ISyE 2030 Summer Semester 2004 June 30, 2004

Transcription

1 ISyE 030 Summer Semester 004 June 30, Every day I must feed my 130 pound dog some combnaton of dry dog food and canned dog food. The cost for the dry dog food s $0.50 per cup, and the cost of a can of dog food s $0.35 per can. Each day, my dog has nutrtonal requrements of at least 1500 calores, 6 grams of fber, 30 grams of proten, and 15 grams of fat. The nutrtonal content of the two foods are shown below. What s the mnmum cost for me to satsfy my dog s detary requrements? Calores fber proten Fat Cost Cup dry dog food $0.50 Can of dog food $0.35 Requrement Defne the decson varables: x1 = number of dry dog food bought x = number of canned dog food bought What s the objectve functon? mn 0.50x x What are the constrants? 400x x 1500 (Calores requrement) 3x1 + x 6 (Fber requrement) 8x1 + 10x 30 (Proten requrement) 3x1 + 11x 15 (Fat requrement) x1, x 0 (Nonnegatvty)

2 Solve the problem graphcally: Wrte the Mosel model: model LPproblem1 uses "mmxprs"! Start a new model! Load the optmzer lbrary declaratons x1: mpvar! Number of dry dog food bought x: mpvar! Number of canned dog food bought end-declaratons objectve:= 0.50*x *x const1:= 400*x *x >= 1500 const:= 3*x1 + x >= 6 const3:= 8*x1 + 10*x >= 30 const4:= 3*x1 + 11*x >= 15 mnmze(objectve)! Objectve: mnmze total cost!calores requrement!fber requrement!proten requrement!fat requrement! Solve the LP-problem! Prnt out the soluton wrteln("soluton:\n Objectve: ", getobjval) wrteln(" x1= ", getsol(x1)) wrteln(" x= ", getsol(x)) end-model

3 Output Soluton: Objectve: 1.77 x1= 0.6 x= 4.. You have been hred by a farmer to help her maxmze proft for next year s crop. She can choose between corn (c) and jalapeños (j). She has a total of 600 acres to plant. Each acre of jalapenos that she plants s worth 5 tmes as much as an acre of corn she plants. However, at least one thrd of her acreage must be planted wth corn. In addton, she s allocated at most 5000 unts of water. An acre of corn takes unts of water, but an acre of jalapenos takes 10 unts of water. Fnally, envronmental restrctons lmt her use of fertlzer to 1000 unts. Each acre of corn takes 1 unt of fertlzer and each acre of jalapenos takes unts. a. Wrte the lnear programmng formulaton for ths problem. Decson varables: c = acre of corn planted j = acre of jalapenos planted max c + 5j subject to: c + j 600 (Total sze constrant) c j 0 (Corn constrant) c + 10j 5000 (Water capacty) c + j 1000 (Fertlzer constrant) c, j 0 (Nonnegatvty) b. On the grd on the next page, plot the constrants and show the feasble regon. Try to be as careful as possble on the plot. See the graph on the next page c. Show the objectve functon (.e., soproft curve) on ths plot at the optmum. See the red lne n the graph. d. What s the value of the optmal proft n ths case? Optmal soluton = (c, j) = (00, 400) Optmal proft = c + 5j = 00

4 Note: Mosel model: model LPproblem uses "mmxprs"! Start a new model! Load the optmzer lbrary declaratons c: mpvar! Acre of corn planted j: mpvar! Acre of jalapenos planted end-declaratons objectve:= c + 5*j const1:= c + j <= 600 const:= *c - j >= 0 const3:= *c + 10*j <= 5000 const4:= c + *j <= 1000 maxmze(objectve)! Objectve: maxmze total proft!sze constrant!corn constrant!water capacty!fertlzer constrant! Solve the LP-problem! Prnt out the soluton wrteln("soluton:\n Objectve: ", getobjval) wrteln(" c= ", getsol(c)) wrteln(" j= ", getsol(j)) end-model

5 Output Soluton: Objectve: 00 c= 00 j= A newsvendor faces the followng dscrete demand dstrbuton: P{Demand=} The newsvendor buys the papers for 70 cents and sells them for $1.00. Leftover papers are sold for 0 cents each. What would the margnal beneft be to the newsvendor of gong from 36 to 37 papers? Gven c (cost) = 0.7, p (sellng prce) = 1.0, and s (salvage value) = 0., we have MP (margnal proft) = p c = 0.3 and ML (margnal loss) = c s = 0.5. So, the crtcal value wll be Pc = ML/(MP + ML) = (c s) /(p s) = From the table below, the margnal beneft to the newsvendor of gong from 36 to 37 papers s $0.10. x Prob. that demand = x P = Prob. of sellng the x th unt (1-P) = Prob. of NOT sellng the x th unt P MP = Expected proft from sellng x th unt (1-P) ML = Expected loss from NOT sellng x th unt Expected NET proft from stockng x th unt $0.3 $0 $ $0.7 $0.05 $ $0.5 $0.15 $ $0.15 $0.5 -$ $0.075 $ $ $0.03 $0.45 -$0.4 > $0 $0.5 -$0.50

6 4. T F Jos A Banks uses a batch producton system. True 5. T F In the newsvendor soluton, the smaller the standard devaton of demand, the further away wll the order sze be from the mean demand. False (For problems 6 and 7, let y = 1 f project ( = 1 to 4) s selected and 0 otherwse) 6. T F y 1 + y = 1 means both projects 1 and must be selected False 7. T F y 1 + y + y 3 + y 4 means that at least projects wll be selected. True 8. For the followng LP: max subject to : 10x + 9x 3 5x x1,x 0 The pont (x 1 =1, x =3) s a: a. boundary pont b. extreme pont c. both a and b d. nether a or b x x 37 If you plot the constrant, the (1,3) t s at the ntersecton of the two constrants. Therefore t s both an extreme pont and on the boundary. The correct answer then s c. 9. Formulate the makng change problem (.e., casher returns as few cons as possble) assumng the casher wll gve back only cons of value 1, 5, 10, and 5, and the change you wll receve s 87. Make sure and clearly defne the varables. Decson varables: x1 = number of 1 cons x = number of 5 cons x3 = number of 10 cons x4 = number of 5 cons mn x1 + x + x3 + x4 subject to: 1x1 + 5x + 10x3 + 5x11 = 87 x1, x, x3, x11 0 and nteger

7 10. Formulate for the followng problem (make sure and clearly defne your varables!). A manufacturer produces engnes, and the goal s to satsfy demand at the least cost. Demand over the next four months s 50, 0, 35, and 80. The maxmum that the manufacturer can make n any month s 55 unts. There are three costs for the manufacturer: producton, nventory holdng, and setup cost. Producton cost s $1500 per engne, and nventory s carred from one perod to the next at a cost of $50 per engne. Fnally, f the manufacturer produces engnes n perod, then they have to pay a setup cost of $3000. Assume that there s an ntal nventory of 0 tems. mn ( I S ) subject to : I1 = 1 50 I = I1 + 0 I 3 = I I 4 = I M S, I 0 1 f produce n perod S = 0 otherwse s producton quantty n perod I s nventory n perod