Operations Management 33:623:386, Spring Solutions to Final Exam Practice Questions. Insurance Advertising. Issuing Bonds.

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1 Operations Management 33:623:386, Spring 2002 Solutions to Final Exam Practice Questions Insurance Advertising x 1 = amount spent on TV ads (in thousands of dollars) x 2 = letters sent min 1000x x 2 ST (25 + 6)x 1 + ( )x (number of new contracts) 6x x [(25 + 6)x 1 + ( )x 2 ] (30% for house insurance) x 1, x 2 0 Issuing Bonds (a) SUMPRODUCT($B13:$B15,B4:B6) (b) Target cell E24 (maximize). Changing cells: B13:B15, B18:E18. Constraints: B13:B15 <= E13:E15, B24:E24 >= B3:E3, Assume nonnegative, assume linear model. Below is an algebraic formulation of the problem. It was not required by the question, but is included here as additional study material. x j = Value of type j = A, B, C bonds issued y i = amount of short term loans taken in year i, i = 1, 2, 3. z i = cash balance at the end of year i, i = 1, 2, 3, 4 All variables express amounts in millions of dollars. Computer Assembly max z 4 z 1 = 55 + y 1 + x A + x B + x C 120 z 2 = 1.045z 1 + y 2 1.1y x A 0.080x B 0.040x C + 20 z 3 = 1.045z 2 + y 3 1.1y x A 0.080x B 0.040x C + 20 z 4 = 1.045z 3 1.1y x A 0.040x C z 1, z 2, z 3, z x A, x B, x C 60 x t = number of computers assembled in Factory 1 in month t = 1, 2, 3, 4 y t = number of computers assembled in Factory 2 in month t = 1, 2, 3, 4 s t = number of computers held in inventory at the end of month t = 1, 2, 3, 4 min 400(x 1 + x 2 + x 3 + x 4 ) + 300(y 1 + y 2 + y 3 + y 4 ) + 100(s 1 + s 2 + s 3 + s 4 ) ST x 1 + y 1 = s s 1 + x 2 + y 2 = s s 2 + x 3 + y 3 = s s 3 + x 4 + y 4 = s x t 800, 3y t 600 for all t = 1, 2, 3, 4 x t 0, y t 0, s t 0, for all t = 1, 2, 3, 4 Note that the nonnegativity constraints on the inventory variables s t ensure that we always satisfy the demand. 1

2 Mixing Fertilizer (a) SUM(E16:E17) SUM(B19:C19) (b) B11*$D16 (c) Target cell E20 (maximize). Changing cells B16:C17. Constraints: B16:C17 >= B23:C24 (this is the key blending constraint), B18:C18 <= B3:C3. Assume nonnegative, assume linear model. Production and Shipment Optimization (a) =D6 + D12/D7 (b) =SUM(C18:C20) (c) =F10*F21 (d) D24 = SUMPRODUCT(B3:B5,F18:F20), D25 = SUM(D12:F12), D26 = SUMPRODUCT(D3:F5,C18:E20) (e) =D9*F21, =F24 D27 (f) target cell F26; minimize; constraints: F18:F20 <= C3:C5, C21:E21 <= F22, D27 <= D10, C21:E21 <= D14:F14, assume linear model, assume nonnegative Goofy Park (a) (b) Additional Variables: s i = Start time (in weeks) of activity i = A,..., H t i = End time (in weeks) of activity i = A,..., H T = Weeks until we declare the project complete min T ST t A = s A + 4 t B = s B + 2 t C = s C + 2 t D = s D + 8 t E = s E + 5 t F = s F + 2 t G = s G + 2 t H = s H + 8 s A 0 s B t A s C t B s D t B s E t B s F t C s G t D s G t E s G t F s H t A T t A,..., t H r A = reduction of time of activity A r C = reduction of time of activity C r D = reduction of time of activity D 2

3 Modified constraints: t A = s A + 4 r A t C = s C + 2 r C t D = s D + 8 r D Additional constraints: T 12 r A, r C, r D 0 New objective function (in thousands of dollars): min 10r A + 5r B + 12r C. (c) Add a new binary variable: { 1 buy the prefabricated court z = 0 do not buy and modify the start/end relation for E as follows: t E = s E + 5 4z. Also, add the constraint z {0, 1}. The objective function becomes min 10r A + 5r B + 12r C + 30z. Note: If this formulation looks unfamiliar, just eliminate the t X variables using the equality constraints. This comment applies to all parts of this question. Either style of answer is acceptable. Memory Manufacturing and Purchasing x 1 = number of Type 1 SIMMs produced per week x 2 = number of Type 2 SIMMs produced per week x 3 = number of DIMMs produced per week x 4 = number of X-DIMMs produced per week y 1 = number of chips purchased from Sleeman y 2 = number of chips purchased from Chang y 3 = number of chips purchased from Malaya 3

4 max (75x x x x 4 ) (14.50y y y 3 ) 20.50(0.03x x x x 4 ) 15.00(0.025x x x x 4 ) ST 0.03x x x x (labor hours) 0.025x x x x (machine hours) 4x 1 + 4x 2 + 8x 3 + 8x 4 = y 1 + y 2 + y 3 (memory chips) y y y y 1 (0.4)(y 1 + y 2 + y 3 ) y 2 (0.4)(y 1 + y 2 + y 3 ) (blending constraints) y 3 (0.4)(y 1 + y 2 + y 3 ) x 1, x 2, x 3, x 4, y 1, y 2, y 3 0 Grading Language Tests Let i = 1, 2, 3 denote senior, junior, and outside raters, respectively. Let j = 1, 2, 3 denote written, computer, and recorded tests, respectively. x ij = number of type j = 1, 2, 3 tests assigned to raters of type i = 1, 2, 3 min 25x x x 33 x 11 + x 21 + x 31 = 800 x 12 + x 22 + x 32 = 800 x 13 + x 23 + x 33 = x x x x x x x (x 11 + x 21 ) x (x 12 + x 22 ) x (x 13 + x 23 ) x 11,..., x 33 0 (written tests) (computer tests) (recorded tests) (time of senior raters) (time of junior raters) (10% senior in house raters for written tests) (10% senior in house raters for computer tests) (10% senior in house raters for recorded tests) (all variables nonnegative) Atlas Valve { 1 make valve i x i = i = 1,..., 5 0 do not make valve i y i = number of type i = 1,..., 5 valves made max 741y y y y y 5 ST 0 y i 100x i, i = 1,..., 5 16x x x 3 + 8x x 5 + 2y y y y 4 + 3y y 1 + 3y 2 + 3y y y y y y y y x 1 + x 2 + x 3 + x 4 + x 5 4 x 2 + x 4 1 x i {0, 1} i = 1,..., 5 4

5 Investment Planning (a) =SUM(B18:B21) (b) = SUMPRODUCT(B3:D3,B17:D17) (c) =SUMPRODUCT(B4:D4,B18:D18) (d) =SUMPRODUCT(E3:F3,E$13:F$13) (e) =G9 + E17 + F17 G3 (f) The formula in G18 is = G17*(1+D$9) + E18 + F18 G4 (g) Target cell G21; maximize; changing cells are B17:D21, E13;F13; constraints: B17:D21 binary, B22:D22 <= B17:D17 (sell assets that you bought), G17:G21 >=0 (keep nonnegative cash balance); assume linear model; assume nonnegative Inventory with Setup Costs (a) =E12 + C13 B2 (b) =B13*E$4 (c) =B13*E$5 (d) =D13*F$5 (e) =C13*D$3 + B13*B$2 + E13*F$3 + D13*F$2 (f) =SUM(I13:I18) (g) target cell I19; minimize; changing cells B13:D18; constraints: B13:B18 binary, D13:D18 binary, C13:C18 <= G13:G18, C13:C18 >= F13:F18, E13:E18 >= 0, E13:E18 <= H13:H18, D13:D18 <= B13:B18; assume linear model; you may, but need not, assume nonnegative Housing Developments x i = { 1 accept offer i 0 do not accept i = 1,..., 7. max (40)(370)x 1 + ( )x 2 + ( )x 3 +(65)(160)x 4 + (35)(420)x 5 + x 6 (75)(130) + x 7 (50)(190) ST x 1 + x x 7 = 3 40x x x x x x x x 2 + x 3 + x 6 1 x 2 + x 3 + x 4 + x 7 1 x 1 + x 5 = 1 x 1,..., x 7 {0, 1} 5

6 Siting Restaurants (a) SUMPRODUCT(B6:E9,B12:E15) (b) SUMPRODUCT($A12:$A15,B12:B15) (c) B5*B18 (d) Target cell B21 (maximize). Changing cells: B12:E15. Constraints: B12:E15 binary, B16:E16=1, F18:F19 <= B1:B2, B18 <= C18. Assigning People We denote projects A, B, C, and D by 1, 2, 3, and 4, respectively. x ij = 1 if we assign exactly j = 1, 2, 3 people to project i = 1, 2, 3, 4, otherwise 0 min 5x x x x x x x x x x x x 43 ST x 11 + x 12 + x 13 = 1 x 21 + x 22 + x 23 = 1 x 31 + x 32 + x 33 = 1 x 41 + x 42 + x 43 = 1 (x 11 + x 21 + x 31 + x 41 ) + 2(x 12 + x 22 + x 32 + x 42 ) + 3(x 13 + x 23 + x 33 + x 43 ) 6 x 11,..., x 43 {0, 1} Locating Restaurants (a) { 1 locate restaurant in town j x j = 0 do not locate j = 1,..., 6. min x 1 + x x 6 ST x 1 + x 3 1 x 2 + x 4 1 x 1 + x 3 + x 4 1 x 2 + x 3 + x 4 + x 5 1 x 4 + x 5 + x 6 1 x 5 + x 6 1 x 2 = x 5 (or use this to eliminate one variable) x 1,..., x 6 {0, 1} (b) Additional variables: y i = { 1 cover town i 0 do not cover i = 1,..., 6. 6

7 max 11y y y 6 ST x 1 + x 3 y 1 x 2 + x 4 y 2 x 1 + x 3 + x 4 y 3 x 2 + x 3 + x 4 + x 5 y 4 x 4 + x 5 + x 6 y 5 x 5 + x 6 y 6 x 1 + x x 6 = 2 x 2 = x 5 x 1,..., x 6 {0, 1} y 1,..., y 6 {0, 1} Constructing a Mall (a) (b) x i = start time (in days) of activity i = A, B,..., H. d = time we declare we are done (in days) min d ST x C x A + 25 x C x B + 10 x D x C + 30 x E x C + 30 x F x D + 22 x F x E + 17 x G x E + 17 x H x F + 10 x H x G + 14 d x H + 5 x A, x B 0 (Note: x C, x D,..., x G, d 0 may also be included, as well as additional (redundant) constraints like d x F + 10.) x i = start time (in days) of activity i = A, B,..., H. d = time we declare we are done (in days) y i = days to reduce activity i = B, C, D, E, F, G z i = 1 if we reduce activity i = B, C, D, E, F, G at all, 0 if we do not 7

8 min ST d x C x A + 25 y A x C x B + 10 y B x D x C + 30 y C x E x C + 30 y C x F x D + 22 y D x F x E + 17 y E x G x E + 17 y E x H x F + 10 y F x H x G + 14 y G d x H + 5 y H y B 5z B y C 12z C y D 6z D y E 8z E y F 4z F y G 6z G 2000z B z C z G + 400y B + 250y C y G x A, x B 0 y B,..., y G 0 z B,..., z G {0, 1} Refining Germanium x 1 = Kilos of raw germanium processed x 2 = Kilos of regular germanium remelted Optional helper variables: y 1 = Kilos of regular germanium produced y 2 = Kilos of premium germanium produced y 3 = Kilos of defective germanium produced min 44x x x y 3 ST y 1 =.45x 1 x 2 y 2 =.3x 1 +.9x 2 y 3 =.25x 1 +.1x 2 y 1 + y 2 50 y 2 35 x 1, x 2 0 y 1 0 Note: the constraint y 3 0 would customarily be included too, but is redundant. The constraint y 1 0 (or, if you don t use the helper variables,.45x 1 x 2 0) is critically important. Writing Modular Software x i = 1 if we write module i = A, B,..., F, otherwise 0 y j = 1 if we market product j = 1, 2,..., 7, otherwise 0 8

9 max (7y y y 7 ) (4x A + 5x B + + 5x F ) ST y 1 x A y 1 x B y 2 x A y 2 x E y 3 x B y 3 x C y 3 x D y 4 x C y 4 x D y 5 x C y 5 x D y 5 x E y 6 x D y 6 x F y 7 x E y 7 x F y 1,..., y 7 {0, 1} x A,..., x F {0, 1} Some students gave alternative solutions in which the constraints above were grouped and added together either by module or product. Grouping by product, the constraints would reduce to Grouping by module, we d get 2y 1 x A + x B 2y 2 x A + x E 3y 3 x B + x C + x D.. y 1 + y 2 2x A y 1 + y 3 2x B y 3 + y 4 + y 5 3x C These solutions are correct and were also given full credit, although for technical reasons having to do with how difficult the problem would be to solve, a professional would probably prefer the first formulation. Pension Fund Planning (a) G18 = SUM(B18:E18) (b) B19 = B17*$G6 (c) B20 = B17*$G2 (d) B24 = SUMPRODUCT(B3:E3,B$18:E$18) (e) C24 = B12*$G$18 (f) B28 = SUMPRODUCT(B8:E8,B18:E18) (g) Target cell is B28, Maximize Changing cells are B17:E18 B18:E18 >= B19:E19 B18:E18 <= B20:E20 B17:E17 binary B24:B26 >= C24:C26.. 9

10 B24:B26 <= D24:D26 Assume linear model Assume nonnegative is not necessary, but may be selected Setting Up Telemarketing Centers Denote the locations as i = 1,..., 7 in the same order as the table. x i = 1 if we open a center at location i = 1,..., 7, otherwise 0. The objective function below is in $100,000 s; any scaling is permissible. min 5x 1 + 8x x 7 ST x 2 + x 3 + x 5 + x 7 1 (cover area 314) x 1 + x 2 + x 6 1 (cover area 316) x 1 + x 3 + x 5 + x 6 1 (cover area 417) x 1 + x 4 + x 6 1 (cover area 501) x 2 + x 4 + x 5 + x 7 1 (cover area 502) x 2 + x 4 + x 6 + x 7 1 (cover area 606) x 2 + x 5 + x 7 1 (cover area 816) x 1 + x 3 1 (cover area 918) x 1 + x 3 1 (cannot open both Dallas and Louisville) x 7 x 1 (if you open St. Louis, you must open Dallas) x 7 x 2 (if you open St. Louis, you must open Atlanta) x 1,..., x 7 {0, 1} Producing Biotech Chemicals Let reagent type i = 1, 2, 3 denote regular, refined, and ultrapure, respectively. Bouquets x i = Liters of type i = 1, 2, 3 reagent made y i = 1 if you sell any reagent of type i = 1, 2, 3, otherwise 0 max 500(x 1 1.1x 2 ) (x 2 1.2x 3 ) x 3 [400x x x 3 ] ST 3x 1 + 2x x x x x x 1 + 1x x y 1 x 1 1.1x 2 100y 1 10y 2 x 2 1.2x 3 50y 2 10y 3 x 3 30y 3 500(x 1 1.1x 2 ) 0.8 (500(x 1 1.1x 2 ) (x 2 1.2x 3 ) x 3 ) 1000(x 2 1.2x 3 ) 0.8 (500(x 1 1.1x 2 ) (x 2 1.2x 3 ) x 3 ) 1500x (500(x 1 1.1x 2 ) (x 2 1.2x 3 ) x 3 ) y 1, y 2, y 3 {0, 1} x 1, x 2, x 3 0 (a) GENBINOMIAL(B9,B1) (because each of B9 bouquets fails with probability B1). (b) GENPOISSON(B2) 10

11 (c) MIN(B11,B12) (because we can t sell more than we have and more than people want) (d) B21 (total profit), that is, B21 should be replaced with SIMOUTPUT(SUM(B18:B20) SUM(B16:B17),A21). Other cells can also have their formulas wrapped in SIMOUTPUT too if you want. (e) PARAMETER(D4:D8,1,A9) Mixing Funds (a) = PARAMETER(G11:G14,1,C10) (b) = SUMPRODUCT(C4:E4,C$11:E$11) (c) E15 = GENNORMAL(H4,I4) E16 = GENNORMAL(H5,I5) E17 = GENUNIFORM(H7,I7) (d) = SIMOUTPUT(SUMPRODUCT(C15:C17,E15:E17),D19) (e) = SIMOUTPUT(IF(E19<H7,1,0),D20) (f) The fourth case, because it has the highest average total return; the probability is (g) The second case since it has the highest average return of the two scenarios with the probability criterion below 10%. Gambling (a) =IF(C12>0,GENTABLE(B$1:D$1,B$2:D$2),0) (b) =C12 + B13 (c) (or roughly $7.19) (d) from the summary statistics for B8 (or one may estimate 40%-45% from the percentile detail statistics for B6). (e) from the summary statistics for B9 (or one may estimate slightly more than 30% since the 70th percentile of B6 is 10.25). Risky Stocks (a) B7 = PARAMETER(C5:E5, 1, A7) (b) B12 = E11 (c) C12 = B$7*B12 (d) D12 = GENBINOMIAL(1, B$1) or GENTABLE(B$2:C$2, B$1:C$1) (e) E12 = IF(D12 = 1, B12 + B$4*C12, B12 C12) Note: in row 23, we would enclose a copied version of this formula in a SIMOUTPUT call. (f) E7 = SIMOUTPUT(IF(E23 >= E1, 1, 0),D7) 11

12 (g) To maximize expected ending capital, use scenario 3, with a 20% investment each month. The expected ending capital is $16,454, but the probability of hitting the target is only 27.6%. (h) To maximize the probability of hitting the target, use scenario 2 instead, with the 10% investment percentage. This raises the probability of hitting the target to 51.6%, although the expected ending capital drops to $12,798. Shuttle Buses (a) B10 = PARAMETER(B6:E6) (b) B17 = GENPOISSON(E$1) (c) C17 = G16 + B17 (d) D17 = MIN(C17,B $10) (e) E17 = C17 D17 (f) F17 = GENBINOMIAL(E17,E$3) (g) G17 = E17 F17 (h) F9 = SIMOUTPUT(AVERAGE(E17:E61, E9) F10 = SIMOUTPUT(E2*SUM(D17:D61), E10) F11 = SIMOUTPUT(F10 B11, E11) (i) Bus for 20 people (Scenario 2); average number left is Temporary Water Supply for a Hotel (a) B13 = PARAMETER(C13:F13,1,A13) B14 = PARAMETER(C14:F14,1,A14) (or copy the formula in B13) (b) B19 = GENBINOMIAL(B3,B7) B20 = GENPOISSON(B4) B21 = MIN(B5,B2+B19+B20) (c) B24 = B10*B21 B25 = B11*SQRT(B21) B26 = GENNORMAL(B24,B25) (d) B28 = SIMOUTPUT(IF(B13<B26,1,0),A28) (e) B29 = IF(B28=1,B16 + B17*B21,0) B20 = SIMOUTPUT(B29 + B14,A30) (f) Option 2 is the cheapest; the probability estimate is about 8.6% 12

13 Managing Inventory (a) D14 = PARAMETER(E14:G14,1,C14) D15 = PARAMETER(E15:G15,3,C15) (Note the second argument: for D14 it is 1, and for D15 it is 3; you could also do it the opposite way the second argument of D14 would be 3, and for D15 it would be 1.) (b) B20 = G19 + C19 (c) C20 = D19 (d) D20 = IF(B20+C20 <= D$14,D$15,0) (e) E20 = GENPOISSON(GENTABLE(D$3:D$4,E$3:E$4)) (f) F20 = MIN(B20,E20) (g) G20 = B20 - F20 (h) H20 = IF(B20>D$11,D$12,0) + D$10*(B20 + G20)/2 + IF(D20>0,D$6 + D20*D$7,0) (i) I20 = IF(F20<E20,1,0) (j) J20 = IF(B20>D$11,1,0) (k) D47 = D8*SUM(F20:F45) D48 = SUM(H20:H45) D49 = D9*(C45 + D45 + G45) D50 = SIMOUTPUT(D47 + D49 - D48,C50) (l) I48 = SIMOUTPUT(SUM(I20:I45), Number of Stockouts ) J48 = SIMOUTPUT(SUM(J20:J45), Number of Overflows ) I49 = SIMOUTPUT(IF(I48>0,1,0), Were there any Stockouts ) J49 = SIMOUTPUT(IF(J49>0,1,0), Were there any Overflows ) (m) The best profit is scenario 2, with R = 225, Q = 200 With this choice of R, Q: We estimate the probability of stockout to be 74.1 % We estimate the probability of overflow to be 0% If we restrict ourselves to (R, Q) giving an expected number of stockouts less than 1, the best choice is scenario 3, R = 250, Q =