL procedure in SAS/OR */
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proc optmodel;
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/* declare sets and parameters, and read input data */
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set <str> ITEMS;
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num weight {ITEMS};
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num value {ITEMS};
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read data mydata into ITEMS=[item] weight value;
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/* declare variables, objective, and constraints */
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var WeightSelected {i in ITEMS} >= 0 <= weight[i];
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max TotalValue = sum {i in ITEMS} (value[i]/weight[i]) * WeightSelected[i];
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con WeightCon:
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sum {i in ITEMS} WeightSelected[i] <= 15;
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/* call linear programming (LP) solver */
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solve;
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/* print optimal solution */
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print TotalValue;
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print {i in ITEMS: WeightSelected[i].sol > 1e-3} WeightSelected;
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quit;
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