orie321sol2 - A“ W L “‘1 0/ 4*- m 0/ e45.» Myra/ax;...

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I 0’ I741 601% :- ~‘V’/JJ"JF‘//‘-£I~ f) V; $00441 ,a/h Q“ GA 1-; ‘IJ dead-{6:1 .1 vi nooéd fr Jrcoép/ ach/pc j'fQ/IJ . MT/( “40'; canrmdZ ran LJJG '/ (.6 ' Jc/wa/‘l O, (acéh/O/o/ fémr‘J/o” (1) (i) . , . -19..-.__-_.._LQ._.-. mm- ._,.___Mh~, mki)=\30-to~z , oz=\z-x, dao...lL (3.4) 1.5: 0-- C: M 11 r. O,_ \L crab) : “mm { \;(-a+\‘\»c,)k [30-033 = Q:o.-\?, awn. {lOQ—13i4239 itou‘z “202 -\3r{ U: o.-e\ a” b. u 0 3&13:1..\L Lu T : 113%) -1 \‘2. «31(5); MW“ i\1L—&+3+QB +\';o- wig :‘ Iva:ng : 2.02 43»; + 3(vGB = \qO—‘Ow‘ ($9-42} «13(50 : Zoi~l31 ()3 ; Q—«K 1:0,- Q "7375).? \CW ' ‘0‘ ;, ,0 _= ,0 . .7 133w)?” Us i-G), : Mm -{ MW i 1\(-1+\‘\Ho} +202 42.93 )mg ‘1\\(~1+\§+ro)+\fio»loq‘§ 3:0“6 V1.41 -_._ wkxmm igxx—mafi MN fi'Eoo—\\q+\\’g7)= =0. e ’5=%. \7. T kg ‘1 wiwv13= = Mm’110g*uq P30}-th i z 305,!“ w mg Iqr-HN’L M Igzlq—k),~ )2 m4 = NW ‘2 mm, {3% -\\«;' I Mm, iloo-\\jy+rx'1fi: i=&-IO)...’G ERA“; MN: V Hm“, E TM "x: : M WU) = sag—nag ,, OR -=— \6—i, 4: 0.- x2”, (33c) ,0— m Wk L, W n13 NS(O\®W Mu M :mkdugwg, «HQ 2 ’mi \OCQH‘U ‘r BOG’HJ‘E : - i:°., _ w M was _ V; m = w n (rm) _ SMWth$ '- IL 0L RL (08" V , 5 O J? Pr \0 = 90 _ H \1 Jor H H. = w - -- a- _ __ 3 G No 0 63.3. = Ow mm-..“ _._ _ 1 0 “I37 H 3 \3 _\o = V50 m mm “M m l __ Lg _ __Q__w__fl Lk 0 -n = 0 Problem 20 We can solve this problem using the general resource allocation model. If we view workers as “resources,” and divisions as “production activities,” we can determine the parameters for the model as follows. 1. B = 10; we have at most ten workers. 2. cl. (xi )= x J. ; we can also identify workers with “units of products,” because the profit we make depends on the numbers of workers allocated to the different divisions, and not on how many units they produce. 3. p j (x I.) is the amount of profitachieved if x j workers are assigned to division j, according to the table given in the problem. 4. b j = 10; we can assign at most ten workers to one division. We are interested in the minimum number of workers needed to achieve 250 units of profit. So we need to compute f (5,0), f (5,1), f (5,2).., and stop once we break the 250 mark. Since units and resources have basically been equated, the recursion for this model can thus be simplified as follows: f(k,y)= maxim (x)+ f(k-1,y-x)} 05x5 y All we need to do now is compute the optimal profit values for various combinations of divisions and workers: f (k,0) = f (0,k) == 0; we can’t accomplish anything with zero workers or divisions. f (1,k) = pl (k); if we only have one division, we assign all of our workers to it. f(2,1) = fig;wa m y -— x)}= max{p2 (1), f(1,1)} = max{p2 (1), p. (1)}= p. (1) = 30 2,2) = max{p,(x)+ f(1,2 - x)}= 55; x = o 2 05x52 f ,3) = 933%{172 (x) + f (1,3 —— x)} = 80; x = 1 f(2,4 = 323945;» f(1,4-— x)}=105; = f (2,5) = Ego, (x)+ f(1,5 -— x)}= 128, = “16) = max{p2 (x = II x 2 x 2 f(1,6-—x)}= 150; x 3 x 3 x 3 x= 4 f(2,10) = niax{p2 (x)+ f(1,10 —- x)}= 223; x = 5 05x51 0 f(3,1) = (x)+ f(2,1- x)}= 35; x =1 f(3,2) = max{p3(x)+ f(2,2 - x)}= 65; x = 1 03x52 f (33) = $33M (x)+ f (2,3 - x)}= 95; x = 2 f(3,4) = max{p3(x)+ f(2,4— x)}= 120; x = 2 03x54 f(3,5) = gagb3(x)+ f(2,5 — x)}= 145; x = 2 OSxS7 = %g@3(x)+f(2,8—x)}= 218; x = 3 = max{p3(x)+ f(2,9 — x)}= 240; x = 3 3,9 05x59 3,10) = max{p3(x)+ f(2,10— x)}= 260; x = 4 OSxSl 0 ) f(3,7) = max{p3(x)+ f(2,7 - x)}= 195; x = 3 ) ) (4.9 (dv.{!!1.(( {a 14(5 It}: xv. . ‘ fF< hr: < I n. . .3. r. o. p v . . ID .Sm mm 02? ESE? 05 Ea . Qmfifid mm 5580:“ 32:30 as? 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orie321sol2 - A“ W L “‘1 0/ 4*- m 0/ e45.» Myra/ax;...

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