114 年 國立中央大學工業管理研究所碩士班《作業研究》

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第 1 題20 分

Solve the following linear programming (LP) problem and report your findings. For example, if the problem has no feasible solution, report "the problem is infeasible;" on the other hand, report an optimal solution if you can find one.
Note: You must show a detailed, step-by-step calculation process to receive full scores.

Minimize x1+1.5x2+2.5x3+x4+1.5x5x_1 + 1.5x_2 + 2.5x_3 + x_4 + 1.5x_5
subject to
3x1+3x2+6x3+3x4+9x5≥123x_1 + 3x_2 + 6x_3 + 3x_4 + 9x_5 \geq 12
x1−x2+1.5x3+0.5x4+0.5x5≥1.5x_1 - x_2 + 1.5x_3 + 0.5x_4 + 0.5x_5 \geq 1.5
x1,x2,x3,x4,x5≥0x_1, x_2, x_3, x_4, x_5 \geq 0

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這一題的完整詳解

核心觀念

本題考查線性規劃的最適解判定,核心工具包括:

  1. 可行解:滿足所有限制式與非負限制。
  2. 基底解:在兩個主要限制式下,通常可由兩個變數為正、其餘變數為 00 求得候選解。
  3. 對偶問題與弱對偶定理:
    原問題為最小化且限制式為「≥\geq」,其對偶為最大化問題。若找到一組原問題可行解與對偶問題可行解,且兩者目標值相同,則兩者皆為最適解。

解題方法

原問題為:

min⁡z=x1+1.5x2+2.5x3+x4+1.5x5s.t.3x1+3x2+6x3+3x4+9x5≥12,x1−x2+1.5x3+0.5x4+0.5x5≥1.5,xi≥0.\begin{aligned} \min \quad & z=x_1+1.5x_2+2.5x_3+x_4+1.5x_5\\ \text{s.t.}\quad &3x_1+3x_2+6x_3+3x_4+9x_5\geq 12,\\ &x_1-x_2+1.5x_3+0.5x_4+0.5x_5\geq 1.5,\\ &x_i\geq 0. \end{aligned}

由於所有變數成本皆為正,最小化時不會任意增加變數;最適解通常會使兩個主要限制式恰好取等號。

為了嚴格驗證最適性,建立其對偶問題。


建立對偶問題

令第一條限制式的對偶變數為 y1y_1,第二條限制式的對偶變數為 y2y_2。

因為原問題是最小化問題,且限制式為「≥\geq」,所以對偶為:

max⁡w=12y1+1.5y2s.t.3y1+y2≤1,3y1−y2≤1.5,6y1+1.5y2≤2.5,3y1+0.5y2≤1,9y1+0.5y2≤1.5,y1,y2≥0.\begin{aligned} \max \quad & w=12y_1+1.5y_2\\ \text{s.t.}\quad &3y_1+y_2\leq 1,\\ &3y_1-y_2\leq 1.5,\\ &6y_1+1.5y_2\leq 2.5,\\ &3y_1+0.5y_2\leq 1,\\ &9y_1+0.5y_2\leq 1.5,\\ &y_1,y_2\geq 0. \end{aligned}

各條對偶限制式分別對應原問題的 x1,x2,x3,x4,x5x_1,x_2,x_3,x_4,x_5。


求對偶問題的候選解

觀察第一條與第五條對偶限制式:

3y1+y2≤1,3y_1+y_2\leq 1, 9y1+0.5y2≤1.5.9y_1+0.5y_2\leq 1.5.

令這兩條限制式取等號:

3y1+y2=1,3y_1+y_2=1, 9y1+0.5y2=1.5.9y_1+0.5y_2=1.5.

由第一式得:

y2=1−3y1.y_2=1-3y_1.

代入第二式:

9y1+0.5(1−3y1)=1.5,9y_1+0.5(1-3y_1)=1.5, 9y1+0.5−1.5y1=1.5,9y_1+0.5-1.5y_1=1.5, 7.5y1=1,7.5y_1=1, y1=215.y_1=\frac{2}{15}.

因此:

y2=1−3(215)=1−615=35.y_2=1-3\left(\frac{2}{15}\right) =1-\frac{6}{15} =\frac{3}{5}.

所以候選對偶解為:

(y1,y2)=(215,35).(y_1,y_2)=\left(\frac{2}{15},\frac{3}{5}\right).

驗證對偶可行性

逐一代入其他限制式。

對應 x1x_1

3y1+y2=3(215)+35=25+35=1.3y_1+y_2 =3\left(\frac{2}{15}\right)+\frac{3}{5} =\frac{2}{5}+\frac{3}{5} =1.

符合:

3y1+y2≤1.3y_1+y_2\leq 1.

對應 x2x_2

3y1−y2=25−35=−15≤1.5.3y_1-y_2 =\frac{2}{5}-\frac{3}{5} =-\frac{1}{5}\leq 1.5.

對應 x3x_3

6y1+1.5y2=6(215)+1.5(35)=45+910=1710=1.7≤2.5.6y_1+1.5y_2 =6\left(\frac{2}{15}\right)+1.5\left(\frac{3}{5}\right) =\frac{4}{5}+\frac{9}{10} =\frac{17}{10} =1.7\leq 2.5.
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第 2 題30 分

The following is a simplex tableau for a minimization LP problem. As you know, such a tableau is simply a convenient way to express (or to represent) a normal LP model.

Zx1x_1x2x_2x3x_3x4x_4x5x_5x6x_6RHS
10-40-10-2-17
01-1/301/30-2/31/3
00200116
002/311/301/313/3

Part 2.1 (15 points)
What is the normal LP model that is expressed by the above simplex tableau?
Note: You must answer the complete LP model to receive full scores.
(請寫出上方表格所表示的LP模型。你必需寫出該模型完整的目標式及限制式,才能得到所有分數。)

Part 2.2 (15 points)
Please determine the status of the LP model that you answered in Part 2.1 and also provide sufficient reasons. For example, your answer may be that "the model can still be improved to lower its objective value,” or “the model has reached optimality," or "the model has no feasible solutions," or any other statuses. (接下頁)
注意:你的答案必須依據你在Part 2.1所回答的LP模型,而非依據題目所給的simplex tableau。同時,你必須給出明確的原因才能得到分數(提示:仔細檢視你在Part 2.1所回答的LP模型,看它是否有透露某些數學方面的訊息,以此來判定該模型的狀態)。

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這一題的完整詳解

核心觀念

單純形表中的每一列代表一條等式。若 x4,x5,x6x_4,x_5,x_6 為三條限制式所對應的剩餘變數(slack variables),則可由表中的等式還原原始 LP 模型。

目前表中之基底變數為 x1,x5,x3x_1,x_5,x_3,非基底變數為 x2,x4,x6x_2,x_4,x_6。


Part 2.1:還原完整 LP 模型

一、由單純形表讀出等式

目標列為

Z−4x2−x4−2x6=−17,Z-4x_2-x_4-2x_6=-17,

因此

Z=−17+4x2+x4+2x6.Z=-17+4x_2+x_4+2x_6.

三條限制列為

x1−13x2+13x4−23x6=13,x_1-\frac13x_2+\frac13x_4-\frac23x_6=\frac13, 2x2+x5+x6=6,2x_2+x_5+x_6=6, 23x2+x3+13x4+13x6=133.\frac23x_2+x_3+\frac13x_4+\frac13x_6=\frac{13}{3}.

二、消去剩餘變數

由第一條等式:

x4=1−3x1+x2+2x6.x_4=1-3x_1+x_2+2x_6.

由第二條等式:

x5=6−2x2−x6.x_5=6-2x_2-x_6.

由第三條等式:

x3=133−23x2−13x4−13x6.x_3=\frac{13}{3}-\frac23x_2-\frac13x_4-\frac13x_6.

將 x4x_4 代入,可得

x3=4+x1−x2−x6.x_3=4+x_1-x_2-x_6.

因此

x6=4+x1−x2−x3.x_6=4+x_1-x_2-x_3.

再將 x6x_6 代入 x4,x5x_4,x_5:

x4=9−x1−x2−2x3,x_4=9-x_1-x_2-2x_3, x5=2−x1−x2+x3.x_5=2-x_1-x_2+x_3.

由於 x4,x5,x6x_4,x_5,x_6 是剩餘變數且必須滿足非負性,因此得到:

x1+x2+2x3+x4=9x_1+x_2+2x_3+x_4=9

對應為

x1+x2+2x3≤9;x_1+x_2+2x_3\le 9; x1+x2−x3+x5=2x_1+x_2-x_3+x_5=2

對應為

x1+x2−x3≤2;x_1+x_2-x_3\le 2; −x1+x2+x3+x6=4-x_1+x_2+x_3+x_6=4

對應為

−x1+x2+x3≤4.-x_1+x_2+x_3\le 4.

三、還原目標函數

將 x4,x6x_4,x_6 代入目標列:

Z=−17+4x2+x4+2x6=−17+4x2+(9−x1−x2−2x3)+2(4+x1−x2−x3)=x1+x2−4x3.\begin{aligned} Z &=-17+4x_2+x_4+2x_6\\ &=-17+4x_2+(9-x_1-x_2-2x_3) +2(4+x_1-x_2-x_3)\\ &=x_1+x_2-4x_3. \end{aligned}

因此完整的原始 LP 模型為

min⁡Z=x1+x2−4x3s.t.x1+x2+2x3≤9,x1+x2−x3≤2,−x1+x2+x3≤4,x1,x2,x3≥0.\boxed{ \begin{aligned} \min\quad & Z=x_1+x_2-4x_3\\ \text{s.t.}\quad &x_1+x_2+2x_3\le 9,\\ &x_1+x_2-x_3\le 2,\\ &-x_1+x_2+x_3\le 4,\\ &x_1,x_2,x_3\ge 0. \end{aligned} }
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第 3 題50 分

【模型背景描述】
A typical international supply chain (ISC) consists of a number of suppliers, manufacturing plants and distribution centers (DCs) which are located in different countries or even continents. To illustrate, the following figure shows an example of ISC which contains three suppliers S1, S2 and S3 located in China, U.S.A. and Mexico, respectively; this ISC has three manufacturing plants PL1, PL2 and PL3 located in Bangladesh, Mexico and Nicaragua, respectively; also, this ISC has two DCs (DC1 and DC2) located in New York and Los Angeles, respectively.

🖼️【此處有附圖,請對照原卷】

The following describes a basic (simplest) way as to how such an ISC may operate to produce products and then distribute them.

  • A number of raw materials are used to make products. For simplicity, we would assume that all products are made with the same raw materials; furthermore, we would assume that all suppliers can supply all raw materials. Of course, there is a maximum quantity that each supplier j can supply each kind of raw material m, and we would denote such a quantity as SCmjSC_{mj}. Also, there is a specific quantity that each kind of raw material m must be used to make each unit of product i, and we would denote such a specific quantity as RmiR_{mi}.

  • For simplicity, assume that every manufacturing plant k can make every kind of product i. Obviously, k has a maximum capacity that it can produce i and we would use "units of finished i's" (i的完成品數量) to express this capacity (denoted as UCikUC_{ik}). Also, k has a minimum quantity that it is obligated to produce i; this minimum quantity is denoted as LCikLC_{ik}.

  • In addition to UCikUC_{ik} and LCikLC_{ik}, every manufacturing plant k has a maximum total capacity that it is allowed to use to make products and we would denote such a capacity as CPkCP_k (所有完成品的數量上限).

  • When production processes are done, manufacturing plants will immediately transport finished products to DCs. Again, for simplicity, we would assume that every manufacturing plant k can transport every product i to every DC l, and we would not consider capacities in these transportation activities.

【模型開發】
There are two key decisions to be made for the above ISC.

  • the amount of each kind of raw material m to be purchased from each supplier j for use of production at each manufacturing plant k (we would denote this decision variable as GmjkG_{mjk})
  • the units of each product i to be made at each manufacturing plant k and then transported to each DC l (we would denote this decision variable as HiklH_{ikl})

Note: When answering the following four questions (i.e., Parts 3.1 to 3.4), you MUST use and only use those decision variables and parameters that are given in each question. You will not receive any points if your answer includes any extra decision variables, parameters, definitions, assumptions, and so on.

Part 3.1: supply limits (10 points)
For each supplier j to supply each kind of raw material m:
Please use GmjkG_{mjk} and SCmjSC_{mj} only to derive a constraint ensuring that j will stay within its limits to supply m.

Part 3.2: production limits/obligations (10 points)
For every manufacturing plant k to make every kind of product i:
Please use HiklH_{ikl}, LCikLC_{ik} and UCikUC_{ik} only to derive constraints ensuring that k will follow its production limits as well as production obligations to produce i.

Part 3.3: plants' maximum total capacities (10 points)
For every manufacturing plant k:
Please use HiklH_{ikl} and CPkCP_k only to derive a constraint ensuring that k will make products according to its maximum total capacity.

Part 3.4: raw material consumption (20 points)
For every manufacturing plant k to consume every kind of raw material m:
Please use HiklH_{ikl}, GmjkG_{mjk} and RmiR_{mi} to derive a constraint ensuring that the purchase of raw material m and the following production and transportation of product i is well governed (or controlled) by RmiR_{mi}.

🖼️ 本題含圖表,以下為原卷對應頁面:
原卷第 2 頁原卷第 3 頁原卷第 4 頁

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這一題的完整詳解

核心觀念

本題屬於作業研究中「線性規劃(Linear Programming, LP)於國際供應鏈網路設計與生產排程」之數學模式建立。解題核心為:

  1. 下標(Indices)與加總維度識別:
    • 原料 mm、供應商 jj、製造廠 kk、產品 ii、物流中心(DC)ll。
  2. 決策變數意義:
    • GmjkG_{mjk}:從供應商 jj 採購原料 mm 並運往製造廠 kk 的數量。
    • HiklH_{ikl}:製造廠 kk 生產產品 ii 並運送至物流中心 ll 的數量。
  3. 限制式本質:
    • 供應上限限制(Supply Capacity Limits):某供應商對某原料的總供應量不可超過其最大供應能力。
    • 個別產能上下限限制(Production Limits & Obligations):某製造廠對特定產品的生產總量須介於規定之下限(義務量)與上限之間。
    • 總產能限制(Total Plant Capacity):某製造廠生產所有產品的總量不可超過其總容許上限。
    • 原料消耗與物料平衡限制(Raw Material Consumption / Bill of Materials Balance):製造廠投入的原料量必須足以支持其生產各產品所需的原料總消耗量。

解題方法與推導

Part 3.1:供應商原料供應上限限制(Supply limits)

  • 物理意義:對於固定的供應商 jj 與特定的原料 mm,供應商 jj 提供給所有工廠 kk 的原料 mm 總量,不能超過該供應商對原料 mm 的最大供應能力 SCmjSC_{mj}。
  • 數學推導:
    固定 mm 與 jj,將製造廠下標 kk 進行加總: ∑kGmjk≤SCmj∀m,∀j\sum_{k} G_{mjk} \le SC_{mj} \quad \forall m, \forall j

Part 3.2:製造廠個別產品生產上下限(Production limits/obligations)

  • 物理意義:工廠 kk 生產產品 ii 後會立即分送到各物流中心 ll。因此,工廠 kk 製造產品 ii 的總量即為對所有物流中心 ll 運送量的總和 ∑lHikl\sum_{l} H_{ikl}。
    此產量必須滿足合約義務(最低生產量 LCikLC_{ik}),且不得超過個別產能上限 UCikUC_{ik}。
  • 數學推導:
    固定工廠 kk 與產品 ii,對物流中心下標 ll 加總: LCik≤∑lHikl≤UCik∀i,∀kLC_{ik} \le \sum_{l} H_{ikl} \le UC_{ik} \quad \forall i, \forall k (亦可拆寫為 ∑lHikl≥LCik\sum_{l} H_{ikl} \ge LC_{ik} 與 ∑lHikl≤UCik\sum_{l} H_{ikl} \le UC_{ik})。

Part 3.3:製造廠整體最大產能限制(Plants' maximum total capacities)

  • 物理意義:工廠 kk 所能製造的所有產品(所有 ii)之總量上限為 CPkCP_k。
  • 數學推導:
    固定工廠 kk,對所有產品 ii 及運往的所有物流中心 ll 進行雙重加總: ∑i∑lHikl≤CPk∀k\sum_{i} \sum_{l} H_{ikl} \le CP_k \quad \forall k

Part 3.4:原料消耗與物料需求平衡(Raw material consumption)

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