AMS210_PreHW-Solutions_Unit_01

AMS 210: Unit 1 Homework Solutions — Fall 2022 Question Sources Most questions are adapted from either “Introduction to Linear Algebra, bth Edition” by Gilbert Strang or from “Introduction to Linear Algebra: Models, Methods and Theory” by Alan Tucker. All questions not labeled as from one of these sources have been created by David Green, Stony Brook University. Any sharing of this material without the express permission of the copyright holder is strictly forbidden. Pre-Homework Practice (Optional) 1. (Tucker 1.1.5) Store A expects to sell three times as many books as store B this month, but next month store B’s sales are expected to double while store A’s sales remain the same. Over the 2-month period, the two stores together are expected to sell a total of 4500 books. How many books would each store sell each month? Solution: Let x be the number of books sold by store B this month. Then 3z is the number sold by store A both this month and next, and 2z is the number sold by store B next month. The equation for the total sales is then: z + 3z 4 2z 4+ 3z = 4500 — 9z = 4500 — =z = 500, 2z = 1000, 3=z = 1500 Store A is expected to sell 1500 books both this month and next, while store B is expected to sell 500 books this month and 1000 next month. 2. (Tucker 1.2.1) Consider a model for the production of heating oil, diesel and gasoline from three refineries: H = 20161 + 4.’L'2 + 4163 D = 10%1 + ].4.2’,‘2 + 5373 G = 5171 + 5.’13'2 + ].22’,'3 where x1, z,, and z3 are the number of barrels of petroleum (in thousands) processed at each refinery in a day, and H, D, and G are the number of gallons (in thousands) of heating oil, diesel and gasoline produced, respectively. (a) If refinery 1 processes 15 thousand barrels of petroleum, refinery 2 processes 20 thousand barrels and refinery 3 processes 60 thousand barrels each day, how much of each product is produced daily? Solution: We have z; = 15, 25 = 20, and z3 = 60, using units of “thousands of barrels” to simplify the arithmetic. H = 20(15) + 4(20) + 4(60) = 620 D = 10(15) + 14(20) + 5(60) =730 G = 5(15) + 5(20) + 12(60) =895 620 thousand gallons of heating oil, 730 thousand gallons of diesel and 895 thousand gallons of gasoline are produced each day. (©)2018-2022. David F. Green, Stony Brook University

(b) Given the same inputs as (a), if there is a daily demand for 500 thousand gallons of heating oil, 850 thousand gallons of diesel and 1 million gallons of gasoline, which products will meet the demand, and which will not? What is the greatest deviation between production and demand? Solution: Only heating oil is produced in a quantity that is able to meet the demand, with a 120 thousand gallon excess. Diesel production leaves a 120 thousand deficit compared to the demand, while gasoline productions leaves a 105 gallon deficit. 3. (Tucker 1.2.9) Consider a linear economic supply-demand model described by: Industrial Demands Consumer Total Demand Ener. Const. Trans. Steel Demand Energy : d = 04z; + 02z + 0223 + 0.2z4 + 100 Construction : d = 03x; + 03z + 0223 + 0.1xy + 50 Transportation : d3 = 0.1x; 4+ 0.1z, + 0224 + 100 Steel : dy = 0.1z, + 0.1zs + 0 where z; is the supply of energy, x5 is the supply of construction, z3 is the supply of transportation, and xz, is the supply of steel, all in arbitrary “units of supply”, and d, ds, ds, and dy are the corresponding total demands for each commodity. If 300 units of energy, 250 units of construction, 160 units of transportation and 40 units of steel are produced, what is the largest deviation between supply and demand among all four commodities? Solution: We have x; = 300, x5 = 250, z3 = 160 and x4, = 40. Thus the total demands are: d = 04(300) + 0.2(250) + 0.2(160) + 0.2(40) + 100 =310 dy = 0.3(300) + 0.3(250) + 0.2(160) + 0.1(40) + 50 =251 ds = 0.1(300) + 0.1(250) + 0.2(40) + 100 =163 dy = 0.1(250) + 0.1(160) + 0=41 The largest deviation is 310 — 300 = 10, for energy. 4. (Tucker 1.3.3) Let p1, p2, ps, P4, s and pg be a set of current probabilities for 6 states of a Markov chain. Let p} be the probabilities of the same states one period in the future, defined by: P, = 05p1 + 0.25p, ph = 0.5p; + 0.50p; + 0.25p3 p, = 0.25p, + 0.50ps + 0.25p, Py = 0.25p3 + 0.50ps + 0.25p5 Py = 0.25p, + 0.50ps + 0.5pg Py = 0.25p5 + 0.5ps What are the probabilities of all six states one period in the future if: (a) po =p3s =0.5 and p; = ps = ps = pg = 07 (©)2018-2022. David F. Green, Stony Brook University

the values to be the actual number of individuals in a specific (small) area, then rounding to whole numbers might be more appropriate. This would giveg;: Ry = 30+ 0.1(30) — 0.15(24) = 29.4 — 29 Fy =24 —0.15(24) + 0.1(30) = 23.4 — 23 After 2 months: Ry =29 + 0.1(29) — 0.15(23) = 28.45 — 28 Fy = 23 — 0.15(23) + 0.1(29) = 22.45 — 22 After 3 months: R; =28 +0.1(28) — 0.15(22) = 27.5 — 28 F3 =22 —0.15(22) + 0.1(28) = 21.5 — 22 Note that rounding leads to apparent stability at R = 28 and F' = 22, at least if we round both 27.5 and 21.5 up. 6. (Tucker 1.4.3) Use trial and error to find an approximate solution to the system of equa- tions with a maximum error of no more than 10% for any of the equations: 10y + 6x2 = 200 7.’L’1 + 7172 = 150 55171 + 4372 = 100 Solution: Note: There are many different guesses that can be made. This is just one possible approach to coming up with guesses. Let’s choose an initial guess of £; = x5 (a balanced guess). Inspection tells us if we set these both to 10, we will get a right-hand side a bit below that desired for all three equations: 10(10) + 6(10) = 160 < 200 7(10) + 7(10) = 140 < 150 5(10) + 4(10) = 90 < 100 The largest error is % = 20%. All values are too low, but the first equation has the greatest error, so let’s try to increase x;, which has the largest coeflicient for z; we’ll try 1 = 13, which should increase each equation’s value by between 15 and 30. 10(13) + 6(10) = 190 < 200 7(13) + 17(10) 161 > 150 5(13) + 4(10) = 105 > 100 The largest error is now % ~ 7.3%. With a maximum error of less than 10%, this is a reasonable place to stop. (©2018-2022. David F. Green, Stony Brook University