Solution (1) Formulate the problem as a LP model.

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1 Benha Unversty Department: Mechancal Engneerng Benha Hgh Insttute of Technology Tme: 3 hr. January 0 -Fall semester 4 th year Eam(Regular) Soluton Subject: Industral Engneerng M Green valley mlls produces carpet at plants n St. Lous and Rchmond. The carpet s then shp to two outlets located n Chcago and Atlanta. The cost per ton of shppng carpet from each of two plants to the two warehouses s as follows: From St. Lous Rchmond Chcago $04 04 To Atlanta $ 04 The plant at St. Lous can supply 0 tons of carpet per week; the plant at Rchmond can supply 400 tons per week. The Chcago outlet has a demand of 300 tons per week, and the outlet at Atlanta demands 30 tons per week. Formulate the problem as a lnear programmng model. The company wants to know the number of tons of carpet to shp from each plant to each outlet n order to mnmze the total shppng cost. Solve ths transportaton problem usng least cost method. Soluton () Formulate the problem as a LP model. Mn Z m n C j j j 40 Subjected to constrans All j From St. Lous Rchmond DEMAND Chcago $ To Atlanta $ SUPPLY Dr. Soher Hussen

2 II. Usng the Least- Cost method, fnd the basc feasble soluton that would mnmze the transportaton cost. Testng the problem s standard. j a b j 0 0 The Least- Cost method a j M M supples A B Demands b j The total cost of the transportaton s gven =0*40+0*70+30*30=4000 Number of basc feasble soluton= m+n-=+-=3 III. Check the optmalty for ths soluton. Used cell v=40 v =0 supples u= u= Demands Cell (,) :u+ v=40 Cell (,) :u+ v=70 Cell (,) :u+ v=30 C C U V for Unused cell The optmalty f 0 j j C3 (0 0) j The soluton optmum. The Prmo Insurance Co. s ntroducng two new product lnes: specal rsk nsurance and mortgages. The epected proft s $ per unt of specal rsk Dr. Soher Hussen 0

3 nsurance and $ per unt on mortgages. Management wshes to establsh sales quotas for the new product lnes to mamze total epected proft. The work requrements are as follows: Department Work-hours per unt Work-hours Specal rsk Mortgage Avalable Underwrtng Admnstraton 0 0 Clams 0 00 a. Formulate a LP model for ths problem. Soluton () Let Specal rsk = Mortgage =y Objectve functon: Mamze Z = + y Subject to the constrants: 3+y 400 Y 0 00 and y 0. Consder the followng lnear programmng problem : a) Ma ( + ) Subject to the constrants: , 0 3. For the Hawkns Company, the monthly percentages of all that were receved on tme over the past months are as shown n table. month shpments 8, 8 8 Dr. Soher Hussen 0

4 . Use a weght of. for the most recent observaton, /3 for the second most recent, and / for the thrd most recent to compute a three month weghted movng average for the tme seres. 3. What s the forecast for month? Soluton (3) I -Three month weghted movng average for the tme seres. month actual weght forecast F4= =8.7 F= =.7 F= =.7 F7= =. II - The forecast for month usng tme trend seres Y Y ^ sum Y=0 sum ^=0 sum =78 sum Y=997 Dr. Soher Hussen 0

5 y m b y y 0 * * 0 b = 8. 0 * 0 78 * 78 n n y y * 0 78 *997 m = * 0 78 * 78 n y 0.03* 8.0 y 4.3 * Soluton (4) a) Ma ( + ) Subject to the constrants:, =0 = =0 = =0 =0 = = M 00 0 M 40 0 M Dr. Soher Hussen

6 M=(0,0) Z=*0+*0=70 M=(70,7) Z= *70 +*7=0 M3=(00,0) Z3 =*00+*0= 00 MA Z at =70, =7, z= 0 b) Mn f() = + Subject to the constrants: =0 = =3 =0 + 4 =0 =4 =4 = M=(0,) Z=*0+*= M=(,) Z= *+*=34 M3=(4,0) Z3 =*4+*0= 308 Dr. Soher Hussen

7 MIN Z at =4, =0, z= Name three advantages and three dsadvantages for these types of layouts wth drawng : A. Product layout. B. Process layout. Soluton () A. Product layout. Materal M M0 M0 M0 Product Advantages: product layout provdes the followng benefts: a) Low cost of materal handlng, due to straght and short route and absence of backtrackng b) Smooth and unnterrupted operatons. c) Contnuous flow of work. Dsadvantages: Product layout suffers from followng drawbacks: a) Heavy overhead charges. b) Breakdown of one machne wll hamper the whole producton process. c) Lesser fleblty as specally lad out for partcular product. B. Process layout. Drllng Plannng Grndng [] [3] [] Mllng Weldng Assembly [] [4] 0 0 [] Dr. Soher Hussen 0

8 Advantages: a) Lower ntal captal nvestment n machnes and equpments. There s hgh degree of machne utlzaton, as a machne s not blocked for a sngle product. b) Breakdown of one machne does not result n complete work stoppage. c) There s a greater fleblty of scope for epanson. Dsadvantages: a. Materal handlng costs are hgh due to backtrackng. b. More sklled labor s requred n hgher cost. c. Tme gap or lag n producton s hgher. C. Compute the shortest path between node and node 7 (and ts length) n the network below. For every lnk of the network, the length of that lnk s gven n the pcture [,] [7,] 8 [0,-] 4 [,] [3,] Dr. Soher Hussen 8

9 Node j Computaton of u j Label u 0 [0,-] u = u +d = 0+ =, from [,] 3 u 3 = u +d 3 = 0+4 =4, from [4,] 4 u 4 = mn { u +d 4, u +d 4, u 3 +d 34 } [,] = mn { 0+, +, 4+3 } = from u = mn { u +d, u 4 +d 4 } [7,] = mn { +, +8 } = 7, from u = mn { u 3 +d 3, u 4 +d 4 } [,3] = mn { 4+, +7} =, from 3 7 u 7 = mn { u +d 7, u +d 7 } = mn { 7+, +9} = 3, from [3,] (7) () () () Dr. Soher Hussen 9