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Unformatted text preview: Space IOE 202: lecture 6 outline Announcements Last time... Other examples of optimization models
EOQ with shortages Cutting Stock problem Radiation treatment planning models IOE 202: Operations Modeling, Fall 2009 Page 1 Space Last time
Using integer variables in models where variables represent nondivisible quantities
Post oﬃce staﬃng problem Course selection: used constraints on the variables to represent logical restrictions Pollution reduction: combining binary and continuous variables; using “forcing” constraints to ensure feasibility and to correctly account for ﬁxed costs Investment with brokerage fees Using binary (0or1) variables to represent yes/no decisions
Important observation: all models we constructed were Linear Integer Programs, that is, objective and constraints involved only linear functions of the variables. This is crucial to our ability to ﬁnd optimal solutions to the models we formulate. IOE 202: Operations Modeling, Fall 2009 Page 2 Space Cutting Stock problem The Better Food Company produces creamﬁlled sponge rolls with a standard width of 20 cm each. Each 20 cm roll costs the company $2.00 to produce. Special customer orders with diﬀerent widths are produced by cutting (slitting) the standard rolls of sponge into shorter lengths. Typical orders are summarized in the following table. These orders need to be met at least cost. Order A B C Desired Width (cm) 5 7 9 Desired Number of Rolls 150 200 300 IOE 202: Operations Modeling, Fall 2009 Page 3 Space How to split the rolls An order is ﬁlled by cutting a standard 20 cm. roll into the desired widths; there are several ways in which a standard roll can be slit to ﬁll a given order. We will consider below three possible ways to cut up a 20cm roll. Note that the shaded area in each diagram represents lengths of sponge that are too short to be used in meeting orders, and so these pieces must be thrown away. Although there are other feasible settings, we limit the discussion for the moment to considering settings 1, 2 and 3 in the ﬁgure. IOE 202: Operations Modeling, Fall 2009 Page 4 Space Three possible knifesetting patterns
20
7 9 4 5 5 20
7 3 Setting 1 20
5 5 9 1 Setting 2 Setting 3 5 cm rolls produced Pattern 1 Pattern 2 Pattern 3
IOE 202: Operations Modeling, Fall 2009 7 cm rolls produced 9 cm rolls produced Page 5 Space Two possible solutions
Number of No of 5 cm No of 7 cm No of 9 cm times pattern rolls rolls rolls is used (150 ordered) (200 ordered) (300 ordered) 300 75 0 Pattern 1 Pattern 2 Pattern 3 Totals: Pattern 1 Pattern 2 Pattern 3 Totals: Number of No of 5 cm No of 7 cm No of 9 cm times pattern rolls rolls rolls is used (150 ordered) (200 ordered) (300 ordered) 200 0 100 IOE 202: Operations Modeling, Fall 2009 Page 6 Space All possible patterns All possible patterns:
Pattern 1 Pattern 2 Pattern 3 Pattern 4 Pattern 5 Pattern 6 5 cm rolls produced 0 2 2 7 cm rolls produced 1 1 0 9 cm rolls produced 1 0 1 IOE 202: Operations Modeling, Fall 2009 Page 7 Space Mathematical model for the Cutting Stock problem Decision variables: Objective function: Constraints: Optimal solution: IOE 202: Operations Modeling, Fall 2009 Page 8 Space Recall the EOQ model for inventory management: Suppose you are facing a uniform demand of a items per unit of time. It costs you $K to place an order with the manufacturer of these items, plus a cost of $c per item you order. Your holding cost is $h per item per unit of time held. We decided to order the items in batches of size Q , and time the orders in such a way so that the inventory is replenished just as we are about to run out. Thus we make sure that we do not experience any shortages. We discussed that the batch size that minimizes the costs of ordering and holding inventory per unit of time, or the Economic Order Quantity, is Q ∗ = 2aK , and the orders should be placed h every t ∗ = 2K units of time. ah
IOE 202: Operations Modeling, Fall 2009 Page 9 Space Ordering With Planned Shortages Planned shortages are now allowed. When a shortage occurs, the aﬀected customers will wait for the product to become available again. Their backorders are ﬁlled immediately when the order quantity arrives to replenish inventory. Additional notation:
p = shortage cost per item short per unit of time short S = inventory level just after a batch of Q items is added to the inventory Q − S = shortage in the inventory just before a batch of Q items is added IOE 202: Operations Modeling, Fall 2009 Page 10 Space Inventory level vs. time ✻ ✲ IOE 202: Operations Modeling, Fall 2009 Page 11 Space One ordering cycle: details
Cycle length = Production/ordering cost per cycle = In each cycle, inventory is held for Average inventory during this time = In each cycle, shortages are encountered for time Average shortage during this time = Total cost per cycle (ordering+inventory+shortage) = units of units of time IOE 202: Operations Modeling, Fall 2009 Page 12 Space Expression of cost for EOQ with shortages Total cost per cycle = K + cQ + and total cost per unit of time is Total cost per cycle aK hS 2 p (Q − S )2 T (Q , S ) = = + ac + + . Length of the cycle Q 2Q 2Q
hS 2 p (Q − S )2 + , 2a 2a Notice that we are facing a nonlinear optimization problem: min T (Q , S ) subject to S ≥ 0, Q ≥ S .
Q ,S Let us try to minimize T (Q , S ) (unconstrained), and check if the solution we obtain satisﬁes the constraints.
Page 13 IOE 202: Operations Modeling, Fall 2009 Space Finding (unconstrained) minimum of T (Q , S ) aK hS 2 p (Q − S )2 T (Q , S ) = + ac + + . Q 2Q 2Q To ﬁnd the minimum, we need to set the partial derivatives of T w.r.t. S and Q to zero simultaneously: ∂T hS p (Q − S ) = − = 0, ∂S Q Q ∂T aK hS 2 p (Q − S ) p (Q − S )2 =− 2 − + − = 0. ∂Q Q 2Q 2 Q 2Q 2 IOE 202: Operations Modeling, Fall 2009 Page 14 Space Optimal solution Solving the above two equations simultaneously leads to 2aK p 2aK p+h S∗ = , Q∗ = h p+h h p
Do these values satisfy the constraints? p +h 2K ∗ = Q ∗ /a = Optimal cycle time: t ah p h Maximum shortage: Q ∗ − S ∗ = 2aK p+h p What happens to the solution when p is much larger than h? What happens to the solution when h is much larger than p ? IOE 202: Operations Modeling, Fall 2009 Page 15 Space Beneﬁts of allowing shortages In the speaker manufacturing example, we had K = 12000, h = 0.30, a = 8000, c = 10. When shortages were not allowed, we determined the EOQ is Q ∗ = 25298. The total cost was T (Q ∗ ) = $87589.47. Suppose now we allow shortages with p = 1.10, then S ∗ = 22424, Q ∗ = 28540, T (Q ∗ , S ∗ ) = $86727.34. IOE 202: Operations Modeling, Fall 2009 Page 16 Space Background on radiation therapy in cancer treatment This year, 1,200,000 Americans will be diagnosed with cancer 600,000+ patients will receive radiation therapy
beam(s) of radiation delivered to the body in order to kill cancer cells High doses of radiation (energy/unit mass) can kill cells and/or prevent them from growing and dividing
true for cancer cells and normal cells Treatment works because the repair mechanisms for cancer cells is less eﬃcient than for normal cells IOE 202: Operations Modeling, Fall 2009 Page 17 Space Treatment delivery illustration A linear accelerator and a treatment couch A view of a collimator (beamshaping device) IOE 202: Operations Modeling, Fall 2009 Page 18 Space Radiation delivery — single beam direction 10 9 8 7 6 tumor Relative Intensity of Dose Delivered
IOE 202: Operations Modeling, Fall 2009 Page 19 Space Radiation delivery — multiple beam directions 5 4 5 4 2 1 9 2 4 2 5 tumor 1 1 Relative Intensity of Dose Delivered
IOE 202: Operations Modeling, Fall 2009 Page 20 Space A schematic representation of a patient’s anatomy 151 T is the target area C is the critical area N is normal tissue S := T ∪ C ∪ N
151 Treatment planning problem
For a given tumor and given critical areas For a given set of possible “beamlet” origins and angles Determine the intensity of each beamlet such that:
dosage over the tumor area will be at least a target level γL dosage over the critical area will be at most a target level γU
Page 21 IOE 202: Operations Modeling, Fall 2009 Space Building a model of the treatment planning problem — I Divide up the treatment region into a 2dimensional (or 3dimensional) grid of pixels
j pixel (i,j) i IOE 202: Operations Modeling, Fall 2009 Page 22 Space Building a model of the treatment planning problem — II Create the beamlet data for each of p = 1, . . . , n possible beamlets. D p is the matrix of unit doses delivered by beam p .
0 0 0 0 0 0 0 0 0 1.0 0 0 0 1.0 1.0 0 0.9 0.9 1.0 0 0.8 0.9 0.9 0 0 0.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 p Dij = unit dose delivered to pixel (i , j ) by beamlet p .
IOE 202: Operations Modeling, Fall 2009 Page 23 Space Building a model of the treatment planning problem — III Decision variables:
wp : intensity of “beamlet” p , p = 1, . . . , n Dij : total dose delivered to pixel (i , j ) for all pixels Relationship (stated as constraints): Dij =
n p =1 p Dij wp for each beamlet (i , j ) IOE 202: Operations Modeling, Fall 2009 Page 24 Space Building a model of the treatment planning problem — III minimizew ,D Dij
n p =1 p Dij wp (i , j ) in S (i ,j ) in S Dij = wp ≥ 0 p = 1, . . . , n (i , j ) in T (i , j ) in C Dij ≥ γL Dij ≤ γU
Unfortunately, this model is typically infeasible. Cannot deliver dose to tumor without some harm to critical area(s). IOE 202: Operations Modeling, Fall 2009 Page 25 Space A linear programming model minimizew ,D s.t. Dij − (Target)ij 
p p =1 Dij wp (i ,j ) in S Dij = wp ≥ 0 This is the same as: minimizew ,D ,∆ s.t. n (i , j ) in S p = 1, . . . , n ∆ij n
p p =1 Dij wp (i ,j ) in S Dij = (i , j ) in S (i , j ) in S p = 1, . . . , n −∆ij ≤ Dij − (Target)ij ≤ ∆ij wp ≥ 0
IOE 202: Operations Modeling, Fall 2009 Page 26 Space A linear programming model 151 (Target)ij (Target)ij (Target)ij
151 = 16, (i , j ) in T = 0, (i , j ) in C = 0, (i , j ) in N minimizew ,D ,∆ 1 · s.t. (i ,j ) in N p Dij = n=1 Dij wp , p ∆ij + 100 · (i , j ) in S (i ,j ) in C ∆ij + 30 · ∆ij (i ,j ) in T −∆ij ≤ Dij − (Target)ij ≤ ∆ij , (i , j ) in S wp ≥ 0, p = 1, . . . , n
IOE 202: Operations Modeling, Fall 2009 Page 27 Space Solution of the linear programming model 200 180 160 140 120 100 80 60 40 20 20 40 60 80 100 120 140 160 180 200 IOE 202: Operations Modeling, Fall 2009 Page 28 Space A nonlinear programming model minimizew ,D ,∆ 1 · s.t. (i ,j ) in N p Dij = n=1 Dij wp , p ∆2 + 100 · ij (i , j ) in S (i ,j ) in C ∆2 + 30 · ij (i ,j ) in T ∆2 ij −∆ij ≤ Dij − (Target)ij ≤ ∆ij , (i , j ) in S wp ≥ 0, p = 1, . . . , n IOE 202: Operations Modeling, Fall 2009 Page 29 Space Solution of the nonlinear programming model 200 180 160 140 120 100 80 60 40 20 20 40 60 80 100 120 140 160 180 200 IOE 202: Operations Modeling, Fall 2009 Page 30 Space Use of integer variables Partial Volume Constraints:
“No more than 20% of the critical region can exceed a dose of 30Gy .” Adding integer variables to incorporate the constraints: Let M be a very large number, Dij ≤ 30 + M · yij ,
(i ,j ) in C yij binary , ≤ C × 0.20 (i , j ) in C yij IOE 202: Operations Modeling, Fall 2009 Page 31 ...
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This note was uploaded on 03/17/2010 for the course IOE 202 taught by Professor Marinaepelman during the Fall '09 term at University of MichiganDearborn.
 Fall '09
 MarinaEpelman

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