quantitative tool in business

MAY 2012 QUANTITATIVE TOOLS IN BUSINESS SOLUTION 2(c)Let x be the project durationThen x N (24.3, 1.1553)P (x > 26) = P x – 24.3 > 26 – 24.31.1553 1.1553= P (Ƶ > 1.47)From tables, P(x > 26) = 0.0708(d)From the Gantt Chart the activities that were curtailed are E, H and ISOLUTION 3(a)The problem can be formulated as follows:Max Z = 25x 1 + 20x 2 + 15x 3S.t. 2x 1 + x 2 + 3x 3 ≤ 400 (Skilled Labour)4x 1 + 2x 2 + x 3 ≤ 600 (Leather)6x 1 + 2x 2 + x 3 ≤ 1200 (Glue)x 1 ≤ 100 (Demand for A)x 1 , x 2 , x 3 ≤ 0 (Non negativity)(b)Initial Simplex TableauBasis X1 X2 X3 S1 S2 S3 S4 Constant ……S1S2S3S4246(1)1222311010000100001000014006001200100200150200100Ƶ -25 -20 -15 0 0 0 0 0(c)(d)From the illustration in the table above, the increase in total revenue is GH¢2500From the final Simplex Tableau;(i) Production-mix is: x 1 = 100x 2 = 80x 3 = 40(ii)(iii)Maximum total revenue = Max Ƶ = GH¢4900Shadow price for skilled labour = GH¢2/hourPage 4 of 12

MAY 2012 QUANTITATIVE TOOLS IN BUSINESS SOLUTION 2Daily Forecast for next weekDay Trend (T) SA1 Sale figureA(Y/GH¢’00)MonTueWedThurFri3.886 + 3.289 (16) = 56.513.886 + 3.289 (17) = 59.7993.886 + 3.289 (18) = 63.0883.886 + 3.289 (19) = 66.3773.886 + 3.289 (20) = 69.6661.2961.0400.9180.9000.84673.2462.1957.9159.7458.94SOLUTION 5(i) Total Cost (C) = x 2 + 16x + 39Average Cost C = x 2 + 16x + 39X x= x + 16 + 39x(ii) Total Cost (C) = x2 + 16x + 39Marginal Cost = dc = 2x + 16…= 2x + 16(iii) Price (P) = x 2 - 24 x + 117Total Revenue (TR) = Px = x 3 - 24x 2 + 117xTotal Revenue (TR) x 3 – 24x 2 + 117x(iv) Total Revenue (TR) = x 3 – 24x 2 + 117xMarginal Revenue = dTR = 3x 2 - 48x + 117= dx= 3x 2 - 48x + 117x 0 1 2 3 4 5 6 7 8 9 10x + 16 + 39/x 00 56 37.5 32 29.9 28.8 28.5 28.6 28.9 29.3 29.92x + 16 16 18 20 22 24 26 28 30 32 34 36x2 – 48x + 117 117 72 33 0 -27 -48 -63 -72 -75 -72 -63Page 7 of 12

MAY 2012 QUANTITATIVE TOOLS IN BUSINESS SOLUTION 2R = x3 – 24x2 + 117Total Revenue (TR) is maximized where MR = cMR = dTR = 3x 2 – 48x + 117 = 0dx3x 2 – 48x + 117 = 03x 2 – 39x – 9x + 117 = 0(x – 13) (3x – 9) = 0x = 13, x = 3d-TR = dMR = 6x - 48dx 2 dxwhen x = 3, dMR = 18 – 48 = - 30dx:. x = 3 maximumP = x 2 – 24x + 117when x = 3P = 3 2 – 24 (3) + 117 = 54P = x – 24x + 117 (Price function)dP/dx – 2x – 24Elasticity (x = 3) = P . 1/dP/dxx= 54 . 1___3 (2x – 24)= 18 . 1 __ = - 1(6 – 24)Elasticity = - 1(v)Profit is maximized at mc = MRFrom the graph at mc = MRx = 2.35Price (x = 2.35) = (2.35) 2 – 24 (2.35) + 117= 66.1225Elasticity (x = 2.35) = 66.1225 . [1/2 (2.35) – 24)]2.35= 28.137-19.3Elasticity = -1.458Page 8 of 12

MAY 2012 QUANTITATIVE TOOLS IN BUSINESS SOLUTION 2SOLUTION 6Random sampling is a sampling technique in which all observations have an equal chance tobeing selected. Example: if there are 10 balls in a box, the selection of any one of random hasequal chance of 1/10 of being selected. Their sampling method is mostly used by Lotteries.Quota sampling is a sampling technique where quotas are allotted to identifiable groups and fromeach group selection is made. The groups are homogenous in character. Example: if arepresentation is required for a university student’s body, quotas can be given to each universitythen within each group another sampling method could be used to select the observations. Thistechnique is used by FIFA to select representations for its tournaments.Systematic sampling is a sampling technique of selecting sample members from a largerpopulation according to a random starting point and a fixed, periodic interval. Typically, every"nth" member is selected from the total population for inclusion in the sample population.For example, if you wanted to select a random group of 1,000 people from a population of50,000 using systematic sampling, you would simply select every 50th person, since50,000/1,000 = 50. Another example is selecting samples from a production line where every10 th product can be selected for inspection.Frequency Distribution TableYields (kg) Tally Frequency10 – 1415 – 1920 – 2425 – 29|| || | | || | | | |125630 – 34| | | | | | |835 – 39| | | | | | | |940 – 44| | | | | |745 – 4950 – 5455 – 5960 - 64| | | || | || | ||5331__Σf50Page 9 of 12

MAY 2012 QUANTITATIVE TOOLS IN BUSINESS SOLUTION 2(ii)Cumulative Frequency Distribution TableLess than9.5-14.519.524.529.534.539.544.549.554.559.564.5Σf01381422313843464950Frequency TableFields- 14- 19- 24- 29- 34- 39- 445– 492– 545 – 592 - 64Frequency(f)Mid- point(x)___ fx x 2 fx 21121214414421734289578522110484 2420627162729 4374832256 1024 8192937333 1369 12 321742294 1764 12 348547235 2209 11 045352156 2704 8 112357171 3249 9 747162623844 3 844Σf50 Σfx1825 Σfx 2 73,125Mean (x) = Σfx = 1825 = 36.5 kgΣf 50____________Std deviation = √ Σfx 2 - Σfx2Σf Σf________________= √ 73125 - 1825250 50_______________ ______= √ 1462.5 - 1332.25 = √ 130.25σ= 11.41 kgPage 10 of 12

MAY 2012 QUANTITATIVE TOOLS IN BUSINESS SOLUTION 2From abovemean = 36.5 kg andσ = 11.41 kgInterval within one std deviation of mean= [25.09, 47.91]on the chart, ((41 – 63) = 35) 35 trees…. Within x + σ, which is 3552 = 70%SOLUTION 7(a)Let L = {worker comes late}M = {worker is male}P(L/M) = 0.08P(M) = 150 = 0.30500Pr (A male late comer) = P(L∩M)= P(M) P(L/M)= 0.30 x 0.08= 0.24(b)Quarterly Production Levels Matrix:P =Q1 Q2 Q3 Q44000 4500 4500 40002000 2600 2400 22005800 6200 6000 6000ABC(c)Unit Production Costs Matrix:C =A B C100 300 150300 400 250100 200 150Raw materialsLabouroverheadsPage 11 of 12

MAY 2012 QUANTITATIVE TOOLS IN BUSINESS SOLUTION 2The matrix CP gives the quarterly costsi.eCP =100300100300400200150250150400020005800450026006200450024006000400022006000= 185000345000167000216000394000190000207000381000183000196000358000174000Hence the following table can be presented at the stockholders’ meeting.QuarterCategory Q1 Q2 Q3 Q4 TotalRaw MaterialLabourOverheads1850003450001670002160003940001900002070003810001830001960003580001740008060001478000714000Total 699000 800000 771000 728000 2998000Page 12 of 12