111 年 國立成功大學製造資訊與系統研究所乙組《生產管理》

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第 1 題25 分

A competitor of Kozar International, Inc. has begun marketing a new instant-developing film project. Kozar has had a similar product under study in its R&D department but has not yet been able to begin production. Because of the competitor's action, top managers have asked for a speedup of R&D activities so that Kozar can produce and market instant film at the earliest possible date. The table below shows the predecessor information and activity time estimates in months, estimates of the optimistic, most probable, and pessimistic times. What is the probability the project will be completed in time for Kozar to begin marketing the new product within 27 months?

ActivityImmediate PredecessorOptimistic TimeMost Probable TimePessimistic Time
A-11.55
BA345
CA123
DB,C3.556.5
EB4512
FC,D,E6.57.511.5
GE5913

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這一題的完整詳解

本題考驗專案管理中的 PERT (Program Evaluation and Review Technique) 方法，旨在計算專案的預計完成時間及其機率。PERT 方法使用三點估計法（樂觀、最可能、悲觀時間）來估計活動持續時間，並計算專案的期望時間與變異數。

步驟一：計算各活動的期望時間 (Expected Time, tet_e) 與變異數 (Variance, σ2\sigma^2)

PERT 使用以下公式計算期望時間：
te=to+4tm+tp6t_e = \frac{t_o + 4t_m + t_p}{6}
其中 tot_o 為樂觀時間，tmt_m 為最可能時間，tpt_p 為悲觀時間。

PERT 使用以下公式計算變異數：
σ2=(tp−to6)2\sigma^2 = (\frac{t_p - t_o}{6})^2

Activitytot_otmt_mtpt_pte=to+4tm+tp6t_e = \frac{t_o + 4t_m + t_p}{6}σ2=(tp−to6)2\sigma^2 = (\frac{t_p - t_o}{6})^2
A11.551+4(1.5)+56=1+6+56=126=2\frac{1 + 4(1.5) + 5}{6} = \frac{1+6+5}{6} = \frac{12}{6} = 2(5−16)2=(46)2=(23)2=49≈0.444(\frac{5-1}{6})^2 = (\frac{4}{6})^2 = (\frac{2}{3})^2 = \frac{4}{9} \approx 0.444
B3453+4(4)+56=3+16+56=246=4\frac{3 + 4(4) + 5}{6} = \frac{3+16+5}{6} = \frac{24}{6} = 4(5−36)2=(26)2=(13)2=19≈0.111(\frac{5-3}{6})^2 = (\frac{2}{6})^2 = (\frac{1}{3})^2 = \frac{1}{9} \approx 0.111
C1231+4(2)+36=1+8+36=126=2\frac{1 + 4(2) + 3}{6} = \frac{1+8+3}{6} = \frac{12}{6} = 2(3−16)2=(26)2=(13)2=19≈0.111(\frac{3-1}{6})^2 = (\frac{2}{6})^2 = (\frac{1}{3})^2 = \frac{1}{9} \approx 0.111
D3.556.53.5+4(5)+6.56=3.5+20+6.56=306=5\frac{3.5 + 4(5) + 6.5}{6} = \frac{3.5+20+6.5}{6} = \frac{30}{6} = 5(6.5−3.56)2=(36)2=(12)2=14=0.25(\frac{6.5-3.5}{6})^2 = (\frac{3}{6})^2 = (\frac{1}{2})^2 = \frac{1}{4} = 0.25
E45124+4(5)+126=4+20+126=366=6\frac{4 + 4(5) + 12}{6} = \frac{4+20+12}{6} = \frac{36}{6} = 6(12−46)2=(86)2=(43)2=169≈1.778(\frac{12-4}{6})^2 = (\frac{8}{6})^2 = (\frac{4}{3})^2 = \frac{16}{9} \approx 1.778
F6.57.511.56.5+4(7.5)+11.56=6.5+30+11.56=486=8\frac{6.5 + 4(7.5) + 11.5}{6} = \frac{6.5+30+11.5}{6} = \frac{48}{6} = 8(11.5−6.56)2=(56)2=2536≈0.694(\frac{11.5-6.5}{6})^2 = (\frac{5}{6})^2 = \frac{25}{36} \approx 0.694
G59135+4(9)+136=5+36+136=546=9\frac{5 + 4(9) + 13}{6} = \frac{5+36+13}{6} = \frac{54}{6} = 9(13−56)2=(86)2=(43)2=169≈1.778(\frac{13-5}{6})^2 = (\frac{8}{6})^2 = (\frac{4}{3})^2 = \frac{16}{9} \approx 1.778

步驟二：繪製 PERT 圖並計算關鍵路徑 (Critical Path)

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第 2 題25 分

Supreme Auto Parts produces components for motorcycle engines. It has plants in Amarillo, Texas, and Charlotte, North Carolina, and supply factories in Detroit and Atlanta. Production and cost data for a major component are as follows.

PlantFreight Costs (Detroit)Freight Costs (Atlanta)CapacityUnit Cost
Amarillo$12$81,200$125
Charlotte$9$33,000$140
DemandDetroitAtlanta
2,000900

(a) Formulate a transportation model to determine the best distribution plan.
(b) Solve the above transportation model to find the optimal objective and optimal solutions.

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這一題的完整詳解

核心觀念

本題考查「運輸問題（Transportation Problem）」：

  • 供應端：Amarillo、Charlotte 兩座工廠。
  • 需求端：Detroit、Atlanta 兩座供應工廠。
  • 每單位總成本：
總成本=單位生產成本+單位運費\text{總成本}=\text{單位生產成本}+\text{單位運費}

總供給量為

1,200+3,000=4,2001,200+3,000=4,200

總需求量為

2,000+900=2,9002,000+900=2,900

供給量大於需求量，因此屬於「供給過剩」的非平衡運輸問題。可加入一個需求量為 1,3001,300 的虛擬目的地 Dummy，代表未使用的產能，該欄運輸成本設為 00。

(a) 運輸模型 formulation

1. 計算單位總成本

來源／目的地DetroitAtlantaDummy
Amarillo125+125+12=$137125+125+8=$133$0
Charlotte140+140+9=$149140+140+3=$143$0

其中 Dummy 的需求量為

4,200−2,900=1,3004,200-2,900=1,300

完整運輸表如下：

工廠／需求地DetroitAtlantaDummy供給量
Amarillo13713301,200
Charlotte14914303,000
需求量2,0009001,3004,200

2. 定義決策變數

令：

  • xADx_{AD}：Amarillo 運往 Detroit 的數量
  • xAAx_{AA}：Amarillo 運往 Atlanta 的數量
  • xAUx_{AU}：Amarillo 運往 Dummy 的數量
  • xCDx_{CD}：Charlotte 運往 Detroit 的數量
  • xCAx_{CA}：Charlotte 運往 Atlanta 的數量
  • xCUx_{CU}：Charlotte 運往 Dummy 的數量

3. 目標函數

最小化總生產與運輸成本：

min⁡Z=137xAD+133xAA+0xAU+149xCD+143xCA+0xCU\min Z = 137x_{AD} +133x_{AA} +0x_{AU} +149x_{CD} +143x_{CA} +0x_{CU}

4. 供給限制式

Amarillo 的供給量為 1,2001,200：

xAD+xAA+xAU=1,200x_{AD}+x_{AA}+x_{AU}=1,200

Charlotte 的供給量為 3,0003,000：

xCD+xCA+xCU=3,000x_{CD}+x_{CA}+x_{CU}=3,000

5. 需求限制式

Detroit 的需求量為 2,0002,000：

xAD+xCD=2,000x_{AD}+x_{CD}=2,000

Atlanta 的需求量為 900900：

xAA+xCA=900x_{AA}+x_{CA}=900

Dummy 的需求量為 1,3001,300：

xAU+xCU=1,300x_{AU}+x_{CU}=1,300

且

xij≥0x_{ij}\geq 0

(b) 求最適配送方案

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第 3 題25 分

The Wilcox Student Health Center has just implemented a new computer system and service process to "improve efficiency." The process flowchart and analysis framework is also provided. As pharmacy manager, you are concerned about waiting time and its potential impact on college students who "get no respect." All prescriptions (Rxs) go through the following process:

Assume that students arrive to drop-off Rxs at a steady rate of two Rxs per minute, with an average of one Rx per student. The average number of students in process (assume waiting and being serviced) at each station is: DROP-OFF-five students, PICK-UP-three students and PAY CASHIER-six students.

🖼️【此處有附圖，請對照原卷】

The fill Rx station typically has 40 Rxs in process and waiting on average. Because of this perceived long wait, 95 percent of the students decide to come back later for pick-up. They come back an average of three hours later. If the students choose to stay, each name is called as soon as the Rx is filled and the student then enters the pick-up line. Assume that the system is operating at a steady state.

(a) What is the average time a student spends in the pharmacy if they stay to pick-up their Rx?
(b) How many minutes does the student spend in the pharmacy if he or she picks-up the Rx 3 hours later (i.e., the student goes home after dropping the Rx off)?
(c) What is the average time in minutes that the Rx spends in the process? Count time from entering the drop-off line to completing payment.

🖼️ 本題含圖表，以下為原卷對應頁面：
原卷第 3 頁

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這一題的完整詳解

核心觀念

本題使用 Little’s Law：

L=λW⇒W=LλL=\lambda W \quad\Rightarrow\quad W=\frac{L}{\lambda}

其中：

  • LL：平均在製數量
  • λ\lambda：流量率
  • WW：平均流程時間

因為平均每位學生持有 11 張 Rx，故學生與 Rx 的流量率皆為：

λ=2 件／分鐘\lambda=2\text{ 件／分鐘}

各站流程時間為：

Tdrop-off=52=2.5 分鐘T_{\text{drop-off}}=\frac{5}{2}=2.5\text{ 分鐘} Tfill Rx=402=20 分鐘T_{\text{fill Rx}}=\frac{40}{2}=20\text{ 分鐘} Tpick-up=32=1.5 分鐘T_{\text{pick-up}}=\frac{3}{2}=1.5\text{ 分鐘} Tcashier=62=3 分鐘T_{\text{cashier}}=\frac{6}{2}=3\text{ 分鐘}

注意：雖然只有 5%5\% 的學生選擇留下，仍有 95%95\% 的學生在三小時後回來，因此穩態下 Pick-up 與 Cashier 的總流量仍是 22 人／分鐘。

解題方法

(a) 學生留下來等候取藥

學生必須依序經過 Drop-off、Fill Rx、Pick-up 及 Cashier：

Tstay=2.5+20+1.5+3=27 分鐘T_{\text{stay}} =2.5+20+1.5+3 =27\text{ 分鐘}

因此，留下來取藥的學生平均在藥局停留：

27 分鐘\boxed{27\text{ 分鐘}}

(b) 學生三小時後回來取藥

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第 4 題25 分

High Tech, Inc. is a virtual store that stocks a variety of calculators in their warehouse. Customer orders are placed, the order is picked and packaged, and then shipped to the customer. A fixed order quantity inventory control system (FQS) helps monitor and control these SKUs. The following information is for one of the calculators that they stock, sell, and ship.

Average demand: 12.5 calculators per week
Lead time: 3 weeks
Order cost: 20/orderUnitcost:20/order Unit cost: 8.00
Carrying charge rate: 0.25
Number of weeks: 52 weeks per year
Standard deviation of weekly demand: 3.75 calculators
SKU service level: 95 percent

Current on-hand inventory: 35 calculators
Scheduled receipts: 20 calculators
Backorders: 2 calculators

(a) What is the Economic Order Quantity?
(b) What is the total annual order and inventory-holding costs for the EOQ?
(c) What is the reorder point without safety stock?
(d) What is the reorder point with safety stock?
(e) Based on the previous information, should a fixed order quantity be placed, and if so, for how many calculators?

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這一題的完整詳解

核心觀念

本題屬於存貨管理中的定量訂購系統（Fixed Order Quantity System, FQS），又稱連續盤點系統（Continuous Review System, (Q,R)(Q, R) 系統）。核心觀念與公式涵蓋：

  1. 基本經濟訂購量模型（EOQ Model）：

    • 年需求量：D=d×週數D = d \times \text{週數}（其中 dd 為平均週需求量）。
    • 單位年持有成本：H=i×CH = i \times C（其中 ii 為年持有成本率，CC 為單位購買成本）。
    • 每次訂購成本：SS。
    • 經濟訂購量公式： EOQ=2DSHEOQ = \sqrt{\frac{2DS}{H}}
    • 年總訂購與持有成本（Total Annual Relevant Cost）： TC=DQS+Q2H=2DSH(當 Q=EOQ)TC = \frac{D}{Q}S + \frac{Q}{2}H = \sqrt{2DSH} \quad (\text{當 } Q = EOQ)

  2. 再訂購點（Reorder Point, ROP）與安全庫存（Safety Stock, SS）：

    • 無安全庫存時之再訂購點：前置時間內的期望需求量。 ROP0=d×LROP_0 = d \times L
    • 需求具隨機性（常態分配）時的前置時間需求標準差： σL=L×σd\sigma_L = \sqrt{L} \times \sigma_d
    • 安全庫存量（配合指定服務水準 ZZ 值）： SS=Z×σL=Z×L×σdSS = Z \times \sigma_L = Z \times \sqrt{L} \times \sigma_d
    • 具安全庫存時之再訂購點： ROP=d×L+SS=d×L+Z×L×σdROP = d \times L + SS = d \times L + Z \times \sqrt{L} \times \sigma_d

  3. 庫存部位（Inventory Position, IP）與訂購決策：

    • 庫存部位定義為現有存貨（On-Hand, OHOH）加上預計到貨量（Scheduled Receipts, SRSR），再扣除延遲交貨／未交訂單量（Backorders, BOBO）： IP=OH+SR−BOIP = OH + SR - BO
    • 連續盤點決策準則：當 IP≤ROPIP \le ROP 時，觸發訂購，訂購批量為 EOQEOQ；若 IP>ROPIP > ROP，則目前不需要發出訂單。

解題方法與步驟推導

基礎參數整理

  • 平均週需求量 d=12.5 台/週d = 12.5\text{ 台/週}
  • 前置時間 L=3 週L = 3\text{ 週}
  • 每次訂購成本 S=S = $20/次20/\text{次}
  • 單位成本 C=C = $8.00/台8.00/\text{台}
  • 年持有成本率 i=0.25i = 0.25
  • 單位年持有成本 H=i×C=0.25×8.00=H = i \times C = 0.25 \times 8.00 = $2.00/台/年2.00/\text{台/年}
  • 一年週數 =52 週= 52\text{ 週}，故年需求量 D=12.5×52=650 台/年D = 12.5 \times 52 = 650\text{ 台/年}
  • 週需求標準差 σd=3.75 台/週\sigma_d = 3.75\text{ 台/週}
  • 服務水準 Service Level =95%= 95\%   ⟹  \implies 標準常態分配對應臨界值 Z0.95≈1.645Z_{0.95} \approx 1.645（若取 Z=1.65Z = 1.65 計算亦可）
  • 現有存貨 OH=35 台OH = 35\text{ 台}，預計到貨 SR=20 台SR = 20\text{ 台}，缺貨待補 BO=2 台BO = 2\text{ 台}

各小題詳細解析

(a) Economic Order Quantity (EOQ)

代入 EOQ 公式：

EOQ=2×650×202=13000≈114.02 台EOQ = \sqrt{\frac{2 \times 650 \times 20}{2}} = \sqrt{13000} \approx 114.02\text{ 台}

實務上可取整數 114 台114\text{ 台}。

(b) Total annual order and inventory-holding costs for the EOQ

於最佳訂購量 $EOQ = \sqrt{13000}

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其他考古題

111 年成功大學的其他科目

統計方法

成功大學《生產管理》其他年度

113 年112 年109 年108 年